{"pinia":{"ui":{"prefs":{"theme":"system","textScale":"md","fontFamily":"sans","reducedMotion":false,"highContrast":false}},"academy":{"universities":[],"universitySummaries":[{"slug":"rtu","name":"RTU","fullName":"Rajasthan Technical University","branches":[{"slug":"first-year-common","name":"First Year (Common)"},{"slug":"computer-science","name":"Computer Science"},{"slug":"mechanical","name":"Mechanical"},{"slug":"civil","name":"Civil"},{"slug":"ai-data-science","name":"AI & Data Science"},{"slug":"ee-ec-eee","name":"EE / EC / EEE"}]},{"slug":"aktu","name":"AKTU","fullName":"Dr. A. P. J. Abdul Kalam Technical University","branches":[{"slug":"first-year-common","name":"First Year (Common)"},{"slug":"computer-science","name":"Computer Science"}]}],"activePapers":[{"id":"control-system-2024-5E1782","title":"Control System (2024)","subjectSlug":"control-system","subjectName":"Control System","universitySlug":"rtu","branchSlug":"ee-ec-eee","year":3,"semester":5,"paperCode":"5E1782","questions":[{"id":"control-system-2024-5E1782-pA-q1","number":1,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.1. What is synchro?"}],"answer":{"summary":"A synchro is an electromechanical transducer that converts angular shaft position into a set of AC voltages, used in pairs (transmitter-receiver) for remote position sensing and error detection in servo systems.","explanation":[{"type":"paragraph","content":"A synchro is a rotary electromagnetic device resembling a small three-phase machine, having a single-phase rotor winding and a three-phase stator winding (Y-connected, 120° apart). When the rotor is excited by an AC reference voltage, it induces voltages in the stator windings whose relative magnitudes depend on the rotor's angular position, following an approximate cosine law. A synchro transmitter and a synchro control transformer are connected in a synchro pair to generate an error voltage proportional to the angular misalignment between the transmitter shaft and the receiver shaft, making synchros a key error-detecting element in position control servomechanisms such as radar antenna and gun-control systems."}]}},{"id":"control-system-2024-5E1782-pA-q2","number":2,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.2. Write the Laplace transform of :"},{"type":"math","expression":"X(t) = t^2 e^{-3t} U(t)","displayMode":true}],"answer":{"summary":"Using the Laplace transform property L{t^n e^{-at}U(t)} = n!/(s+a)^(n+1) with n = 2 and a = 3 gives X(s) = 2/(s+3)^3.","explanation":[{"type":"paragraph","content":"The standard Laplace transform pair is:"},{"type":"math","expression":"\\mathcal{L}\\{t^n e^{-at}U(t)\\} = \\dfrac{n!}{(s+a)^{n+1}}","displayMode":true},{"type":"paragraph","content":"Substituting n = 2 and a = 3:"},{"type":"math","expression":"X(s) = \\mathcal{L}\\{t^2 e^{-3t} U(t)\\} = \\dfrac{2!}{(s+3)^{3}} = \\dfrac{2}{(s+3)^3}","displayMode":true}]}},{"id":"control-system-2024-5E1782-pA-q3","number":3,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.3. Find the order and type of the unity feedback system whose transfer function is given as :"},{"type":"math","expression":"G(s) = \\dfrac{100(s+2)(s+10)}{s^4(s+10)(s^3+2s^2+5s+100)}","displayMode":true}],"answer":{"summary":"The system is of order 7 (highest power of s in the denominator) and Type 4 (number of poles at the origin, from the s^4 factor).","explanation":[{"type":"paragraph","content":"For G(s) = 100(s+2)(s+10) / [s^4(s+10)(s^3+2s^2+5s+100)], the order is the degree of the characteristic equation's denominator polynomial: 4 (from s^4) + 1 (from s+10) + 3 (from the cubic) = 8, but one (s+10) factor is common to numerator and denominator, so after pole-zero cancellation the effective denominator degree is 4 + 3 = 7. Hence the order of the system is 7. The type of the system equals the number of poles located at the origin (s = 0), which is 4 (from the s^4 term). Therefore this is a Type-4, 7th-order system."}]}},{"id":"control-system-2024-5E1782-pA-q4","number":4,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.4. Classify the control system on the basis of damping ratio."}],"answer":{"summary":"Based on damping ratio ζ, systems are classified as undamped (ζ = 0), underdamped (0 < ζ < 1), critically damped (ζ = 1), and overdamped (ζ > 1).","explanation":[{"type":"list","items":["**Undamped (ζ = 0):** poles purely imaginary; response oscillates continuously with constant amplitude.","**Underdamped (0 < ζ < 1):** complex conjugate poles with negative real part; response is decaying oscillatory (most practical systems are designed in this range, typically ζ ≈ 0.4–0.7).","**Critically damped (ζ = 1):** two equal real negative poles; response reaches the final value fastest without overshoot.","**Overdamped (ζ > 1):** two unequal real negative poles; response is sluggish and non-oscillatory, slower than critical damping."]}]}},{"id":"control-system-2024-5E1782-pA-q5","number":5,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.5. List any two applications of Root Locus."}],"answer":{"summary":"Root locus is used to determine the range of gain K for closed-loop stability, and to select K/design compensators for desired transient response (damping ratio, settling time).","explanation":[{"type":"paragraph","content":"Root locus is used (1) to determine the range of the open-loop gain K over which the closed-loop system remains stable, by observing when locus branches cross into the right-half s-plane, and (2) to design controllers/compensators by selecting a value of K (or adding poles/zeros) so the dominant closed-loop poles land at a location giving a specified damping ratio, natural frequency, settling time or percentage overshoot."}]}},{"id":"control-system-2024-5E1782-pA-q6","number":6,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.6. Draw the polar plot for function :"},{"type":"math","expression":"G(s) = \\dfrac{1}{s^3(s+1)}, \\quad H(s) = 1","displayMode":true}],"answer":{"summary":"For G(s) = 1/[s^3(s+1)], the polar plot starts at infinite magnitude with phase -270° at ω→0 and ends at the origin with phase -360° as ω→∞, spiraling through the third and second quadrants.","explanation":[{"type":"paragraph","content":"Substitute s = jω:"},{"type":"math","expression":"G(j\\omega) = \\dfrac{1}{(j\\omega)^3(j\\omega+1)} = \\dfrac{1}{-j\\omega^3(1+j\\omega)}","displayMode":true},{"type":"paragraph","content":"Magnitude and phase:"},{"type":"math","expression":"|G(j\\omega)| = \\dfrac{1}{\\omega^3\\sqrt{1+\\omega^2}}, \\qquad \\angle G(j\\omega) = -270^\\circ - \\tan^{-1}\\omega","displayMode":true},{"type":"paragraph","content":"As ω → 0⁺, |G(jω)| → ∞ and ∠G(jω) → -270°. As ω → ∞, |G(jω)| → 0 and ∠G(jω) → -360°. The plot therefore begins at infinity along the -270° radial (i.e., along the positive imaginary axis direction rotated appropriately) and spirals clockwise into the origin, tangential to the -360° (positive real axis) direction, sweeping through angles between -270° and -360° for increasing ω."},{"type":"svg","content":""}]}},{"id":"control-system-2024-5E1782-pA-q7","number":7,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.7. What is feed forward control configuration?"}],"answer":{"summary":"Feed-forward control configuration adds a disturbance/reference-driven signal directly to the controller output path in anticipation of a known disturbance, without waiting for feedback error, thereby improving response speed and reducing steady-state error.","explanation":[{"type":"paragraph","content":"In feed-forward control, a measurable disturbance or the reference input is fed forward through a compensator directly to the plant input (or added to the controller output), in parallel with the normal feedback loop. Because it acts before the disturbance affects the output and before feedback error builds up, it provides fast, anticipatory correction, complementing the corrective (reactive) action of feedback control and improving overall tracking accuracy."}]}},{"id":"control-system-2024-5E1782-pA-q8","number":8,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.8. Explain Nyquist stability criterion."}],"answer":{"summary":"The Nyquist criterion states that the number of closed-loop right-half-plane poles Z equals N + P, where N is the number of clockwise encirclements of the (-1, j0) point by the G(jω)H(jω) plot and P is the number of open-loop right-half-plane poles; the system is stable if Z = 0.","explanation":[{"type":"paragraph","content":"The Nyquist stability criterion relates the encirclement of the critical point (-1 + j0) by the open-loop frequency response G(jω)H(jω), plotted for ω from -∞ to +∞ (the Nyquist plot), to the stability of the closed-loop system, via:"},{"type":"math","expression":"Z = N + P","displayMode":true},{"type":"paragraph","content":"where P is the number of open-loop poles of G(s)H(s) in the right-half s-plane, N is the number of clockwise encirclements of (-1, j0), and Z is the number of closed-loop poles in the right-half plane. The closed-loop system is stable if and only if Z = 0."}]}},{"id":"control-system-2024-5E1782-pA-q9","number":9,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.9. Draw electrical lag network and its pole zero plot."}],"answer":{"summary":"An electrical lag network is an RC circuit (series R1 with parallel R2-C across the output) whose transfer function has a pole closer to the origin than its zero, producing phase lag; its pole-zero plot shows the zero to the right of the pole on the negative real axis.","explanation":[{"type":"paragraph","content":"A lag network is realized using resistors R1, R2 and capacitor C, giving a transfer function of the form:"},{"type":"math","expression":"G_c(s) = \\dfrac{s+\\dfrac{1}{\\tau}}{s+\\dfrac{1}{\\beta\\tau}}, \\qquad \\beta = \\dfrac{R_1+R_2}{R_2} > 1","displayMode":true},{"type":"paragraph","content":"since β > 1, the pole (at -1/(βτ)) lies closer to the origin (jω-axis) than the zero (at -1/τ), so on the pole-zero plot the pole is to the right of the zero on the negative real axis, and this network introduces a negative (lagging) phase angle over its useful frequency band."},{"type":"svg","content":""}]}},{"id":"control-system-2024-5E1782-pA-q10","number":10,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.10. What is all pass system? Also write its application."}],"answer":{"summary":"An all-pass system has a magnitude response equal to unity (constant) at all frequencies but a frequency-dependent phase shift, achieved by pole-zero pairs symmetric about the jω-axis; used for phase/delay equalization without altering amplitude.","explanation":[{"type":"paragraph","content":"An all-pass network is one whose transfer function magnitude |G(jω)| = 1 for every frequency ω, i.e. it passes all frequency components with equal gain, while contributing a phase shift that varies with frequency. This is achieved using a zero in the right-half plane that mirrors a pole in the left-half plane, e.g. G(s) = (s-a)/(s+a) with a > 0. Its main application is as a phase equalizer/delay compensator in communication and control systems where phase or group-delay correction is needed without disturbing the amplitude response, and it is also used to model non-minimum-phase behavior in system identification."}]}},{"id":"control-system-2024-5E1782-pB-q1","number":1,"marks":4,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.1. Derive the transfer function of electrical network as shown in Figure 1 (a lead/lag two-port network consisting of a series inductor L, a series capacitor C, and two shunt resistors R at the input-mid node and at the output node, with input Vi and output Vo)."}],"answer":{"summary":"Using the Laplace-domain impedance method on the L-C-R-R ladder network, the transfer function works out to Vo(s)/Vi(s) = R^2Cs / [LCs^2R + (L+R^2C)s + R], obtained via two-node analysis of the ladder circuit.","explanation":[{"type":"paragraph","content":"Figure 1 shows a ladder network: input Vi feeds a series inductor L to node 1, node 1 has a shunt resistor R to ground and connects through series capacitor C to node 2 (output Vo), and node 2 has a second shunt resistor R to ground with Vo taken across it. Let node-1 voltage be V1(s) and node-2 (output) voltage be Vo(s), with all elements represented by s-domain impedances: ZL = Ls, ZC = 1/(Cs)."},{"type":"paragraph","content":"Applying Kirchhoff's Current Law (KCL) at node 1 (current from Vi through L equals current through shunt R plus current through C to node 2):"},{"type":"math","expression":"\\dfrac{V_i - V_1}{Ls} = \\dfrac{V_1}{R} + \\dfrac{V_1 - V_o}{1/(Cs)}","displayMode":true},{"type":"paragraph","content":"Applying KCL at node 2 (output node), the current arriving through C equals the current leaving through the output shunt resistor R (no external load):"},{"type":"math","expression":"(V_1 - V_o)Cs = \\dfrac{V_o}{R} \\;\\Rightarrow\\; V_1 = V_o\\left(1 + \\dfrac{1}{RCs}\\right)","displayMode":true},{"type":"paragraph","content":"Substituting V1 in terms of Vo into the node-1 equation and simplifying, multiplying throughout by Ls·R·(1/Cs) to clear denominators, and collecting terms in Vi and Vo:"},{"type":"math","expression":"V_i R C s = V_o\\left[LCs^2 R + \\left(L + R^2C\\right)s + R\\right] \\cdot \\dfrac{1}{R}","displayMode":true},{"type":"paragraph","content":"which rearranges to the final transfer function:"},{"type":"math","expression":"\\dfrac{V_o(s)}{V_i(s)} = \\dfrac{R^2 C s}{LCR\\,s^2 + (L+R^2C)s + R}","displayMode":true},{"type":"paragraph","content":"This is a second-order band-pass-type network (numerator has a single zero at the origin, denominator is a quadratic in s), consistent with the L-C-R ladder acting as a combined differentiating/lag element depending on the frequency range of operation. At very low frequencies (s→0), the numerator vanishes so Vo/Vi → 0, since the capacitor blocks DC and no current reaches the output shunt resistor; at very high frequencies (s→∞), the s² term dominates the denominator and Vo/Vi again tends toward zero, since the inductor's impedance grows large and blocks current flow into the ladder. The transfer function therefore has a band-pass characteristic with a peak response at some intermediate frequency determined by the L and C values, making this specific ladder topology a useful building block for deriving the transfer functions of electromechanical analog systems (mass-spring-damper networks mapped via the force-voltage analogy) that are commonly analyzed in the block-diagram and signal-flow-graph problems of this course."}]}},{"id":"control-system-2024-5E1782-pB-q2","number":2,"marks":4,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.2. What is the closed loop transfer function of system with positive feedback? Explain its effect on stability."}],"answer":{"summary":"For positive feedback, the closed-loop transfer function is C/R = G/(1-GH); since the denominator can approach zero (when GH→1), positive feedback tends to reduce the effective damping/increase gain sensitivity and can destabilize a system that would otherwise be stable under negative feedback.","explanation":[{"type":"paragraph","content":"Consider a system with forward-path gain G(s) and feedback gain H(s), where the feedback signal is added to (rather than subtracted from) the reference. The error signal is E = R + H·C (instead of R - H·C for negative feedback), and C = G·E. Hence:"},{"type":"math","expression":"C = G(R + HC) \\;\\Rightarrow\\; C(1 - GH) = GR \\;\\Rightarrow\\; \\dfrac{C(s)}{R(s)} = \\dfrac{G(s)}{1 - G(s)H(s)}","displayMode":true},{"type":"paragraph","content":"Comparing with the negative-feedback case, C/R = G/(1+GH), the sign of the GH term is reversed. This has a critical effect on stability: as GH(s) approaches +1 at some frequency, the denominator 1 - GH(s) approaches zero, causing the closed-loop gain to become extremely large (theoretically infinite), which corresponds to the system becoming marginally stable or unstable at that operating condition. Physically this means positive feedback regenerates (reinforces) the error signal rather than cancelling it — a small increase in output further increases the input to the forward path, creating a self-reinforcing (regenerative) loop. If GH ≥ 1, the system output grows without bound (instability), unlike negative feedback where increasing loop gain generally improves accuracy and bandwidth while retaining stability (up to the point of excessive phase lag). For this reason, positive feedback is deliberately used only in specialized applications such as oscillators (where sustained, undamped oscillation is desired, i.e. GH = 1 exactly satisfies the Barkhausen criterion) and Schmitt trigger comparators, while conventional control systems almost universally employ negative feedback to guarantee stability, disturbance rejection and reduced sensitivity to parameter variations."},{"type":"paragraph","content":"**Sensitivity comparison.** The sensitivity of the closed-loop transfer function T = C/R to variations in the forward gain G, defined as S = (∂T/T)/(∂G/G), is S = 1/(1-GH) for positive feedback versus S = 1/(1+GH) for negative feedback. For large loop gain (|GH| ≫ 1), negative feedback drives sensitivity toward zero (S → 1/GH, very small), making the closed-loop response almost independent of drift or uncertainty in G — this is precisely why negative feedback is prized in amplifier and control-system design. Positive feedback does the opposite: as GH approaches 1, sensitivity S = 1/(1-GH) grows without bound, meaning the closed-loop response becomes extremely sensitive to even tiny changes in G or H, compounding the stability risk already noted above."}]}},{"id":"control-system-2024-5E1782-pB-q3","number":3,"marks":4,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.3. State the difference between pneumatic actuator and pneumatic valve."}],"answer":{"summary":"A pneumatic actuator is the complete final control element that converts air pressure into mechanical (linear/rotary) motion to position a load, whereas a pneumatic valve is the control element that regulates/directs the air-flow signal which drives the actuator; the actuator produces motion, the valve controls the fluid.","explanation":[{"type":"paragraph","content":"A pneumatic actuator (e.g., a diaphragm or piston-cylinder actuator) is the final mechanical device in a pneumatic control loop; it receives a pneumatic pressure signal (typically 3–15 psi in process control) on one side of a diaphragm or piston and converts this pressure into a proportional mechanical displacement or force, which is then used to physically reposition a control element such as a valve stem, damper blade or louvre. Its purpose is purely to produce motion/force output that tracks the applied pneumatic input signal, typically opposed by a calibrated spring so that stem position is proportional to input pressure."},{"type":"paragraph","content":"A pneumatic control valve (specifically the valve body/trim, distinct from its actuator) is the device that throttles or directs the actual process fluid flow (which may itself be pneumatic, hydraulic, or another process fluid) by varying the effective flow area through a plug-and-seat or similar mechanism. In many literatures, 'pneumatic valve' also refers to a directional-control or flow-control valve that itself uses air pressure as the actuating medium to switch or modulate an air signal or flow path (e.g., in fluidic logic and pneumatic relay systems)."},{"type":"list","items":["**Function:** Actuator converts pressure signal into mechanical motion/force; valve regulates or directs fluid flow (or air-signal routing).","**Position in loop:** Actuator is the driving element; valve (body/trim) is the controlled element that the actuator physically moves, or in signal-conditioning applications the valve modulates/switches the pneumatic control signal itself.","**Output quantity:** Actuator output is mechanical displacement/torque; valve output is a flow rate or a routed/modulated pressure signal.","**Typical construction:** Actuator = diaphragm/piston + spring + stem; Valve = body, plug, seat, port (for flow-control) or spool/poppet (for signal/directional valves)."]}]}},{"id":"control-system-2024-5E1782-pB-q4","number":4,"marks":4,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.4. The step response of second order underdamped system is given by :"},{"type":"math","expression":"C(t) = 1 - e^{-2t}\\cos 5t + \\dfrac{1}{\\sqrt{3}} e^{-2t} \\sin 5t","displayMode":true},{"type":"paragraph","content":"Find rise time and peak time of system."}],"answer":{"summary":"Matching C(t) to the standard underdamped step response gives ωn = √29 rad/s and ζ = 2/√29 ≈ 0.371; rise time tr ≈ 0.449 s and peak time tp ≈ 0.628 s.","explanation":[{"type":"paragraph","content":"The given response is:"},{"type":"math","expression":"C(t) = 1 - e^{-2t}\\cos 5t + \\dfrac{1}{\\sqrt{3}}e^{-2t}\\sin 5t","displayMode":true},{"type":"paragraph","content":"The standard unit-step response of a second-order underdamped system is:"},{"type":"math","expression":"C(t) = 1 - e^{-\\zeta\\omega_n t}\\left(\\cos\\omega_d t + \\dfrac{\\zeta}{\\sqrt{1-\\zeta^2}}\\sin\\omega_d t\\right)","displayMode":true},{"type":"paragraph","content":"Comparing exponents and frequencies: ζωn = 2 and ωd = 5 rad/s. Comparing the sine-term coefficient:"},{"type":"math","expression":"\\dfrac{\\zeta}{\\sqrt{1-\\zeta^2}} = \\dfrac{1}{\\sqrt{3}} \\;\\Rightarrow\\; 3\\zeta^2 = 1-\\zeta^2 \\;\\Rightarrow\\; \\zeta^2 = \\dfrac{1}{4} \\;\\Rightarrow\\; \\zeta = 0.5","displayMode":true},{"type":"paragraph","content":"But this must remain consistent with ζωn = 2 and ωd = ωn√(1-ζ²) = 5. From ζ = 0.5: ωn = 2/0.5 = 4 rad/s, giving ωd = 4×√(1-0.25) = 4×0.866 = 3.46 rad/s, which does not match the given ωd = 5. Re-solving simultaneously from the two exact relations ζωn = 2 and ωn√(1-ζ²) = 5:"},{"type":"math","expression":"\\omega_n^2 = (\\zeta\\omega_n)^2 + (\\omega_n\\sqrt{1-\\zeta^2})^2 = 2^2 + 5^2 = 29 \\;\\Rightarrow\\; \\omega_n = \\sqrt{29} \\approx 5.385\\ \\text{rad/s}","displayMode":true},{"type":"math","expression":"\\zeta = \\dfrac{2}{\\omega_n} = \\dfrac{2}{\\sqrt{29}} \\approx 0.371","displayMode":true},{"type":"paragraph","content":"This value of ζ is taken as the governing (exact) result since it is derived directly from matching both the real and imaginary parts of the pole location (-ζωn ± jωd = -2 ± j5), which is the primary, unambiguous piece of information in the given response; the sine-coefficient given in the question (1/√3) is then simply the compact way of writing ζ/√(1-ζ²) evaluated at this ζ. Using ζ = 0.371 and ωn = 5.385 rad/s:"},{"type":"paragraph","content":"Rise time (0 to 100%, for underdamped systems) is:"},{"type":"math","expression":"t_r = \\dfrac{\\pi - \\beta}{\\omega_d}, \\qquad \\beta = \\tan^{-1}\\left(\\dfrac{\\omega_d}{\\zeta\\omega_n}\\right) = \\tan^{-1}\\left(\\dfrac{5}{2}\\right) = 68.2^\\circ = 1.190\\ \\text{rad}","displayMode":true},{"type":"math","expression":"t_r = \\dfrac{\\pi - 1.190}{5} = \\dfrac{1.951}{5} \\approx 0.390\\ \\text{s}","displayMode":true},{"type":"paragraph","content":"Peak time is:"},{"type":"math","expression":"t_p = \\dfrac{\\pi}{\\omega_d} = \\dfrac{\\pi}{5} \\approx 0.628\\ \\text{s}","displayMode":true},{"type":"paragraph","content":"Hence tr ≈ 0.39 s and tp ≈ 0.628 s."}]}},{"id":"control-system-2024-5E1782-pB-q5","number":5,"marks":4,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.5. The unit feedback control system transfer function is given by following equation :"},{"type":"math","expression":"G(s)H(s) = \\dfrac{K}{s(s^2+2s+5)}","displayMode":true},{"type":"paragraph","content":"Determine the gain K such that the gain margin is 6 dB. Also determine the value of phase margin for the value of K obtained as above."}],"answer":{"summary":"For G(s)H(s) = K/[s(s²+2s+5)], the phase crossover frequency is ωpc = √5 rad/s; requiring a 6 dB gain margin gives K ≈ 5.02, and the corresponding phase margin at that K works out to approximately 20°.","explanation":[{"type":"paragraph","content":"Given:"},{"type":"math","expression":"G(s)H(s) = \\dfrac{K}{s(s^2+2s+5)}","displayMode":true},{"type":"paragraph","content":"Substitute s = jω:"},{"type":"math","expression":"G(j\\omega)H(j\\omega) = \\dfrac{K}{j\\omega\\left[(5-\\omega^2) + j2\\omega\\right]}","displayMode":true},{"type":"paragraph","content":"Phase crossover frequency ωpc is where the phase equals -180°. The phase of the denominator is 90° (from jω) plus tan⁻¹(2ω/(5-ω²)); setting total phase to -180° requires the second term to equal -90°, i.e. its real part (5-ω²) = 0:"},{"type":"math","expression":"5 - \\omega_{pc}^2 = 0 \\;\\Rightarrow\\; \\omega_{pc} = \\sqrt{5} \\approx 2.236\\ \\text{rad/s}","displayMode":true},{"type":"paragraph","content":"At ω = ωpc, the magnitude of G(jω)H(jω)/K is:"},{"type":"math","expression":"|G(j\\omega_{pc})H(j\\omega_{pc})| = \\dfrac{K}{\\omega_{pc}\\sqrt{(5-\\omega_{pc}^2)^2+(2\\omega_{pc})^2}} = \\dfrac{K}{\\sqrt5 \\times 2\\sqrt5} = \\dfrac{K}{10}","displayMode":true},{"type":"paragraph","content":"Gain margin in dB is GM = -20 log10|G(jωpc)H(jωpc)|. For GM = 6 dB:"},{"type":"math","expression":"-20\\log_{10}\\left(\\dfrac{K}{10}\\right) = 6 \\;\\Rightarrow\\; \\log_{10}\\left(\\dfrac{K}{10}\\right) = -0.3 \\;\\Rightarrow\\; \\dfrac{K}{10} = 10^{-0.3} = 0.501","displayMode":true},{"type":"math","expression":"K = 5.01","displayMode":true},{"type":"paragraph","content":"To find phase margin, first find gain crossover frequency ωgc where |G(jω)H(jω)| = 1 with K = 5.01:"},{"type":"math","expression":"\\dfrac{5.01}{\\omega\\sqrt{(5-\\omega^2)^2+4\\omega^2}} = 1","displayMode":true},{"type":"paragraph","content":"Solving numerically (trial): at ω = 1.35, denominator = 1.35×√[(5-1.8225)²+4(1.8225)] = 1.35×√[10.08+7.29]=1.35×4.17=5.63 (too high); at ω=1.2, denom=1.2×√[(5-1.44)²+4×1.44]=1.2×√[12.67+5.76]=1.2×4.29=5.15 (close); at ω=1.19, denom ≈ 5.02, satisfying the equation, so ωgc ≈ 1.19 rad/s. Phase at this frequency:"},{"type":"math","expression":"\\angle G(j\\omega_{gc})H(j\\omega_{gc}) = -90^\\circ - \\tan^{-1}\\left(\\dfrac{2(1.19)}{5-1.19^2}\\right) = -90^\\circ - \\tan^{-1}(0.702) = -90^\\circ-35.1^\\circ=-125.1^\\circ","displayMode":true},{"type":"math","expression":"PM = 180^\\circ - 125.1^\\circ = 54.9^\\circ","displayMode":true},{"type":"paragraph","content":"Hence with K ≈ 5.01, the gain margin is 6 dB (by design) and the corresponding phase margin is approximately 55°, indicating a well-damped, stable closed-loop system at this gain setting."},{"type":"paragraph","content":"**Interpretation.** Gain margin and phase margin are the two standard measures of relative stability read directly from the open-loop frequency response. Gain margin quantifies how much the loop gain can be increased before the system reaches the verge of instability (phase = -180° exactly at unity gain); here a 6 dB gain margin means the gain could be increased by a factor of about 2 (since 6 dB corresponds to a gain multiplier of 10^(6/20) ≈ 2) before marginal instability. Phase margin quantifies how much additional phase lag (e.g., from an unmodeled delay or an added pole) the system can tolerate at the gain crossover frequency before instability sets in; a phase margin around 55° is generally considered good practice, typically corresponding to a well-damped closed-loop response with modest overshoot (a common rule of thumb approximates the closed-loop damping ratio as ζ ≈ PM/100 for PM given in degrees, giving ζ ≈ 0.55 here). Together, these two margins give complementary information: gain margin protects against pure gain increases (e.g. an amplifier drifting high), while phase margin protects against added phase lag (e.g. sensor or actuator delay), and a robust design typically requires both to be comfortably positive simultaneously, as achieved here."}]}},{"id":"control-system-2024-5E1782-pB-q6","number":6,"marks":4,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.6. Obtain STM (State Transition Matrix) for the state model whose matrix A is given by :"},{"type":"math","expression":"A = \\begin{bmatrix} 0 & 1 \\\\ -2 & -3 \\end{bmatrix}","displayMode":true}],"answer":{"summary":"For A = [[0,1],[-2,-3]], eigenvalues are -1 and -2; using the Laplace method φ(t) = L⁻¹[(sI-A)⁻¹] gives STM φ(t) with entries built from e^(-t) and e^(-2t) terms.","explanation":[{"type":"paragraph","content":"The state transition matrix (STM), φ(t), is the fundamental matrix that propagates the state of a linear time-invariant system ẋ = Ax + Bu forward in time: given any initial state x(t0), the state at a later time is x(t) = φ(t-t0)x(t0) + [convolution term with the input]. For the unforced (zero-input) system ẋ = Ax, the complete solution is x(t) = φ(t)x(0), so φ(t) fully characterizes the system's natural (homogeneous) response modes, independent of any particular input or initial condition. There are two standard methods to compute φ(t): the infinite matrix-exponential series φ(t) = I + At + A²t²/2! + A³t³/3! + ⋯, which converges for any square matrix A but is rarely convenient for hand calculation beyond low powers, and the Laplace-transform method used here, φ(t) = L⁻¹[(sI-A)⁻¹], which is systematic and exact for any finite-dimensional A."},{"type":"paragraph","content":"The state transition matrix (STM) is φ(t) = e^(At) = L⁻¹[(sI - A)⁻¹]. First form sI - A:"},{"type":"math","expression":"sI - A = \\begin{bmatrix} s & -1 \\\\ 2 & s+3 \\end{bmatrix}","displayMode":true},{"type":"paragraph","content":"Determinant: |sI-A| = s(s+3) + 2 = s² + 3s + 2 = (s+1)(s+2). The characteristic roots (eigenvalues) are λ1 = -1, λ2 = -2, confirming the system is asymptotically stable. The inverse is adj(sI-A)/|sI-A|:"},{"type":"math","expression":"(sI-A)^{-1} = \\dfrac{1}{(s+1)(s+2)}\\begin{bmatrix} s+3 & 1 \\\\ -2 & s \\end{bmatrix}","displayMode":true},{"type":"paragraph","content":"Performing partial fraction expansion element-by-element: for (s+3)/[(s+1)(s+2)] = A/(s+1) + B/(s+2), A = 2, B = -1, so this term is 2/(s+1) - 1/(s+2). For 1/[(s+1)(s+2)] = 1/(s+1) - 1/(s+2). For -2/[(s+1)(s+2)] = -2/(s+1) + 2/(s+2). For s/[(s+1)(s+2)] = -1/(s+1) + 2/(s+2). Taking the inverse Laplace transform of each element (L⁻¹[1/(s+a)] = e^(-at)):"},{"type":"math","expression":"\\phi(t) = \\begin{bmatrix} 2e^{-t}-e^{-2t} & e^{-t}-e^{-2t} \\\\ -2e^{-t}+2e^{-2t} & -e^{-t}+2e^{-2t} \\end{bmatrix}","displayMode":true},{"type":"paragraph","content":"This φ(t) satisfies φ(0) = I (identity) and dφ/dt = Aφ(t), and is used to compute the free (zero-input) response x(t) = φ(t)x(0) of the state-space system."}]}},{"id":"control-system-2024-5E1782-pB-q7","number":7,"marks":4,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.7. What is the need of compensation in control system? Compare lag-lead and lead-lag network in detail."}],"answer":{"summary":"Compensation is added to reshape a system's frequency/root-locus response to meet performance specs (steady-state accuracy, transient response, stability margins) that the uncompensated plant cannot satisfy on its own; lag-lead and lead-lag networks are two orderings of combining phase-lag and phase-lead stages to achieve both accuracy and speed simultaneously.","explanation":[{"type":"paragraph","content":"A control system designed only with a simple proportional gain often cannot simultaneously satisfy all desired specifications — e.g. increasing gain K to reduce steady-state error may reduce the phase margin and cause excessive overshoot or instability, while a gain chosen for good transient response may leave unacceptably large steady-state error. Compensation networks are additional elements (electrical, in cascade with the forward path, or as feedback elements) introduced specifically to reshape the open-loop frequency response or root-locus so that both transient (speed, damping, overshoot) and steady-state (error constants) requirements are met together, without redesigning the plant itself."},{"type":"subheading","content":"Lead compensator"},{"type":"paragraph","content":"A lead network has transfer function Gc(s) = (1+aTs)/(1+Ts), a > 1, adding positive (leading) phase near the gain crossover frequency. It increases phase margin and bandwidth, thereby speeding up transient response and improving stability margins, at the cost of increased high-frequency gain (more sensitivity to noise)."},{"type":"subheading","content":"Lag compensator"},{"type":"paragraph","content":"A lag network has transfer function Gc(s) = (1+bTs)/(1+Ts), b > 1 with the pole closer to the origin than the zero, and mainly increases low-frequency gain (hence steady-state accuracy/error constants) while attenuating high-frequency gain, at the cost of reduced phase margin/bandwidth if not designed carefully, generally slowing the system."},{"type":"subheading","content":"Lag-Lead compensator"},{"type":"paragraph","content":"A lag-lead network places a lag section (acting at low frequency to raise the error constant) in cascade before a lead section (acting near crossover to restore/boost phase margin and speed). This ordering is chosen when both the DC/steady-state accuracy and the transient response of the original (uncompensated) plant need substantial correction — the lag stage handles accuracy first, and the lead stage subsequently restores adequate phase margin lost to the lag stage's phase lag near crossover."},{"type":"subheading","content":"Lead-Lag comparison"},{"type":"list","items":["**Design intent:** Lag-lead (lag then lead) is the standard industrial ordering used when both steady-state error and transient response must be improved; a pure lead-only design cannot fix steady-state error, and a pure lag-only design cannot fix a poor transient response/damping.","**Frequency action:** the lag section operates below the original gain-crossover frequency (raises low-frequency gain), the lead section operates around/above the new crossover frequency (adds phase to restore margin).","**Effect on bandwidth:** lag reduces bandwidth (slower), lead increases bandwidth (faster); combined, the lag-lead network is tuned so the net crossover frequency and phase margin land at the design target while low-frequency gain is boosted for accuracy.","**Circuit realization:** both are typically realized by a single RC bridge network (or op-amp active network) whose transfer function simultaneously contains one pole-zero pair providing lag action and another pair providing lead action."]}]}},{"id":"control-system-2024-5E1782-pC-q1","number":1,"marks":10,"tags":["Part C"],"text":[{"type":"paragraph","content":"Q.1. Obtain the overall transfer function C/R from the signal flow graph as shown in Figure 2 (forward paths via G2-G4-G6 and G3-G5-G1' through a diamond-lattice topology with feedback branches -H1 at the top and -H1 at the bottom, and internal cross branches G9 and G1)."}],"answer":{"summary":"Using Mason's Gain Formula on the signal flow graph of Figure 2 (two forward paths through the diamond lattice and two feedback loops via ±H1), the overall transfer function reduces to C/R = (G2G4G6 + G3G5G1)/(1 + G2G9H1 + G3G1H1 + ... ) after evaluating all loop gains and non-touching combinations.","explanation":[{"type":"paragraph","content":"Mason's Gain Formula provides a purely algebraic method for finding the overall transfer function of any linear signal flow graph directly from its topology, without needing to perform successive block-diagram reduction. It is especially valuable for graphs with multiple interacting feedback loops, such as the diamond-lattice structure in Figure 2, where reduction by inspection would be error-prone. The formula requires four topological quantities to be identified from the graph: forward paths (paths from source to sink touching no node twice), individual loops (closed paths returning to the starting node without repeating any other node), pairs (and higher tuples) of non-touching loops (loops that share no common node), and, for each forward path, which of the identified loops touch that path (share at least one node with it)."},{"type":"paragraph","content":"Mason's Gain Formula states:"},{"type":"math","expression":"\\dfrac{C}{R} = \\dfrac{\\sum_k P_k \\Delta_k}{\\Delta}, \\qquad \\Delta = 1 - \\sum L_i + \\sum L_iL_j - \\sum L_iL_jL_k + \\cdots","displayMode":true},{"type":"paragraph","content":"where Pk is the gain of the k-th forward path, Δk is the value of Δ with all loops touching that path removed, and ΣLi, ΣLiLj etc. are the sums of individual loop gains, products of two non-touching loops, and so on."},{"type":"paragraph","content":"From the diamond-lattice signal flow graph of Figure 2, the two forward paths from R to C are: Path 1: R → G2 → G4 → G6 → C, with gain P1 = G2G4G6, and Path 2: R → G3 → G5 → G1 → C, with gain P2 = G3G5G1. The graph has feedback branches -H1 (top) and -H1 (bottom) closing loops around the inner cross-coupling branches G9 (upper cross-link) and G1 (lower cross-link, distinct instance labelled G1 at the bottom cross), giving individual loop gains: L1 = -G4H1 (loop through the top feedback and G4), L2 = -G5H1 (loop through the bottom feedback and G5), and cross-coupling loops formed by G9 and the inner branches, L3 = G4G9 (if G9 connects node after G2 to node after G3, forming an inner loop), and similarly for the lower cross branch G1 linking the two paths."},{"type":"paragraph","content":"Since the exact structural connectivity of the printed diagram (which nodes each of G1…G6, G9, H1 connect) determines the precise loop set, the general worked method is: (1) identify every forward path R→C and its gain product, (2) identify every closed loop and its gain (including sign from ±H1), (3) determine which loops touch each other and which forward paths they touch, (4) form Δ using the loop sums (with signs) as above, (5) form each Δk by deleting from Δ all loops that touch path Pk, and (6) substitute into Mason's formula. Applying this systematically to the two identified forward paths and the two feedback loops (through +H1 and -H1) plus the internal cross-loops through G9 and the second G1 branch, and noting no two loops here are non-touching (they all share at least one common node in the compact diamond topology), Δ reduces to a first-order sum of loop gains only:"},{"type":"math","expression":"\\Delta = 1 - (L_1+L_2+L_3+L_4)","displayMode":true},{"type":"paragraph","content":"and since both forward paths touch all loops, Δ1 = Δ2 = 1, giving the final result:"},{"type":"math","expression":"\\dfrac{C(s)}{R(s)} = \\dfrac{G_2G_4G_6 + G_3G_5G_1}{1-(L_1+L_2+L_3+L_4)}","displayMode":true},{"type":"paragraph","content":"with the loop gains L1 through L4 substituted from the -H1 top/bottom feedback paths and the G9-based cross-coupling branches as read off the given figure."}]}},{"id":"control-system-2024-5E1782-pC-q2","number":2,"marks":10,"tags":["Part C"],"text":[{"type":"paragraph","content":"Q.2. The open loop transfer function of a unity feedback control system is :"},{"type":"math","expression":"G(s) = \\dfrac{K}{s(1+sT)}","displayMode":true},{"type":"paragraph","content":"By what factor should the gain K be multiplied so that damping ratio increases from 0.3 to 0.69?"}],"answer":{"summary":"For G(s)=K/[s(1+sT)], damping ratio ζ is inversely proportional to √K, so raising ζ from 0.3 to 0.69 requires multiplying K by (0.3/0.69)² ≈ 0.189, i.e. the gain must be reduced to about 18.9% of its original value.","explanation":[{"type":"paragraph","content":"The closed-loop transfer function for unity feedback with G(s) = K/[s(1+sT)] is:"},{"type":"math","expression":"\\dfrac{C(s)}{R(s)} = \\dfrac{G(s)}{1+G(s)} = \\dfrac{K/T}{s^2 + \\dfrac{1}{T}s + \\dfrac{K}{T}}","displayMode":true},{"type":"paragraph","content":"Comparing with the standard second-order form s² + 2ζωn s + ωn²:"},{"type":"math","expression":"\\omega_n^2 = \\dfrac{K}{T}, \\qquad 2\\zeta\\omega_n = \\dfrac{1}{T}","displayMode":true},{"type":"paragraph","content":"From these, eliminating ωn:"},{"type":"math","expression":"\\omega_n = \\sqrt{\\dfrac{K}{T}} \\;\\Rightarrow\\; 2\\zeta\\sqrt{\\dfrac{K}{T}} = \\dfrac{1}{T} \\;\\Rightarrow\\; \\zeta = \\dfrac{1}{2\\sqrt{KT}}","displayMode":true},{"type":"paragraph","content":"This shows ζ ∝ 1/√K for fixed T, i.e. K ∝ 1/ζ². If K1 corresponds to ζ1 = 0.3 and K2 corresponds to ζ2 = 0.69, then:"},{"type":"math","expression":"\\dfrac{K_2}{K_1} = \\left(\\dfrac{\\zeta_1}{\\zeta_2}\\right)^2 = \\left(\\dfrac{0.3}{0.69}\\right)^2 = (0.4348)^2 = 0.189","displayMode":true},{"type":"paragraph","content":"Hence the gain K must be multiplied by a factor of approximately 0.189 (i.e., reduced to about 18.9% of its original value) to raise the damping ratio from 0.3 (lightly damped, large overshoot) to 0.69 (close to optimal damping with minimal overshoot). This inverse-square relationship between K and ζ is a standard and important result for Type-1 second-order servo systems: increasing gain always reduces damping (and vice versa), which is why gain and transient response cannot be tuned independently without additional compensation."},{"type":"paragraph","content":"**Effect on transient performance.** To appreciate why this gain reduction is desirable, consider the percentage peak overshoot formula for a unit-step input to a standard underdamped second-order system:"},{"type":"math","expression":"\\%OS = e^{-\\pi\\zeta/\\sqrt{1-\\zeta^2}} \\times 100","displayMode":true},{"type":"paragraph","content":"At the original damping ratio ζ1 = 0.3, the exponent is -π(0.3)/√(1-0.09) = -0.943/0.954 = -0.988, giving %OS = e^(-0.988)×100 ≈ 37.2% — a large, often unacceptable overshoot for a precision positioning servo. At the improved damping ratio ζ2 = 0.69, the exponent is -π(0.69)/√(1-0.4761) = -2.168/0.7238 = -2.995, giving %OS = e^(-2.995)×100 ≈ 5.0% — a much more acceptable, lightly-oscillatory response close to the commonly cited 'optimal' engineering damping range of ζ ≈ 0.6–0.7, which balances a reasonably fast response against minimal overshoot. This concretely demonstrates why a designer would deliberately choose to reduce the loop gain K to about 18.9% of its original value: the resulting seven-fold reduction in peak overshoot (from ≈37% down to ≈5%) is a dramatic improvement in transient quality, illustrating the fundamental gain-versus-damping trade-off inherent to simple proportional-gain (uncompensated) Type-1 servo systems, and motivating the use of dedicated compensators (as discussed elsewhere in this paper) when both high steady-state accuracy (which favors high K) and low overshoot (which favors low K) are simultaneously required."},{"type":"paragraph","content":"**Settling time consideration.** It is also worth checking the settling time impact of this gain change, since reducing K reduces ωn (as ωn = √(K/T)) even as ζ increases. Using the 2% settling time approximation ts ≈ 4/(ζωn), and noting that the product ζωn = 1/(2T) is actually independent of K (from the earlier relation 2ζωn = 1/T, which holds regardless of the specific K chosen), the settling time ts ≈ 8T remains unchanged by this gain adjustment. This is a subtle but important point: for this particular Type-1 configuration, reducing the gain to improve damping and overshoot does not come at the cost of a slower settling time, because the real part of the closed-loop poles (which governs the settling time via its reciprocal) is fixed by T alone and does not depend on K. What does change is the damped natural frequency ωd = ωn√(1-ζ²) and hence the peak time tp = π/ωd, which increases somewhat as ζ increases (fewer, slower oscillations before settling), but the overall envelope of decay to within 2% of the final value is unaffected — making this an unusually 'free' improvement in transient quality achievable simply by re-tuning the proportional gain."}]}},{"id":"control-system-2024-5E1782-pC-q3","number":3,"marks":10,"tags":["Part C"],"text":[{"type":"paragraph","content":"Q.3. A unit feedback system is described by :"},{"type":"math","expression":"G(s)H(s) = \\dfrac{10}{s(1+0.2s)(1+0.01s)}","displayMode":true},{"type":"paragraph","content":"Construct a Bode plot and find : (a) Gain and Phase crossover frequency (b) Gain and Phase margin (c) Stability of closed loop system"}],"answer":{"summary":"For G(s)H(s) = 10/[s(1+0.2s)(1+0.01s)], the Bode plot gives gain crossover frequency ≈ 6.9 rad/s, phase crossover frequency = √(1/0.002) ≈ 22.4 rad/s, gain margin ≈ 12.2 dB, phase margin ≈ 42°, and the closed-loop system is stable.","explanation":[{"type":"paragraph","content":"This transfer function is already in standard Bode time-constant form, with an overall DC gain K = 10 acting through a single integrator (pole at the origin, giving Type-1 behavior) followed by two simple real poles. To sketch the asymptotic Bode magnitude plot, each factor contributes independently: the 1/(jω) term contributes a straight line of slope -20 dB/decade passing through 0 dB at ω = K = 10 rad/s (if it acted alone); each simple pole 1/(1+jωτ) contributes 0 dB/decade below its corner frequency ω = 1/τ and -20 dB/decade above it. The overall asymptotic magnitude curve is obtained by summing these individual straight-line segments at every frequency, while the overall phase curve is obtained similarly by summing each factor's phase contribution (-90° constant for the integrator, and a smooth -0° to -90° arctangent transition centered at each pole's corner frequency for the two simple poles)."},{"type":"paragraph","content":"The open-loop transfer function has corner frequencies at ω1 = 1/0.2 = 5 rad/s and ω2 = 1/0.01 = 100 rad/s, plus a pole at the origin (Type-1 system):"},{"type":"math","expression":"G(j\\omega)H(j\\omega) = \\dfrac{10}{j\\omega(1+j0.2\\omega)(1+j0.01\\omega)}","displayMode":true},{"type":"paragraph","content":"The asymptotic magnitude plot starts at -20 dB/decade (due to the 1/jω pole) from a 0 dB crossing determined by K = 10, then breaks to -40 dB/decade at ω = 5 rad/s, and to -60 dB/decade at ω = 100 rad/s."},{"type":"paragraph","content":"**Gain crossover frequency (ωgc):** where |G(jω)H(jω)| = 1 (0 dB). In the region between the two corner frequencies, magnitude ≈ 10/(ω × 0.2ω) = 50/ω² (using the -40 dB/decade asymptote approximation beyond ω1 = 5). Setting this to 1: ω² = 50, ωgc ≈ 7.07 rad/s (checking this lies between 5 and 100, valid). Refining with the exact magnitude expression at ω = 6.9 rad/s gives |GH| ≈ 1.0, confirming ωgc ≈ 6.9 rad/s."},{"type":"paragraph","content":"**Phase crossover frequency (ωpc):** where phase = -180°. Total phase is:"},{"type":"math","expression":"\\angle GH = -90^\\circ - \\tan^{-1}(0.2\\omega) - \\tan^{-1}(0.01\\omega)","displayMode":true},{"type":"paragraph","content":"Setting this equal to -180° requires tan⁻¹(0.2ω) + tan⁻¹(0.01ω) = 90°, which (using the tangent addition identity, this occurs when the product of the two arguments approaches the point where the combined arctangents sum to 90°) solves numerically to ωpc ≈ 22.4 rad/s (verified: at ω=22.4, tan⁻¹(4.48)=77.4°, tan⁻¹(0.224)=12.6°, sum=90.0° ✓)."},{"type":"paragraph","content":"**Gain margin:** magnitude at ωpc = 22.4:"},{"type":"math","expression":"|GH(j22.4)| = \\dfrac{10}{22.4\\sqrt{1+(0.2\\times22.4)^2}\\sqrt{1+(0.01\\times22.4)^2}} = \\dfrac{10}{22.4\\times4.55\\times1.025} \\approx 0.0993","displayMode":true},{"type":"math","expression":"GM = -20\\log_{10}(0.0993) \\approx 20.06\\ \\text{dB}","displayMode":true},{"type":"paragraph","content":"**Phase margin:** phase at ωgc = 6.9 rad/s:"},{"type":"math","expression":"\\angle GH(j6.9) = -90^\\circ - \\tan^{-1}(1.38) - \\tan^{-1}(0.069) = -90^\\circ-54.1^\\circ-3.9^\\circ=-148^\\circ","displayMode":true},{"type":"math","expression":"PM = 180^\\circ - 148^\\circ = 32^\\circ","displayMode":true},{"type":"paragraph","content":"**Stability:** Since both the gain margin (≈ 20 dB, positive) and phase margin (≈ 32°, positive) are positive, the closed-loop system is stable, though the phase margin of 32° indicates a moderately, not highly, damped response (some overshoot expected, roughly ζ ≈ PM/100 = 0.32)."},{"type":"paragraph","content":"**Practical significance.** The large gap between ωgc (≈6.9 rad/s) and ωpc (≈22.4 rad/s) in this design reflects a comfortably stable system with substantial margin before instability, which is desirable in practice because real plants always carry unmodeled dynamics (extra small time constants in sensors, actuators, or structural resonances) that add further phase lag beyond what the nominal model predicts. A gain margin of roughly 20 dB means the loop gain could be increased by a factor of about 10 before the system reaches the stability boundary, providing generous tolerance for component aging, temperature drift, or manufacturing variation in the actual hardware gain. Similarly, the phase margin of about 32° — while positive and hence formally stable — sits below the commonly recommended design target of 45°-60° for well-damped industrial servo loops, so a designer reviewing this Bode plot would likely still consider adding a mild lead compensator to push the phase margin higher, trading some of the generous gain margin for improved transient damping and reduced overshoot, since gain margin and phase margin are not independent — reshaping the loop to improve one often affects the other."}]}},{"id":"control-system-2024-5E1782-pC-q4","number":4,"marks":10,"tags":["Part C"],"text":[{"type":"paragraph","content":"Q.4. Given the unity feedback system, make an accurate plot of the root locus for the following system :"},{"type":"math","expression":"G(s) = \\dfrac{K^2(s^2 - 2s + 2)}{(s+1)(s+2)}","displayMode":true},{"type":"paragraph","content":"Calibrate the gain for at least four points. Also find the breakaway points, the jω axis crossing and range of gain for stability. Find angle of arrival also."}],"answer":{"summary":"For G(s)=K²(s²-2s+2)/[(s+1)(s+2)], the root locus has 2 poles (-1,-2), 2 zeros (1±j1); breakaway/entry points occur on the real axis between the poles, the loci cross the jω-axis at specific points found via the Routh array, and the angle of arrival at the complex zero is computed from the angle condition.","explanation":[{"type":"paragraph","content":"Open-loop poles are at s = -1 and s = -2 (on the real axis). Open-loop zeros are the roots of s² - 2s + 2 = 0, i.e. s = 1 ± j1 (complex, in the right-half plane, since s²-2s+2=0 gives s=(2±√(4-8))/2=1±j1)."},{"type":"paragraph","content":"**Number of branches and asymptotes:** number of poles n = 2, zeros m = 2, so n - m = 0; there are no asymptotes going to infinity — both branches terminate on the finite zeros at 1±j1."},{"type":"paragraph","content":"**Real-axis segments:** the segment between s = -1 and s = -2 lies on the root locus (odd number of poles/zeros to its right, namely one pole at -1)."},{"type":"paragraph","content":"**Breakaway/break-in points:** found from dK/ds = 0 where the characteristic equation is 1 + K²(s²-2s+2)/[(s+1)(s+2)] = 0, i.e."},{"type":"math","expression":"K^2 = -\\dfrac{(s+1)(s+2)}{s^2-2s+2}","displayMode":true},{"type":"paragraph","content":"Differentiating the right side with respect to s and setting the derivative to zero (standard breakaway condition d/ds[(s+1)(s+2)/(s²-2s+2)] = 0) and solving the resulting polynomial numerically over the real-axis segment (-2,-1) gives a breakaway point at approximately s ≈ -1.45 (the unique real root of the derivative equation lying within the valid real-axis locus segment)."},{"type":"paragraph","content":"**Calibration of gain at four points:** using the magnitude condition |K²(s²-2s+2)| = |(s+1)(s+2)| at test points along the locus. At s = -1.45 (breakaway): numerator |(-1.45)²-2(-1.45)+2| = |2.1+2.9+2| = 7.0, denominator |(-0.45)(0.55)| = 0.2475, so K² = 0.2475/7.0 = 0.0354, K ≈ 0.188. At s = -1.2: numerator |1.44+2.4+2|=5.84, denom |(-0.2)(0.8)|=0.16, K²=0.0274, K≈0.166. At s=-1.7: numerator |2.89+3.4+2|=8.29, denom|(-0.7)(0.3)|=0.21,K²=0.0253,K≈0.159. At s=-1.0 (pole, K=0) and s=-2 (pole, K=0) mark the start points of the two loci branches on the real axis, each moving toward the breakaway point before departing into the complex plane toward the zeros at 1±j1."},{"type":"paragraph","content":"**jω-axis crossing:** substitute s = jω into the characteristic equation 1 + K²(s²-2s+2)/[(s+1)(s+2)] = 0, i.e. (s+1)(s+2) + K²(s²-2s+2) = 0:"},{"type":"math","expression":"(j\\omega+1)(j\\omega+2) + K^2\\left[(j\\omega)^2-2j\\omega+2\\right] = 0","displayMode":true},{"type":"math","expression":"(-\\omega^2+3j\\omega+2) + K^2(-\\omega^2-2j\\omega+2) = 0","displayMode":true},{"type":"paragraph","content":"Separating real and imaginary parts: Real: -ω²+2 + K²(-ω²+2) = 0 ⟹ (2-ω²)(1+K²) = 0. Since (1+K²) ≠ 0, this requires ω² = 2, i.e. ω = ±√2. Imaginary: 3ω - 2K²ω = 0 ⟹ ω(3-2K²)=0. For ω = √2 ≠ 0: K² = 1.5, K = 1.225. So the locus crosses the imaginary axis at ω = ±√2 ≈ ±1.414 rad/s when K ≈ 1.225, and this is also the gain range boundary for stability."},{"type":"paragraph","content":"**Range of gain for stability:** since the loci originate at the stable poles (-1, -2) and move toward the zeros (1±j1, which lie in the right-half plane), the closed-loop poles cross into the RHP once K exceeds the value found above. Hence the system is stable for 0 ≤ K < 1.225 and becomes unstable for K > 1.225 (marginally stable exactly at K = 1.225)."},{"type":"paragraph","content":"**Angle of arrival at the complex zero (1+j1):** using the angle condition, angle of arrival at a zero = 180° + Σ(angles from poles to zero) - Σ(angles from other zeros to this zero). Angles from poles -1 and -2 to the zero (1+j1): from -1: angle = tan⁻¹(1/2) = 26.57°; from -2: angle = tan⁻¹(1/3) = 18.43°. Angle from the other zero (1-j1) to (1+j1): this is a vertical line, angle = 90°. Angle of arrival:"},{"type":"math","expression":"\\theta_{arr} = 180^\\circ + (26.57^\\circ+18.43^\\circ) - 90^\\circ = 180^\\circ+45^\\circ-90^\\circ=135^\\circ","displayMode":true},{"type":"svg","content":""}]}},{"id":"control-system-2024-5E1782-pC-q5","number":5,"marks":10,"tags":["Part C"],"text":[{"type":"paragraph","content":"Q.5. Write short notes on the following : (a) Optimal control (b) Non-linear control"}],"answer":{"summary":"Optimal control designs the input that minimizes a defined performance index (e.g., quadratic cost, LQR) subject to system dynamics and constraints; non-linear control deals with systems violating superposition/homogeneity, using techniques like feedback linearization, sliding mode, and Lyapunov-based design instead of standard linear tools.","explanation":[{"type":"subheading","content":"(a) Optimal Control"},{"type":"paragraph","content":"Optimal control theory addresses the problem of determining the control input u(t) for a dynamic system, described typically by state equations ẋ = f(x,u,t), that minimizes (or maximizes) a specified performance index (cost functional) J, subject to the system dynamics as equality constraints and possibly additional inequality constraints on states/inputs. A widely used performance index in linear systems is the quadratic cost:"},{"type":"math","expression":"J = \\int_0^{\\infty} \\left(x^T Q x + u^T R u\\right)\\,dt","displayMode":true},{"type":"paragraph","content":"where Q and R are positive semi-definite/positive-definite weighting matrices that penalize state deviation and control effort respectively. For a linear time-invariant plant ẋ = Ax + Bu, minimizing this cost leads to the Linear Quadratic Regulator (LQR), whose optimal control law is a state feedback u = -Kx, where K = R⁻¹BᵀP and P is the unique positive-definite solution of the Algebraic Riccati Equation:"},{"type":"math","expression":"A^TP + PA - PBR^{-1}B^TP + Q = 0","displayMode":true},{"type":"paragraph","content":"Optimal control formalizes the trade-off between fast/accurate regulation (large Q) and conservative, low-effort actuation (large R), and is derived using variational calculus (Pontryagin's Minimum Principle) or dynamic programming (Bellman's Principle of Optimality/Hamilton-Jacobi-Bellman equation). It is applied extensively in aerospace guidance, robotics trajectory optimization, and economic/process control where minimizing energy, time, or fuel consumption is as important as achieving the desired state."},{"type":"subheading","content":"(b) Non-linear Control"},{"type":"paragraph","content":"Non-linear control deals with systems whose governing differential equations do not satisfy the superposition and homogeneity properties assumed by classical linear control theory (transfer functions, Laplace transforms, Bode/root-locus techniques do not directly apply). Common sources of non-linearity include saturation, dead-zone, backlash, hysteresis, and Coulomb friction in actuators, as well as intrinsically non-linear plant dynamics (e.g. pendulum dynamics involving sin θ, or bilinear/product terms between states). Because closed-form transfer functions generally do not exist, non-linear systems are analyzed and controlled using dedicated techniques such as: phase-plane analysis and the describing function method (for approximate frequency-domain analysis of mild non-linearities and limit-cycle prediction); Lyapunov's direct method, which establishes stability by finding an energy-like function V(x) that is positive definite with a negative semi-definite derivative along system trajectories, without needing to solve the non-linear differential equations explicitly; feedback linearization, which uses a non-linear state transformation and non-linear inner-loop control law to cancel the plant's non-linear terms and render the closed-loop dynamics linear, so that standard linear design can then be applied to the outer loop; and sliding-mode control, which drives system trajectories onto a designed switching surface and keeps them there despite model uncertainty or external disturbances, offering strong robustness at the cost of high-frequency chattering in the control signal. Non-linear control is essential in robotics, aircraft flight control, power electronics converters, and any system where linearized small-signal models fail to capture large-signal or highly non-linear plant behavior."},{"type":"paragraph","content":"A key qualitative distinction from linear systems is that non-linear systems can exhibit behaviors with no linear counterpart, such as finite escape time (the state diverging to infinity in finite time), multiple isolated equilibrium points, and self-sustained limit cycles (bounded periodic oscillations that persist independent of initial conditions, unlike the marginally stable oscillations of a linear system which depend entirely on the initial energy injected). Because linear stability tools (root locus, Bode, Nyquist) are built on the assumption of a single, globally valid transfer function, they can only be applied to a non-linear plant after linearizing about a specific operating point (using a Jacobian/small-signal approximation), and such a linearized model is only locally valid near that operating point — this is precisely why genuinely non-linear analysis and design techniques, as summarized above, remain indispensable wherever a system must operate correctly across a wide range of conditions rather than in the small neighborhood of a single set-point."}]}}],"examYear":2024},{"id":"control-system-2022-5E1393","title":"Control System (2022)","subjectSlug":"control-system","subjectName":"Control System","universitySlug":"rtu","branchSlug":"ee-ec-eee","year":3,"semester":5,"paperCode":"5E1393","questions":[{"id":"control-system-2022-5E1393-pA-q1","number":1,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.1. Define forward path and forward path gain."}],"answer":{"summary":"A forward path in a signal flow graph is a path from the input (source) node to the output (sink) node along which no node is traversed more than once; forward path gain is the product of the branch gains along that path.","explanation":[{"type":"paragraph","content":"In a signal flow graph, a forward path is any continuous path traced from the input (source) node to the output (sink) node, following the direction of the arrows, such that no node is encountered more than once along the path. The forward path gain Pk is obtained by multiplying together all the individual branch transmittances (gains) that lie along that particular path, and this quantity is a key term used directly in the numerator of Mason's Gain Formula when computing the overall input-output transfer function of the system."}]}},{"id":"control-system-2022-5E1393-pA-q2","number":2,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.2. Differentiate closed loop and open loop control system."}],"answer":{"summary":"A closed loop (feedback) control system measures its output and feeds it back to compare with the reference, correcting error automatically; an open loop system has no feedback path and its output does not influence the control action.","explanation":[{"type":"paragraph","content":"An open-loop control system generates its output purely as a function of the given input and a predetermined control action, with no path to measure the actual output and feed it back for comparison; hence it cannot correct for disturbances or parameter drift (e.g. a simple washing machine timer). A closed-loop control system, by contrast, continuously measures the output via a sensor and feeds it back to be compared with the reference input, generating an error signal that drives the controller; this makes closed-loop systems self-correcting, more accurate, and less sensitive to disturbances and parameter variations, but at the cost of added complexity and the possibility of instability if not properly designed."}]}},{"id":"control-system-2022-5E1393-pA-q3","number":3,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.3. State Mason's gain formula."}],"answer":{"summary":"Mason's gain formula gives the overall transfer function as T = (1/Δ)·ΣPkΔk, where Pk is each forward path gain, Δ = 1 − Σ(loop gains) + Σ(products of non-touching loop pairs) − ..., and Δk is Δ evaluated after removing loops touching path k.","explanation":[{"type":"paragraph","content":"Mason's gain formula computes the overall gain (transfer function) between an input and output node of a signal flow graph as:"},{"type":"math","expression":"T = \\dfrac{C}{R} = \\dfrac{1}{\\Delta}\\sum_{k} P_k \\Delta_k","displayMode":true},{"type":"paragraph","content":"where Pk is the gain of the k-th forward path, Δ (the graph determinant) = 1 − (sum of all individual loop gains) + (sum of gain products of all possible pairs of non-touching loops) − (sum of gain products of all possible triples of non-touching loops) + ⋯, and Δk is the value of Δ computed by deleting from the graph all loops that touch (share a node with) the k-th forward path."}]}},{"id":"control-system-2022-5E1393-pA-q4","number":4,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.4. What is meant by steady state error?"}],"answer":{"summary":"Steady state error is the difference between the reference input and the actual output as time approaches infinity, i.e. ess = lim(t→∞)[r(t)-c(t)], reflecting the system's inherent tracking inaccuracy after all transients have decayed.","explanation":[{"type":"paragraph","content":"Steady state error ess is defined as the value that the error signal e(t) = r(t) - c(t) settles to as time t tends to infinity, once all transient components of the response have decayed to zero. It is computed conveniently using the final value theorem applied to the error transform: ess = lim(s→0) s·E(s), and depends on the system type (number of poles at the origin of the open-loop transfer function) and the type of input applied (step, ramp, or parabolic)."}]}},{"id":"control-system-2022-5E1393-pA-q5","number":5,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.5. List the main properties of a state transition matrix."}],"answer":{"summary":"The state transition matrix φ(t) satisfies φ(0)=I, is non-singular with φ⁻¹(t)=φ(-t), obeys the semigroup property φ(t2-t1)φ(t1-t0)=φ(t2-t0), and satisfies the differential equation dφ/dt = Aφ(t).","explanation":[{"type":"list","items":["**Identity at t = 0:** φ(0) = I, the identity matrix.","**Non-singularity/invertibility:** φ(t) is always invertible, and φ⁻¹(t) = φ(-t).","**Semigroup (transition) property:** φ(t2 - t0) = φ(t2 - t1)·φ(t1 - t0) for any t0 < t1 < t2, meaning transitions can be composed.","**Differential equation:** φ(t) satisfies dφ(t)/dt = A φ(t), and by definition φ(t) = e^{At} = L⁻¹[(sI-A)⁻¹]."]}]}},{"id":"control-system-2022-5E1393-pA-q6","number":6,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.6. State the necessary condition for stability."}],"answer":{"summary":"The necessary (but not sufficient) condition for stability is that all coefficients of the characteristic equation must be present (non-zero) and must have the same sign; if any coefficient is missing or of different sign, the system is definitely unstable.","explanation":[{"type":"paragraph","content":"For a characteristic equation a0sⁿ + a1sⁿ⁻¹ + ⋯ + an = 0, a necessary condition for all roots to lie in the left-half s-plane (stability) is that all the coefficients a0, a1, …, an must be real, non-zero (no missing/zero terms), and of the same sign. This condition is necessary but not sufficient — satisfying it does not guarantee stability, but violating it (a missing term or a sign change) immediately guarantees instability, without needing to construct the full Routh array."}]}},{"id":"control-system-2022-5E1393-pA-q7","number":7,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.7. What are the characteristics of phase-lead network?"}],"answer":{"summary":"A phase-lead network provides positive (leading) phase angle over a specific frequency range, has its zero closer to the origin than its pole, increases bandwidth and speeds up transient response, but amplifies high-frequency noise.","explanation":[{"type":"paragraph","content":"Key characteristics of a phase-lead network: it introduces a positive phase shift (phase lead) at frequencies between its zero and pole locations; on the pole-zero plot, its zero lies closer to the origin than its pole (i.e., |zero| < |pole|); it increases the gain crossover frequency and hence the closed-loop bandwidth, speeding up the transient response and reducing rise/settling time; it improves the phase margin, improving relative stability; and it acts approximately as a high-pass filter, so it increases susceptibility to high-frequency measurement noise."}]}},{"id":"control-system-2022-5E1393-pA-q8","number":8,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.8. What is dominant pole?"}],"answer":{"summary":"A dominant pole is the closed-loop pole (or complex-conjugate pole pair) closest to the imaginary axis, whose contribution decays slowest and therefore dominates the overall transient response, allowing a higher-order system to be approximated by a lower-order (often second-order) model.","explanation":[{"type":"paragraph","content":"In a system with multiple closed-loop poles, the dominant pole(s) are those located nearest to the jω-axis (i.e., having the smallest real-part magnitude) compared to the other poles, which are typically located much farther to the left (at least 5 times the real part of the dominant pole, by common rule of thumb). Because the transient response terms associated with far-left poles decay much faster (e^(-large·t)) than those from the dominant pole(s), the dominant pole's exponential/oscillatory term persists longest and effectively governs the overall shape, settling time, and overshoot of the system response, permitting an accurate approximation of a high-order system by a simpler first- or second-order model based on the dominant pole(s) alone."}]}},{"id":"control-system-2022-5E1393-pA-q9","number":9,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.9. What are constant M and N circle?"}],"answer":{"summary":"Constant M and N circles are loci on the G(jω) plane representing constant closed-loop magnitude (M circles) and constant closed-loop phase (N circles) for a unity feedback system, used in the Nichols/Nyquist-based design of closed-loop frequency response.","explanation":[{"type":"paragraph","content":"For a unity-feedback system, constant-M circles are loci of points in the G(jω)-plane (real vs imaginary axes) along which the magnitude of the closed-loop transfer function |C/R| = |G/(1+G)| remains constant at a specified value M; each value of M traces a distinct circle (centered on the real axis, except M=1 which is a straight vertical line). Similarly, constant-N circles are loci along which the closed-loop phase angle ∠(C/R) remains constant at a specified value N; these circles pass through the origin and the point (-1, j0). Overlaying the open-loop polar plot G(jω) on this M-N circle grid (the basis of the Nichols chart) allows direct graphical readout of closed-loop magnitude and phase response, and hence resonance peak Mr, resonant frequency ωr, and bandwidth, without algebraically deriving the closed-loop transfer function."}]}},{"id":"control-system-2022-5E1393-pA-q10","number":10,"marks":2,"tags":["Part A"],"text":[{"type":"paragraph","content":"Q.10. What is the value of gain K at any given point on the root locus?"}],"answer":{"summary":"At any point on the root locus, the gain K is found from the magnitude condition K = (product of distances from that point to all open-loop poles) / (product of distances from that point to all open-loop zeros).","explanation":[{"type":"paragraph","content":"For a unity-feedback characteristic equation 1 + KG(s)H(s) = 0, at any test point s1 that lies on the root locus (i.e., satisfying the angle condition), the corresponding value of gain K is obtained from the magnitude condition:"},{"type":"math","expression":"K = \\dfrac{\\prod_{i=1}^{n}|s_1 - p_i|}{\\prod_{j=1}^{m}|s_1 - z_j|}","displayMode":true},{"type":"paragraph","content":"i.e., K equals the product of the vector lengths from s1 to each open-loop pole pi, divided by the product of the vector lengths from s1 to each open-loop zero zj, measured directly on the s-plane (graphically, using a ruler/protractor on the root locus plot, or analytically by direct substitution)."}]}},{"id":"control-system-2022-5E1393-pB-q1","number":1,"marks":8,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.1. Find the transfer function of the system shown in fig. using block diagram reduction technique (a multi-loop diagram with forward blocks G1, G2, G3 in cascade from R(s) to C(s), an inner positive feedback via H2 around G2-G3, an outer feedback via H3 from after G2G3 back to before G2, and an outermost negative feedback via H1 around the entire forward path back to the input summing junction)."}],"answer":{"summary":"Reducing the block diagram step-by-step (combining cascaded G2G3, absorbing the inner H2 feedback loop, moving/absorbing the H3 feedback around G2G3, and finally reducing the outer H1 feedback loop) yields the overall transfer function C(s)/R(s) = G1G2G3 / [1 + G2G3H2 + G2G3H3 + G1G2G3H1].","explanation":[{"type":"paragraph","content":"The block diagram consists of a forward path R(s) → G1 → G2 → G3 → C(s), with three feedback paths: an inner positive feedback H2 taken from after G3 back to before G2 (around G2-G3), a feedback H3 taken from after G2G3 back to a summing point before G2 as well, and an outer negative feedback H1 taken from the output C(s) all the way back to the primary input summing junction (around the entire forward path G1G2G3)."},{"type":"paragraph","content":"**Step 1 — Combine the inner feedback loops around G2-G3:** Since both H2 (positive/regenerative, based on the arrow into the summing junction before G2) and H3 are feedback paths around the block G2G3 acting at the same summing point before G2, they combine into an equivalent single feedback path (H2 + H3) or (H2 − H3) around G2G3 depending on their signs at the summer, feeding into the point after G1's output. Taking the diagram's H2 as positive feedback (regenerative) and H3 as negative feedback (as is conventional at that junction), the combined inner loop reduces G2G3 to:"},{"type":"math","expression":"G_{23,eq}(s) = \\dfrac{G_2G_3}{1 - G_2G_3H_2 + G_2G_3H_3} = \\dfrac{G_2G_3}{1+G_2G_3(H_3-H_2)}","displayMode":true},{"type":"paragraph","content":"**Step 2 — Combine with G1 in cascade:** The equivalent block G23,eq is in cascade with G1, giving a combined forward-path gain:"},{"type":"math","expression":"G_{eq}(s) = G_1 \\cdot G_{23,eq}(s) = \\dfrac{G_1G_2G_3}{1+G_2G_3(H_3-H_2)}","displayMode":true},{"type":"paragraph","content":"**Step 3 — Reduce the outer feedback loop H1:** This combined forward block is now in a standard single negative-feedback loop with H1 around the entire path from R(s) to C(s):"},{"type":"math","expression":"\\dfrac{C(s)}{R(s)} = \\dfrac{G_{eq}(s)}{1+G_{eq}(s)H_1} = \\dfrac{\\dfrac{G_1G_2G_3}{1+G_2G_3(H_3-H_2)}}{1+\\dfrac{G_1G_2G_3H_1}{1+G_2G_3(H_3-H_2)}}","displayMode":true},{"type":"paragraph","content":"Multiplying numerator and denominator by [1 + G2G3(H3−H2)] to clear the compound fraction, the final simplified transfer function is:"},{"type":"math","expression":"\\dfrac{C(s)}{R(s)} = \\dfrac{G_1G_2G_3}{1+G_2G_3(H_3-H_2)+G_1G_2G_3H_1}","displayMode":true},{"type":"paragraph","content":"This result illustrates the systematic block-diagram-reduction technique: (i) collapse cascaded blocks by multiplying their gains, (ii) collapse an inner feedback loop using G/(1∓GH), and (iii) repeat outward until only a single forward block remains in a single feedback loop with the outermost feedback element, at which point the standard closed-loop formula G/(1+GH) yields the final answer."},{"type":"paragraph","content":"**Alternative check via Mason's Gain Formula.** As a cross-check, treating the block diagram as a signal flow graph gives one forward path with gain G1G2G3, and loop gains L1 = G2G3H2 (positive feedback, hence entering Δ with opposite sign), L2 = -G2G3H3, and L3 = -G1G2G3H1 (all sharing common nodes with the forward path and with each other, so no non-touching loop products exist). Substituting into Δ = 1 - ΣLi gives Δ = 1 - G2G3H2 + G2G3H3 + G1G2G3H1, and with Δ1 = 1 (the single forward path touches every loop), Mason's formula directly returns C/R = G1G2G3/[1 + G2G3(H3-H2) + G1G2G3H1], matching the block-diagram-reduction result exactly and confirming the derivation."}]}},{"id":"control-system-2022-5E1393-pB-q2","number":2,"marks":8,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.2. Obtain the differential equations describing the complete dynamics of the mechanical system shown in figure (a two-mass translational spring-mass-damper system: mass M1 with applied force F(t) and displacement X1 connected to a fixed wall through spring K1 and damper B1, and to mass M2 (displacement X2) through the same K1-B1 pair; M2 is connected to a fixed ceiling through spring K2 and damper B2)."}],"answer":{"summary":"Applying Newton's second law to each mass in the two-mass mechanical network gives two coupled second-order differential equations: M1ẍ1 + B1(ẋ1-ẋ2) + K1(x1-x2) = F(t), and M2ẍ2 + B2ẋ2 + K2x2 + B1(ẋ2-ẋ1) + K1(x2-x1) = 0.","explanation":[{"type":"paragraph","content":"The mechanical system consists of mass M1 (displacement x1, driven by external force F(t)) connected to a fixed wall through spring K1 and damper B1, with M1 also connected through the same K1 and B1 to mass M2 (displacement x2); M2 in turn is connected to a fixed upper support through a separate spring K2 and damper B2."},{"type":"paragraph","content":"**Free body diagram for M1:** The forces acting on M1 are the applied force F(t), the spring force from K1 (proportional to the relative displacement between M1 and M2, since K1 connects them: force = K1(x1 - x2), opposing extension), and the damping force from B1 (proportional to relative velocity: B1(ẋ1 - ẋ2)). Applying Newton's second law (ΣF = Mẍ):"},{"type":"math","expression":"M_1\\ddot{x}_1 + B_1(\\dot{x}_1-\\dot{x}_2) + K_1(x_1-x_2) = F(t)","displayMode":true},{"type":"paragraph","content":"**Free body diagram for M2:** M2 experiences the reaction spring/damper forces from K1, B1 (pulling it toward M1, i.e., proportional to (x2-x1) and (ẋ2-ẋ1)) plus its own spring K2 and damper B2 connecting it to the fixed upper support (proportional to its absolute displacement x2 and velocity ẋ2, since the far end is fixed). No external force acts directly on M2. Applying Newton's second law:"},{"type":"math","expression":"M_2\\ddot{x}_2 + B_2\\dot{x}_2 + K_2x_2 + B_1(\\dot{x}_2-\\dot{x}_1) + K_1(x_2-x_1) = 0","displayMode":true},{"type":"paragraph","content":"These two coupled linear second-order ODEs completely describe the dynamics of the two-mass, two-spring, two-damper mechanical translational system. They form a direct electrical-mechanical analogy (force-voltage or force-current analogy) with a two-loop RLC electrical network, where mass M corresponds to inductance L (or capacitance, depending on analogy), damping B corresponds to resistance R, and stiffness K corresponds to the reciprocal of capacitance 1/C (or inductance), which is the standard basis for representing mechanical systems using equivalent electrical circuit analysis and transfer function techniques in control system design."},{"type":"paragraph","content":"**Transfer function form.** Taking the Laplace transform of both equations (assuming zero initial conditions) converts them into an algebraic pair in terms of X1(s), X2(s) and F(s):"},{"type":"math","expression":"\\left[M_1s^2+B_1s+K_1\\right]X_1(s) - \\left[B_1s+K_1\\right]X_2(s) = F(s)","displayMode":true},{"type":"math","expression":"-\\left[B_1s+K_1\\right]X_1(s) + \\left[M_2s^2+(B_1+B_2)s+(K_1+K_2)\\right]X_2(s) = 0","displayMode":true},{"type":"paragraph","content":"These two simultaneous algebraic equations can be solved for X1(s)/F(s) or X2(s)/F(s) using Cramer's rule, giving the transfer function relating the applied force to either mass's displacement as a ratio of polynomials in s — this is the standard procedure by which the physical mechanical system is converted into the s-domain block-diagram or signal-flow-graph representation used throughout the rest of control system analysis and design."}]}},{"id":"control-system-2022-5E1393-pB-q3","number":3,"marks":8,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.3. A servomechanism is represented by the equation :"},{"type":"math","expression":"\\dfrac{d^2\\theta}{dt^2} + 10\\dfrac{d\\theta}{dt} = 150E","displayMode":true},{"type":"paragraph","content":"Where E = (r - θ) is the actuating signal. Calculate the value of damping ratio, undamped frequency of oscillations."}],"answer":{"summary":"Rewriting the servo equation in standard second-order form gives ωn = √150 ≈ 12.25 rad/s and ζ = 10/(2×12.25) ≈ 0.408, indicating an underdamped response.","explanation":[{"type":"paragraph","content":"A servomechanism is a feedback control system in which the controlled variable is a mechanical position, velocity, or acceleration (here, angular position θ), and the actuating signal E is generated by comparing this output against a reference command r. This particular equation is the classic model of a DC position servo (e.g., an armature-controlled DC motor driving a load through a gear train, with a potentiometer or synchro pair providing position feedback): the term 10(dθ/dt) represents the combined viscous friction and back-EMF damping torque of the motor-load system (normalized by inertia), while the term 150E represents the net driving torque, proportional to the actuating error signal through the combined motor-torque-constant and amplifier gain (again normalized by inertia)."},{"type":"paragraph","content":"The servomechanism equation is:"},{"type":"math","expression":"\\dfrac{d^2\\theta}{dt^2} + 10\\dfrac{d\\theta}{dt} = 150E, \\qquad E = r - \\theta","displayMode":true},{"type":"paragraph","content":"Substituting E = r - θ:"},{"type":"math","expression":"\\ddot{\\theta} + 10\\dot{\\theta} + 150\\theta = 150r","displayMode":true},{"type":"paragraph","content":"Comparing the homogeneous (characteristic) part with the standard second-order form s² + 2ζωn s + ωn² = 0, i.e. matching coefficients ẍ + 2ζωnẋ + ωn²x:"},{"type":"math","expression":"\\omega_n^2 = 150 \\;\\Rightarrow\\; \\omega_n = \\sqrt{150} \\approx 12.247\\ \\text{rad/s}","displayMode":true},{"type":"math","expression":"2\\zeta\\omega_n = 10 \\;\\Rightarrow\\; \\zeta = \\dfrac{10}{2\\times 12.247} = \\dfrac{10}{24.495} \\approx 0.408","displayMode":true},{"type":"paragraph","content":"Since 0 < ζ < 1, the system is underdamped, meaning θ(t) will exhibit a decaying oscillatory response to a step change in reference r(t), settling to the final value with some overshoot (approximately 100·e^(-πζ/√(1-ζ²)) % ≈ 100·e^(-1.485) ≈ 22.6% peak overshoot) before stabilizing, with natural (undamped) frequency of oscillation ωn ≈ 12.25 rad/s and damped oscillation frequency ωd = ωn√(1-ζ²) = 12.247×√(1-0.1665) ≈ 11.19 rad/s."},{"type":"paragraph","content":"**Practical relevance.** A damping ratio of ζ ≈ 0.41 places this servo in the moderately underdamped range — noticeably livelier than the 'optimal' engineering range of ζ ≈ 0.6–0.7 typically targeted for position servos, and closer to the more oscillatory end of acceptable designs. In practice, such a system would exhibit a visible, if not excessive, overshoot (≈23%) on step commands, along with several visible oscillation cycles before settling, since the ratio ωd/(ζωn) = 1/tan(cos⁻¹ζ) is relatively large at this damping level. A control engineer reviewing this design would likely recommend increasing the damping term (e.g., by adding tachometer/velocity feedback, which directly adds to the effective 10(dθ/dt) coefficient without altering ωn as strongly as changing the forward gain would) to push ζ closer to the 0.6–0.7 range, reducing overshoot and oscillatory ringing while retaining a fast enough response — a design trade-off that recurs throughout this course whenever a raw, uncompensated servo model is evaluated against desired transient-response specifications."}]}},{"id":"control-system-2022-5E1393-pB-q4","number":4,"marks":8,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.4. Comment on the stability of the system whose characteristic equation is given below :"},{"type":"math","expression":"s^5 + s^4 + 2s^3 + 2s + 4 = 0","displayMode":true}],"answer":{"summary":"Routh's array for s⁵+s⁴+2s³+2s+4=0 produces a zero in the first column at the s³ row (with all-zero replacement needed) and subsequent sign changes, indicating the system is unstable with poles in the right-half plane.","explanation":[{"type":"paragraph","content":"The Routh-Hurwitz criterion determines the number of roots of a polynomial that lie in the right-half s-plane purely from its coefficients, without needing to factor or numerically solve the polynomial. It is constructed as a triangular array: the first two rows are formed directly from the characteristic equation's coefficients (alternating terms in each row), and every subsequent row is computed from the two rows immediately above it using a standard cross-multiplication rule, continuing until the array has as many rows as the polynomial has degree, plus one. The key theorem is that the number of sign changes encountered while reading down the first column of the completed array exactly equals the number of characteristic roots in the right-half plane — zero sign changes confirms stability, while any sign change (or a special case such as a zero appearing in the first column) signals instability or requires special handling."},{"type":"paragraph","content":"Characteristic equation: s⁵ + s⁴ + 2s³ + 0·s² + 2s + 4 = 0 (note the s² coefficient is 0, already a warning sign per the necessary condition for stability). Constructing the Routh array:"},{"type":"math","expression":"\\begin{array}{c|cccc} s^5 & 1 & 2 & 2 \\\\ s^4 & 1 & 0 & 4 \\\\ s^3 & & & \\end{array}","displayMode":true},{"type":"paragraph","content":"Row s³ coefficients: b1 = (1×2 - 1×0)/1 = 2, b2 = (1×2 - 1×4)/1 = -2. Row s³: [2, -2]. Row s² coefficients: c1 = (2×0 - 1×(-2))/2 = (0+2)/2 = 1, c2 = (2×4 - 1×0)/2 = 4. Row s²: [1, 4]. Row s¹: d1 = (1×(-2) - 2×4)/1 = (-2-8)/1 = -10. Row s¹: [-10]. Row s⁰: e1 = 4 (from cross-multiplication with the s² row's second term and zero, effectively carrying down 4)."},{"type":"math","expression":"\\begin{array}{c|cc} s^5 & 1 & 2 \\\\ s^4 & 1 & 4 \\\\ s^3 & 2 & -2 \\\\ s^2 & 1 & 4 \\\\ s^1 & -10 & 0 \\\\ s^0 & 4 & \\end{array}","displayMode":true},{"type":"paragraph","content":"The first column is: 1, 1, 2, 1, -10, 4. There are two sign changes in this column (from +1 at the s² row to -10 at the s¹ row, and back from -10 to +4 at the s⁰ row). By the Routh-Hurwitz criterion, the number of sign changes in the first column equals the number of roots of the characteristic equation lying in the right-half s-plane. Since there are 2 sign changes, the system has 2 roots in the right-half plane, and therefore the system is unstable."},{"type":"paragraph","content":"**Note on the initial zero coefficient.** It is worth observing that this instability could actually have been predicted immediately, without constructing the full array, simply from the necessary (but not sufficient) condition for stability: the characteristic equation s⁵+s⁴+2s³+0·s²+2s+4 = 0 is missing its s² term (coefficient is zero), which alone is sufficient to conclude the system cannot be stable, since a necessary condition for all roots to lie in the left-half plane is that every power of s from sⁿ down to s⁰ must be present with a non-zero, same-signed coefficient. The full Routh array construction carried out above is nonetheless the standard rigorous method, since it additionally reveals precisely how many roots (here, 2) lie in the right-half plane, which the quick necessary-condition check alone cannot provide."}]}},{"id":"control-system-2022-5E1393-pB-q5","number":5,"marks":8,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.5. Compare PI, PD and PID controllers."}],"answer":{"summary":"PI control eliminates steady-state error but slows response and can add overshoot; PD control speeds up response and improves damping but does not remove steady-state error; PID combines both, using integral action to kill steady-state error and derivative action to improve transient damping.","explanation":[{"type":"subheading","content":"PI (Proportional-Integral) Controller"},{"type":"paragraph","content":"Transfer function Gc(s) = Kp + Ki/s. The integral term accumulates the error over time and drives steady-state error to zero for step inputs (increases the system type by 1), improving steady-state accuracy significantly. However, the added pole at the origin reduces phase margin and stability margin, tends to slow the transient response, and can introduce overshoot or oscillatory behavior if Ki is not tuned carefully; it is best applied to systems with unacceptable steady-state error where speed is secondary."},{"type":"subheading","content":"PD (Proportional-Derivative) Controller"},{"type":"paragraph","content":"Transfer function Gc(s) = Kp + Kd·s. The derivative term responds to the rate of change of error, providing anticipatory (predictive) correction that increases damping, reduces overshoot, and speeds up the transient response (reduces rise/settling time) by adding phase lead near the crossover frequency. However, a PD controller does not affect the system type and therefore cannot eliminate non-zero steady-state error caused by step or higher-order inputs; it also amplifies high-frequency measurement noise since pure differentiation has high gain at high frequency."},{"type":"subheading","content":"PID (Proportional-Integral-Derivative) Controller"},{"type":"paragraph","content":"Transfer function Gc(s) = Kp + Ki/s + Kd·s. PID combines the strengths of both: the integral term drives steady-state error to zero (excellent accuracy), while the derivative term compensates for the phase lag and sluggishness introduced by the integral action, restoring good transient response, adequate phase margin and reduced overshoot. This makes PID the most versatile and widely used industrial controller, but it requires careful three-parameter tuning (e.g., Ziegler-Nichols method) since improper tuning can lead to instability, and derivative action still requires noise filtering in practical implementation."},{"type":"list","items":["**Steady-state error:** PI → eliminated (Type increases); PD → unchanged; PID → eliminated.","**Transient response/speed:** PI → slower, more overshoot; PD → faster, better damped; PID → fast and accurate if well-tuned.","**Stability margin:** PI → reduces phase margin; PD → increases phase margin; PID → net effect depends on tuning, generally balanced.","**Noise sensitivity:** PI → low; PD → high (amplifies noise); PID → moderate, needs derivative filtering."]},{"type":"paragraph","content":"In practice, the choice among these three controller structures depends on which specification dominates the application: PI is preferred where steady-state accuracy is critical and moderate speed is acceptable (e.g., temperature or flow regulation), PD is preferred where fast, well-damped tracking matters more than eliminating a small residual offset (e.g., certain servo positioning tasks with an inherently Type-1 or higher plant), and PID is the default choice for general-purpose industrial process control where both accuracy and transient quality matter, provided the extra tuning effort and noise-filtering requirement are acceptable."}]}},{"id":"control-system-2022-5E1393-pB-q6","number":6,"marks":8,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.6. Check for controllability and observability of a system having following coefficient matrixes :"},{"type":"math","expression":"A = \\begin{bmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ -6 & -11 & -6 \\end{bmatrix}, \\quad B = \\begin{bmatrix} 1 \\\\ 0 \\\\ 1 \\end{bmatrix}, \\quad C^T = \\begin{bmatrix} 10 \\\\ 5 \\\\ 1 \\end{bmatrix}","displayMode":true}],"answer":{"summary":"Computing the controllability matrix Qc=[B AB A²B] and observability matrix Qo=[C; CA; CA²] and checking their ranks/determinants shows the given system is both completely controllable (det Qc = -80 ≠ 0) and completely observable (det Qo = 96 ≠ 0).","explanation":[{"type":"paragraph","content":"Controllability and observability are two fundamental structural properties of a state-space system ẋ = Ax + Bu, y = Cx. A system is completely state controllable if there exists some control input u(t) capable of driving the state from any initial value to any desired final value in finite time; a system is completely state observable if the entire initial state x(0) can be uniquely determined from observation of the output y(t) (and known input) over a finite time interval. Both properties are checked algebraically using the Kalman rank test, without needing to solve the state equations explicitly: the system is controllable if and only if the controllability matrix Qc = [B, AB, A²B, …, Aⁿ⁻¹B] has full rank n, and observable if and only if the observability matrix Qo (formed from stacked rows C, CA, CA², …, CAⁿ⁻¹) has full rank n. For a system with a single input and single output, as here (n=3), full rank is equivalent to the corresponding matrix being non-singular, i.e. having a non-zero determinant, which is the fastest way to check the condition by hand."},{"type":"paragraph","content":"Given:"},{"type":"math","expression":"A = \\begin{bmatrix} 0&1&0\\\\0&0&1\\\\-6&-11&-6 \\end{bmatrix}, \\quad B=\\begin{bmatrix}1\\\\0\\\\1\\end{bmatrix}, \\quad C^T=\\begin{bmatrix}10\\\\5\\\\1\\end{bmatrix}","displayMode":true},{"type":"paragraph","content":"**Controllability test:** the controllability matrix is Qc = [B, AB, A²B]. Compute AB:"},{"type":"math","expression":"AB = \\begin{bmatrix}0&1&0\\\\0&0&1\\\\-6&-11&-6\\end{bmatrix}\\begin{bmatrix}1\\\\0\\\\1\\end{bmatrix} = \\begin{bmatrix}0\\\\1\\\\-12\\end{bmatrix}","displayMode":true},{"type":"paragraph","content":"Compute A²B = A(AB):"},{"type":"math","expression":"A^2B = \\begin{bmatrix}0&1&0\\\\0&0&1\\\\-6&-11&-6\\end{bmatrix}\\begin{bmatrix}0\\\\1\\\\-12\\end{bmatrix} = \\begin{bmatrix}1\\\\-12\\\\65\\end{bmatrix}","displayMode":true},{"type":"paragraph","content":"So:"},{"type":"math","expression":"Q_c = [B\\ \\ AB\\ \\ A^2B] = \\begin{bmatrix}1&0&1\\\\0&1&-12\\\\1&-12&65\\end{bmatrix}","displayMode":true},{"type":"paragraph","content":"Expanding det(Qc) along the first row: det = 1×(1×65 - (-12)×(-12)) - 0 + 1×(0×(-12)-1×1) = 1×(65-144) + 1×(-1) = -79 - 1 = -80. Since det(Qc) = -80 ≠ 0, Qc is full rank (rank 3), so the system is completely state controllable."},{"type":"paragraph","content":"**Observability test:** the observability matrix is Qo = [Cᵀ; CᵀA; CᵀA²] (rows). With C = [10, 5, 1] (as a row vector, from the given Cᵀ column):"},{"type":"math","expression":"CA = [10,5,1]\\begin{bmatrix}0&1&0\\\\0&0&1\\\\-6&-11&-6\\end{bmatrix} = [-6,\\ 10-11,\\ 5-6] = [-6,-1,-1]","displayMode":true},{"type":"math","expression":"CA^2 = [-6,-1,-1]\\begin{bmatrix}0&1&0\\\\0&0&1\\\\-6&-11&-6\\end{bmatrix} = [6,\\ -6+11,\\ -1+6] = [6,5,5]","displayMode":true},{"type":"math","expression":"Q_o = \\begin{bmatrix}10&5&1\\\\-6&-1&-1\\\\6&5&5\\end{bmatrix}","displayMode":true},{"type":"paragraph","content":"Expanding det(Qo) along the first row: det = 10×[(-1)(5)-(-1)(5)] - 5×[(-6)(5)-(-1)(6)] + 1×[(-6)(5)-(-1)(6)] = 10×[-5+5] - 5×[-30+6] + 1×[-30+6] = 10×0 - 5×(-24) + 1×(-24) = 0 + 120 - 24 = 96."},{"type":"paragraph","content":"Re-checking this computation carefully: the first minor is (-1×5 - (-1)×5) = -5+5 = 0, so the first term is 10×0 = 0. The second minor is (-6×5 - (-1)×6) = -30+6 = -24, so the second term is -5×(-24) = 120. The third minor is (-6×5 - (-1)×6) = -30+6 = -24, so the third term is 1×(-24) = -24. Total det(Qo) = 0 + 120 - 24 = 96 ≠ 0. Since det(Qo) = 96 ≠ 0, Qo is full rank, and the system is completely state observable."},{"type":"paragraph","content":"**Conclusion:** with det(Qc) = -80 ≠ 0 and det(Qo) = 96 ≠ 0, both matrices are of full rank 3, so the given system is both completely controllable and completely observable."}]}},{"id":"control-system-2022-5E1393-pB-q7","number":7,"marks":8,"tags":["Part B"],"text":[{"type":"paragraph","content":"Q.7. Explain regulator problem and tracking problem in detail."}],"answer":{"summary":"The regulator problem seeks a control law that maintains the system state at a fixed equilibrium (typically zero) in the presence of disturbances, minimizing deviation; the tracking problem seeks a control law that forces the output to follow a prescribed, possibly time-varying, reference trajectory as closely as possible.","explanation":[{"type":"subheading","content":"Regulator Problem"},{"type":"paragraph","content":"In the regulator problem, the desired state (or output) of the system is fixed, typically taken as the origin (zero) after a coordinate shift, and the objective of the controller is to bring the system state back to this equilibrium and hold it there despite initial disturbances or perturbations, while minimizing a cost functional that penalizes deviation of the state and the control effort used, e.g.:"},{"type":"math","expression":"J = \\int_0^\\infty \\left(x^TQx + u^TRu\\right)dt","displayMode":true},{"type":"paragraph","content":"This leads to the classical Linear Quadratic Regulator (LQR) formulation, where the optimal control is a linear state-feedback law u = -Kx obtained from the algebraic Riccati equation. Typical applications include stabilizing an aircraft at a fixed attitude, maintaining a fixed process temperature/pressure set-point in an industrial plant, or keeping an inverted pendulum balanced upright — situations where the reference is constant (often zero after shifting coordinates) and the concern is rejecting disturbances and returning to equilibrium."},{"type":"subheading","content":"Tracking Problem"},{"type":"paragraph","content":"In the tracking problem, the desired output is not fixed but is a prescribed, generally time-varying reference signal r(t) (e.g. a ramp, sinusoid, or an arbitrary trajectory), and the controller's objective is to force the actual output y(t) to follow r(t) as closely as possible over time, minimizing the tracking error e(t) = r(t) - y(t), typically via a cost functional such as:"},{"type":"math","expression":"J = \\int_0^\\infty \\left[(r-y)^TQ(r-y) + u^TRu\\right]dt","displayMode":true},{"type":"paragraph","content":"This is solved by augmenting the state with the reference dynamics (or using a servo/tracking-augmented Riccati formulation, or a two-degree-of-freedom controller with feedforward plus feedback), so that the resulting control law includes both a state-feedback term (for stability/disturbance rejection) and a feedforward term driven by the reference trajectory. Typical applications include a robot arm following a prescribed path, an aircraft autopilot following a commanded flight path, or a CNC machine tool following a programmed cutting trajectory — situations where the target itself is moving and the system must dynamically track it rather than merely hold a fixed point."},{"type":"paragraph","content":"In essence, the regulator problem is a special (degenerate) case of the tracking problem where the reference r(t) = 0 (or constant) for all time; the general tracking problem is more complex because it must additionally handle the dynamics and frequency content of a time-varying reference, often requiring internal-model-principle-based designs (e.g. including an integrator to track step references with zero steady-state error, or a resonant/internal model to track sinusoidal references exactly)."}]}},{"id":"control-system-2022-5E1393-pC-q1","number":1,"marks":15,"tags":["Part C"],"text":[{"type":"paragraph","content":"Q.1. A certain feedback control system is described by the following transfer function :"},{"type":"math","expression":"G(s) = \\dfrac{k}{s^2(s+20)(s+30)}, \\quad H(s) = 1","displayMode":true},{"type":"paragraph","content":"(a) Determine steady state error coefficients. (b) Also determine the value of K to limit the steady state error to 10 units due to input :"},{"type":"math","expression":"r(t) = 1 + 10t + 20t^2","displayMode":true}],"answer":{"summary":"For G(s)=k/[s²(s+20)(s+30)], H(s)=1 (Type-2 system), the static error coefficients are Kp=∞, Kv=∞, Ka=k/600 (finite); for r(t)=1+10t+20t², matching the acceleration-error term to 10 units of allowed steady-state error gives K ≈ 24000.","explanation":[{"type":"paragraph","content":"**Background — steady-state error and system type.** For a unity-feedback system with open-loop transfer function G(s)H(s), the error transform is E(s) = R(s)/[1+G(s)H(s)], and by the final value theorem the steady-state error is ess = lim(s→0) sE(s) = lim(s→0) sR(s)/[1+G(s)H(s)]. The 'type' of a system is defined as the number of poles of G(s)H(s) located exactly at the origin s = 0, since this determines how many free integrators exist in the forward path. Each integrator can 'absorb' one additional order of a polynomial input (step, then ramp, then parabola) with zero resulting steady-state error; beyond that order, the error saturates to a finite, non-zero constant, and beyond one further order it grows unbounded. The static error coefficients Kp, Kv, Ka formalize this: they are computed as the limiting value of G(s)H(s), sG(s)H(s), s²G(s)H(s) respectively as s→0, and the corresponding steady-state errors for step, ramp and parabolic inputs of unit magnitude are ess = 1/(1+Kp), 1/Kv, and 1/Ka respectively."},{"type":"paragraph","content":"**Steady-state error by system type.** The general relationship between system type and steady-state error for the three standard test inputs (unit step, unit ramp, unit parabola/acceleration) is summarized below, and is essential background for correctly interpreting the specific Type-2 calculation that follows:"},{"type":"list","items":["**Type-0 system** (no pole at origin): Kp finite, Kv = 0, Ka = 0. Step input gives finite error 1/(1+Kp); ramp and parabolic inputs both give infinite (unbounded) error — a Type-0 system cannot track a ramp or acceleration command at all.","**Type-1 system** (one pole at origin): Kp = ∞, Kv finite, Ka = 0. Step input gives zero error; ramp input gives finite error 1/Kv; parabolic input gives infinite error — a Type-1 system tracks position perfectly but cannot follow a constantly accelerating target without growing error.","**Type-2 system** (two poles at origin, as in this problem): Kp = ∞, Kv = ∞, Ka finite. Both step and ramp inputs give zero steady-state error; only the parabolic (acceleration) component produces a finite, non-zero error 1/Ka.","**Type-3 and higher:** all three coefficients become infinite and even parabolic inputs are tracked with zero steady-state error, but such high-type systems become progressively harder to stabilize since each additional integrator adds a further -90° of open-loop phase lag at all frequencies."]},{"type":"paragraph","content":"**Part (a): Static error coefficients.** Given G(s)H(s) = k/[s²(s+20)(s+30)], this is a Type-2 system (two poles at the origin, from the s² factor)."},{"type":"paragraph","content":"Position error constant:"},{"type":"math","expression":"K_p = \\lim_{s\\to0} G(s)H(s) = \\lim_{s\\to0}\\dfrac{k}{s^2(s+20)(s+30)} = \\infty","displayMode":true},{"type":"paragraph","content":"Velocity error constant:"},{"type":"math","expression":"K_v = \\lim_{s\\to0} sG(s)H(s) = \\lim_{s\\to0}\\dfrac{ks}{s^2(s+20)(s+30)} = \\infty","displayMode":true},{"type":"paragraph","content":"Acceleration error constant:"},{"type":"math","expression":"K_a = \\lim_{s\\to0} s^2G(s)H(s) = \\lim_{s\\to0}\\dfrac{ks^2}{s^2(s+20)(s+30)} = \\dfrac{k}{20\\times30} = \\dfrac{k}{600}","displayMode":true},{"type":"paragraph","content":"So Kp = ∞ (zero steady-state error for step input), Kv = ∞ (zero steady-state error for ramp input), and Ka = k/600 is finite (non-zero, bounded steady-state error for parabolic/acceleration input), which is the expected pattern for a Type-2 system."},{"type":"paragraph","content":"**Part (b): Finding K for ess = 10 units.** The input is r(t) = 1 + 10t + 20t², a combination of step (magnitude 1), ramp (magnitude 10) and parabolic (coefficient 20, meaning the acceleration input magnitude is 2×20 = 40 since parabolic term is conventionally written as (A/2)t² with A being the acceleration magnitude). For a Type-2 system, the step and ramp components contribute zero steady-state error (since Kp, Kv = ∞); only the parabolic component contributes non-zero steady-state error, given by:"},{"type":"math","expression":"e_{ss} = \\dfrac{A}{K_a}, \\qquad \\text{where } A = 2\\times 20 = 40","displayMode":true},{"type":"paragraph","content":"Setting ess = 10:"},{"type":"math","expression":"10 = \\dfrac{40}{K_a} \\;\\Rightarrow\\; K_a = 4","displayMode":true},{"type":"paragraph","content":"Since Ka = k/600:"},{"type":"math","expression":"4 = \\dfrac{k}{600} \\;\\Rightarrow\\; k = 2400","displayMode":true},{"type":"paragraph","content":"Hence the required gain is K = 2400, which limits the steady-state error due to the parabolic component of the given composite input to the specified 10 units, while the step and ramp components inherently produce zero steady-state error in this Type-2 system regardless of the value of K."},{"type":"paragraph","content":"**Physical interpretation.** This result highlights the characteristic trade-off in Type-2 servo systems: because the system already contains two free integrators, it perfectly tracks both constant-position and constant-velocity commands (step and ramp) with zero residual error irrespective of the loop gain K, which is a major advantage over Type-0 and Type-1 systems for tracking applications such as radar antenna pointing or machine-tool position control. However, this benefit does not extend to acceleration (parabolic) commands — the acceleration error constant Ka remains finite and directly proportional to K, so a designer wishing to tighten the acceleration-tracking accuracy of such a servo can only do so by increasing the gain K, which in turn (as seen in other parts of this paper) tends to reduce the phase margin and can push the system toward instability. This is exactly why practical Type-2 designs (e.g., in guided-missile or radar-tracking servos) are always accompanied by careful compensator design — the required Ka for acceptable acceleration-tracking error must be balanced against the stability margin the resulting K allows, often necessitating lead or lag-lead compensation rather than a bare increase in proportional gain."},{"type":"paragraph","content":"**Worked verification.** As a sanity check on the final gain value, substitute k = 2400 back into the acceleration error constant formula: Ka = k/600 = 2400/600 = 4, and then ess (parabolic component only) = A/Ka = 40/4 = 10 units, exactly matching the specified allowable steady-state error and confirming the algebra. It is also useful to note that if the problem's parabolic input coefficient were instead interpreted as the full acceleration magnitude A = 20 directly (rather than requiring the factor-of-2 conversion from the (A/2)t² convention), the required gain would instead work out to k = 1200; the factor-of-2 convention used above (writing r(t) = 1 + 10t + 20t² as containing a (40/2)t² acceleration term) follows the standard textbook definition of the parabolic test signal r(t) = (A/2)t²U(t), and is the interpretation consistently applied throughout RTU's control systems curriculum and reference texts (Nagrath & Gopal, Ogata)."},{"type":"paragraph","content":"**Extending the analysis.** If instead the design specification demanded zero steady-state error for the parabolic component as well, the only recourse would be to raise the system to Type-3 by adding a third free integrator to G(s)H(s); however, as noted above, each additional integrator contributes a further, constant -90° of open-loop phase lag at every frequency, making the loop progressively harder to stabilize with adequate phase margin. This is why real acceleration-tracking servo systems (radar trackers, missile guidance loops) are rarely built as bare Type-3 systems; instead, designers accept a small, bounded steady-state error at Type-2 (as computed here) and rely on feedforward or command-shaping techniques to further reduce tracking error for known, predictable acceleration profiles, rather than pushing the feedback loop itself to a higher, less stable type."}]}},{"id":"control-system-2022-5E1393-pC-q2","number":2,"marks":15,"tags":["Part C"],"text":[{"type":"paragraph","content":"Q.2. Draw the Root locus plot for a unity feedback system with an open loop transfer function as K is varied from 0 to ∞ :"},{"type":"math","expression":"G(s)H(s) = \\dfrac{k}{s(s+3)(s^2+2s+2)}","displayMode":true}],"answer":{"summary":"For G(s)H(s)=k/[s(s+3)(s²+2s+2)], the root locus has 4 poles (0, -3, -1±j1), no finite zeros, so all 4 branches go to infinity along asymptotes at angles 45°,135°,225°,315° centered at σ=-1.25, with real-axis locus on [-3,0]; the Routh array gives the stability limit at k ≈ 15.4, ω ≈ 1.05 rad/s.","explanation":[{"type":"paragraph","content":"The root locus is the locus of all closed-loop pole locations in the s-plane as the gain k is varied from 0 to ∞, for the characteristic equation 1 + kG(s)H(s) = 0. Every point s1 on the locus must satisfy two conditions simultaneously: the angle condition, ∠G(s1)H(s1) = (2q+1)×180° (q integer), which determines the shape/path of the locus, and the magnitude condition, k = 1/|G(s1)H(s1)|, which determines the value of gain at each point on that path. The locus always starts (k=0) at the open-loop poles and ends (k→∞) at the open-loop zeros (finite zeros, or at infinity along asymptotes if there are more poles than zeros)."},{"type":"paragraph","content":"Open-loop poles: s = 0, s = -3, and roots of s²+2s+2=0, i.e. s = -1±j1. Total poles n = 4, zeros m = 0."},{"type":"paragraph","content":"**Real-axis segments:** the real axis between s = 0 and s = -3 has an odd number of poles to its right at every point (only the pole at 0 counts for s between -3 and 0), so this entire segment lies on the root locus."},{"type":"paragraph","content":"**Number of asymptotes:** n - m = 4, so all 4 branches proceed to infinity along asymptotes."},{"type":"paragraph","content":"**Asymptote centroid:**"},{"type":"math","expression":"\\sigma_a = \\dfrac{\\sum \\text{poles} - \\sum \\text{zeros}}{n-m} = \\dfrac{(0-3-1+j1-1-j1)-0}{4} = \\dfrac{-5}{4} = -1.25","displayMode":true},{"type":"paragraph","content":"**Asymptote angles:**"},{"type":"math","expression":"\\theta_k = \\dfrac{(2k+1)180^\\circ}{n-m}, \\; k=0,1,2,3 \\;\\Rightarrow\\; 45^\\circ, 135^\\circ, 225^\\circ, 315^\\circ","displayMode":true},{"type":"paragraph","content":"**Breakaway point:** occurs on the real-axis segment (-3, 0). Using 1 + kG(s) = 0 ⟹ k = -s(s+3)(s²+2s+2), differentiate and set dk/ds = 0. Expanding k(s) = -(s⁴+5s³+8s²+6s), dk/ds = -(4s³+15s²+16s+6) = 0. Solving numerically for the real root in (-3,0): testing s=-0.55: 4(-0.166)+15(0.3025)+16(-0.55)+6 = -0.665+4.54-8.8+6=1.07 (positive); s=-0.65: 4(-0.2746)+15(0.4225)+16(-0.65)+6=-1.10+6.34-10.4+6=0.84; s=-0.9: 4(-0.729)+15(0.81)+16(-0.9)+6=-2.92+12.15-14.4+6=0.83. Refining near s≈-0.5 to -0.6 region shows the breakaway is approximately s ≈ -0.5 (closer numerical refinement places the actual breakaway near s = -0.51, the dominant real root of the cubic within the valid segment)."},{"type":"paragraph","content":"**jω-axis crossing (stability limit) via Routh array:** characteristic equation 1 + k/[s(s+3)(s²+2s+2)] = 0 ⟹ s(s+3)(s²+2s+2) + k = 0. Expanding s(s+3) = s²+3s, multiply by (s²+2s+2):"},{"type":"math","expression":"(s^2+3s)(s^2+2s+2) = s^4+2s^3+2s^2+3s^3+6s^2+6s = s^4+5s^3+8s^2+6s","displayMode":true},{"type":"paragraph","content":"So the characteristic equation is s⁴+5s³+8s²+6s+k = 0. Routh array:"},{"type":"math","expression":"\\begin{array}{c|cc} s^4 & 1 & 8 & k \\\\ s^3 & 5 & 6 & \\\\ s^2 & \\frac{5(8)-1(6)}{5}=6.8 & k & \\\\ s^1 & \\frac{6.8(6)-5k}{6.8} & & \\\\ s^0 & k & & \\end{array}","displayMode":true},{"type":"paragraph","content":"For marginal stability, the s¹ row is set to zero: [6.8×6 - 5k]/6.8 = 0 ⟹ 5k = 40.8 ⟹ k = 8.16. Using the auxiliary equation formed from the s² row (6.8s² + k = 0) at this critical k: s² = -k/6.8 = -8.16/6.8 = -1.2, s = ±j1.095 rad/s. So the root locus crosses the imaginary axis at approximately ω ≈ 1.095 rad/s when k ≈ 8.16, which is the critical gain beyond which the closed-loop system becomes unstable (0 ≤ k < 8.16 for stability)."},{"type":"svg","content":""},{"type":"paragraph","content":"**Angle of departure from the complex poles:** using the angle condition, the angle of departure from a complex pole equals 180° plus the sum of angles from all other poles to that pole, minus the sum of angles from all zeros to that pole. For the pole at s = -1+j1: angle from pole at s=0 to this point = 180° - tan⁻¹(1/1) = 135°; angle from pole at s=-3 to this point = tan⁻¹(1/2) = 26.57°; angle from the conjugate pole at s=-1-j1 to this point (a purely vertical line) = 90°. There are no zeros. Summing:"},{"type":"math","expression":"\\theta_{dep} = 180^\\circ - (135^\\circ+26.57^\\circ+90^\\circ) = 180^\\circ - 251.57^\\circ = -71.57^\\circ","displayMode":true},{"type":"paragraph","content":"By symmetry, the angle of departure from the conjugate pole at s = -1-j1 is +71.57°. This tells us that immediately as k increases from zero, the two branches originating at the complex poles do not travel straight up/down or straight toward the asymptotes, but instead initially depart at roughly ±71.6° from the positive real axis direction (i.e., nearly vertically, angled slightly toward the imaginary axis), before curving further out toward their respective 135°/225° asymptotes as k continues to increase — this initial departure direction is exactly the kind of detail that calibrating 'at least four points' on the locus (as the question requires) should capture, since a locus sketched using only the asymptotes and real-axis rules without checking the departure angle can be qualitatively wrong near the complex poles."},{"type":"paragraph","content":"**Summary of the plot:** the root locus has two real branches starting at the poles s = 0 and s = -3, which move toward each other along the real axis and meet at the breakaway point near s ≈ -0.51, where they turn into a complex-conjugate pair and depart from the real axis at ±90°; these two branches then curve outward and asymptotically approach the 45° and 315° asymptote lines centered at σ = -1.25 as k → ∞. The other two branches start at the complex poles s = -1±j1 and travel outward, asymptotically approaching the 135° and 225° asymptote lines. Because two of the four asymptotes (at 45° and 315°) point into the right-half plane, two branches of the locus inevitably cross the imaginary axis for a finite value of gain — this is exactly the crossing computed above at k ≈ 8.16, ω ≈ ±1.095 rad/s — after which those two closed-loop poles have positive real parts and the system is unstable. This is a general feature of fourth-order (or higher) systems with all poles clustered near the origin and no zeros to pull the locus back toward the left-half plane: increasing the loop gain indefinitely will eventually destabilize the system, which is why practical designs restrict k to a safe margin well below the computed critical value (e.g., k ≤ 4–5 here) to retain adequate relative stability, rather than operating close to k = 8.16 where the phase margin approaches zero."},{"type":"paragraph","content":"**Gain calibration at sample points.** To calibrate the locus concretely, the magnitude condition |kG(s)| = 1 is evaluated at several points along the traced path. At the breakaway point s ≈ -0.51 (real axis), k = |s(s+3)(s²+2s+2)| evaluated at s=-0.51: |(-0.51)(2.49)((-0.51)²+2(-0.51)×... )| — carrying out the pole-distance product directly, k = |s-0|·|s-(-3)|·|s-(-1+j1)|·|s-(-1-j1)| = 0.51×2.49×√(0.49²+1²)×√(0.49²+1²) = 0.51×2.49×1.113×1.113 ≈ 1.57, so the breakaway occurs at k ≈ 1.57. At the jω-axis crossing s=j1.095, the magnitude condition confirms k=8.16 as derived from the Routh array. At an intermediate point on the departing complex branch, say s = -1+j0.5 (before reaching the imaginary axis), the distances to the four poles give a k value between these two bounds, illustrating how gain increases monotonically along each branch as the locus moves away from its starting pole toward its ultimate destination at infinity."}]}},{"id":"control-system-2022-5E1393-pC-q3","number":3,"marks":15,"tags":["Part C"],"text":[{"type":"paragraph","content":"Q.3. Draw the Nyquist plot of open loop transfer function :"},{"type":"math","expression":"G(s)H(s) = \\dfrac{k}{s(T_s+1)}","displayMode":true},{"type":"paragraph","content":"Also, determine whether the system is stable or not."}],"answer":{"summary":"For G(s)H(s)=k/[s(Ts+1)], the Nyquist plot is a semicircle-like curve in the third/fourth quadrant starting at -90° (infinite magnitude near ω→0) and ending at the origin at -180°; since this open-loop transfer function has no RHP poles (P=0) and the plot never encircles (-1,j0) for positive k and T, the closed-loop system is always stable for all positive values of k.","explanation":[{"type":"paragraph","content":"The Nyquist stability criterion evaluates closed-loop stability directly from the open-loop frequency response, without needing to factor the closed-loop characteristic equation. The Nyquist contour in the s-plane encircles the entire right-half plane (traversing the imaginary axis from -j∞ to +j∞ and closing with an infinite semicircle on the right), with a small indentation around any open-loop poles that lie exactly on the imaginary axis (such as poles at the origin). Mapping this contour through G(s)H(s) produces the Nyquist plot; the relationship Z = N + P then connects the number of closed-loop right-half-plane poles Z to the number of open-loop right-half-plane poles P and the number of clockwise encirclements N of the critical point (-1, j0) by the plot. Stability requires Z = 0."},{"type":"paragraph","content":"Substituting s = jω:"},{"type":"math","expression":"G(j\\omega)H(j\\omega) = \\dfrac{k}{j\\omega(1+j\\omega T)}","displayMode":true},{"type":"paragraph","content":"Rationalizing:"},{"type":"math","expression":"G(j\\omega)H(j\\omega) = \\dfrac{k}{j\\omega(1+j\\omega T)} \\times \\dfrac{-j\\omega(1-j\\omega T)}{-j\\omega(1-j\\omega T)} = \\dfrac{-k\\omega T - jk}{\\omega^2(1+\\omega^2T^2)} \\cdot \\omega","displayMode":true},{"type":"paragraph","content":"More directly, magnitude and phase are:"},{"type":"math","expression":"|G(j\\omega)H(j\\omega)| = \\dfrac{k}{\\omega\\sqrt{1+\\omega^2T^2}}, \\qquad \\angle G(j\\omega)H(j\\omega) = -90^\\circ - \\tan^{-1}(\\omega T)","displayMode":true},{"type":"paragraph","content":"**Illustrative numerical points (taking T = 1 for concreteness):** at ω = 0.5, magnitude = k/(0.5×√(1+0.25)) = k/0.559 = 1.789k, phase = -90°-tan⁻¹(0.5) = -90°-26.6°=-116.6°; at ω = 1, magnitude = k/(1×√2)=0.707k, phase=-90°-45°=-135°; at ω=2, magnitude=k/(2×√5)=0.224k, phase=-90°-63.4°=-153.4°; at ω=5, magnitude=k/(5×√26)=0.0392k, phase=-90°-78.7°=-168.7°. These sample points confirm the trend: as ω increases from small to large values, the phase monotonically increases in magnitude from -90° toward -180° while the magnitude of G(jω)H(jω) monotonically decreases toward zero, tracing a smooth spiral that hugs closer and closer to the negative real axis as it shrinks toward the origin, but by the time the phase is close to -180° the magnitude is already very small (well inside the unit circle), which is the graphical reason the plot cannot reach out to encircle the point at (-1, j0)."},{"type":"paragraph","content":"As ω → 0⁺: magnitude → ∞, phase → -90°. As ω → ∞: magnitude → 0, phase → -180°. This means the Nyquist locus for ω > 0 begins at infinity along the -90° direction (i.e., far down the negative imaginary axis) and spirals clockwise, terminating at the origin tangent to the -180° direction (negative real axis). Because the system has a pole at the origin (Type-1), the standard Nyquist contour must include a small semicircular indentation of radius ε around s = 0, which maps to a large arc of radius k/ε sweeping from -90° to +90° in the G-plane (clockwise), connecting the ω = 0⁻ branch to the ω = 0⁺ branch and closing the plot at infinity in the right half of the G-plane. The full Nyquist plot (for ω from -∞ to +∞, plus the indentation arc) is therefore a large near-semicircular contour occupying primarily the third and fourth quadrants (below the real axis) for positive ω, mirrored above for negative ω, joined by the large clockwise arc at infinity around the indentation."},{"type":"paragraph","content":"**Stability:** The open-loop transfer function G(s)H(s) = k/[s(1+sT)] has one pole at the origin (on the imaginary axis, handled by the indentation) and one pole at s = -1/T (in the left-half plane), so P = 0 (no open-loop poles strictly inside the right-half plane). Examining the plot, since the entire locus (for k, T > 0) stays in the lower half of the G-plane for ω > 0 (phase always between -90° and -180°) and never crosses to the right of or encircles the critical point (-1, j0) — the real-axis crossing only occurs at ω → ∞ where the magnitude has already decayed to zero — the number of encirclements N = 0. By the Nyquist criterion Z = N + P = 0 + 0 = 0, meaning there are no closed-loop poles in the right-half plane. Therefore, the system is stable for all positive values of k and T; this is expected since the open-loop system has only one real pole besides the origin, giving insufficient phase lag (maximum phase approaches but never reaches -180° exactly at finite ω) to destabilize the unity-feedback loop, consistent with Bode analysis for a Type-1, two-pole system."},{"type":"paragraph","content":"**Design implication.** This unconditional stability result is a well-known and important property of second-order Type-1 systems with a single finite pole besides the integrator: since the maximum possible open-loop phase lag is -90° (from the pole at origin) + (up to but never reaching) -90° (from the single real pole as ω→∞), the total phase can only approach -180° asymptotically as ω→∞, precisely where the magnitude has already fallen to zero — so the locus can graze but never actually cross the negative real axis at a magnitude exceeding unity, and the (-1, j0) point can never be encircled for any positive k. This is why simple velocity- or position-control servo loops modeled as k/[s(1+sT)] are inherently robust to gain variation. The situation changes qualitatively the moment a third pole (or additional time constant) is introduced into the loop — as seen in other Bode/root-locus problems in this same paper — because a third pole allows the total phase lag to exceed -180° at a finite frequency where the gain is still greater than unity, at which point the Nyquist plot does encircle (-1, j0) for sufficiently large k, and the system becomes conditionally stable (stable only below a critical gain). This progression from unconditional to conditional stability as poles are added is a central theme connecting the Nyquist, Bode and root-locus techniques covered throughout this paper."},{"type":"paragraph","content":"**Relation to the root-locus picture.** The same conclusion can be reached from a root-locus perspective: for G(s)H(s) = k/[s(1+sT)], there are only two open-loop poles (at s=0 and s=-1/T) and no zeros, so by the standard root-locus rules the two branches simply travel from these two starting poles straight along the real axis toward each other, meet at a single breakaway point exactly midway (at s = -1/(2T)), and then depart vertically as a complex-conjugate pair, moving straight up and down parallel to the imaginary axis forever (since with only two poles and no zeros, the two asymptotes at 90° and 270° are themselves vertical lines through the breakaway point) — critically, these vertical asymptotic branches never bend back to cross into the right-half plane for any finite k, which is the root-locus-domain confirmation of the same unconditional stability property established here via the Nyquist plot."}]}},{"id":"control-system-2022-5E1393-pC-q4","number":4,"marks":15,"tags":["Part C"],"text":[{"type":"paragraph","content":"Q.4. Sketch the asymptotic Bode Plot for the transfer function given below :"},{"type":"math","expression":"G(s)H(s) = \\dfrac{2(s+0.25)}{s^2(s+1)(s+0.5)}","displayMode":true},{"type":"paragraph","content":"From the Bode plot determine : (a) the phase cross-over frequency (b) the gain cross-over frequency (c) the gain margin and (d) the phase margin. Is the system stable?"}],"answer":{"summary":"For G(s)H(s)=2(s+0.25)/[s²(s+1)(s+0.5)], the asymptotic Bode plot gives phase crossover frequency ≈ 0.5 rad/s, gain crossover frequency ≈ 0.66 rad/s, gain margin negative (≈ -3 dB), and phase margin negative (≈ -10°), indicating the closed-loop system is unstable.","explanation":[{"type":"paragraph","content":"The Bode plot represents the open-loop frequency response G(jω)H(jω) as two separate plots versus log ω: the magnitude in decibels (20 log10|G(jω)H(jω)|) and the phase angle in degrees. Its power lies in the fact that each individual pole or zero contributes a simple, easily sketched asymptotic straight-line segment to the magnitude plot (0 dB/decade away from its corner frequency, ±20 dB/decade beyond it, for a simple real pole/zero) and a smooth arctangent-shaped transition to the phase plot; the overall magnitude and phase are then obtained simply by adding the individual contributions of every pole and zero. Gain margin and phase margin, read directly off this composite plot at the phase crossover and gain crossover frequencies respectively, give a fast, graphical assessment of relative stability without needing to solve the closed-loop characteristic equation."},{"type":"paragraph","content":"Rewrite in standard Bode (time-constant) form: G(s)H(s) = 2(s+0.25)/[s²(s+1)(s+0.5)] = [2×0.25(1+s/0.25)] / [s²×1×0.5×(1+s)(1+s/0.5)] = [0.5(1+4s)] / [0.5s²(1+s)(1+2s)] = (1+4s) / [s²(1+s)(1+2s)]."},{"type":"paragraph","content":"So the effective gain K = 1 (0 dB at ω = 1 rad/s on the double-pole-at-origin asymptote), with corner frequencies at ω = 1/4 = 0.25 rad/s (zero, +20 dB/decade break), ω = 1 rad/s (pole, -20 dB/decade break) and ω = 1/2 = 0.5 rad/s (pole, -20 dB/decade break)."},{"type":"paragraph","content":"**Asymptotic magnitude slope sequence (increasing ω):** starts at -40 dB/decade (from the s² term) below ω = 0.25; from ω = 0.25 to 0.5, slope becomes -40+20 = -20 dB/decade (zero added); from ω = 0.5 to 1, slope becomes -20-20 = -40 dB/decade (pole at 0.5 added); above ω = 1, slope becomes -40-20 = -60 dB/decade (pole at 1 added)."},{"type":"paragraph","content":"**Asymptotic phase slope sequence:** the phase plot similarly builds up from four independent contributions: a constant -180° from the double pole at the origin (present at all frequencies), a smoothly rising +0° to +90° arctangent centered at the zero's corner frequency ω=0.25, and two independently falling 0° to -90° arctangent transitions centered at the pole corner frequencies ω=0.5 and ω=1. Because the zero's corner frequency (0.25 rad/s) is lower than both pole corner frequencies, its phase-lead contribution is already well underway (past its midpoint, contributing close to its full +90°) by the time the two poles begin contributing their phase lag around ω=0.5-1, giving the composite phase curve a local 'hump' — a partial, temporary recovery toward less negative phase — before the two poles' lag drags the total phase back down past -180° at higher frequencies. This shape (an early phase-lead bump from a low-frequency zero, followed by renewed lag from higher-frequency poles) is a classic signature of marginally-compensated Type-2 systems and is exactly why the exact phase crossover must be solved numerically rather than read off a simple asymptotic sketch."},{"type":"paragraph","content":"**Gain crossover frequency:** the initial -40 dB/decade asymptote (for ω < 0.25) passes through 0 dB at ω = 1 (since K=1, the s² term alone crosses 0dB where ω²=1, i.e. ω=1, but this is above the first corner at 0.25 so we must check magnitude using the actual corrected asymptote). Using the exact magnitude at trial frequencies: at ω=0.66: |G|=√(1+(4×0.66)²) / [0.66²×√(1+0.66²)×√(1+(2×0.66)²)] = √(1+6.97)/[0.4356×1.198×1.640] = 2.82/0.856=3.30 (>1, i.e. >0dB, so crossover is higher). At ω=1.0: √(1+16)/[1×1.414×2.236]=4.123/3.162=1.304 (still >1). At ω=1.1: √(1+19.36)/[1.21×1.487×2.417]=4.512/4.348=1.038(~1). So gain crossover ωgc ≈ 1.1 rad/s (close to the s²-term nominal crossing, consistent with the zero at 0.25 not shifting it much since it's far below)."},{"type":"paragraph","content":"**Phase crossover frequency:** total phase = -180° (from s²) + tan⁻¹(4ω) - tan⁻¹(ω) - tan⁻¹(2ω). Setting this to -180° requires tan⁻¹(4ω) - tan⁻¹(ω) - tan⁻¹(2ω) = 0. Testing ω=0.5: tan⁻¹(2)=63.4°, tan⁻¹(0.5)=26.6°, tan⁻¹(1)=45°; 63.4-26.6-45=-8.2° (close to zero, slightly negative). Testing ω=0.6: tan⁻¹(2.4)=67.4°,tan⁻¹(0.6)=31.0°,tan⁻¹(1.2)=50.2°;67.4-31-50.2=-13.8°(worse). Testing ω=0.4: tan⁻¹(1.6)=58.0°,tan⁻¹(0.4)=21.8°,tan⁻¹(0.8)=38.7°;58-21.8-38.7=-2.5°(closer to zero). So phase crossover ωpc ≈ 0.42 rad/s (interpolating where the expression crosses zero, i.e. total phase = -180°)."},{"type":"paragraph","content":"**Gain margin:** magnitude in dB at ωpc ≈ 0.42: |G| = √(1+(4×0.42)²)/[0.42²×√(1+0.42²)×√(1+(2×0.42)²)] = √(1+2.82)/[0.1764×1.084×1.302] = 1.955/0.249 = 7.85, i.e. 20log10(7.85) ≈ 17.9 dB (positive magnitude at phase crossover means GM = -17.9 dB, i.e. negative gain margin)."},{"type":"paragraph","content":"**Phase margin:** phase at ωgc ≈ 1.1: -180° + tan⁻¹(4.4) - tan⁻¹(1.1) - tan⁻¹(2.2) = -180°+77.2°-47.7°-65.6°=-216.1°. PM = 180° + (-216.1°) = -36.1° (negative)."},{"type":"paragraph","content":"**Stability:** Both the gain margin (≈ -17.9 dB) and phase margin (≈ -36°) are negative, which conclusively indicates that the closed-loop system is unstable. This is consistent with the system being Type-2 (two poles at the origin contribute -180° phase immediately), which inherently struggles to maintain adequate phase margin unless the zero (here at ω=0.25) is positioned aggressively enough to add sufficient phase lead before the gain crosses 0 dB; in this case the poles at ω=0.5 and ω=1 add further lag that pushes the phase past -180° before crossover, confirming instability."},{"type":"paragraph","content":"**Design remedy.** This result is instructive because it shows a Type-2 system's characteristic weakness: with two integrators, the -180° phase contribution is present at every frequency from the outset, leaving only a narrow margin (contributed by the single zero at ω=0.25 partially cancelling the lag of the two poles at ω=0.5 and ω=1) before the phase drops below -180°. Since the calculated phase margin is already negative at the current gain, simply reducing the gain K will not by itself rescue stability here in the way it often does for Type-0/Type-1 systems, because the phase crossover frequency ωpc (≈0.42 rad/s) is already below the gain crossover frequency ωgc (≈1.1 rad/s) — i.e., the magnitude is still above 0 dB at the frequency where phase has already reached -180°. The standard remedy is to add a lead compensator whose maximum phase boost is placed near ωgc to pull the phase back above -180° before the magnitude falls to 0 dB, effectively increasing the phase margin to a positive, adequate value (typically 30°-60°) without needing to reduce the loop gain (and hence without sacrificing the excellent steady-state accuracy that this Type-2 configuration otherwise provides for both step and ramp inputs)."},{"type":"paragraph","content":"**Cross-check via Nyquist reasoning.** The conclusion of instability can be independently sanity-checked using the encirclement interpretation of the Nyquist criterion: with P = 0 open-loop right-half-plane poles (both poles at the origin lie on, not inside, the imaginary axis, and are handled by the standard indentation), stability requires zero clockwise encirclements of the (-1, j0) point. A negative gain margin at the phase-crossover frequency (magnitude greater than 0 dB, i.e. greater than unity, exactly where the phase first reaches -180°) means the open-loop locus crosses the negative real axis to the left of the critical point (-1,j0) before its magnitude has decayed below unity — geometrically, this is precisely the condition under which the Nyquist plot does encircle (-1, j0), giving N > 0 and hence Z = N + P > 0 closed-loop right-half-plane poles, consistent with the instability already established from the Bode-domain margins."}]}},{"id":"control-system-2022-5E1393-pC-q5","number":5,"marks":15,"tags":["Part C"],"text":[{"type":"paragraph","content":"Q.5. Consider a unity feedback system having open loop transfer function :"},{"type":"math","expression":"G(s) = \\dfrac{1}{s^2+10s+15}","displayMode":true},{"type":"paragraph","content":"Calculate rise time, peak time, peak overshoot and settling time."}],"answer":{"summary":"For G(s)=1/(s²+10s+15) as a unity-feedback open-loop transfer function, the closed-loop system is actually overdamped (ζ>1) since ωn=√16=4 and ζ=10/8=1.25, so classical underdamped time-response formulas do not apply; instead the response is found from the two real closed-loop poles, giving no overshoot, and settling/rise times are computed from the dominant real pole.","explanation":[{"type":"paragraph","content":"The time-domain performance specifications — rise time, peak time, peak overshoot and settling time — are conventionally defined for the standard second-order underdamped step response, C(t) = 1 - [e^(-ζωnt)/√(1-ζ²)]·sin(ωdt+φ), which applies only when the closed-loop poles are a complex-conjugate pair (0 < ζ < 1). Before applying any of the standard formulas, the damping ratio of the actual closed-loop system must always be checked first, since a system with ζ ≥ 1 has purely real closed-loop poles and an entirely different, non-oscillatory step response for which the peak-time and percentage-overshoot formulas are meaningless (there being no peak to exceed the final value)."},{"type":"paragraph","content":"For unity feedback, the closed-loop transfer function is:"},{"type":"math","expression":"\\dfrac{C(s)}{R(s)} = \\dfrac{G(s)}{1+G(s)} = \\dfrac{1}{s^2+10s+16}","displayMode":true},{"type":"paragraph","content":"Comparing with standard form s²+2ζωns+ωn²:"},{"type":"math","expression":"\\omega_n^2 = 16 \\Rightarrow \\omega_n=4\\ \\text{rad/s}, \\qquad 2\\zeta\\omega_n=10 \\Rightarrow \\zeta = \\dfrac{10}{8}=1.25","displayMode":true},{"type":"paragraph","content":"Since ζ = 1.25 > 1, the closed-loop system is overdamped, not underdamped — the standard formulas for peak time, peak overshoot (which assume complex conjugate poles) do not apply, since there is no oscillation or overshoot in an overdamped response. This must be handled using the actual real, unequal closed-loop poles, found from the characteristic equation s²+10s+16=0:"},{"type":"paragraph","content":"**Why this arises here.** It is instructive to see why this particular problem lands in the overdamped regime even though it superficially looks like a routine second-order time-response question. The open-loop transfer function G(s) = 1/(s²+10s+15) itself already has a large damping ratio: comparing s²+10s+15 to s²+2ζ'ωn's+ωn'², the open-loop values are ωn' = √15 ≈ 3.87 rad/s and ζ' = 10/(2×3.87) ≈ 1.29, so the open-loop plant is already overdamped even before feedback is applied. Closing the unity feedback loop only shifts the constant term from 15 to 16 (since the DC loop gain here is unity), a very small change, so it is unsurprising that the closed-loop system remains overdamped (ζ = 1.25) rather than being pulled into the underdamped region — this would only happen if the feedback loop gain were substantially higher, which is not the case for this unity-feedback configuration."},{"type":"math","expression":"s = \\dfrac{-10\\pm\\sqrt{100-64}}{2} = \\dfrac{-10\\pm6}{2} \\;\\Rightarrow\\; s_1=-2,\\ s_2=-8","displayMode":true},{"type":"paragraph","content":"The closed-loop step response is a sum of two decaying exponentials (no oscillation): C(t) = 1 + Ae^{-2t} + Be^{-8t}, with A and B determined from initial conditions C(0)=0, Ċ(0)=0, giving (standard overdamped step-response formula for unity DC gain, since C(∞) = 16/16 = 1):"},{"type":"math","expression":"C(t) = 1 - \\dfrac{s_2}{s_2-s_1}e^{s_1t} + \\dfrac{s_1}{s_2-s_1}e^{s_2t} = 1 - \\dfrac{-8}{-6}e^{-2t}+\\dfrac{-2}{-6}e^{-8t} = 1-\\dfrac{4}{3}e^{-2t}+\\dfrac{1}{3}e^{-8t}","displayMode":true},{"type":"paragraph","content":"**Peak overshoot:** for an overdamped response there is no overshoot; peak overshoot = 0% (the response rises monotonically to the final value without exceeding it)."},{"type":"paragraph","content":"**Peak time:** since there is no overshoot peak, 'peak time' in the classical oscillatory sense does not exist; the response asymptotically approaches C(∞) = 1 without a distinct peak."},{"type":"paragraph","content":"**Rise time (10% to 90%, commonly used for overdamped systems since 0-100% rise time is undefined without a peak crossing):** solving C(t1) = 0.1 and C(t2) = 0.9 numerically from C(t) = 1 - 1.333e^{-2t} + 0.333e^{-8t}. Since the -8t term decays about 4× faster than the -2t term, for t beyond ~0.3s the response is dominated by the slower pole: C(t) ≈ 1 - 1.333e^{-2t}. Setting 1-1.333e^{-2t2}=0.9 ⟹ e^{-2t2}=0.075 ⟹ t2 = ln(13.33)/2 ≈ 1.294 s. Setting 1-1.333e^{-2t1}=0.1 (approximate, ignoring fast term at small t introduces some error, but as a working estimate) ⟹ e^{-2t1}=0.675 ⟹ t1 ≈ 0.196 s. So rise time tr = t2 - t1 ≈ 1.294 - 0.196 ≈ 1.10 s."},{"type":"paragraph","content":"**Settling time (2% criterion):** dominated by the slower (dominant) pole at s1 = -2, using the standard approximation ts ≈ 4/|dominant pole real part| = 4/2 = 2 s (using the 2% settling criterion applied to the dominant time constant of the slower exponential term)."},{"type":"paragraph","content":"This question illustrates an important conceptual point: applying the standard underdamped formulas (tp = π/ωd, %OS = e^{-πζ/√(1-ζ²)}×100) blindly to any second-order system without first checking ζ is a common error — here ζ = 1.25 > 1 confirms the system is overdamped, so peak time and percentage overshoot are not applicable, and rise/settling time must instead be computed from the actual real exponential pole locations."},{"type":"paragraph","content":"**Exact rise-time refinement.** The estimate above used the dominant-pole approximation (dropping the fast e^{-8t} term) when solving for t1 at the 10% crossing, which introduces some error since the fast term is not yet fully negligible at small t. Solving the full equation 1 - 1.333e^{-2t1} + 0.333e^{-8t1} = 0.1 numerically (e.g. by iteration): at t1=0.15, C=1-1.333×0.741+0.333×0.301=1-0.988+0.100=0.112 (slightly above 0.1); at t1=0.17, C=1-1.333×0.712+0.333×0.254=1-0.949+0.0846=0.135 (moving the wrong way, indicating the approximation crosses non-monotonically at very small t due to the competing fast term); at t1=0.13, C=1-1.333×0.771+0.333×0.352=1-1.028+0.117=0.089 (below 0.1). Interpolating between t1=0.13 (C=0.089) and t1=0.15 (C=0.112), the 10% crossing occurs at approximately t1 ≈ 0.14 s, close to but slightly refined from the earlier dominant-pole estimate of 0.196 s, illustrating that for high-accuracy work the full two-exponential expression should be solved rather than relying purely on the dominant-pole simplification, especially near the start of the response where the faster pole's contribution has not yet fully decayed."},{"type":"paragraph","content":"**Physical interpretation.** An overdamped response, while free of overshoot and oscillation (a desirable property in applications where overshoot could be damaging or unsafe, such as valve or elevator positioning), is inherently slower to reach the vicinity of its final value than a well-tuned underdamped response, because it lacks the 'overshoot-then-settle' behavior that a slightly underdamped system uses to approach the target faster from below. In this problem, the very large effective damping ratio (ζ = 1.25) arises because the plant's own characteristic polynomial (s²+10s+15) already has widely separated real poles (s = -1.5 and s = -10, using the open-loop values before closing the loop) and unity feedback here adds relatively little corrective dynamics (the closed-loop DC gain moves from 1/15 open-loop to 1/16 closed-loop, a minor change), so the closed-loop pole locations remain close to the open-loop ones and stay real rather than becoming complex. If faster response with retained non-oscillatory behavior were desired, a designer would typically add a compensator (e.g., a PD or lead controller) to reposition the dominant closed-loop pole further from the origin (reducing its time constant) while carefully avoiding excessive gain that would drive ζ below 1 and reintroduce overshoot — illustrating the general trade-off in second-order response shaping between speed, damping and overshoot that runs through this entire question paper."}]}}],"examYear":2022}],"loading":false,"error":null}}}

Q2Control System

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