Question 1

Q.1. Define forward path and forward path gain.

Accepted Answer

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.

Question 2

Q.2. Differentiate closed loop and open loop control system.

Accepted Answer

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.

Question 3

Q.3. State Mason's gain formula.

Accepted Answer

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.

Question 4

Q.4. What is meant by steady state error?

Accepted Answer

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.

Question 5

Q.5. List the main properties of a state transition matrix.

Accepted Answer

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).

Question 6

Q.6. State the necessary condition for stability.

Accepted Answer

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.

Question 7

Q.7. What are the characteristics of phase-lead network?

Accepted Answer

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.

Question 8

Q.8. What is dominant pole?

Accepted Answer

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.

Question 9

Q.9. What are constant M and N circle?

Accepted Answer

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.

Question 10

Q.10. What is the value of gain K at any given point on the root locus?

Accepted Answer

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).

Question 11

Q.1. Find the transfer function of the system shown in fig. using block diagr…

Accepted Answer

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].

Question 12

Q.2. Obtain the differential equations describing the complete dynamics of th…

Accepted Answer

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.

Question 13

Q.3. A servomechanism is represented by the equation :

Accepted Answer

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.

Question 14

Q.4. Comment on the stability of the system whose characteristic equation is …

Accepted Answer

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.

Question 15

Q.5. Compare PI, PD and PID controllers.

Accepted Answer

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.

Question 16

Q.6. Check for controllability and observability of a system having following…

Accepted Answer

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).

Question 17

Q.7. Explain regulator problem and tracking problem in detail.

Accepted Answer

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.

Question 18

Q.1. A certain feedback control system is described by the following transfer…

Accepted Answer

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.

Question 19

Q.2. Draw the Root locus plot for a unity feedback system with an open loop t…

Accepted Answer

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.

Question 20

Q.3. Draw the Nyquist plot of open loop transfer function :

Accepted Answer

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.

Question 21

Q.4. Sketch the asymptotic Bode Plot for the transfer function given below :

Accepted Answer

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.

Question 22

Q.5. Consider a unity feedback system having open loop transfer function :

Accepted Answer

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.

Control System