RTUEE / EC / EEEYr 2020 · Sem 62020

Q3Control System

Question

16 marks

Q.2 (a) Derive the expression for time response of second order control system subjected to unit step input function. [8]

(b) Consider a unity feedback system having open loop transfer function G(s) = 1/(s²+10s+15). Calculate rise time, peak time, peak overshoot and settling time. [8]

Answer

The time response of a standard second-order system to a unit step input is derived by taking the inverse Laplace transform of C(s) = ωn²/[s(s²+2ζωn·s+ωn²)], yielding c(t) = 1 - [e^(-ζωnt)/√(1-ζ²)]·sin(ωd·t + φ) for the underdamped case; applying this to G(s)=1/(s²+10s+15) with unity feedback gives a closed-loop system with ζ=1.25 (overdamped), so peak time and overshoot do not exist in the classical sense, with rise time (10-90%) ≈ 1.16s and 2% settling time (dominant pole) ≈ 2s.

(a) Time Response of Second-Order System to Unit Step Input

A standard second-order control system has the closed-loop transfer function:

where ωn is the undamped natural frequency and ζ is the damping ratio. For a unit step input, R(s)=1/s, so:

For the underdamped case (0<ζ<1), the denominator's complex conjugate roots are at s = -ζωn ± jωn√(1-ζ²) = -ζωn ± jωd, where ωd = ωn√(1-ζ²) is the damped natural frequency. Performing partial fraction expansion and taking the inverse Laplace transform yields the standard time-response expression:

This response consists of a steady-state value of 1 (the final value, matching the unit step input, confirming the system has unity DC gain) plus a decaying sinusoidal oscillation term, whose amplitude decays exponentially with time constant 1/(ζωn) and whose oscillation frequency is the damped natural frequency ωd. From this general expression, the standard time-domain performance specifications are derived: rise time tr = (π-φ)/ωd (time to first reach the final value); peak time tp = π/ωd (time to reach the first, maximum overshoot peak); peak overshoot Mp = e^(-πζ/√(1-ζ²)) (the fractional amount by which the response exceeds the final value at the peak); and settling time ts ≈ 4/(ζωn) (2% criterion) or 3/(ζωn) (5% criterion), the time for the response to remain within the specified tolerance band around the final value.

(b) Time-Domain Specifications for G(s) = 1/(s²+10s+15)

Given: unity feedback system with open-loop transfer function G(s) = 1/(s²+10s+15).

The closed-loop transfer function for a unity feedback system is C(s)/R(s) = G(s)/(1+G(s)):

Comparing with the standard form ωn²/(s²+2ζωns+ωn²): ωn² = 16, so ωn = 4 rad/s; and 2ζωn = 10, so ζ = 10/(2×4) = 1.25.

Critical observation: since ζ = 1.25 > 1, this closed-loop system is overdamped (not underdamped), meaning the standard underdamped formulas for peak time and peak overshoot given in part (a) do NOT apply — an overdamped second-order system exhibits a purely exponential-type (non-oscillatory) step response with no overshoot at all, so peak time and peak overshoot, as classically defined for underdamped systems, do not exist for this particular system.

Exact overdamped step response: the characteristic equation s²+10s+16=0 factors as (s+2)(s+8)=0, giving real poles at s=-2 and s=-8 (both real and distinct, confirming the overdamped condition). The unit-step response is found via partial fractions of C(s) = 1/[s(s+2)(s+8)]:

Rise time (10%-90%): solving c(t) = 0.1×(1/16) and c(t) = 0.9×(1/16) numerically for the respective crossing times gives approximately t10% ≈ 0.14 s and t90% ≈ 1.30 s, so:

Peak time and peak overshoot: since the response is purely overdamped (monotonically approaching its final value without ever exceeding it), there is no overshoot at all (Mp = 0%) and hence no meaningful 'peak time' in the classical sense — the response simply rises monotonically and asymptotically approaches its final value of 1/16 = 0.0625 as t→∞.

Settling time (2% criterion): for an overdamped system, settling time is dominated by the slower (less negative, here s=-2) of the two real poles, giving an approximate settling time of:

Important conclusion: this problem illustrates a valuable practical lesson — before blindly applying the standard underdamped second-order formulas (peak time, overshoot) to any given transfer function, one must first verify that ζ<1 (underdamped condition); when ζ≥1 (critically damped or overdamped, as found here with ζ=1.25), those particular formulas are not applicable, and the actual time-domain response must instead be found through direct partial-fraction inversion of the overdamped (real-pole) closed-loop transfer function, as carried out above.

Cross-Check via the Alternative Overdamped Formula

As an independent check, the general overdamped step response of a second-order system can also be written directly in terms of ζ and ωn (rather than first factoring into real poles) using the standard formula:

Substituting ζ=1.25 and ωn=4 rad/s gives √(ζ²-1)=√(1.5625-1)=√0.5625=0.75, so the two poles are at s=-ζωn±ωn√(ζ²-1) = -5±3, i.e., s=-2 and s=-8, exactly matching the factorization (s+2)(s+8) obtained earlier from the characteristic equation s²+10s+16=0, confirming that both derivation routes (direct factoring of the characteristic polynomial, and the general ζ,ωn-parametrized overdamped formula) are consistent with one another.

Final value and steady-state error check: applying the final value theorem directly to C(s)=1/[s(s+2)(s+8)] gives lim(t→∞)c(t) = lim(s→0)[s·C(s)] = 1/(2×8) = 1/16 = 0.0625, matching the constant term obtained in the partial-fraction expansion above; since the reference input is a unit step (final value 1), the steady-state error is ess = 1-0.0625 = 0.9375, a very large steady-state error. This large residual error arises because the given open-loop transfer function G(s)=1/(s²+10s+15) is Type-0 (no free integrator), so under unity feedback it necessarily produces a finite, nonzero steady-state error for a step input — a general property of Type-0 systems that can be independently confirmed using the position error constant Kp = lim(s→0)G(s) = 1/15, giving ess = 1/(1+Kp) = 1/(1+1/15) = 15/16 = 0.9375, exactly matching the value computed above from the explicit time-response expression.

Physical and Design Interpretation

The heavy overdamping (ζ=1.25) found here means the closed-loop response, while completely free of overshoot and oscillation, is noticeably sluggish compared to the fastest possible non-oscillatory response (which occurs at critical damping, ζ=1). This is evident from the two widely-separated real poles at s=-2 and s=-8: the pole closer to the origin (s=-2, corresponding to the slower of the two exponential decay modes) dominates the settling behavior, since its associated term e^(-2t) decays roughly four times more slowly than the e^(-8t) term contributed by the faster pole, and it is this slower (dominant) pole that governs both the rise time and settling time estimates computed above. In a practical design context, if faster settling were required while still avoiding overshoot entirely, a designer would need to either increase the loop gain (moving both poles closer together toward critical damping, ζ→1, at which point the response time is minimized for a still non-oscillatory response) or apply a derivative/lead compensator to reshape the closed-loop pole locations — illustrating the classical trade-off in second-order system design between response speed and oscillatory overshoot that recurs throughout root-locus and compensator-design problems in this subject. It is also worth noting that, since both closed-loop poles (s=-2, s=-8) are real and negative, the Routh-Hurwitz criterion applied to s²+10s+16=0 trivially confirms stability (both coefficients 1, 10, 16 are positive, and being only a second-order polynomial with all-positive coefficients guarantees stability), consistent with the fact that this system's step response was found to settle smoothly to a finite bounded value rather than growing or oscillating indefinitely.

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