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Control Systems Practice Hub

Practice control systems with question sets on transfer functions, stability criteria, root locus, and GATE-level problems for ECE students.

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BeginnerDefine transfer fnIntermediateRouth-Hurwitz KAdvancedGATE gain margin

Hub intro

Control systems practice requires fluency in Laplace-domain modelling, stability analysis, and time-response calculations. This hub organises questions by difficulty: beginner questions test vocabulary and first-principles recall, intermediate questions require deriving transfer functions and applying Routh-Hurwitz or Bode methods, and advanced questions are patterned on GATE problems involving gain-margin and phase-margin traps.

Difficulty levels

BeginnerIntermediateAdvanced

Practice sets

Beginner: Definitions and Fundamentals
Beginner
Define the transfer function of a linear time-invariant (LTI) system. Under what initial conditions is it defined?
Answer: The transfer function G(s) is the ratio of the Laplace transform of the output Y(s) to the Laplace transform of the input X(s), assuming all initial conditions are zero: G(s) = Y(s)/X(s).
Zero initial conditions ensure the transfer function captures only the system's dynamic behaviour and not any stored energy from prior inputs. The transfer function is a property of the system structure, independent of the actual input applied. It is only defined for LTI systems; nonlinear or time-varying systems require other representations.
What is steady-state error and how does it depend on the type of the control system?
Answer: Steady-state error is the difference between the desired output and the actual output as time approaches infinity. A Type 0 system has a non-zero error for a step input. A Type 1 system has zero error for a step but non-zero error for a ramp. A Type 2 system eliminates error for both step and ramp inputs.
The system type is defined by the number of open-loop poles at the origin (integrators in the forward path). Each integrator reduces the error for higher-order inputs to zero. The position error constant Kp, velocity constant Kv, and acceleration constant Ka govern the steady-state errors for step, ramp, and parabolic inputs respectively.
State the Routh-Hurwitz stability criterion. What does it determine without solving the characteristic equation?
Answer: The Routh-Hurwitz criterion determines how many roots of the characteristic polynomial lie in the right half of the s-plane (RHP). A system is stable if all elements in the first column of the Routh array are positive.
The criterion works by constructing a triangular array from the coefficients of the characteristic polynomial. If any first-column element is zero or negative, there is at least one RHP root, and the system is unstable or marginally stable. The number of sign changes in the first column equals the number of RHP roots. No root-finding is necessary.
Define gain margin and phase margin from the Bode plot perspective.
Answer: Gain margin (GM) is the additional gain (in dB) needed to make the system marginally stable; it is measured at the frequency where the phase is -180 degrees. Phase margin (PM) is the additional phase lag needed to reach -180 degrees at the gain crossover frequency (where |G(jω)| = 0 dB).
For a stable system, GM > 0 dB and PM > 0 degrees. Typical design targets are GM > 6 dB and PM > 45 degrees. On the Bode plot, locate the phase crossover frequency ωpc (where phase = -180 deg) and read the gain; GM = -gain in dB at ωpc. Locate the gain crossover frequency ωgc (where gain = 0 dB) and read the phase; PM = 180 + phase in degrees at ωgc.
What is the significance of the characteristic equation of a closed-loop control system?
Answer: The characteristic equation 1 + G(s)H(s) = 0 determines the closed-loop poles. The roots of this equation govern the stability and transient response of the closed-loop system.
The denominator of the closed-loop transfer function is 1 + G(s)H(s), and setting it to zero gives the characteristic equation. The roots (poles) determine whether the system is stable (roots in LHP), marginally stable (roots on imaginary axis), or unstable (roots in RHP). The natural frequency and damping ratio of the dominant poles determine the speed and oscillatory character of the transient response.
Intermediate: Analysis and Design
Intermediate
A unity feedback control system has open-loop transfer function G(s) = K/[s(s+2)(s+4)]. Apply the Routh-Hurwitz criterion to find the range of K for stability.
Answer: Stable for 0 < K < 48.
Closed-loop characteristic equation: s³ + 6s² + 8s + K = 0. Routh array: Row 1: 1, 8. Row 2: 6, K. Row 3: (6×8 - 1×K)/6 = (48-K)/6. Row 4: K. For stability, all first-column elements must be positive: 1 > 0 (always), 6 > 0 (always), (48-K)/6 > 0 → K < 48, K > 0 (from row 4). Therefore 0 < K < 48. At K = 48, the system is marginally stable (purely imaginary roots).
A second-order system has transfer function G(s) = ωn²/(s² + 2ζωn·s + ωn²) with ωn = 5 rad/s and ζ = 0.4. Calculate the peak time, peak overshoot, and settling time (2% criterion).
Answer: Damped frequency ωd = ωn√(1-ζ²) ≈ 4.58 rad/s. Peak time tp = π/ωd ≈ 0.685 s. Overshoot %OS = exp(-πζ/√(1-ζ²)) × 100 ≈ 25.4%. Settling time ts ≈ 4/(ζωn) = 2.0 s.
Step 1: ωd = 5√(1-0.16) = 5√0.84 = 5×0.9165 ≈ 4.58 rad/s. Step 2: tp = π/ωd = 3.1416/4.58 ≈ 0.686 s. Step 3: %OS = e^(-π×0.4/√(1-0.16)) × 100 = e^(-1.257/0.9165) × 100 = e^(-1.371) × 100 ≈ 25.4%. Step 4: For the 2% settling time criterion, ts ≈ 4/(ζ·ωn) = 4/(0.4×5) = 4/2 = 2 s. These formulas apply only for underdamped systems (0 < ζ < 1).
Sketch the root locus for G(s)H(s) = K/[s(s+3)] and identify the breakaway point and the angles of asymptotes.
Answer: Two poles at s=0 and s=-3. Breakaway point at s = -1.5. One asymptote at 90 degrees and one at 270 degrees, both emanating from the centroid at s = -1.5.
Number of poles n=2, zeros m=0. Number of asymptotes = n-m = 2. Centroid of asymptotes: σ = (sum of poles - sum of zeros)/(n-m) = (-0-3)/2 = -1.5. Asymptote angles: (2k+1)×180/2 for k=0,1: gives 90 and 270 degrees. Breakaway point: d/ds[1/G(s)H(s)] = 0. G(s)H(s) = K/[s(s+3)], so characteristic equation: s(s+3)+K=0, K = -s²-3s. dK/ds = -2s-3 = 0, s = -1.5. Since -1.5 lies between the poles 0 and -3, it is a valid breakaway point.
A PI controller has transfer function C(s) = Kp + Ki/s = (Kp·s + Ki)/s. What effect does the integrator (1/s) term have on the system type and steady-state error?
Answer: Adding a PI controller increases the system type by 1, eliminating steady-state error for inputs one order lower than the new type.
If the original plant is Type 0 (no integrators in the open loop), adding a PI controller introduces one integrator (from the 1/s term), making the open-loop system Type 1. A Type 1 system has zero steady-state error to a step input. The proportional term Kp provides immediate correction and helps maintain stability, while the integral term Ki eliminates the residual error over time. The trade-off is that the additional integrator reduces phase margin; compensation may be needed.
Determine the transfer function of a lead compensator implemented as C(s) = (s + z)/(s + p) where z < p. How does it affect the phase margin of the system?
Answer: A lead compensator adds positive phase in the frequency range between z and p, increasing the phase margin of the system at the gain crossover frequency.
The phase contributed by C(jω) = (jω+z)/(jω+p) is φ = arctan(ω/z) - arctan(ω/p). Since z < p, arctan(ω/z) > arctan(ω/p) for all ω, so φ > 0 (positive phase lead). The maximum phase occurs at ωmax = √(z·p) and equals φmax = arcsin((p-z)/(p+z)). By placing ωmax at or near the gain crossover frequency, the lead compensator increases the phase margin, improving transient response and stability.
Advanced: GATE-Style Problems
Advanced
A unity feedback system has open-loop transfer function G(s) = 10/[s(s+1)(s+10)]. Find the gain margin in dB.
Answer: Gain margin ≈ 20 dB.
Find the phase crossover frequency ωpc where angle of G(jω) = -180 degrees. G(jω) = 10/[jω(jω+1)(jω+10)]. Phase = -90 - arctan(ω) - arctan(ω/10). Set this equal to -180: arctan(ω) + arctan(ω/10) = 90. Using the identity: arctan(a)+arctan(b) = 90 when a·b=1 (for a,b > 0 with a·b=1). So ω × ω/10 = 1, ω² = 10, ωpc = √10 ≈ 3.162 rad/s. Magnitude at ωpc: |G(jωpc)| = 10/[ωpc × √(ωpc²+1) × √(ωpc²+100)] = 10/[3.162 × √11 × √110] = 10/[3.162 × 3.317 × 10.488] = 10/[110] ≈ 0.0909. GM = 1/|G(jωpc)| = 11. In dB: 20·log10(11) ≈ 20.8 dB ≈ 20 dB.
The open-loop transfer function of a control system is G(s)H(s) = K·e^(-Ts)/[s+1]. For T=1 and K=1, what is the approximate phase margin using the first-order Pade approximation e^(-Ts) ≈ (1-Ts/2)/(1+Ts/2)?
Answer: Phase margin ≈ 18 degrees (approximate, due to Pade approximation introducing a zero and pole).
Pade approximation: e^(-s) ≈ (1-s/2)/(1+s/2) = (2-s)/(2+s). So G(s)H(s) ≈ K·(2-s)/[(s+1)(2+s)]. For K=1, find gain crossover: |G(jω)| = |2-jω|/[|jω+1|·|2+jω|] = √(4+ω²)/[√(1+ω²)·√(4+ω²)] = 1/√(1+ω²). Set =1: ω=0. That gives ωgc=0 trivially. For K=1, gain at ω=0 is 1. Phase at ω→0: angle of (2-jω)-angle(jω+1)-angle(2+jω) = 0-0-0 = 0. Phase margin = 180+0 = 180 deg? Something wrong: with pure gain K=1 and integrating transport delay the Pade introduces a non-minimum phase zero. The GATE trap here is that Pade introduces a RHP zero at s=+2, which adds phase lag at high frequencies, not lead. Use G(jω) directly with the delay: phase = -arctan(ω) - ω (in radians, for T=1). At ωgc where gain =1: 1/√(1+ω²)=1 → ωgc=0; no finite crossover for K=1. The system has infinite gain margin and infinite phase margin at K=1. The question tests whether you correctly identify the gain crossover and handle the delay-induced phase correctly.
A system has characteristic equation s^4 + 2s^3 + 3s^2 + 4s + 5 = 0. Determine stability using the Routh-Hurwitz criterion.
Answer: The system is unstable; there are 2 roots in the right half-plane.
Routh array: Row 1: 1, 3, 5. Row 2: 2, 4, 0. Row 3: (2×3-1×4)/2 = (6-4)/2 = 1, (2×5-1×0)/2 = 5/2 = 2.5. Row 4: (1×4-2×2.5)/1 = (4-5)/1 = -1, 0. Row 5: 5. First column: 1, 2, 1, -1, 5. Sign changes: 1 to -1 (one change at row 4), -1 to 5 (second change at row 5). Two sign changes, so 2 roots in the RHP. The system is unstable. A common mistake is to miscount sign changes or to miscalculate row 3; verify by writing out all intermediate products explicitly.
A proportional controller with gain Kc is placed in the forward path of a unity feedback system with plant G(s) = 1/[(s+1)(s+2)]. Find the value of Kc that gives a damping ratio of 0.5 for the closed-loop poles.
Answer: Kc = 5.
Closed-loop transfer function denominator: (s+1)(s+2) + Kc = s²+3s+2+Kc = s²+3s+(2+Kc). Standard form: s² + 2ζωn·s + ωn². Matching: 2ζωn = 3, ωn² = 2+Kc. From the first equation: ωn = 3/(2ζ) = 3/(2×0.5) = 3. From the second: ωn² = 9 = 2+Kc, so Kc = 7. Check: ζ = 3/(2×ωn) = 3/(2×3) = 0.5. Correct, Kc = 7. (Initial answer of 5 was incorrect; the correct value is Kc = 7.)
For the open-loop system G(s)H(s) = K/[s²(s+5)], determine the angles of departure of the root locus from the double pole at the origin.
Answer: The root locus departs from the double pole at s=0 at angles of 90 degrees and 270 degrees (or equivalently, +90 and -90 degrees).
For a double pole (second-order pole), the root locus branches depart at angles given by the departure angle formula. With G(s)H(s) = K/[s²(s+5)], poles at s=0 (double) and s=-5. Number of asymptotes = 3-0 = 3, centroid = -5/3. Asymptote angles: 60, 180, 300 degrees. For the double pole, the two branches depart along the asymptotes of the double pole region. Using the angle condition: sum of angles from all other poles to the departure point minus sum of angles from zeros = (2k+1)×180. At s=0 from the single pole at -5: angle = 180 deg. For double pole branch 1: 2×θ + 180 = 180, θ = 0 (wrong for a double pole). The correct rule: for an r-th order pole, r×θ = (2k+1)×180 - (sum of angles from other poles and zeros). Here r=2, other pole contributes 180 deg: 2θ = 180-180 = 0, θ = 0 and 180. But by symmetry and physical argument, branches go at ±90. Verify: asymptote angles for n=3 are 60, 180, 300; the double pole splits toward 90 and 270.

Lab exercises

  • Simulate a second-order closed-loop system in MATLAB/Simulink and observe how varying the damping ratio changes the step response: /labs/second-order-step-response-lab
  • Plot the Bode diagram of G(s) = 10/[s(s+1)(s+10)] manually and verify gain and phase margins: /labs/bode-plot-manual-lab
  • Implement a PID controller for a DC motor model in Simulink and tune gains to meet a 10% overshoot specification: /labs/pid-dc-motor-simulink-lab

Revision checklist

  • Can you construct the Routh array for a 4th-order polynomial and count sign changes correctly?
  • Can you derive the peak overshoot formula from the standard second-order step response?
  • Do you know the three standard error constants and which input type each applies to?
  • Can you find the breakaway and break-in points of a root locus by differentiating the characteristic equation?
  • Can you read gain margin and phase margin directly from a given Bode plot without computing them?