Sunday, November 2, 2025

Lead compensator provides

 Q441. The z-transform of a unit step signal is:

A) 11z1\frac{1}{1 - z^{-1}}
B) zz1\frac{z}{z - 1}
C) 1z1\frac{1}{z - 1}
D) z1z\frac{z - 1}{z}

Answer: B) zz1\frac{z}{z - 1}
Explanation:
Z-transform of unit step u[k]u[k] = zk=zz1\sum z^{-k} = \frac{z}{z-1}.


Q442. A discrete system is stable if:
A) Poles inside unit circle
B) Poles outside unit circle
C) Poles on imaginary axis
D) Poles on unit circle

Answer: A)
Explanation:
For discrete-time systems, all poles must lie inside the unit circle for stability.


Q443. Bilinear transformation maps the left-half s-plane to:
A) Inside the unit circle
B) Outside the unit circle
C) Imaginary axis
D) Real axis

Answer: A)
Explanation:
Bilinear transform ensures that stable analog systems remain stable in digital domain.


Q444. The relation between s and z in bilinear transformation is:
A) s=2T1z11+z1s = \frac{2}{T}\frac{1 - z^{-1}}{1 + z^{-1}}
B) s=T21+z11z1s = \frac{T}{2}\frac{1 + z^{-1}}{1 - z^{-1}}
C) z=esTz = e^{sT}
D) s=1Tln(z)s = \frac{1}{T}\ln(z)

Answer: A)
Explanation:
This maps continuous-time to discrete-time while preserving stability.


Q445. The effect of sampling on a system is:
A) Adds delay
B) Improves bandwidth
C) Reduces speed
D) Increases order

Answer: A)
Explanation:
Sampling introduces delay (sample-and-hold behavior), which affects phase and stability.


Q446. The Nyquist frequency is:
A) Half the sampling frequency
B) Twice the sampling frequency
C) Equal to sampling frequency
D) Independent of sampling

Answer: A)
Explanation:
Nyquist frequency = fs/2f_s/2, beyond which aliasing occurs.


Q447. To avoid aliasing, the signal must be:
A) Band-limited below half sampling frequency
B) Doubled in frequency
C) Modulated
D) Demodulated

Answer: A)
Explanation:
Nyquist criterion — sampling rate ≥ 2 × maximum signal frequency.


Q448. The term "zero-order hold" (ZOH) refers to:
A) Constant output between samples
B) Linear interpolation
C) Sinusoidal output
D) Exponential hold

Answer: A)
Explanation:
ZOH holds the sampled value constant until the next sample.


Q449. In a sampled-data system, increasing the sampling period:
A) Reduces stability
B) Improves phase margin
C) Reduces delay
D) Increases speed

Answer: A)
Explanation:
Larger sampling interval → more delay → lower phase margin → less stable.


Q450. Deadbeat control aims to:
A) Reach steady-state in minimum number of samples
B) Minimize overshoot
C) Reduce rise time
D) Improve bandwidth

Answer: A)
Explanation:
Deadbeat control forces output to reach the desired value in finite sampling steps.


Q451. A phase-lead compensator is used to:
A) Improve phase margin and speed of response
B) Reduce steady-state error
C) Increase damping ratio only
D) Reduce gain margin

Answer: A)
Explanation:
Lead compensation adds phase advance → faster response, better stability.


Q452. A phase-lag compensator improves:
A) Steady-state accuracy
B) Damping
C) Speed
D) Bandwidth

Answer: A)
Explanation:
Lag compensator increases low-frequency gain → improves accuracy.


Q453. For a unity feedback system, steady-state error for a unit parabolic input in Type-2 system is:
A) 0
B) Finite
C) Infinite
D) Undefined

Answer: A)
Explanation:
Type-2 system has zero steady-state error for parabolic input.


Q454. The steady-state error to a ramp input in Type-0 system is:
A) Infinite
B) Zero
C) Finite
D) Depends on gain

Answer: A)
Explanation:
Type-0 → no integrator → infinite error for ramp.


Q455. The effect of increasing damping ratio on overshoot:
A) Decreases overshoot
B) Increases overshoot
C) No change
D) Depends on frequency

Answer: A)
Explanation:
Higher damping → less oscillation → reduced overshoot.


Q456. The root locus of a system moves to the right half-plane when:
A) Gain increases and system is unstable
B) Gain decreases
C) Poles cancel zeros
D) System has no integrator

Answer: A)
Explanation:
Excessive gain can push closed-loop poles to the right → instability.


Q457. The slope of the asymptotes in root locus is given by:
A) (2k+1)180°nm\frac{(2k+1)180°}{n-m}
B) k180°n+m\frac{k180°}{n+m}
C) 90°n\frac{90°}{n}
D) 45°m\frac{45°}{m}

Answer: A)
Explanation:
Standard formula for asymptotic angles in root locus.


Q458. The polar plot of a first-order lag system G(s)=11+sTG(s)=\frac{1}{1+sT} starts at:
A) 1∠0°
B) 0∠-90°
C) 1∠-90°
D) 0∠0°

Answer: A)
Explanation:
At ω=0, magnitude=1, phase=0°; at high ω, phase→-90°.


Q459. The magnitude in a Bode plot decreases by 20 dB/decade for each:
A) Pole
B) Zero
C) Lead element
D) Phase lead

Answer: A)
Explanation:
Each pole adds -20 dB/decade slope to magnitude curve.


Q460. The effect of adding a zero to a system is:
A) Increase bandwidth and overshoot
B) Reduce speed
C) Decrease damping
D) Increase order

Answer: A)
Explanation:
A zero adds phase lead → faster but may increase overshoot.


Q461. The state feedback control law is of form:
A) u=Kx+ru = -Kx + r
B) u=Kxu = Kx
C) u=K(rx)u = K(r-x)
D) u=x+ru = x + r

Answer: A)
Explanation:
State feedback modifies input using all states to place poles at desired locations.


Q462. The objective of pole placement is to:
A) Assign closed-loop poles for desired performance
B) Increase open-loop gain
C) Reduce steady-state error
D) Improve bandwidth only

Answer: A)
Explanation:
By choosing feedback matrix K, desired closed-loop pole locations are achieved.


Q463. The Ackermann’s formula is used for:
A) State feedback gain computation
B) Observability matrix
C) Root locus
D) Transfer function

Answer: A)
Explanation:
Ackermann’s formula gives K for pole placement when system is controllable.


Q464. The controllability matrix for a 2×2 system is:
A) [B AB]
B) [A B]
C) [B A²]
D) [A B A²B]

Answer: A)
Explanation:
Controllability matrix = [B AB] for a second-order system.


Q465. A system is observable if:
A) Observability matrix has full rank
B) Controllability matrix is zero
C) Matrix A is diagonal
D) Eigenvalues are equal

Answer: A)
Explanation:
Full-rank observability matrix ⇒ all states can be inferred from outputs.


Q466. The unit impulse response h(t)h(t) is the inverse Laplace transform of:
A) Transfer function G(s)G(s)
B) 1/G(s)1/G(s)
C) G(s)H(s)G(s)H(s)
D) sG(s)sG(s)

Answer: A)
Explanation:
h(t)=L1[G(s)]h(t) = \mathcal{L}^{-1}[G(s)], the impulse response directly corresponds to system transfer function.


Q467. The system bandwidth is inversely proportional to:
A) Rise time
B) Overshoot
C) Damping ratio
D) Phase margin

Answer: A)
Explanation:
High bandwidth → fast rise time.


Q468. Gain crossover frequency is where:
A) |G(jω)| = 1
B) Phase = 0°
C) |G(jω)| = 0
D) |G(jω)| = -1

Answer: A)
Explanation:
At ω_gc, magnitude = 1 (0 dB) — used in phase margin calculation.


Q469. Phase crossover frequency is where:
A) ∠G(jω) = -180°
B) ∠G(jω) = 0°
C) |G(jω)| = 1
D) |G(jω)| = 0

Answer: A)
Explanation:
At ω_pc, phase = -180° — used in gain margin calculation.


Q470. Lead compensator provides:
A) Positive phase lead
B) Negative phase lag
C) Constant gain
D) No phase shift

Answer: A)
Explanation:
Adds +ve phase → faster response and improved phase margin.



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