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Control Systems & Signal Processing (Electronics Engineering) Solved Questions & Notes (2026) - Apex Rankers

Engineering & Technology > Electronics Engineering > Control Systems & Signal Processing

35 Total Solved Questions
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Control Systems & Signal Processing

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Q. 1 Electronics Engineering
Difficulty: Easy (1 Mark)
A linear time-invariant (LTI) continuous-time system is BIBO (Bounded-Input Bounded-Output) stable if and only if:
A
All closed-loop poles lie strictly in the left half of the s-plane (LHP)
✓ Correct
B
All poles lie on the imaginary axis
C
All zeros lie in the right half of the s-plane
D
The system impulse response diverges to infinity as t approaches infinity
💡 Step-by-Step Explanation & Concept Rationale
For an LTI continuous system to be BIBO stable, all poles of the transfer function must have negative real parts, meaning they must lie strictly in the open left-half of the s-plane.
Q. 2 Electronics Engineering
Difficulty: Easy (1 Mark)
In a discrete-time LTI system with transfer function H(z), BIBO stability requires that all system poles lie:
A
Strictly inside the unit circle (|z| < 1) in the z-plane
✓ Correct
B
Strictly outside the unit circle (|z| > 1)
C
On the real axis only
D
Along the imaginary axis
💡 Step-by-Step Explanation & Concept Rationale
In discrete-time systems, the stable region in the s-plane (Re(s) < 0) maps into the interior of the unit circle in the z-plane (|z| < 1).
Q. 3 Electronics Engineering
Difficulty: Medium (1 Mark)
In the Routh-Hurwitz stability criterion, the number of sign changes in the first column of the Routh array equals:
A
The number of closed-loop poles in the left-half s-plane
B
The number of closed-loop poles with positive real parts (in the right-half s-plane)
✓ Correct
C
The total number of system zeros
D
The system damping ratio
💡 Step-by-Step Explanation & Concept Rationale
According to the Routh-Hurwitz theorem, the number of roots of the characteristic equation with positive real parts (RHP poles) is exactly equal to the number of sign changes in the first column of the Routh array.
Q. 4 Electronics Engineering
Difficulty: Easy (1 Mark)
For a standard second-order system with transfer function G(s) = omega_n^2 / (s^2 + 2*zeta*omega_n*s + omega_n^2), what is the system response when damping ratio zeta = 0?
A
Overdamped (non-oscillatory)
B
Critically damped
C
Undamped with sustained sinusoidal oscillations at frequency omega_n
✓ Correct
D
Exponentially decaying with zero overshoot
💡 Step-by-Step Explanation & Concept Rationale
When zeta = 0, the roots are s = +-j*omega_n (purely imaginary), causing undamped, perpetual sinusoidal oscillation at the natural frequency omega_n.
Q. 5 Electronics Engineering
Difficulty: Easy (1 Mark)
A standard second-order control system is critically damped when the damping ratio zeta is equal to:
A
zeta = 0
B
0 < zeta < 1
C
zeta = 1
✓ Correct
D
zeta > 1
💡 Step-by-Step Explanation & Concept Rationale
When zeta = 1, the system is critically damped: two repeated real roots at s = -omega_n, giving the fastest response possible without overshoot.
Q. 6 Electronics Engineering
Difficulty: Medium (1 Mark)
What is the steady-state error of a Type-1 control system subjected to a unit step input?
A
Zero
✓ Correct
B
Constant finite value 1 / (1 + Kp)
C
Infinity
D
Undetermined
💡 Step-by-Step Explanation & Concept Rationale
A Type-1 system has one integrator (pole at origin) in open loop, giving an infinite position error constant (Kp = infinity). Therefore, ess = 1 / (1 + Kp) = 0.
Q. 7 Electronics Engineering
Difficulty: Medium (1 Mark)
In a Bode plot, Phase Margin (PM) is defined as:
A
180 degrees + Phase angle at the gain crossover frequency (where |G(j*omega)| = 1)
✓ Correct
B
Gain at the phase crossover frequency
C
Phase angle at omega = 0
D
180 degrees - Gain crossover frequency
💡 Step-by-Step Explanation & Concept Rationale
Phase Margin PM = 180° + phi_gc, where phi_gc is the phase of the open-loop frequency response at the gain crossover frequency (where |G(j*omega)| = 1 or 0 dB).
Q. 8 Electronics Engineering
Difficulty: Medium (1 Mark)
Gain Margin (GM) in decibels is measured at which frequency on a Bode diagram?
A
At the gain crossover frequency
B
At the phase crossover frequency (where phase angle is -180 degrees)
✓ Correct
C
At resonance peak frequency
D
At the origin (omega = 0)
💡 Step-by-Step Explanation & Concept Rationale
Gain Margin GM = -20 * log10|G(j*omega_pc)|, measured at the phase crossover frequency omega_pc where the phase crosses -180 degrees.
Q. 9 Electronics Engineering
Difficulty: Hard (1 Mark)
According to the Nyquist stability criterion, the number of unstable closed-loop poles Z in the RHP is given by:
A
Z = N + P (where N is clockwise encirclements of -1+j0, P is open-loop RHP poles)
✓ Correct
B
Z = N - P
C
Z = P - N
D
Z = N * P
💡 Step-by-Step Explanation & Concept Rationale
Nyquist criterion: N = Z - P, where N is the number of clockwise encirclements of the critical point (-1+j0). Therefore, the number of unstable closed-loop poles is Z = N + P. For stability, Z must be 0, requiring N = -P (P counter-clockwise encirclements).
Q. 10 Electronics Engineering
Difficulty: Medium (1 Mark)
A lead compensator is primarily used in control system design to:
A
Improve steady-state accuracy by adding open-loop gain at low frequencies
B
Improve transient response, increase phase margin, and enhance system stability/speed
✓ Correct
C
Filter out high-frequency sensor noise
D
Eliminate DC offset voltage
💡 Step-by-Step Explanation & Concept Rationale
A lead compensator adds positive phase (lead angle) near the gain crossover frequency, increasing the phase margin and bandwidth, which speeds up transient response and improves stability.
Q. 11 Electronics Engineering
Difficulty: Medium (1 Mark)
What is the role of a lag compensator in control systems?
A
Increases bandwidth and rise time
B
Increases low-frequency gain to reduce steady-state error without significantly degrading stability
✓ Correct
C
Shifts system poles into the right half of the s-plane
D
Acts as an ideal differentiator
💡 Step-by-Step Explanation & Concept Rationale
A lag compensator provides high gain at low frequencies to minimize steady-state error while attenuating high frequencies, maintaining phase margin without sacrificing stability.
Q. 12 Electronics Engineering
Difficulty: Easy (1 Mark)
In PID controllers, adding the Derivative (D) control action provides:
A
Elimination of steady-state offset
B
Anticipatory control, increasing damping and reducing overshoot
✓ Correct
C
Infinite gain at DC
D
Immunity to high-frequency noise
💡 Step-by-Step Explanation & Concept Rationale
The derivative action responds to the rate of change of error, acting as an anticipatory brake that adds damping, reduces overshoot, and speeds settling time.
Q. 13 Electronics Engineering
Difficulty: Easy (1 Mark)
What is the primary drawback of adding Derivative (D) control in practical closed-loop loops?
A
It slows down system response
B
It amplifies high-frequency measurement noise
✓ Correct
C
It introduces a steady-state offset error
D
It prevents the integrator from winding up
💡 Step-by-Step Explanation & Concept Rationale
Because differentiation multiplies amplitude by frequency (omega), high-frequency sensor noise is heavily amplified, often requiring a low-pass filter on the derivative term.
Q. 14 Electronics Engineering
Difficulty: Easy (1 Mark)
The Nyquist sampling theorem states that an analog signal with highest frequency component fmax can be uniquely reconstructed from its samples if the sampling frequency fs satisfies:
A
fs >= 2 * fmax
✓ Correct
B
fs <= fmax / 2
C
fs = fmax
D
fs >= 4 * fmax^2
💡 Step-by-Step Explanation & Concept Rationale
Nyquist-Shannon sampling theorem requires the sampling rate fs to be at least twice the maximum frequency component fmax (fs >= 2*fmax) to prevent aliasing.
Q. 15 Electronics Engineering
Difficulty: Easy (1 Mark)
What type of filter must precede an Analog-to-Digital Converter (ADC) to prevent frequency components above fs / 2 from folding into the baseband?
A
High-pass filter
B
Anti-aliasing low-pass filter
✓ Correct
C
All-pass phase delay filter
D
Notch reject filter at DC
💡 Step-by-Step Explanation & Concept Rationale
An analog anti-aliasing low-pass filter attenuates all frequencies above the Nyquist frequency (fs/2) before sampling, preventing aliased distortion.
Q. 16 Electronics Engineering
Difficulty: Medium (1 Mark)
What is the fundamental difference between an FIR (Finite Impulse Response) filter and an IIR (Infinite Impulse Response) digital filter?
A
FIR filters are always unstable, whereas IIR filters are unconditionally stable
B
FIR filters have no feedback (all-zero) and can have strictly linear phase, while IIR filters use feedback (poles and zeros)
✓ Correct
C
IIR filters cannot be implemented using digital signal processors
D
FIR filters require infinite computation memory
💡 Step-by-Step Explanation & Concept Rationale
FIR filters rely solely on present and past inputs (feedforward only), ensuring unconditional stability and exact linear phase. IIR filters utilize recursive feedback, offering sharper cutoff with fewer coefficients but risking phase distortion or instability.
Q. 17 Electronics Engineering
Difficulty: Easy (1 Mark)
The Discrete Fourier Transform (DFT) of an N-point sequence requires O(N^2) complex multiplications. Using the Fast Fourier Transform (FFT) Cooley-Tukey algorithm, the computational complexity is reduced to:
A
O(N)
B
O(N * log2(N))
✓ Correct
C
O(log2(N))
D
O(N^(1/2))
💡 Step-by-Step Explanation & Concept Rationale
The Cooley-Tukey radix-2 FFT reduces the computational complexity from O(N^2) to O(N * log2(N)), making real-time digital spectral analysis feasible.
Q. 18 Electronics Engineering
Difficulty: Hard (1 Mark)
The bilinear transform method maps an analog s-plane transfer function to a digital z-plane transfer function using which substitution?
A
s = (2 / T) * (z - 1) / (z + 1)
✓ Correct
B
s = (T / 2) * (z + 1) / (z - 1)
C
s = (z - 1) / T
D
z = (s - 1) / (s + 1)
💡 Step-by-Step Explanation & Concept Rationale
The bilinear transformation utilizes the trapezoidal rule approximation: s = (2/T) * (1 - z^(-1)) / (1 + z^(-1)) = (2/T) * (z - 1) / (z + 1), mapping the entire left-half s-plane inside the unit circle.
Q. 19 Electronics Engineering
Difficulty: Hard (1 Mark)
What undesirable phenomenon occurs when using the bilinear transformation to design digital filters, necessitating frequency pre-warping?
A
Aliasing of high frequencies into DC
B
Nonlinear frequency compression (frequency warping) between analog and digital domains
✓ Correct
C
Filter coefficients becoming complex numbers
D
Instability of all poles regardless of location
💡 Step-by-Step Explanation & Concept Rationale
The bilinear transform maps the infinite analog frequency range (-infinity, +infinity) onto the finite digital frequency range (-pi, pi) via omega_d = 2 * arctan(omega_a * T / 2). This nonlinear warping requires pre-warping critical cutoff frequencies.
Q. 20 Electronics Engineering
Difficulty: Easy (1 Mark)
A Butterworth low-pass filter is characterized by which frequency response property?
A
Equiripple in both passband and stopband
B
Maximally flat magnitude response in the passband with no ripple
✓ Correct
C
Fastest possible roll-off rate at the expense of severe passband ripple
D
Linear phase response in the stopband only
💡 Step-by-Step Explanation & Concept Rationale
Butterworth filters feature a maximally flat passband magnitude response with no ripples. Chebyshev filters have passband or stopband ripple, and Elliptic filters have ripples in both.
Q. 21 Electronics Engineering
Difficulty: Medium (1 Mark)
Which filter approximation provides the most linear phase response and constant group delay in the passband?
A
Chebyshev Type I
B
Bessel (Thomson) filter
✓ Correct
C
Elliptic (Cauer) filter
D
Butterworth filter
💡 Step-by-Step Explanation & Concept Rationale
Bessel filters are designed for maximally flat group delay (linear phase response), preserving pulse shapes without ringing or overshoot, though with a gentler amplitude roll-off.
Q. 22 Electronics Engineering
Difficulty: Easy (1 Mark)
The convolution of an input signal x(t) with the unit impulse response h(t) of an LTI system yields:
A
The system transfer function H(s)
B
The system output y(t)
✓ Correct
C
The frequency spectrum X(omega)
D
The derivative of the input dx/dt
💡 Step-by-Step Explanation & Concept Rationale
The response of any continuous-time LTI system to an arbitrary input x(t) is given by the convolution integral: y(t) = integral from -infinity to +infinity of x(tau) * h(t - tau) dtau.
Q. 23 Electronics Engineering
Difficulty: Easy (1 Mark)
What is the Laplace transform of the unit step function u(t)?
A
1
B
1 / s
✓ Correct
C
1 / s^2
D
s
💡 Step-by-Step Explanation & Concept Rationale
The unilateral Laplace transform of u(t) is L{u(t)} = integral from 0 to infinity of e^(-st) dt = 1/s for Re(s) > 0.
Q. 24 Electronics Engineering
Difficulty: Easy (1 Mark)
What is the Laplace transform of the Dirac delta function delta(t)?
A
1
✓ Correct
B
1 / s
C
s
D
0
💡 Step-by-Step Explanation & Concept Rationale
By definition, L{delta(t)} = integral from 0- to 0+ of delta(t) * e^(-st) dt = 1.
Q. 25 Electronics Engineering
Difficulty: Easy (1 Mark)
In the z-transform, what does a unit delay operation x[n - 1] correspond to in the frequency/transform domain?
A
z * X(z)
B
z^(-1) * X(z)
✓ Correct
C
X(z - 1)
D
-z * X(z)
💡 Step-by-Step Explanation & Concept Rationale
By the time-shifting property of the z-transform, shifting by k samples corresponds to multiplication by z^(-k). Thus, a 1-sample delay is z^(-1) * X(z).
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