Electronics (ECE) - MCQ Practice Questions

Time shifting gives X(s)e^(-st₀), not e^(st₀). The sign must be negative.

Nyquist sampling rate = 2 × maximum frequency = 2 × 1000 Hz = 2000 Hz = 2 kHz.

This is the superposition property of LTI systems, fundamental to linear systems theory.

For causal exponential signals, X(z) = z/(z-a) with ROC |z| > |a|.

H(z) = (1 + z^(-1))/2. Setting numerator to zero: 1 + z^(-1) = 0 gives z = -1.

For a signal, power P = R(0) = autocorrelation at lag 0.

At ω = ωc (cutoff frequency), |H(jωc)| = 1/√(1+1) = 1/√2. This is the -3dB point.

For poles at σ ± jωd, ωn = √(σ² + ωd²) = √(1 + 4) = √5.

(s+3)/((s+1)(s+2)) = A/(s+1) + B/(s+2). Solving: A=2, B=-1. h(t) = 2e^(-t)u(t) - e^(-2t)u(t).

Frequency resolution Δf = fs/N, so fs = Δf × N = 0.1 × 1024 = 102.4 Hz.

The signal contains frequencies at 500 Hz and 1500 Hz. Maximum frequency is 1500 Hz, so Nyquist frequency required = 3 kHz. But sampling at 4 kHz gives Nyquist frequency = 2 kHz. Since 1500 Hz < 2 kHz, no aliasing occurs. However, check: for the 1500 Hz component sampled at 4 kHz, it aliases to 4000 - 1500 = 2500 Hz which folds to 4000 - 2500 = 1500 Hz. Actually, Nyquist = fs/2 = 2 kHz. The 1500 Hz signal is below 2 kHz, so no aliasing. Correction: Maximum frequency in signal is 1500 Hz, Nyquist minimum needed = 3 kHz. Sampling at 4 kHz gives Nyquist = 2 kHz < 3 kHz, so aliasing WILL occur. Option B is correct.

Using partial fractions or standard Z-transform tables: X(z) = z/(z-0.5) corresponds to x[n] = 0.5^n × u[n] where u[n] is the unit step function. The ROC |z| > 0.5 confirms a causal right-sided sequence.

At DC (ω=0): |H(j0)| = 1. At ω = ωc: |H(jωc)| = ωc/√(ωc² + ωc²) = 1/√2. This is the -3dB cutoff frequency, a fundamental property of first-order filters.

X(s) = 1/(s+2). Y(s) = H(s)×X(s) = 1/[(s+2)²]. This is a repeated pole, inverse Laplace gives y(t) = t×e^(-2t)×u(t). This is a resonance condition in the system.

The autocorrelation at τ=0 gives the total average power: R(0) = 10 + 8cos(0) = 10 + 8 = 18 W. Wait—rechecking: Power = R(0) = 10 + 8 = 18 W. But if the question implies periodic component, the DC power is R(∞) or the constant term = 10 W. Standard definition: total power = R(0) = 18 W. If asking for average power excluding periodic oscillation, answer is 10 W.

Understanding:

We need to find $K$ such that the closed-loop system is critically damped (damping ratio $\zeta =1$).

Formula:

The closed-loop transfer function is:

For a standard second-order system $\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$, critical damping requires:

Step 1: Identify parameters from the characteristic equation.

Comparing $s^2+4s+K$ with $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$:

Step 2: Apply critical damping condition $\zeta =1$.

Step 3: Solve for $K$.

Answer:

The value of $K$ for critical damping is 4.

Quick Tip:

For a second-order system, the condition $\zeta =1$ gives repeated real poles. Any $K>4$ makes it underdamped and $K<4$ makes it overdamped for this system.

Understanding:

We need to find the damping ratio $\zeta$ from the percentage peak overshoot (% PO).

Formula:

The relationship between peak overshoot and damping ratio is:

Step 1: Set up the equation.

Step 2: Take natural logarithm of both sides.

Step 3: Solve for $\zeta$.

Answer:

The damping ratio corresponding to $16.3\%$ peak overshoot is $0.5$.

Quick Tip:

The standard result $\%PO\approx 16.3\%$ corresponds to $\zeta =0.5$ and is a frequently tested value. Memorise: $\zeta =0.707$ gives $\approx 4.3\%$ overshoot (Butterworth), and $\zeta =1$ gives $0\%$ overshoot.

Understanding:

We need to find the settling time $T_s$ using the $2\%$ criterion.

Formula:

Comparing with the standard form $\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$:

Step 1: Extract system parameters.

Step 2: Compute the product $\zeta {\omega }_n$.

Step 3: Apply the settling time formula.

Answer:

The settling time using the $2\%$ criterion is approximately $1.33\text{s}$.

Quick Tip:

For the $5\%$ criterion, use $T_s\approx 3/(\zeta {\omega }_n)$. The product $\sigma =\zeta {\omega }_n$ is the real part of the closed-loop poles and directly controls the settling speed.

Understanding:

We need to find the steady-state error $e_{ss}$ for a unit ramp input $r(t)=tu(t)$ with a Type-1 plant and proportional controller.

Formula:

For a unit ramp, the steady-state error is given by the velocity error constant $K_v$:

where $G_{OL}(s)=K_p\cdot G(s)$.

Step 1: Compute $K_v$.

Step 2: Compute steady-state error.

Answer:

The steady-state error to a unit ramp input is $\frac{2}{K_p}$.

Quick Tip:

A Type-1 system has zero steady-state error to a step input but a finite error to a ramp. A Type-2 system would give zero error to a ramp. Adding an integrator (I-controller) increases system type by 1.

Understanding:

We need to determine the range of $K$ for which all roots of the characteristic equation lie in the left-half $s$-plane.

Formula:

The Routh array for $a{s}^3+b{s}^2+cs+d$ is:

Stability requires all first-column elements to be positive.

Step 1: Build the Routh array.

Step 2: Simplify the $s^1$ row entry.

Step 3: Apply stability conditions (all first-column elements $>0$).

Answer:

The system is stable for $K>2$.

Quick Tip:

A sign change in the first column of the Routh array indicates unstable roots. The number of sign changes equals the number of roots in the right-half $s$-plane.