Electronics (ECE) - MCQ Practice Questions

X[0] = Σ x[n] (sum of all samples in one period). Average = X[0]/N = $\frac{8}{4}$ = 2.

Nyquist theorem: fs ≥ 2 × fmax = 2 × 15 kHz = 30 kHz.

DTFT{δ[n-n₀]} = e^(-jωn₀). With n₀=3, X(e^(jω)) = e^(-j3ω).

Using partial fractions and standard pairs: L^(-1){K/(s+a)} = Ke^(-at)u(t). Here K=5, a=3, so h(t) = 5e^(-3t)u(t).

Linear convolution of signals of length M and N produces output of length M+N-1 = 5+5-1 = 9.

Denominator s²+3s+2 = (s+1)(s+2) = 0, giving poles at s = -1 and s = -2.

The maximum frequency in the signal is 50 Hz (from 100π = 2πf). Nyquist frequency = 2 × 50 = 100 Hz.

The modulation theorem (or frequency convolution property) states that x(t)×y(t) ↔ X(f)*Y(f).

∫₀^∞ e^(-3t) dt = $\frac{1}{3}$ < ∞. For stability, BIBO condition requires absolute integrability, which is satisfied.

X(z) = Σ($\frac{1}{2}$)^n z^(-n) = z/(z - $\frac{1}{2}$) = z/(z - 0.5) with ROC |z| > 0.5.

For an even signal, the Fourier transform is real and even: X(-f) = X(f) and X*(f) = X(f).

Energy = 1² + 2² + (-1)² + 0² = 1 + 4 + 1 + 0 = 6.

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.