Electronics (ECE) - MCQ Practice Questions

Zero at z=0 (DC) causes attenuation at low frequencies. Poles near z=1 allow high frequencies to pass, characteristic of a high-pass filter.

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.

IFT{δ(f-f₀)} = e^(j2πf₀t) and IFT{δ(f+f₀)} = e^(-j2πf₀t). Sum = 2cos(2πf₀t), but accounting for delta properties, result is cos(2πf₀t).

Linear phase FIR filters require symmetric (even) or antisymmetric (odd) impulse responses about the center. For length 5, center is n=2.

h(t) = ds(t)/dt = d/dt[1 - e^(-2t)u(t)] = 2e^(-2t)u(t).

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).

Frequency resolution Δf = fs/N, where N=16. Therefore, fs = Δf × N = 1 × 16 = 16 kHz.

Transfer function H(z) = 0.2/(1-0.8z⁻¹). DC gain H(1) = 0.2/(1-0.8) = 0.$\frac{2}{0}$.2 = 1.

For x[n] = cos(ω₀n), period N = 2π/ω₀ = 2π/(π/4) = 8.

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.

Aliased frequency = |f₀ - kfs| = |3 - 10| = 7 kHz (for k=1). The signal appears as a 7 kHz component in the baseband.

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.