Electronics (ECE) - MCQ Practice Questions

From x(t) = 5cos(2πt + π/4), ω = 2π, so f = ω/2π = 1 Hz

A causal system does not depend on future values. Only option B satisfies this condition.

Z-transform: X(z) = ∑aⁿu[n]z⁻ⁿ = z/(z-a) for |z| > |a|

Nyquist sampling rate = 2 × (maximum frequency) = 2 × 5 = 10 kHz

Convolution is commutative: x[n] * h[n] = h[n] * x[n]

Numerator gives zeros: z=0.5. Denominator gives poles: z=0.8

For x(t) = e^(-at)u(t), the Laplace transform is 1/(s+a) with ROC Re(s) > -a. Here a=2, so X(s) = 1/(s+2).

X[k] = Σ x[n]e^(-j2πkn/N). At k=0: X[0] = x[0] + x[1] + x[2] + x[3] = 1+2+3+2 = 8.

According to Nyquist sampling theorem, minimum sampling rate = 2×(maximum frequency) = 2B Hz.

Nyquist frequency = 2 × maximum frequency = 2 × 100 = 200 Hz. Since sampling rate (250 Hz) > Nyquist frequency, no aliasing occurs.

Pole is at z = 0.5, which lies inside the unit circle. For causality and stability, all poles must be inside |z| < 1.

Energy = Σ|x[n]|² = Σ(0.25)^n = 1/(1-0.25) = $\frac{4}{3}$ ≈ 1.33

The DTFT X(e^jω) is periodic with period 2π because e^j(ω+2π)n = e^jωn.

By definition, X(z) = Σx[n]z^-n. Comparing coefficients: x[0]=1, x[1]=2, x[2]=3.

Length of convolution = M + N - 1 = 3 + 2 - 1 = 4.

By Nyquist theorem, f_s ≥ 2×f_max = 2×2 = 4 kHz.

DFT length equals the input signal length unless zero-padding is applied. For N-point signal, DFT yields N frequency bins.

Using the property that Z{n·a^n·u[n]} = az/(z-a)², with a=0.8, we get X(z) = 0.8z/(z-0.8)².

Nyquist rate = 2 × maximum frequency = 2f₀. The given sampling rate (5f₀) exceeds this, ensuring no aliasing.

Energy E = Σ|x[n]|² = 1² + 2² + 1² + (-1)² = 1 + 4 + 1 + 1 = 7.