Electronics (ECE) - MCQ Practice Questions

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.

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.

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.