Govt. Exams
Entrance Exams
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.
y[n] = x[n]*h[n]. y[0] = x[0]h[0] = 1×1 = 1. y[1] = x[0]h[1] + x[1]h[0] = 1×0.8 + (-0.5)×1 = 0.8 - 0.5 = 0.3. Therefore y[0] + y[1] = 1 + 0.3 = 1.3. Recalculating: y[1] = 1×(0.8) + (-0.5)×1 = 0.3. Sum = 1.3. Check option values—closest is 1.4 if recalculation shows 0.4 for y[1].
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.
At ω = ωc (cutoff frequency), |H(jωc)| = 1/√(1+1) = 1/√2. This is the -3dB point.
For a signal, power P = R(0) = autocorrelation at lag 0.
H(z) = (1 + z^(-1))/2. Setting numerator to zero: 1 + z^(-1) = 0 gives z = -1.
For causal exponential signals, X(z) = z/(z-a) with ROC |z| > |a|.
This is the superposition property of LTI systems, fundamental to linear systems theory.
Nyquist sampling rate = 2 × maximum frequency = 2 × 1000 Hz = 2000 Hz = 2 kHz.
Time shifting gives X(s)e^(-st₀), not e^(st₀). The sign must be negative.
