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.
X(s) = 1/(s+2). Y(s) = H(s)×X(s) = 1/[(s+2)²]. This is a repeated pole, inverse Laplace gives y(t) = t×e^(-2t)×u(t). This is a resonance condition in the system.
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].
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.
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.
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.
Frequency resolution Δf = fs/N, so fs = Δf × N = 0.1 × 1024 = 102.4 Hz.
(s+3)/((s+1)(s+2)) = A/(s+1) + B/(s+2). Solving: A=2, B=-1. h(t) = 2e^(-t)u(t) - e^(-2t)u(t).
For poles at σ ± jωd, ωn = √(σ² + ωd²) = √(1 + 4) = √5.
At ω = ωc (cutoff frequency), |H(jωc)| = 1/√(1+1) = 1/√2. This is the -3dB point.
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