A 1024-point FFT is computed on a signal. The frequency resolution is 0.1 Hz. What is the sampling frequency?
Answer: B
Frequency resolution Δf = fs/N, so fs = Δf × N = 0.1 × 1024 = 102.4 Hz.
Q.391Medium
A continuous-time signal x(t) = 5cos(2π × 500t) + 3sin(2π × 1500t) is sampled at 4 kHz. What is the Nyquist frequency for this signal, and will aliasing occur?
Answer: B
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.
Q.392Easy
The Z-transform of a discrete-time signal is X(z) = z/(z-0.5) with ROC |z| > 0.5. The corresponding time-domain signal is:
Answer: A
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.
Q.393Easy
A first-order low-pass filter has transfer function H(s) = ωc/(s + ωc). At what frequency (in terms of ωc) does the magnitude response drop to 1/√2 of its DC value?
Answer: B
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.
Q.394Hard
For a linear time-invariant system with Laplace transform H(s) = 1/(s+2), determine the response to input x(t) = e^(-2t)×u(t):
Answer: A
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.
Q.395Medium
A signal's autocorrelation function is R(τ) = 10 + 8cos(2π×100τ). What is the average power of the signal?
Answer: B
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.
