The energy of signal x[n] = {1, 2, 1, -1} over the given support is:
A 5 B 7 C 8 D 9
Energy E = Σ|x[n]|² = 1² + 2² + 1² + (-1)² = 1 + 4 + 1 + 1 = 7.
For a periodic signal x[n] with period N=4, the DFT X[k] has X[0]=8. The average value of the signal is:
A 1 B 2 C 4 D 8
X[0] = Σ x[n] (sum of all samples in one period). Average = X[0]/N = 4 8
A continuous-time signal x(t) is band-limited to 15 kHz. Using ideal reconstruction from samples, the minimum sampling frequency required is:
A 7.5 kHz B 15 kHz C 30 kHz D 60 kHz
Nyquist theorem: fs ≥ 2 × fmax = 2 × 15 kHz = 30 kHz.
The DTFT of x[n] = δ[n-3] is:
A e^(-j3ω) B e^(j3ω) C cos(3ω) D 1
DTFT{δ[n-n₀]} = e^(-jωn₀). With n₀=3, X(e^(jω)) = e^(-j3ω).
For a Laplace transform H(s) = 5/(s+3), the system impulse response is:
A 5e^(-3t)u(t) B 5e^(3t)u(t) C -5e^(-3t)u(t) D e^(-3t)u(t)
Using partial fractions and standard pairs: L^(-1){K/(s+a)} = Ke^(-at)u(t). Here K=5, a=3, so h(t) = 5e^(-3t)u(t).
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The convolution of two signals each of length 5 using linear convolution yields an output of length:
A 5 B 9 C 10 D 25
Linear convolution of signals of length M and N produces output of length M+N-1 = 5+5-1 = 9.
A system with transfer function H(s) = (s+2)/(s²+3s+2) has poles at:
A s = -1, -2 B s = 1, 2 C s = -1 ± j D s = 2, -3
Denominator s²+3s+2 = (s+1)(s+2) = 0, giving poles at s = -1 and s = -2.
A continuous-time signal x(t) = 5sin(100πt) is sampled at 150 Hz. What is the Nyquist frequency required to avoid aliasing?
A 50 Hz B 100 Hz C 150 Hz D 200 Hz
The maximum frequency in the signal is 50 Hz (from 100π = 2πf). Nyquist frequency = 2 × 50 = 100 Hz.
Which property of the Fourier transform states that multiplication in time domain equals convolution in frequency domain?
A Linearity B Modulation theorem C Parseval's theorem D Time-shifting property
The modulation theorem (or frequency convolution property) states that x(t)×y(t) ↔ X(f)*Y(f).
A causal LTI system has impulse response h(t) = e^(-3t)u(t). Is this system stable?
A Yes, because the impulse response is absolutely integrable B No, because the pole is at s = 3 C No, because the ROC doesn't include the jω axis D Yes, but only for t > 0
∫₀^∞ e^(-3t) dt = 3 1
The Z-transform of x[n] = (2 1
A z/(z - 0.5), |z| > 0.5 B 1/(1 - 0.5z^(-1)), |z| > 0.5 C z/(z - 2), |z| > 2 D 0.5z/(z - 1), |z| > 1
X(z) = Σ(2 1 2 1
An even signal x(t) has Fourier transform X(f). What can be said about X(f)?
A X(f) is purely imaginary B X(f) is real and even C X(f) is complex D X(f) is purely real but odd
For an even signal, the Fourier transform is real and even: X(-f) = X(f) and X*(f) = X(f).
The energy of a signal x[n] = {1, 2, -1, 0} is:
A 4 B 6 C 5 D 3
Energy = 1² + 2² + (-1)² + 0² = 1 + 4 + 1 + 0 = 6.
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:
A x[n] = 0.5^n × u[n] B x[n] = δ[n] + 0.5^n × u[n] C x[n] = 0.5^(n-1) × u[n-1] D x[n] = δ[n-1] × 0.5^n
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.
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?
A ω = ωc/2 B ω = ωc C ω = 2ωc D ω = ωc/√2
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.
