What is the bandwidth of a first-order low-pass filter with cutoff frequency fc = 1 kHz?
A 500 Hz B 1 kHz C 2 kHz D 4 kHz
Bandwidth of a low-pass filter is defined by its cutoff frequency (-3dB point). At f = fc, magnitude drops to 0.707 of DC gain. Bandwidth = fc = 1 kHz.
A continuous-time signal x(t) = 5cos(2πt + π/4) has a fundamental frequency of:
A 1 Hz B 2 Hz C 0.5 Hz D 4 Hz
From x(t) = 5cos(2πt + π/4), ω = 2π, so f = ω/2π = 1 Hz
Which of the following is a causal system?
A y[n] = x[n+1] + x[n] B y[n] = x[n] + x[n-1] C y[n] = x[-n] D y[n] = x[n] + x[n+2]
A causal system does not depend on future values. Only option B satisfies this condition.
The Z-transform of x[n] = aⁿu[n] is:
A z/(z-a) B 1/(z-a) C z/(z+a) D a/(z-1)
Z-transform: X(z) = ∑aⁿu[n]z⁻ⁿ = z/(z-a) for |z| > |a|
Nyquist sampling rate for a signal with maximum frequency 5 kHz is:
A 2.5 kHz B 5 kHz C 10 kHz D 20 kHz
Nyquist sampling rate = 2 × (maximum frequency) = 2 × 5 = 10 kHz
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The convolution of two sequences is commutative, meaning:
A x[n] * h[n] ≠ h[n] * x[n] B x[n] * h[n] = h[n] * x[n] C Convolution is only associative D Convolution is only distributive
Convolution is commutative: x[n] * h[n] = h[n] * x[n]
A system with H(z) = (z-0.5)/(z-0.8) has:
A One zero at z=0.5 and one pole at z=0.8 B One pole at z=0.5 and one zero at z=0.8 C Two poles at z=0.5 and z=0.8 D Pole-zero cancellation
Numerator gives zeros: z=0.5. Denominator gives poles: z=0.8
A continuous-time signal x(t) = e^(-2t)u(t) is applied to a system. What is the Laplace transform X(s)?
A 1/(s+2), Re(s) > -2 B 1/(s-2), Re(s) > 2 C 2/(s+2), Re(s) > -2 D 1/(s+2)², Re(s) > -2
For x(t) = e^(-at)u(t), the Laplace transform is 1/(s+a) with ROC Re(s) > -a. Here a=2, so X(s) = 1/(s+2).
A discrete signal x[n] = {1, 2, 3, 2} has a length of 4 samples. What is the 4-point DFT at k=0?
A 6 B 8 C 12 D 16
X[k] = Σ x[n]e^(-j2πkn/N). At k=0: X[0] = x[0] + x[1] + x[2] + x[3] = 1+2+3+2 = 8.
A signal x(t) is time-limited to 0 to T seconds. What is the minimum sampling rate required to avoid aliasing if its bandwidth is B Hz?
A B Hz B 2B Hz C B/2 Hz D T×B Hz
According to Nyquist sampling theorem, minimum sampling rate = 2×(maximum frequency) = 2B Hz.
A continuous-time signal x(t) = 5cos(2π×100t) is sampled at 250 Hz. What is the Nyquist frequency required?
A 100 Hz B 200 Hz C 250 Hz D 500 Hz
Nyquist frequency = 2 × maximum frequency = 2 × 100 = 200 Hz. Since sampling rate (250 Hz) > Nyquist frequency, no aliasing occurs.
For a causal LTI system with transfer function H(z) = 1/(1-0.5z^-1), the system is:
A Unstable B Marginally stable C Stable D Critically stable
Pole is at z = 0.5, which lies inside the unit circle. For causality and stability, all poles must be inside |z| < 1.
The energy of a discrete-time signal x[n] = (0.5)^n u[n] is:
A 0.75 B 1.0 C 1.33 D 2.0
Energy = Σ|x[n]|² = Σ(0.25)^n = 1/(1-0.25) = 3 4
Which of the following is a property of the Discrete-Time Fourier Transform (DTFT)?
A DTFT is periodic with period π B DTFT is periodic with period 2π C DTFT is periodic with period 4π D DTFT is non-periodic
The DTFT X(e^jω) is periodic with period 2π because e^j(ω+2π)n = e^jωn.
A signal x[n] has Z-transform X(z) = 1 + 2z^-1 + 3z^-2 for |z| > 0. The signal values are:
A x[0]=1, x[1]=2, x[2]=3 B x[-2]=1, x[-1]=2, x[0]=3 C x[0]=3, x[1]=2, x[2]=1 D x[1]=1, x[2]=2, x[3]=3
By definition, X(z) = Σx[n]z^-n. Comparing coefficients: x[0]=1, x[1]=2, x[2]=3.
The convolution of x[n] = {1, 2, 3} and h[n] = {1, 1} yields a sequence of length:
A 2 B 3 C 4 D 5
Length of convolution = M + N - 1 = 3 + 2 - 1 = 4.
A signal has power spectral density S_x(f) that is nonzero only for |f| ≤ 2 kHz. The minimum sampling frequency is:
A 2 kHz B 4 kHz C 8 kHz D 16 kHz
By Nyquist theorem, f_s ≥ 2×f_max = 2×2 = 4 kHz.
A finite-duration signal x[n] has length N=8. Its DFT X[k] has length:
A 4 B 8 C 16 D Depends on zero-padding
DFT length equals the input signal length unless zero-padding is applied. For N-point signal, DFT yields N frequency bins.
The Z-transform of x[n] = n·(0.8)^n u[n] is:
A 0.8z/(z-0.8)² B 0.8z/(z-0.8) C z/(z-0.8)² D 0.8/(z-0.8)²
Using the property that Z{n·a^n·u[n]} = az/(z-a)², with a=0.8, we get X(z) = 0.8z/(z-0.8)².
A signal x(t) = sin(2πf₀t) is sampled at fs = 5f₀. What is the Nyquist sampling rate required?
A f₀ B 2f₀ C 2.5f₀ D 5f₀
Nyquist rate = 2 × maximum frequency = 2f₀. The given sampling rate (5f₀) exceeds this, ensuring no aliasing.
