A continuous-time signal x(t) = 5cos(2πt + π/4) has a fundamental frequency of:
From x(t) = 5cos(2πt + π/4), ω = 2π, so f = ω/2π = 1 Hz
Which of the following is a causal system?
A causal system does not depend on future values. Only option B satisfies this condition.
The Z-transform of x[n] = aⁿu[n] is:
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:
Nyquist sampling rate = 2 × (maximum frequency) = 2 × 5 = 10 kHz
The convolution of two sequences is commutative, meaning:
Convolution is commutative: x[n] * h[n] = h[n] * x[n]
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A system with H(z) = (z-0.5)/(z-0.8) has:
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)?
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?
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?
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?
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:
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:
Energy = Σ|x[n]|² = Σ(0.25)^n = 1/(1-0.25) = 34 ≈ 1.33
Which of the following is a property of the Discrete-Time Fourier Transform (DTFT)?
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:
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:
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:
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:
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:
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?
Nyquist rate = 2 × maximum frequency = 2f₀. The given sampling rate (5f₀) exceeds this, ensuring no aliasing.
The energy of signal x[n] = {1, 2, 1, -1} over the given support is:
Energy E = Σ|x[n]|² = 1² + 2² + 1² + (-1)² = 1 + 4 + 1 + 1 = 7.
