Govt. Exams
Entrance Exams
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).
DTFT{δ[n-n₀]} = e^(-jωn₀). With n₀=3, X(e^(jω)) = e^(-j3ω).
Nyquist theorem: fs ≥ 2 × fmax = 2 × 15 kHz = 30 kHz.
X[0] = Σ x[n] (sum of all samples in one period). Average = X[0]/N = 8/4 = 2.
Energy E = Σ|x[n]|² = 1² + 2² + 1² + (-1)² = 1 + 4 + 1 + 1 = 7.
Nyquist rate = 2 × maximum frequency = 2f₀. The given sampling rate (5f₀) exceeds this, ensuring no aliasing.
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.
DFT length equals the input signal length unless zero-padding is applied. For N-point signal, DFT yields N frequency bins.
By Nyquist theorem, f_s ≥ 2×f_max = 2×2 = 4 kHz.
Length of convolution = M + N - 1 = 3 + 2 - 1 = 4.
