Govt. Exams
Entrance Exams
For x[n] = cos(2πk n/N), the period N = 6/(1/2) = 12 samples.
For x[n] = cos(ω₀n), period N = 2π/ω₀ = 2π/(π/4) = 8.
Transfer function H(z) = 0.2/(1-0.8z⁻¹). DC gain H(1) = 0.2/(1-0.8) = 0.2/0.2 = 1.
Frequency resolution Δf = fs/N, where N=16. Therefore, fs = Δf × N = 1 × 16 = 16 kHz.
h(t) = ds(t)/dt = d/dt[1 - e^(-2t)u(t)] = 2e^(-2t)u(t).
Linear phase FIR filters require symmetric (even) or antisymmetric (odd) impulse responses about the center. For length 5, center is n=2.
IFT{δ(f-f₀)} = e^(j2πf₀t) and IFT{δ(f+f₀)} = e^(-j2πf₀t). Sum = 2cos(2πf₀t), but accounting for delta properties, result is cos(2πf₀t).
Zero at z=0 (DC) causes attenuation at low frequencies. Poles near z=1 allow high frequencies to pass, characteristic of a high-pass filter.
For a real signal, X(f) exhibits Hermitian symmetry: X(-f) = X*(f). Magnitude symmetry |X(-f)| = |X(f)| indicates even symmetry in x(t).
Pole at z=0.5 (inside unit circle) makes the system stable. The denominator has z⁻¹ term, making it IIR (infinite impulse response).
