Govt. Exams
Entrance Exams
By definition, X(z) = Σx[n]z^-n. Comparing coefficients: x[0]=1, x[1]=2, x[2]=3.
The DTFT X(e^jω) is periodic with period 2π because e^j(ω+2π)n = e^jωn.
Energy = Σ|x[n]|² = Σ(0.25)^n = 1/(1-0.25) = 4/3 ≈ 1.33
Pole is at z = 0.5, which lies inside the unit circle. For causality and stability, all poles must be inside |z| < 1.
Nyquist frequency = 2 × maximum frequency = 2 × 100 = 200 Hz. Since sampling rate (250 Hz) > Nyquist frequency, no aliasing occurs.
According to Nyquist sampling theorem, minimum sampling rate = 2×(maximum frequency) = 2B Hz.
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.
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).
Numerator gives zeros: z=0.5. Denominator gives poles: z=0.8
Convolution is commutative: x[n] * h[n] = h[n] * x[n]
