Question

Given a series RLC circuit with V = 5V, R = 200 kΩ, C = 10µF. Sampling frequency of the circuit is 10 Hz. The expression and the ROC of the z-transform of the sampled signal are ____________

(a) \(\frac{5z}{z-e^{-5′}}\), |z|<e^-5

(b) \(\frac{5z}{z-e^{-0.05′}}\), |z|<e^-0.05

(c) \(\frac{5z}{z-e^{-0.05′}}\), |z|>e^-0.05

(d) \(\frac{5z}{z-e^{-5′}}\), |z|>e^-5

The question was posed to me in an online interview.

The question is from Discrete Fourier Transform in chapter Fourier Transform of Signals and Systems

Answer 1

The correct choice is (c) \(\frac{5z}{z-e^{-0.05′}}\), |z|>e^-0.05

For explanation: I (t) = \(\frac{V}{R}  e^{-t/RC}\)

Voltage across resistor = R I (t)

= V e^-t/RC = 5 e^-t/RC

= 5 \(e{\frac{-t}{200 × 10 × 10^{-6} × 10^3}}\)  = 5 e^-t/2

Given that, the Sampling frequency of the circuit = 10 Hz

Hence, x (n) = 5e^-n/2 X 10 = 5e^-0.05n

Now, X (z) = \(∑_{n=-∞}^∞ x[n]z^{-n}\)

= 5 \(∑_{n=-∞}^∞  (e^{-0.05} z^{-n})^n\)

= 5. \(\frac{1}{1-e^{-0.05} Z^{-1′}}\), ROC |z|>e^-0.05

= \(\frac{5z}{z-e^{-0.05′}}\), |z|>e^-0.05.