Question

A continuous time, band limited signal \(x(t)\) has its Fourier transform described by \( X(f)= \begin{cases} 1-\dfrac{|f|}{200}, & |f|\le 200\text{ Hz}\\[4pt] 0, & |f|>200\text{ Hz} \end{cases} \). The signal is uniformly sampled at a sampling rate of \(600\) Hz. The Fourier transform of the sampled signal is \(X_s(f)\). What is the value of \(\dfrac{X_s(600)}{X_s(500)}\)? (Round off the answer to one decimal place.)

Hint:

The spectrum of a sampled signal is the periodic replication of the original spectrum with period \(f_s\). For ratios at specific frequencies, the global scaling cancels.

Answer 1

Correct Answer: 1.9

Step 1: For uniform sampling with rate \(f_s=600\) Hz,
\( X_s(f)=\dfrac{1}{T}\sum_{k=-\infty}^{\infty} X(f-k f_s) \)
(with \(T=1/f_s\)). For ratios the common factor \(1/T\) cancels.
Step 2: Evaluate the nonzero terms:
- \(X_s(600)\): only \(k=1\) contributes since \(X(600-600)=X(0)=1\). Thus \(X_s(600)\propto 1\).
- \(X_s(500)\): only \(k=1\) contributes with \(X(500-600)=X(-100)=1-\frac{100}{200}=0.5\).
Hence \(X_s(500)\propto 0.5\).
Step 3: Therefore, \( \dfrac{X_s(600)}{X_s(500)}=\dfrac{1}{0.5}=2.0. \)