Question

A 1 kHz signal is flat-top sampled at the rate of 1800 samples/sec and the samples are applied to an ideal rectangular low pass filter with a cut-off frequency of 1100 Hz. Then the output of the filter contains:

• A. 800 Hz and 1000 Hz components.
• B. 800 Hz and 900 Hz components.
• C. 800 Hz, 900 Hz and 1000 Hz components.
• D. only 800 Hz component.

Hint:

When sampling, if the input frequency \(f_{in}\) is greater than half the sampling frequency (\(f_s/2\)), an aliased frequency will appear at \(f_{alias} = |f_{in} - f_s|\). The output of a low-pass filter will contain all frequency components (original and aliased) that are below its cutoff frequency.

Answer 1

The Correct Option is A

Step 1: Identify the signal parameters. Original signal frequency: \( f_{in} = 1 \text{ kHz} = 1000 \text{ Hz} \). Sampling frequency: \( f_s = 1800 \text{ samples/sec} = 1800 \text{ Hz} \).
Step 2: Check for aliasing. The Nyquist rate required for perfect reconstruction is \( 2f_{in} = 2 \times 1000 = 2000 \text{ Hz} \). The Nyquist frequency of the sampling process is \( f_s/2 = 1800/2 = 900 \text{ Hz} \). Since the sampling frequency \( f_s = 1800 \text{ Hz} \) is less than the Nyquist rate of 2000 Hz (or equivalently, \(f_{in}>f_s/2\)), aliasing will occur.
Step 3: Determine the frequencies present after sampling. When a signal of frequency \(f_{in}\) is sampled at \(f_s\), the resulting spectrum contains components at frequencies \( |f_{in} \pm k f_s| \) for all integers \(k\). The main frequencies we are concerned with are the ones that fall into the low-frequency range. For k=0: \( f = f_{in} = 1000 \text{ Hz} \). For k=1: \( f = |f_{in} - f_s| = |1000 - 1800| = 800 \text{ Hz} \). Also \(f_{in}+f_s = 2800\) Hz, etc. So, after sampling, the baseband will contain frequency components at 1000 Hz and an aliased component at 800 Hz.
Step 4: Apply the low-pass filter. The signal is passed through an ideal low-pass filter with a cutoff frequency \( f_{cutoff} = 1100 \text{ Hz} \). This filter will pass all frequencies below 1100 Hz and block all frequencies above it. The frequencies present in the signal are 800 Hz, 1000 Hz, 2800 Hz, etc. The filter will pass the 800 Hz and 1000 Hz components, as both are less than 1100 Hz. The output of the filter will contain both 800 Hz and 1000 Hz components.