4
$\begingroup$

This is question 4.26 from the third edition of Alan Oppenheim's textbook "Discrete-time Signal Processing". It is stated as follows: enter image description here

Firstly, the answer is the input signal should be bandlimited to 12 kHz. I'm having trouble seeing why it is the case.

My reasoning:

The cutoff frequency of the low pass filter is obviously 8 kHz, so any frequency component of the discretized input signal beyond 8 kHz will be removed. But since we do not want aliasing to occur, we want the input signal to be bandlimited such that after passing the sampled signal through the low pass filter we don't get aliasing below 8 kHz. But in the case of input signal bandlimited to 12 kHz, we get the following (a quick drawing): enter image description here

which causes aliasing, making the overall system no longer LTI.

So where did I go wrong? Any help is greatly appreciated!

$\endgroup$

2 Answers 2

4
$\begingroup$

This question, and a number of similar others, from DTSP book can be a little tricky to recognise the fact that it's not actually asking alias free operation (which would require 8 khz cutoff signal as you expected), rather it is actually asking how much aliasing (due to initial sampling block) is tolerable: you can allow aliasing in those regions of the DT filter $H(e^{j\omega})$ whose stop band will help you get rid of the aliased spectrum, hence enable an LTI processing possible, but not with the samples of the complete original fullband signal $x_c(t)$ but with that of the reduced bandwidth input signal.

To see this you shall better consult to the spectral figure below: aliase sample The mathematical condition on the maximum allowable cutoff frequency $\Omega_c$ of the input signal, whose aliased spectra will be deleted by the ideal lowpass filter with cutoff frequency $\pi/2$, is given as:

$$\Omega_s - \Omega_c > \frac{\pi/2}{T_s}$$ $$\frac{2\pi}{T_s} - \Omega_c > \frac{\pi/2}{T_s}$$ $$\Omega_c < \frac{2\pi}{T_s} - \frac{\pi/2}{T_s}$$ $$\Omega_c < \frac{3\pi/2}{T_s}$$

Thats stated in hertz frequency as: $$f_c < \frac{3}{4}f_s$$

which gives a maximum allowable input signal bandwidth as $f_c=12,000$ Hz when the sampling rate is $f_s=16,000$ Hz.

$\endgroup$
0
2
$\begingroup$

Let's first consider the $H(e^{j\omega})$. The sampling frequency is $16$ kHz. The cutoff frequency of the low-pass filter is $$\frac{\pi/2}{\pi}=\frac{f_c}{16/2}\Rightarrow f_c=4 \text{ KHz}$$ Now consider the whole picture. The spectrum is replicated at a higher frequency seperated $16$ KHz ahead. So the next replica is placed at $16$ KHz. At which bandwidth of that replica at $16$ KHz you will get aliasing in the $4$ KHz passband of the spectrum at the origin?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.