# Minimum possible sampling frequency for continuous time signal

A continuous-time signal $x(t)$ has the magnitude spectrum $\lvert X(F)\rvert$ shown in Figure 6. The signal is sampled to obtain $x(n)$ using the sampling frequency $F_s$. Determine the minimum sampling frequency, which ensures that the information in $[0; 1000 \ \rm{Hz}]$ is not distorted.

Immediately, I would think that $2000 \ \rm Hz$ is the correct answer.

Can someone explain to me why it is not and what formulas/intution is required to answer the above task?

• why does it say $jX(F)j$? Shouldn't that just be identical to $-X(F)$? Or is $X$ some kind of matrix and $j$ not the usual imaginary unit? – Marcus Müller Dec 15 '17 at 19:58
• I think OP meant $\lvert \cdot \rvert$ with the $j \cdot j$. – Gilles Dec 15 '17 at 20:00
• ah, Im stupid. You just copy&pasted... not even fixing hyphenation. For some reason, that converted $|$ to $j$, and you didn't even bother to fix... – Marcus Müller Dec 15 '17 at 20:00
• @Gilles: aaah, simultaneous comments :) – Marcus Müller Dec 15 '17 at 20:00

This is one of those problems which ask for finding the sampling rate that avoids aliasing in the band of interest but don't care aliasing for the rest of the spectrum.

In this problem the real, baseband analog signal has a bandwith of $B = 1800$ Hz, which would require a minimum Nyquist sampling rate of $F_s = 2 B = 3600$ Hz when we are to avoid aliasing on the complete band of the signal; $-1800< f < 1800$ Hz which is depicted as the condition on figure-a below: (without amplitude scaling)

On the other hand the condition on the figure-b tells that if you want to preserve the signal only in the band of $-1000< f < 1000$ , then you can relax the aliasing condition to lower the sampling rate so that aliasing is prevented only in the band of ineterest; $-1000< f < 1000$ . That sampling rate you look is found by requiring on the shifted spectrum that, the leftmost cutoff frequency of right shifted replica is above $1000$ Hz of the unshifted replica, which is dispalyed on the figure-b.

$$F_s - 1800 > 1000 \implies F_s > 2800$$

Hence the minimum sampling rate that will avoid aliasing in the band $-1000< f < 1000$ but otherwise will create aliasing in the range $1000< |f| < 1800$ is $$F_s = 2800$$

Note that what you considered would be true if the signal was lowpass filtered to $1000$ Hz before being sampled. But in the problem no such antialiasing bandlimiting filtering is mentioned.

• Thank you Fat32. Can you explain to me what you mean by " leftmost cutoff frequency of right shifted spectrum". I still don't get the intuition about why we need to add the extra 800 Hz.. – Peter Alexander Dec 16 '17 at 9:00
• @PeterAlexander I've added a figure so that it's crystal clear :-)) , isn't it? – Fat32 Dec 16 '17 at 16:49

Hint #1 (As this looks like homework): draw the resulting spectrum for a few different sampling frequencies, say 1000Hz, 1500Hz, 2000 Hz.

Hint #2: make sure you include aliasing in the exercise of hint #1