I want to clear my thinking on how to approach problems of the type characterized below.

I guess what we want to do is select a sample rate of a sufficient size such that this signal can be fully captured therein but also small enough so as not to waste all the space from 0 to, 156? (77*2).

I've seen in some examples that if we select a sampling rate that is some integer multiple of this rate then we can use one that is much lower than the nyquist frequency, taking advantage of the property of replication.

So maybe for part (a), this would just have the same thing repeated over and over every 56 MHz? and How about specral inversion, would every other sample be inverted?

For part (b) maybe it's 56/2 = 28? or maybe it's 77/2?

For part (c) are we meant to think about the

fs' < 2fc-B/m

or maybe the

2fc-B/m > fs > 2fc+B/m+1

I guess m in this situation is 56? And B is 14? But how about the transition of B, it doesn't drop down straight but it flares out at the bottom, how to think about that? And I guess fc = 70?

enter image description here

EDIT: (reaction to Matt L.'s answer)

Is the image below what you had in mind for part (a)?

But why don't we take the bandwidth into account? Furthermore, in the image it seems that the 14MHz indicated there as the bandwidth doesnt completely describe the size of B since it seems to flare out at the ends, how about that?

So can we generalize that equation to be that for any time we are performing that undersampling we can get the location of all the replications by:

'Highest frequency component for which we must account' - ('number of K term' * 'sampling rate')

and sampling rate must be > 2B

Why did you only calculate for two terms? could be go on? can we do k = 3, 4, 5, 6... ?

I guess you mean the answer to part (b) is 14 because it's the smallest positive spectral component right?

& for (c) you mean f1 = 14 MHz = 70MHz - fs

enter image description here

  • $\begingroup$ Your figure is right. The bandwidth of the signal is the region where the spectrum is non-zero, or where it has significant components. It doesn't really matter that much in your example. Sampling always generates replicas of the original spectrum at multiples of the sampling frequency. I only calculated a few terms because that's the frequency range of interest according to part (a). And for (b) and (c) I mean exactly what I wrote, which it seems you understood well. $\endgroup$
    – Matt L.
    May 14, 2014 at 16:13

1 Answer 1


For (a) you just need to know that sampling will create replicas of the original spectrum shifted by multiples of the sampling frequency. But don't forget the negative frequencies. Since $s(t)$ is real-valued, you have $|S(f)|=|S(-f)|$, and you also have to shift the contribution at negative frequencies (which is a mirror image of the part of the spectrum shown in the figure). Considering the frequency range [-70MHz,70MHz], replicas of the positive part of the spectrum (as shown in the figure) occur at $70MHz-k\cdot56MHz$, $k=1,2$:

$$f_1=14MHz,\quad f_2=-42MHz$$

Replicas of the negative part of the original spectrum in the range [-70MHz,70MHz] occur at $-70MHz+kf_s$, $k=1,2$:

$$f_3=-14MHz,\quad f_4=42MHz$$

So the answer to (b) is $f_1=14MHz=70Mhz-f_s$, with the last equality being the answer to (c).

Check this link for more information on bandpass sampling.

  • $\begingroup$ Thank you very much for responding to my inquiry, I really like your answer and I've read that wiki page, but I'm got some subsequent questions about what you've said above, i wrote it as an answer as opposed to a question so i can post an image, would you mind to consider it a bit? please forgive my ignorance! $\endgroup$
    – user8769
    May 14, 2014 at 14:26

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