Nyquist Rate of cosine modulated function

Question

Here's my understanding:

$$y(t) = x(t)~ \cos(\Omega_0 t)$$

I take the Fourier transform of y(t) and I get this result:

$$Y(\Omega) = \frac{1}{2}X(\Omega - \Omega_0) + \frac{1}{2}X(\Omega + \Omega_0)$$

If the Highest frequency of $X(\Omega)$ is $\Omega_{N_X}$ ,shouldn't the highest frequency of $Y(\Omega)$ just be:

$$\Omega_{N_Y} = \Omega_0 + \Omega_{N_X}$$

instead of:

$$\Omega_{N_Y} = \Omega_{N_X} + 2 \Omega_0$$

as the textbook claims?

I suppose if you defined $\Omega_N$ to be the frequency width of $X(\Omega)$ from max negative frequency to max positive frequenct ...instead of frequency width from 0 to max positive frequency... then you would have a bandpass sampling rate of: $\Omega_s = 2(\frac{1}{2}\Omega_N + \Omega_0) = \Omega_N + 2\Omega_0$ that meets nyquists minimum sampling criteria. — pico, May 03 '20 at 12:42
No, Nyquist rate means "minimum" sampling rate required for a bandlimited signal to be reconstructed from it's samples, this is not the minimum sampling rate that you would require. That is why I bring up the context of difference in sampling of a bandpass vs baseband signal. Please refer to the link in my answer. This sampling rate would also work to recover the spectrum of "x(t)" but it is not the minimum sampling rate and hence not nyquist — Dsp guy sam, May 03 '20 at 12:53
but I'm a little but confused why you would define $\Omega_N$ as being a width from negative to positive max frequency of $X(\Omega)$, instead of from 0 to max positive $X(\Omega0)$ — pico, May 03 '20 at 12:54
Sampling creates shifted replicas of the original spectrum after sampling. Now if this shift is greater than duration in frqeuency from negative to positive these copies will not overlap otherwise they will. Double of 0 to max positive is also the same as from max negative to max positive (for a symmetric signal, "centred around DC", meaning baseband signal) — Dsp guy sam, May 03 '20 at 13:00
The idea is that sampling creates shifted replicas of the original spectrum, and these should not overlap, consider now that the signal extends in negative direction to $-f_o$ and in positive to $f_1$ ,which are not equal. Then minimum sampling rate is the sweep from negative to positive of the signal which $f_1 -(-f_o)$ which is $f_1+f_o$, so in this case we just can't look at positive frequencies and more general approach is to look that negative to positive — Dsp guy sam, May 03 '20 at 13:10
@pico The question says Nyquist Rate of $x_a(t)$ is $\Omega_N$, that means the complete baseband Bandwidth of $x_a(t)$ is $\Omega_N$, from negative to positive. Which in turn means, spectrum of $x_a(t)$ cannot exceed $\frac{\Omega_N}{2}$. — DSP Rookie, May 03 '20 at 14:06

jithin · Answer 1 · 2020-05-03T11:56:48.320

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Update: I feel there is no need to bring concept of Bandpass sampling (as mentioned in comments below) because this is specifically regarding a simple misunderstanding that OP has. The text referenced asks about Nyquist Rate (which wrongly mentions it as Nyquist Frequency) but OP asks about highest Frequency in $y(t)$.

For the component $\frac{1}{2}X(\Omega-\Omega_0)$ the highest frequency content is at $\Omega_0+\Omega_N/2$. The lowest (starting frequency) is at $\Omega_0-\Omega_N/2$. Similary for the other component $\frac{1}{2}X(\Omega+\Omega_0)$, the starting frequency is at $-\Omega_0-\Omega_N/2$ and ending frequency is at $-\Omega_0+\Omega_N/2$.

So if you see the overall frequency content of $y(t)$, lowest = $-\Omega_0-\Omega_N/2$ and highest = $\Omega_0+\Omega_N/2$. This is your signal $y(t)$ whose highest frequency is $\Omega_0+\Omega_N/2$.

The Nyquist rate for this is $2\Omega_0+\Omega_N$ if you consider $y(t)$ as baseband signal in the picture below. But the minimum sampling rate will be much lesser than this if you consider it as Passband signal (concept of bandpass sampling as mentioned in another answer)

If the Highest frequency of $X(\Omega)$ is $\Omega_N$

No. The highest frequency content in this case is $\Omega_N/2$ which is half the Nyquist rate. Otherwise this would cause aliasing.

edited May 03 '20 at 11:56

answered May 03 '20 at 05:12

jithin

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The signal $y(t)$ is a bandpass signal, if you have a Khz signal around a MHz band, we don't require to sample at Mhz – Dsp guy sam May 03 '20 at 10:18
@Dspguysam I know all that but the scenario here is different. The OPs question just asks why Nyquist Frequency is $2\Omega_0+\Omega_N$ for $y(t)$. I showed why. Whoever has downvoted, I hope you will revert it. It may have nothing to do with Bandpass sampling. – jithin May 03 '20 at 10:30
There are applications where signal has to be converted to IF and then use a single DAC to transmit. This is exactly the scenario depicted in question. Why bring concepts like bandpass sampling into this? – jithin May 03 '20 at 10:35
Nyquist frequency, is half the sampling rate, now in this case what is the sampling rate? Only when we talk about baseband signals the Nyquist frqeuency equals the maximum frqeuency of the signal – Dsp guy sam May 03 '20 at 10:36
Because it is a bandpass signal so you will have the concept of bandpass sampling.that is my understanding, please correct me – Dsp guy sam May 03 '20 at 10:40
"Only when you talk about baseband signals" not necessary. I might have loosely interchanged the terms "Nyquist Frequency" and "Nyquist Sampling rate" but that does not make the answer incorrect. $y(t)$ is also a valid baseband because its highest frequency content is $\pm (\Omega_0+\Omega_N/2)$. I have worked on Waveform Generator instrument which generates signal like $y(t)$ and use DAC to transmit. The sampling frequency of DAC is $F_s \ge 2 \Omega_0+\Omega_N$. Bandpass sampling doesn't even come into picture there. – jithin May 03 '20 at 10:42
Let me ask you this question now, let's take numerical values, $\Omega_N=30$Khz and $\Omega_o = 1$ GHz, what is the minimum sampling rate required for this signal according to your answer? – Dsp guy sam May 03 '20 at 10:47
Let us continue this discussion in chat. – jithin May 03 '20 at 10:48
you state the point that OP has asked about highest frqeuency, yes he has, but look at his context he is asking the highest frequency because he has to find the sampling rate of the signal, that's the question in his book and there is an answer also given in his book with about the rate – Dsp guy sam May 03 '20 at 12:09
Also, Nyquist rate: minimum required sampling rate of a "bandlimited" signal – Dsp guy sam May 03 '20 at 12:14
Yes, I have given the justification why the answer in text is correct because $y(t)$ can be considered as baseband signal too. I have also clarified the misunderstanding of OP regarding $\Omega_N/2$. Your misunderstanding bandlimited means it should always have non-zero magnitude within the band. That may not be true always as in the case of $y(t)$. – jithin May 03 '20 at 12:37
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the question title is? Cosine modulated function? Why do you modulate by a cosine, to make it a bandpass signal, if you modulate by a cosine a baseband signal it will always be bandlimited around the frequency of cosine and will always have a zero spectrum around baseband/DC (it's not an assumption), making it bandpass. You are not looking at the question in it's details. – Dsp guy sam May 03 '20 at 12:46

Dsp guy sam · Answer 2 · 2020-05-03T12:29:49.720

Firstly, your book uses the term Nyquist frqeuency as $\Omega_N + 2\Omega_o$, this is incorrect, this is the Nyquist rate (minimum sampling rate) if we consider the signal to be baseband. The maximum frequency content is then $\Omega_o + \frac{\Omega_N}{2}$, since you have defined maximum frequency of $x(t)$ as $\Omega_{N_x}$ this is nothing but $\frac{\Omega_{N}}{2}$ i.e. half of critical sampling rate or nyqusit rate. So it's essentially the same thing, Nyquist rate means sampling at the minimum rate which means the maximum frqeuency in the signal is half of it.

Secondly,

The signal $y(t)$ is a bandpass signal, if you have a Khz signal around a MHz band, we don't require to sample at Mhz.

Note: As soon as the term Nyquist frequency is aksed it is always in the context of sampling, because by definition Nyquist frqeuency is half the sampling rate.

This being a bandpass signal this has to be looked in context of bandpass sampling.

The book should be mentioning the maximum frequency or maximum bandwidth taking into account the signal is baseband. It is asking about Nyquist frqeuency which is half of sampling rate

Otherwise consider this example:

$\Omega_N=30$Khz and $\Omega_o = 1$ GHz, what is the minimum sampling rate required for this signal. Well if you define the Nyquist frequency by the term aksed in the questions this would be in GHz and minimum sampling rate (which is double of nyqusit frqeuency) would also be in GhZ, which is incorrect. minimum sampling rate for this example would be in Khz.

There is a good discussion available in this question by @Matt L.

Is there a condition for bandpass sampling?

DSP Rookie · Answer 3 · 2020-05-03T13:57:02.317

Answer : What you are considering as $\Omega_{N_x}$ is equal to $\frac{\Omega_N}{2}$ according to question. So, what you are saying is same as what answer mentions given we are considering Baseband Samping of $y_a(t)$.

I think you are confused because of the terms Nyquist rate and Nyquist Frequency.

Nyquist rate and Nyquist frequency are two different things. Nyquist rate is sampling rate satisfying the Nyquist criterion.

But Nyquist frequency is maximum frequency in the sampled signal. Nyquist frequency is always half of the sampling rate. It doesnot matter whether sampling rate is Nyquist rate or not. Nyquist frequency is the property of sampling itself.

But all sampling rates are not Nyquist rates. Only those sampling rates are Nyquist rates which satisfy Nyquist sampling criterion. Note that Nyquist sampling criterion is satisfied by both baseband and bandpass sampling where $f_s \ge 2B$.

The Question says Nyquist Rate of the baseband signal $x_a(t)$ is $\Omega_N$, meaning that the highest frequency in baseband $x_a(t)$ before modulation by cosine is $\frac{\Omega_N}{2}$. And then the question asks for Nyquist Rate for the derived signal $y_a(t)$.

And the answer given mentions Nyquist Frequency but then it gives Nyquist Rate for Baseband Sampling of $y_a(t)$. The answer $2\Omega_o + \Omega_N$ is correct for Nyquist Rate with Baseband Sampling of $y_a(t)$. So, the only wrong thing in the answer is mentioning "Frequency" instead of "Rate".

Nyquist Rate of cosine modulated function

3 Answers3