# Nyquist Rate of cosine modulated function 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:

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

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

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 3 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 3 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 3 at 12:54
• whoop...change bandpass to baseband in my first comment... – pico May 3 at 12:57
• 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 3 at 13:00

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. • 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 3 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 3 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 3 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 3 at 10:36
• 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 3 at 12:46

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?

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".