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So, we had a very brief introduction to the Nyquist-Shannon sampling theorem (the discrete time version).

While discussing this, we have seen that multiplying a discrete time signal $x[n]$ by an impulse train $p[n]$, we get a sampled signal $x_p[n]$, and if certain conditions are fulfilled, then it is possible to recover the original signal.

In the frequency domain, what it means (informally) is that we are taking $X(e^{j\omega})$ and shifting it by integer multiples of $2\pi/N$ (i.e the sampling frequency) and "adding" all of this to the plot of $X(e^{j\omega})$ giving us the plot of $X_p(e^{j\omega})$ (in other words we have replicas of $X(e^{j\omega})$)

After looking on the internet and doing a couple of drawings, I've seen that one important condition to recover the initial signal $x[n]$ is that $\omega_s > 2\cdot \omega_m$, where $\omega_s$ is the sampling frequency and $\omega_m$ is the frequency of the original signal (i.e the frequency of $x[n]$).

However, the teacher, in his lecture only mentioned the following condition:

"If the original signal satisfies the following condition: $$X(e^{j\omega}) = 0, \text{ for } \frac{\pi}{N} < |\omega| \leq \pi$$ then the sampled signal $x_p[n]$ contains all the information about the original signal and we can easily recover $x[n]$ from $x_p[n]$".

This makes a little sense to me, as I understand that if $X(e^{j\omega}) \neq 0 $ for all $\omega$, then we do, indeed, have overlapping (i.e aliasing) and it is not possible to recover the signal. But I do not understand why $X(e^{j\omega})$ has to be $0$ specifically for $\frac{\pi}{N} < |\omega| \leq \pi$.

Also, how come the teacher did not speak of the other condition (namely that $\omega_s > 2\cdot \omega_m$) ? Has he forgotten it, or is there something I am missing ?

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Those two conditions are just the same. Note that $\omega_m$ is the highest frequency of $x[n]$, i.e., $X(e^{j\omega})=0$ for $\omega_m<|\omega|\le\pi$. Now we require $\omega_m<\omega_s/2$, where $\omega_s$ is the sampling frequency. In your case you have $\omega_s=2\pi/N$, i.e., $\omega_m<\pi/N$, or, in other words, $X(e^{j\omega})$ must be zero for $\pi/N\le|\omega|\le\pi$.

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  • $\begingroup$ Thank you very much, just one question though... You said, that since $\omega_m$ is the highest frequency of $x[n]$, then we have that $X(e^{j\omega}) = 0$ for $\omega_m < |\omega| \leq \pi$. I do understand the left hand side of this inequality, but not the right hand side. Why do we have that $|\omega| \leq \pi $ $\endgroup$ – Skyris Jul 1 '18 at 12:05
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Lets have a higher level of the idea of signal reconstruction from samples.

When you try to reconstruct something from partial information it is important to know what you know about the result required.

For instance, if you know the signal is a Line if $ \mathbb{R}^{2} $ then all you need is 2 samples, any samples of it.
You can even generalize that to any polynomial with having enough points samples from it.

You can even find more interesting rules (For instance, can you think about the case of having single Sine Function with Zero Phase ad Unit Amplitude, how many point will you need?).

So when dealing with reconstruction we need assumption of the model and then derive the requirements from the samples.

If we assume the signal has limited bandwidth, namely from a given frequency and above it vanishes in the Fourier domain then the Nyquist Shannon Sampling Theorem states a simple rule, sample the signal in a uniform manner with sampling rate which is larger then twice the bandwidth of the signal.
Once you do that, you have enough information to reconstruct it.

So in your case, saying that the signal vanishes from certain frequency and above is just stating it obeys the assumption of the model.

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