Sampling Theorem

English Definition

a signal $ x(t) $ can be uniquely recovered from its samples if the samples within $ x(nT) $ satisfy the nyquist rate, or $ T < \frac{1}{2}\frac{2\pi}{\omega_m} $ for all integers n.

Test for sampling

Instead of using the T based definition, use instead that $ \omega_s>2\omega_m $, where

$ \omega_s $ is the Sampling frequency

$ \omega_m $ is the maximum, or band limited frequency, and

$ 2\omega_m $ is the actual nyquist rate

Proof

The sampling theorem will still only yield an approximation of the original signal, as will be shown by its proof, but the nyquist rate does give a very reliable approximation.

The representation of many samples is represented as $ x_p(t) $. This signal is as follows:

$ x_p(t)=x(t)p(t) $ where p(t) is an impulse train, or mathematically:

$ x_p(t)=\sum_{n=-\infty}^\infty x(t)\delta(t-nT) $

Since $ x(t) $ only has values at intervals of the sampling period $ nT $, the equation can be simplified to

$ x_p(t)=\sum_{n=-\infty}^\infty x(nT)\delta(t-nT) $

This essentially looks like a string of impulses with heights at the same values as the original signal

Our reconstruction, which completes the proof, is done however in the frequency domain:

A multiplication in the time domain becomes a convolution in the frequency domain

$ \mathcal{X}_p(\omega)=\frac{1}{2\pi}\mathcal{X}(\omega)*\sum_{k=-\infty}^\infty \frac{2\pi}{T}\delta(\omega-k\omega_s) $

$ \mathcal{X}_p(\omega)=\sum_{k=-\infty}^\infty \frac{1}{T}\mathcal{X}(\omega-k\omega_s) $

So this represents a lot of shifted frequency representations of the original signal, shifted by $ k\omega_s $ and each with a height of $ \frac{1}{T} $.

To recover the approximate original signal, or what will be called the reconstructed signal $ x_r(t) $, we must use a low pass filter with a gain of T, and a frequency in the range of $ (\omega_m,\omega_s-\omega_m) $. It needs to be greater than the maximum frequency so that it collects all the information, but less than the space in between the end of the first cycle and offset of the second.

We can call this frequency $ \omega_c $, and it is safely within the nyquist range at $ \omega_c=\frac{\omega_s}{2} $

We can use a transfer function to approximate this filter, and it is:

$ \mathcal{H}(\omega)=T(u(\omega+\omega_c)-u(\omega-\omega_c)) $

Therefore we can multiply these in the frequency domain:

$ \mathcal{X}_r(\omega)=\mathcal{H}(\omega)\mathcal{X}_p(\omega) $

$ \mathcal{X}_r(\omega)=T(u(\omega+\omega_c)-u(\omega-\omega_c))\sum_{k=-\infty}^\infty \frac{1}{T}\mathcal{X}(\omega-k\omega_s) $

This isolates the range, and if the convolution was ideal, this function should only have a value when k=0, which would yield:

$ \mathcal{X}_r(\omega)=\frac{1}{T}T\mathcal{X}(\omega-0\omega_s)=\mathcal{X}(\omega) $

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