NyquistRozolis - Rhea

The Nyquist theorem

This theorem applies to a signal that is band limited.

When sampling a signal, it is generally done by multiplying the original signal by an impulse train modeled as:
$ \sum_{k=-\infty}^{\infty} \delta ( t - kTs ) $

This is done to extract data at equidistant points Ts (the sampling period) apart. The Nyquist theorem shows that in order to perfectly reconstruct a signal, then the frequency of the impulses must be at least 2 times the frequency of the initial signal.

Derivation

To sample a signal, one must multiply the above impulse train by the signal:

$ x_s(t) = x(t) \sum_{k=-\infty}^{\infty} \delta ( t - kT_s ) $

Since a $ \delta $ will always result in a zero anywhere other than $ kT_s $, then this can be further simplified:

$ x_s(t) = \sum_{k=-\infty}^{\infty} x(kT_s) \delta ( t - kT_s ) $

Then, this can be brought into the radial frequency domain. Multiplication is convolution in the frequency domain, therefore:

$ X_s(w_s) = \frac{1}{T_s} \sum_{k=-\infty}^{\infty} X(w) \delta ( w - kw_s ) $

The same simplification can be made in the radial frequency domain as used before:

$ X_s(w_s) = \frac{1}{T_s} \sum_{k=-\infty}^{\infty} X(w-kw_s) $

In order to prevent aliasing, all components within the sum must not overlap. In other words, when sampling for each integer k, there cannot be another sample that falls within the same frequency.

The k=0 sample spans from the frequency [-w,w], thus to prevent aliasing other samples must not overlap.

Therefore $ w-kw_s \leq -w $ for all 2 unique integers k between $ -\infty \ $ and $ \ \infty $

Assuming k to be 1:

$ 2w \leq w_s $

This can be converted to frequency in hertz:

$ 2 * 2 \pi f \leq 2 \pi f_s $

$ 2f \leq f_s $

Back to ECE438 slectures, Fall 2014