Nyquist Miguel Castellanos ECE438 slecture - Rhea

Outline

1. Introduction
2. Proof
3. Example
4. Conclusion
5. References

Introduction

The Nyquist theorem provides a sufficient condition for perfect reconstruction of a signal from its sampling. In this slecture I will give a proof of the Nyquist theorem and give an example demonstrating how perfect reconstruction is possible even when violating the Nyquist condition.

Proof

Let the ideal sampling $ x_s(t) \text{ of } x(t) $ be defined as

$ x_s(t) := x(t)p_{\frac{1}{f_s}}(t) $,

where

$ p_{\frac{1}{f_s}}(t) = \sum_{k = -\infty}^\infty \delta(t-\frac{k}{f_s}) $.

Nyquist Theorem: A signal $ x(t) $ that has the property $ X(f) = 0 $ for $ |f| \ge f_M $ can be perfectly reconstructed from its sampling $ x_s(t) $ if sampled at a rate $ f_s > 2f_M $.

To prove that perfect reconstruction is possible, we must find an expression for $ x(t) $ in terms of $ x_s(t) $.

Given that $ \mathcal{F}(x(t)) = X(f) $, we can find $ X_s(f) $ using the convolution property.

Without loss of generality, we can assume that the signal $ x(t) $ has the spectrum shown in the figure below. The shape of the graph of $ X(f) $ does not matter because the only important feature of $ X(f) $ is that $ X(f) = 0 $ for $ |f| \ge f_M $.

We would like to determine what $ X_s(f) $ looks like in order to find a way to reconstruct $ x(t) $.

Since we have sampled at a rate $ f_s > 2f_M $, the following inequalities hold:

$ f_s > 2f_M \iff f_s - f_M > f_M \iff -f_s + f_M < f_M $.

Together, these inequalities, the graph of $ X(f) $, and the expression for $ X_s(f) $ in terms of $ X(f) $ imply that $ X_s(f) $ will have the spectrum shown in the figure below.

Notice that the spectrum of the ideal sampling of a signal is an amplitude scaled periodic repetition of the original spectrum. Since $ x(t) $ is bandlimited and we have sampled at a rate $ f_s > 2f_M $, the periodic repetitions of $ X(f) $ do not overlap.

All the information needed to reconstruct $ X(f) $ can be found in the portion of $ X_s(f) $ that corresponds to $ X(f) $ (shown in red). Therefore we can use a simple lowpass filter with gain $ \tfrac{1}{f_s} $ and cutoff frequency $ \tfrac{f_s}{2} $ to recover $ X(f) $ from $ X_s(f) $.

$ X(f) = X_s(f)\left\{ \begin{array}{ll} \frac{1}{f_s}, & |f| \le \frac{f_s}{2}\\ 0, & \text{else} \end{array} \right. $

$ \iff x(t) = x_s(t)*\text{sinc}(f_st) $

$ \therefore $ We can perfectly reconstruct $ x(t) $ from $ x_s(t) $.

Example

Though the Nyquist theorem states that perfect reconstruction is possible if we satisfy the Nyquist condition $ (f_s > 2f_M) $, it is important to note that this condition is not necessary. The following example demonstrates how perfect reconstruction is sometimes possible even when undersampling.

Let the signal $ x(t) $ have a spectrum $ X(f) $ as seen in the figure below.

The Nyquist condition states that we should sample at a rate $ f_s > 2(2a) = 4a $. Instead, let us sample at $ f_s = 2a $.

As before, we have $ x_s(t) = x(t)p_{\frac{1}{f_s}}(t) $ and $ X_s(f) = f_s\sum_{k = -\infty}^\infty X(f-kf_s) $

$ \implies X_s(f) = 2a\sum_{k = -\infty}^\infty X(f-2ka) $.

Therefore, $ X_s(f) $ will have the spectrum shown in the figure below.

Notice that there is no aliasing in $ X_s(f) $ even though $ f_s < 4a $. In addition, the portion of $ X_s(f) $ that corresponds to $ X(f) $ (shown in red) can be recovered using a bandpass filter with gain $ \tfrac{1}{2a} $ and cutoff frequencies $ a \text{ and } 2a $.

Conclusion

To summarize, the Nyquist theorem states that any bandlimited signal can be perfectly reconstructed from its sampling if sampled at a rate greater than twice its bandwidth $ (f_s > 2f_M) $. However, the Nyquist condition is not necessary for perfect reconstruction as shown in the example above.