Difference between revisions of "Text slecture" - Rhea

Outline

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

Introduction

This slecture explains definition of downsampling and demonstrates downsampled signal in frequency domain. Also, it explains process of decimation and why it needs a low-pass filter.

Derivation

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.