Difference between revisions of "Text slecture" - Rhea

Outline

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

Introduction

This slecture demonstrates downsampled signal in frequency domain. Also, it explains process of decimation and why it needs

a low-pass filter.

Derivation

Let x[n] be a digital-time signal shown below:

GRAPH

 then y[n] is produced by downsampling x [n] by factor D = 3. So, y [n] = x[Dn].

GRAPH

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 and 2a.

Conclusion