Difference between revisions of "Text Slecture" - Rhea

Revision as of 16:31, 2 October 2014

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Outline

• Introduction
• Useful Background
• DTFT Example of a Cosine Sampled Above the Nyquist Rate
• DTFT Example of a Cosine Sampled Below the Nyquist Rate
• Conclusion
• References

Introduction

In this Slecture, I will walk you through taking the DTFT of a pure frequency sampled above and below the Nyquist Rate. Then I will compare the differences between them.

Useful Background

Nyquist Condition: $ f_s = 2f_{max} $</font> DTFT of a Cosine: $ x_d[n] = cos(2\pi nT){\leftrightarrow}X(\omega) = \pi(\delta(\omega-\omega_o) + \delta(\omega+\omega_o)){ ,for\ } \omega \in [-\pi,\pi] $

The DTFT of a sampled signal is periodic with 2π.

DTFT of a Cosine Sampled Above the Nyquist Rate

For our original pure frequency, let’s choose the E below middle C. The E occurs at 330Hz $ x(t) = cos(2\pi * 330t) $. Now let’s sample this pure cosine at a frequency above the Nyquist Rate. The Nyquist Rate is:
 f_s = 2 * f_m'a'x = 2 * (330H'z) = 660Hz.
Let’s sample at 990Hz.

$ \begin{align} \\ x_d[n] & = x(n*\frac{1}{990Hz})\\ & = cos(2\pi n *\frac{330}{990}) = \frac{e^{j2\pi n \frac{330}{990}} + e^{-j2\pi n \frac{330}{990}}}{2}\\ & = cos(\frac{2\pi n}{3}) \end{align} $

Because $ \left | \frac{2\pi}{3}\right | < \pi $, there is no aliasing occurring in the DTFT, and it can be written as follows:

$ \begin{align} \\ X(\omega) & = \frac{1}{2}(2\pi\delta(\omega - 2\pi \frac{330}{990}) + 2\pi\delta(\omega + 2\pi \frac{330}{990})) , \ \omega \in\ [-\pi,\pi]\\ & = \frac{990}{2}(\delta(\frac{990}{2\pi}\omega - 330) + \delta(\frac{990}{2\pi}\omega + 330)) , \ \omega \in\ [-\pi,\pi]\\ & = rep_{2\pi}(\frac{990}{2}(\delta(\frac{990}{2\pi}\omega - 330) + \delta(\frac{990}{2\pi}\omega + 330))), \forall \omega \end{align} $

DTFT Of a Cosine Sampled Below the Nyquist Rate

Let’s use the same pure frequency as above.

x(t) = cos(2π * 330t)

Now let’s sample this pure cosine at a frequency below the Nyquist Rate. From above, the Nyquist Rate is 660Hz. Let’s sample at 550Hz.

$ \begin{align} \\ x_d[n] & = x(n*\frac{1}{550Hz})\\ & = cos(2\pi n *\frac{330}{550}) = \frac{e^{j2\pi n \frac{330}{550}} + e^{-j2\pi n \frac{330}{550}}}{2}\\ \\ \end{align} $

Because $ \pi < \frac{2\pi 330}{550} < 2\pi $, aliasing occurs in the DTFT. The DTFT should be calculated with ω ∈ [ − π, π], so we will use the periodicity of cosine to shift $ x_d[n] $ into an appropriate range.
>$ \begin{align}\\ x_{d}[n] & = cos(2\pi n*\frac{330}{550})\\ & = cos(2\pi n*\frac{330}{550} - 2\pi n)\\ & = cos(2\pi n*(\frac{330}{550} - \frac{550}{550}))\\ & = cos(2\pi n*(\frac{-220}{550}))\\ & = cos(2\pi n*\frac{220Hz}{550Hz}) \end{align} $

Now that the argument of the cosine $ \left | 2\pi \frac{220}{550}\right | < \pi $, we can take the DTFT of x_d[n], and the initial value will fall into a desired range for ω.

$ \begin{align} X(\omega) & = \frac{1}{2}(2\pi\delta(\omega - 2\pi \frac{220}{550}) + 2\pi\delta(\omega + 2\pi \frac{220}{550})) , \ \omega \in\ [-\pi,\pi]\\ & = \frac{550}{2}(\delta(\frac{550}{2\pi}\omega - 220) + \delta(\frac{550}{2\pi}\omega + 220)) , \ \omega \in\ [-\pi,\pi]\\ & = rep_{2\pi}(\frac{550}{2}(\delta(\frac{550}{2\pi}\omega - 220) + \delta(\frac{550}{2\pi}\omega + 220))), \forall \omega \end{align} $

Conclusion

Looking at Figure 3, you can see the comparison between a cosine sampled above and below the Nyquist Rate. The cosine sampled below the Nyquist Rate exhibits aliasing, which in this case causes a frequency shift and a magnitude scaling of the original frequency response.

References

[1] Mireille Boutin, "ECE 438 Digital Signal Processing with Applications," Purdue University. September 9, 2014.

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