Line 38: Line 38:
 
== DTFT of a Cosine Sampled Above the Nyquist Rate  ==
 
== 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 330''Hz'' <math>x(t) = cos(2π * 330t)</math> &lt;p&gt;Now let’s sample this pure cosine at a frequency above the Nyquist Rate. The Nyquist Rate is: <br><span class="math">&nbsp;''f''<sub>''s''</sub> = 2 * ''f''<sub>''m''''a''''x''</sub> = 2 * (330''H''''z'') = 660''H''''z''</span>. <br>Let’s sample at 990<span class="math">''H''''z''</span>.  
+
For our original pure frequency, let’s choose the E below middle C. The E occurs at 330''Hz'' <math>x(t) = cos(2π * 330t)</math> &lt;p&gt;Now let’s sample this pure cosine at a frequency above the Nyquist Rate. The Nyquist Rate is: <br><span class="math">&nbsp;''f''<sub>''s''</sub> = 2 * ''f''<sub>''m''''a''''x''</sub> = 2 * (330''H''''z'') = 660''Hz''</span>. <br>Let’s sample at 990<span class="math">''Hz''</span>.  
  
 
<math> \begin{align} \\  
 
<math> \begin{align} \\  
 
  x_d[n] & = x(n*\frac{1}{990Hz})\\
 
  x_d[n] & = x(n*\frac{1}{990Hz})\\
& = cos(2\pi n *\frac{330}{990}) = \frac{e^{j2\pi n \frac{330}{550}} + e^{-j2\pi n \frac{330}{550}}}{2}\\
+
& = 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})
 
& = cos(\frac{2\pi n}{3})
 
\end{align}  
 
\end{align}  
Line 59: Line 59:
  
 
== DTFT Of a Cosine Sampled Below the Nyquist Rate ==
 
== DTFT Of a Cosine Sampled Below the Nyquist Rate ==
<p>Let’s use the same pure frequency as above.<br /><br /></p>
+
<p>Let’s use the same pure frequency as above.</p>
<p><br /><span class="math"><em>x</em>(<em>t</em>) = <em>c</em><em>o</em><em>s</em>(2<em>π</em> * 330<em>t</em>)</span><br /></p>
+
<p><span class="math"><em>x</em>(<em>t</em>) = <em>c</em><em>o</em><em>s</em>(2<em>π</em> * 330<em>t</em>)</span><br /></p>
 
<p>Now let’s sample this pure cosine at a frequency below the Nyquist Rate. From above, the Nyquist Rate is 660<span class="math"><em>H</em><em>z</em></span>. Let’s sample at 550<span class="math"><em>H</em><em>z</em></span>.</p>
 
<p>Now let’s sample this pure cosine at a frequency below the Nyquist Rate. From above, the Nyquist Rate is 660<span class="math"><em>H</em><em>z</em></span>. Let’s sample at 550<span class="math"><em>H</em><em>z</em></span>.</p>
<p><br /><span class="math">$$\begin{split}
+
<p><br />
  x_{d}[n] &amp; = x(n*\frac{1}{550Hz})\\
+
 
&amp; = cos(2\pi n*\frac{330Hz}{550Hz}) = \frac{e^{j2\pi n \frac{330}{550}} + e^{-j2\pi n \frac{330}{550}}}{2}\\
+
<math> \begin{align} \\
\end{split}$$</span><br /></p>
+
  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}  
 +
</math> <br /><br />
 +
 
 +
Because <font style="vertical-align:-125%;"><math>\pi < \frac{2\pi 330}{550} < 2\pi</math></font>, aliasing occurs in the DTFT. The DTFT should be calculated with <font style="vertical-align:-80%;"><math>\omega \in [-\pi,\pi]</math></font>, so we will use the periodicity of cosine to shift <math>x_d[n]</math> into an appropriate range.
 +
 
 +
<br /></p>
 
<p><br /></p>
 
<p><br /></p>
 
<p>Because <span class="math">$ \pi &lt; \frac{2\pi 330}{550} &lt; 2\pi$</span>, aliasing occurring in the DTFT. The DTFT should be calculated with <span class="math"><em>ω</em> ∈ [ − <em>π</em>, <em>π</em>]</span>, so we will use the periodicity of cosine to shift <span class="math"><em>x</em><sub><em>d</em></sub>[<em>n</em>]</span> into an appropriate range.</p>
 
<p>Because <span class="math">$ \pi &lt; \frac{2\pi 330}{550} &lt; 2\pi$</span>, aliasing occurring in the DTFT. The DTFT should be calculated with <span class="math"><em>ω</em> ∈ [ − <em>π</em>, <em>π</em>]</span>, so we will use the periodicity of cosine to shift <span class="math"><em>x</em><sub><em>d</em></sub>[<em>n</em>]</span> into an appropriate range.</p>
<p><br /><span class="math">$$\begin{split}
+
<p><br /
     x_{d}[n] &amp; = cos(2\pi n*\frac{330}{550})\\
+
><math> \begin{align}\\
     &amp; = cos(2\pi n*\frac{330}{550} - 2\pi n)\\
+
     x_{d}[n] & = cos(2\pi n*\frac{330}{550})\\
     &amp; = cos(2\pi n*(\frac{330}{550} - \frac{550}{550}))\\
+
     & = cos(2\pi n*\frac{330}{550} - 2\pi n)\\
     &amp; = cos(2\pi n*(\frac{-220}{550}))\\
+
     & = cos(2\pi n*(\frac{330}{550} - \frac{550}{550}))\\
     &amp; = cos(2\pi n*\frac{220Hz}{550Hz})
+
     & = cos(2\pi n*(\frac{-220}{550}))\\
     \end{split}$$</span><br /></p>
+
     & = cos(2\pi n*\frac{220Hz}{550Hz})
 +
     \end{align}
 +
    </math><br /></p>
 
<p>Now that <span class="math">$\ |2\pi\frac{220}{550}| &lt; \pi$</span>, we can take the DTFT of <span class="math"><em>x</em><sub><em>d</em></sub>[<em>n</em>]</span>, and the initial value will fall into a desired range for <span class="math"><em>ω</em></span>.</p>
 
<p>Now that <span class="math">$\ |2\pi\frac{220}{550}| &lt; \pi$</span>, we can take the DTFT of <span class="math"><em>x</em><sub><em>d</em></sub>[<em>n</em>]</span>, and the initial value will fall into a desired range for <span class="math"><em>ω</em></span>.</p>
 
<p><br /><span class="math">$$\begin{split}
 
<p><br /><span class="math">$$\begin{split}

Revision as of 10:12, 2 October 2014

Link title

DTFT of a Cosine Sampled Above and Below the Nyquist Rate

A slecture by ECE student Sahil Sanghani

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.


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 .

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(2π * 330t) $ <p>Now let’s sample this pure cosine at a frequency above the Nyquist Rate. The Nyquist Rate is:
 fs = 2 * fm'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} $

AboveNyquist.jpg

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 $ \omega \in [-\pi,\pi] $, so we will use the periodicity of cosine to shift $ x_d[n] $ into an appropriate range.


Because $ \pi < \frac{2\pi 330}{550} < 2\pi$, aliasing occurring in the DTFT. The DTFT should be calculated with ω ∈ [ − π, π], so we will use the periodicity of cosine to shift xd[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 $\ |2\pi\frac{220}{550}| < \pi$, we can take the DTFT of xd[n], and the initial value will fall into a desired range for ω.


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

Back to ECE438, Fall 2014

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010