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Question 3

POI also calculates the frequency response H(z) with z along the unit circle $ z = e^{j\hat{\omega}} = e^{j2\pi f/f_s}$ for |f| less than fs/2. The sampling rate fs is not defined in Assignment 1, the POI software lets you choose the sampling rate. For fs = 44,100 Hz, calculate the frequency response for each of the 3 filters of Assignment 1. Do this two ways:

1) Use the POI software to place the poles and zeros in the complex plane, and let the program calculate the frequency response.

2) Work out the algebraic expression for the frequency response and plot using Matlab.

Compare the results from POI and Matlab.

Solution.

1) Results from POI software

Frequency responses calculated from POI software are shown in Fig.2.


  
Figure 2: Frequency responses of the three filters obtained by POI

2) Matlab plot for the algebraic expression

(a) The frequency response of an LTI system is given by the transfer function evaluated on the unit circle


\begin{eqnarray*}H(f) = H(z)\vert _{z=e^{j\hat \omega} } = H(z)\vert _{z=e^{j2\pi f/f_s}}
\end{eqnarray*}


where $\hat \omega = 2 \pi f/f_s $ is the normalized radian frequency, and fs is the sampling frequency.

For the first filter,


\begin{eqnarray*}H(f) & = & H(z)_{z=e^{j\hat \omega} }\\
& = & H(z)\vert _{z=e...
...1}}\\
& = & \frac{1}{2} \frac{r_1}{d_1} e^{j\phi_1 - \theta_1}
\end{eqnarray*}


The magnitude response is

\begin{eqnarray*}\vert H(f)\vert & = & \frac{1}{2} \frac{r_1}{d_1}
\end{eqnarray*}


and phase response is


\begin{eqnarray*}arg(H(f)) & = & \phi_1 - \theta_1
\end{eqnarray*}


where


\begin{eqnarray*}r_1 & = & (cos(2\pi f/f_s)+1)^2 + sin^2(2\pi f/f_s)\\
d_1 & = ...
...n(2\pi f/f_s)}{cos(2\pi f/f_s)+1} )\\
\theta_1 & = & 2\pi f/f_s
\end{eqnarray*}


(b)

For the second filter,


\begin{eqnarray*}H(f) & = & H(z)\vert _{z=e^{j2\pi f/f_s} }\\
& = & \frac{(e^{...
... & \frac{cos(2\pi f/f_s)+jsin(2\pi f/f_s)-1}{e^{j2\pi f/f_s}}\\
\end{eqnarray*}


The magnitude response is

\begin{eqnarray*}\vert H(f)\vert & = & \sqrt{(cos(2\pi f/f_s)-1)^2 + sin^2(2\pi f/f_s)}
\end{eqnarray*}


and phase response is

\begin{eqnarray*}arg(H(f)) & = & \arctan(\frac{sin(2\pi f/f_s)}{cos(2\pi f/f_s)-1} )-(2\pi f/f_s)
\end{eqnarray*}


(c)

For the third filter,


\begin{eqnarray*}H(f) & = & H(z)\vert _{z=e^{j2\pi f/f_s} }\\
& = & \frac{1}{e...
..._s}-a}\\
& = & \frac{1}{cos(2\pi f/f_s)+jsin(2\pi f/f_s)-a}\\
\end{eqnarray*}


The magnitude response is

\begin{eqnarray*}\vert H(f)\vert & = & \frac{1}{\sqrt{(cos(2\pi f/f_s)-a)^2 + sin^2(2\pi f/f_s)}}
\end{eqnarray*}


and phase response is

\begin{eqnarray*}arg(H(f)) & = & -\arctan(\frac{sin(2\pi f/f_s)}{cos(2\pi f/f_s)-a} )
\end{eqnarray*}


Magnitude responses and phase responses calculated from Matlab are shown in Fig.3 and Fi.4.


  
Figure 3: Magnitude responses of the three filters obtained by Matlab


  
Figure 4: Phase responses of the three filters obtained by Matlab
\begin{figure}
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We can observe that POI's results are the same as Matlab's results.

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Next: Question 4 Up: Solution to Assignment 3 Previous: Question 2
Hyun Ho Jeon
2001-02-05