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Question 4

A linear time invariant filter is described by the difference equation y[n]=0.8y[n-1]-0.8x[n]+x[n-1].

1) Find H(z) and identify the poles and zeros.

2) Find an expression for the frequency response, assuming fs=44,100 Hz.

3) Show that the magnitude of the frequency response is 1 for all values of f.

Solution.

1) The z-transform of y[n] is

\begin{eqnarray*}Y \left (z \right) & = & \sum_{ k= - \infty }^{ \infty } y[k]z^...
...-k}+x[k-1]z^{-k} }\\
& = & 0.8z^{-1}Y(z)-0.8X(z)+z^{-1}X(z)\\
\end{eqnarray*}


Thus

\begin{eqnarray*}H(z) & = & \frac{Y(z)}{X(z)}\\
& = & \frac{-0.8+z^-1}{1-0.8z^-1}\\
& = & -0.8 \frac{z-0.8^{-1}}{z-0.8}
\end{eqnarray*}


H(z) has pole at z=0.8 and zero at z=0.8-1=1.25.

2)

\begin{eqnarray*}H(f) & = & H(z)\vert _{z=e^{j\hat{\omega}} } \\
& = & -0.8 \frac{e^{j\hat{\omega}}-0.8^{-1}}{e^{j\hat{\omega}}-0.8}
\end{eqnarray*}


3)


\begin{eqnarray*}\vert H(e^{j\hat{\omega}})\vert^2 & = & H(e^{j\hat{\omega}}) H(...
...}{-0.8e^{j\hat{\omega}}-0.8e^{-j\hat{\omega}}+1.64 }\\
& = & 1
\end{eqnarray*}


Frequency response of the filter is shown in Fig.5.

  
Figure 5: Frequency response of H(z)



Hyun Ho Jeon
2001-02-05