Question 1

For the three filters of assignment 1, find the transfer function of H(z). Write H(z) as a ratio of polynomials and identify the poles and zeros. Choose a a = 0.9.

Solution.

(a) For the impulse response

$$\begin{eqnarray*}h \left [ k \right ] & = & 0.5 \delta \left [ k \right ] + \delta \left [ k-1 \right ] + 0.5 \delta \left [ k-2 \right ] \end{eqnarray*}$$

its z-transform can be expressed as:

$$\begin{eqnarray*}H \left (z \right) & = & \sum_{ k= - \infty }^{ \infty } h[k]z^... ...& = & 0.5 + z^{-1} + 0.5z^{-2}\\ & = & \frac{0.5(z+1)^2}{z^2} \end{eqnarray*}$$

The above equation is equivalent to

$$\begin{eqnarray*}H \left (z \right) & = & b_{0} \frac{(z-z_1)(z-z_2)}{(z-p_1)(z-p_2)} \end{eqnarray*}$$

where z₁= z₂=-1 and p₁=p₂=0 and b₀=0.5.

(b) For the impulse response

$$\begin{eqnarray*}h \left [ k \right ] & = & \delta \left [ k \right ] - \delta \left [ k-1 \right ] \end{eqnarray*}$$

its z-transform can be expressed as:

$$\begin{eqnarray*}H \left (z \right) & = & \sum_{ k= - \infty }^{ \infty } h[k]z^... ...1 \right ]z^{-1} \\ & = & 1 - z^{-1} \\ & = & \frac{z-1}{z} \end{eqnarray*}$$

The above equation is equivalent to

$$\begin{eqnarray*}H \left (z \right) & = & b_{0} \frac{z-z_1}{z-p_1} \end{eqnarray*}$$

where z₁=1 and p₁=0 and b₀=1.

(c) For the output sequence

$$\begin{eqnarray*}y \left [ n \right ] & = & ay \left [ n-1 \right ] + x\left [ n-1 \right ] \end{eqnarray*}$$

its z-transform can be expressed as:

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

Using the linearity and time-shifting properties of the z-transform , we can write

$$\begin{eqnarray*}Y \left (z \right) & = & aY(z)z^{-1} + X(z)z^{-1} \end{eqnarray*}$$

The above equation is equivalent to (1-az^-1)Y(z) & = & X(z)z^-1

Thus

$$\begin{eqnarray*}H(z) & = & \frac{Y(z)}{X(z)}\\ & = & \frac{z^{-1}}{1-az^{-1}}\\ & = & \frac{1}{z-a}\\ & = & \frac{1}{z-p_1} \end{eqnarray*}$$

where p₁=a=0.9.