Question 1

1.1 For a transfer function with one pole and one zero, where (approximately) in the complex z-plane wolud you place the pole $\gamma_1$ and zero z₁ to obtain a high pass filter characteristic with zero response at f=0 and maximum response at f=f_s/2 ?

1.2 Draw a sketch to show the pole and zero locations.

1.3 Draw a sketch of the approximate frequency (magnitude) response.

1.4 Derive the diference equation to implement this high pass filter in terms of $\gamma_1$ and z₁.

1.5 Draw a diagram of an impulse response of the filter.

Solution.

1.1

For a filter with one pole and one zero,

$$b_0 \frac{z-z_1}{z-\gamma_1}$$ H[z] = (1)

The frequency response becomes

$$H(z)_{z=e^{j\hat \omega} }$$ H(f) = (2)

$$H(z)\vert _{z=e^{j2\pi f/f_s} }$$ = (3)

$$b_0 \frac{z=e^{j2\pi f/f_s-z_1}}{z=e^{j2\pi f/f_s-\gamma_1}}$$ = (4)

$$b_0 \frac{r_1}{d_1} e^{j(\phi_1 - \theta_1)}$$ = (5)

where

$$r_1e^{j\phi_1} e^{j2\pi f/f_s}-z_1$$ = (6)

$$d_1e^{j\theta_1} e^{j2\pi f/f_s}-\gamma_1$$ = (7)

Assume b₀=1. From the table on the notes page 49, we can see that the pole $\gamma_1$ should be located on the negative real axis, whereas the zero z₁ on the positive real axis. If we leave the pole $\gamma_1$ on the negative real axis within the unit circle and move the zero z₁ to 1, H(f) becomes zero at f=0. For these z₁ and $\gamma_1$, H(f) becomes maximum at f=f_s/2.

1.2

The pole and zero locations on the z-plane are shown in Fig.1(a).

1.3

Magnitude response of the H(f) is shown in Fig.1(b).

  

Figure 1: Pole-zero locations and frequency response of the filter

1.4

From H(z) in Eq.(1),

$$Y(z)(z-\gamma_1)$$ = b₀ X(z)(z-z₁) (8)

$$Y(z)(1-\gamma_1 z^{-1})$$ = b₀ X(z)(1-z₁z^-1) (9)

$$\gamma_1 z^{-1} Y(z) + b_0X(z) - b_0 z_1 z^{-1} X(z)$$ Y(z) = (10)

By applying inverse z-transform,

$$\gamma_1 y[n-1] + b_0 x[n] - b_0 z_1 x[n-1]$$ y[n] = (11)

1.5

The block diagram of the filter is shown in Fig.2.

  

Figure 2: Block diagram of the filter