Question 3

Design a simple digital filter whose trasnfer function corresponds to a delay of two sample times, i.e.

y[k] = x[k-2] (22)

Assume a sampling frequency f_s = 48 kHz.

3.1 Specify H(z).

3.2 Find the frequency response (amplitude and phase) of this filter.

3.3 Find the impulse response h[k] of this filter by taking the inverse z-transform of H(z).

3.4 Show by convolution that the output of the filter y[k] is x[k-2] as expected.

Solution.

3.1

Z-transform of y[k] is

$$Y \left (z \right) \sum_{ k= - \infty }^{ \infty } y[k]z^{-k}$$ = (23)

$$\sum_{ k= - \infty }^{ \infty } x[k-2]z^{-k}$$ = (24)

= z^-2X(z) (25)

Thus H(z) is

$$\frac{1}{z^2}$$ H(z) = Y(z)/X(z) = (26)

3.2

The frequency response H(f) is

$$H(z)_{z=e^{j\hat \omega} }$$ H(f) = (27)

$$H(z)\vert _{z=e^{j2\pi f/f_s} }$$ = (28)

$$\frac{1}{(e^{j2\pi f/f_s})^2}$$ = (29)

$$\frac{1}{e^{j4\pi f/f_s}}$$ = (30)

$$\frac{1}{cos(4\pi f/f_s)+jsin(4\pi f/f_s)}$$ = (31)

$$\frac{1}{d_1} e^{-j\theta_1}$$ = (32)

The magnitude response is

$$\frac{1}{d_1}$$ |H(f)| = (33)

and phase response is

$$- \theta_1$$ arg(H(f)) = (34)

where

$$\sqrt{cos^2(4\pi f/f_s) + sin^2(4\pi f/f_s)}$$ d₁ = (35)

$$\theta_1 4\pi f/f_s$$ = (36)

3.3

From the z-transform pair

$$z^{-k} \leftrightarrow \delta[n-k]$$ (37)

The impulse response of the filter is

$$h[n] = \delta[n-2]$$ (38)

3.4

Given filter impulse response h[k] and filter input sequence x[n], the filter out is

y[k] = x[k]*h[k] (39)

$$\sum_{ n= 0 }^{ \infty }h[n]x[k-n]$$ = (40)

$$\sum_{ n= 0 }^{ \infty }\delta[n-2]x[k-n]$$ = (41)

$$\delta[2-2]x[k-2]$$ = (42)

= x[k-2] (43)