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

= | (23) | ||

= | (24) | ||

= | z^{-2}X(z) |
(25) |

Thus *H*(*z*) is

H(z) = Y(z)/X(z) |
= | (26) |

3.2

The frequency response *H*(*f*) is

H(f) |
= | (27) | |

= | (28) | ||

= | (29) | ||

= | (30) | ||

= | (31) | ||

= | (32) |

The magnitude response is

|H(f)| |
= | (33) |

and phase response is

arg(H(f)) |
= | (34) |

where

d_{1} |
= | (35) | |

= | (36) |

3.3

From the z-transform pair

(37) |

The impulse response of the filter is

(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) |

= | (40) | ||

= | (41) | ||

= | (42) | ||

= | x[k-2] |
(43) |