<< Chapter < Page Chapter >> Page >
H ( z ) = n = 0 M b ( n ) z - n n = 0 N a ( n ) z - n = B ( z ) A ( z )

The frequency response of the filter is found by setting z = e j ω , which gives [link] the form

H ( ω ) = n = 0 h ( n ) e - j ω n

It should be recalled that this form assumes a sampling rate of T = 1 . To simplify notation, H ( ω ) is used to denote the frequency response rather than H ( e j ω ) .

This frequency-response function is complex-valued and consists of a magnitude and phase. Even though the impulse responseis a function of the discrete variable n, the frequency response is a function of the continuous-frequency variable ω and is periodic with period 2 π .

Unlike the FIR filter case, exactly linear phase is impossible for the IIR filter. It has been shown that linear phase isequivalent to symmetry of the impulse response. This is clearly impossible for the IIR filter with an impulse response that is zerofor n < 0 and nonzero for n going to infinity.

The FIR linear-phase filter allowed removing the phase from the design process. The resulting problem was a real-valuedapproximation problem requiring the solution of linear equations. The IIR filter design problem is more complicated. Linear phase isnot possible, and the equations to be solved are generally nonlinear. The most common technique is to approximate themagnitude of the transfer function and let the phase take care of itself. If the phase is important, it becomes part of theapproximation problem, which then is often difficult to solve.

Calculation of the iir filter frequency response

As shown in another module, L equally spaced samples of H ( ω ) can be approximately calculated by taking an L -length DFT of h ( n ) given in [link] . However, unlike for the FIR filter, this requires that the infinitely long impulse response betruncated to at least length-L. A more satisfactory alternative is to use the DFT to evaluate the numerator and denominator of [link] separately rather than to approximately evaluate [link] . This is accomplished by appending L - N zeros to the a ( n ) and L - M zeros to the b ( n ) from [link] , and taking length-L DFTs of both to give

H ( 2 π k / L ) = DFT { b ( n ) } DFT { a ( n ) }

where the division is a term-wise division of each of the L values of the DFTs as a function of k . This direct method of calculation is a straightforward and flexible technique that does not involvetruncation of h ( n ) and the resulting error. Even nonuniform spacing of the frequency samples can be achieved by altering the DFTas was suggested for the FIR filter. Because IIR filters are generally lower in order than FIR filters, direct use of the DFT isusually efficient enough and use of the FFT is not necessary. Since the a ( n ) and b ( n ) do not generally have the symmetries of the FIR h ( n ) , the DFTs cannot be made real and, therefore, the shifting and stretching techniques of other modules are notapplicable.

As an example, the frequency-response plot of a third-order elliptic-function lowpass filter with a transfer function of

H ( z ) = 0 . 1335 z 3 + 0 . 056 z 2 + 0 . 056 z + 0 . 1335 z 3 - 1 . 507 z 2 + 1 . 2646 z - 0 . 3786

is given in [link] a. The details for designing this filter are discussed in elsewhere. A similar performance for themagnitude response would require a length of 18 for a linear-phase FIR filter.

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Digital signal processing and digital filter design (draft). OpenStax CNX. Nov 17, 2012 Download for free at http://cnx.org/content/col10598/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Digital signal processing and digital filter design (draft)' conversation and receive update notifications?

Ask