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IIR Filter Design - IIR Filter Design A common method of...

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IIR Filter Design A common method of designing IIR filter is to transform a continuous-time filter into a discrete-time filter. The reasons for this approach are (1) many useful continuous-time IIR filter design methods have relatively simple closed form design formula; (2) techniques for continuous-time IIR filter are not easily applied to discrete-time filters; (3) the transformation from the continuous-time filter to discrete-time filter is totally unrelated to the continuous-time filter design. The design procedure is firstly to design a continuous- time lowpass prototype and then transform to a discrete-time transfer function. Other selective filter can be obtained through frequency transformation. Continuous-time filter design In this section, we discuss the use of Butterworth function as the squared magnitude frequency response function of lowpass filter. The frequencies for continuous systems and digital systems are denoted by and ϖ , respectively. (A) Butterworth filters The squared magnitude response of Butterworth filter is the rational function defined as 1
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N p N c j H j H j H 2 2 2 2 ) / ( 1 1 ) / ( 1 1 ) ( ) ( | ) ( | ε + = + = - = (1) where N is the order of the filter, c is the –3dB frequency (usually called the cutoff frequency, where the squared magnitude is equal to 1/2), p is the passband edge frequency, and ) 1 /( 1 2 ε + is the passband edge value of 2 | ) ( | p j H . The frequency response of a Butterworth filter is shown in the following figure, where s is the stopband edge frequency with the gain frequency gain s δ . The frequency responses of Butterworth filters for different N are shown in the following diagram 2 2 s δ p s 0.5 Ω c
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It is noted that the higher the order, the steeper and narrower is the transition band. The design parameters N and c can be obtained from the two conditions that arise at the edges of the transition band N c p 2 2 ] / [ 1 1 1 1 ε + + (2) N c s s 2 2 ] / [ 1 1 δ + (3) where s δ and s are the stopband ripple and stopband edge frequency, respectively. 3 Fig.1
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We have 2 2 ] / [ ε N c p 1 ] / [ 2 2 - - s N c s δ (4) By the above two equations, we divide one equation by the other ) 1 /( ] / [ 2 2 2 - - s N s p δ ε (5) The order of the filter is obtained as } / log{ / } 1 / log{ 2 s p s s N δ εδ - . (6) The cutoff frequency can be obtained as N p c / 1 ] /[ ε = (7) Once the Butterworth function is fixed, the transfer function is related to the frequency response as ) ( ) ( | ) ( ) ( j H j H s H s H j s - = - = = N c s ) / ( 1 1 2 2 - + (8) The transfer function ) ( s H of the filter can be obtained by finding the poles from the Butterworth function. Setting the denominator to zero, the poles can be obtained as 1 / ) 1 ( 2 2 - = - N c N N s ) 1 ( ) 1 ( 2 2 - - = N N c N s } ) 1 2 ( exp{ } ) 1 2 ( exp{ 2 π π + + = k j N m j N c (9) The poles are given by 4
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} 2 / ) 1 2 ( exp{ } 2 / ) 1 2 ( exp{ N k j m j s c π π + + = , 1 , , 0 - = N k , 1 , 0 = m The poles of ) ( ) ( s H s H - are located on a circle of radius c at equal spaced points. Those poles with m= 0 are located on the left
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