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1、Comp30291 DMP 4.17 21 Oct 08 BMGCUNIVERSITY of MANCHESTERSchool of Computer ScienceComp30291: Digital Media Processing 2009-10Section 4: Design of FIR digital filters4.1. IntroductionAn FIR digital filter of order M may be implemented by programming the signal-flow-graph shown below. Its difference
2、equation is:yn = a0xn + a1xn-1 + a2xn-2 + . + aMxn-Mz-1z-1z-1z-1xnyna0a1.aM-1aMFig. 4.1Its impulse-response is ., 0, ., a0, a1, a2,., aM, 0, . and its frequency-response is the DTFT of the impulse-response, i.e.Now consider the problem of choosing the multiplier coefficients. a0, a1,., aM such that
3、H( ejW ) is close to some desired or target frequency-response H(ejW) say. The inverse DTFT of H(ejW) gives the required impulse-response : The methodology is to use the inverse DTFT to get an impulse-response hn & then realise some approximation to it Note that the DTFT formula is an integral, it h
4、as complex numbers and the range of integration is from -p to p, so it involves negative frequencies.Reminders about integrationabt (Have +ve & -ve areas)Reminder about complex numbers:Let x = a + j.b, j = -1Modulus: |x| = a2 + b2 Arg(x)=tan-1(b/a) + p.sign(b) if a 0 = tan2(b,a) = angle(a + j.b) : r
5、ange -p to pPolar: x = Rejf where R = |x| & f = Arg(x)De Moivre: ejf = cos(f) + j.sin(f) , e-jf = cos(f) - j.sin(f) ejf + e-jf = 2cos(f) & ejf - e-jf = 2 j.sin(f)Complex conjugate: x* = a - j.b = Re-jf. What about the negative frequencies?Examine the DTFT formula for H(ejW).If hn real then hnejW is
6、complex-conjugate of hne-jW. Adding up terms gives H(e-jW ) as complex conj of H(ejW). G(W) = G(-W) since G(W) = |H(ejW)| & G(-W) = H(e-jW)| (Mod of a complex no. is Mod of its complex conj.)Because of the range of integration (-p to p) of the DTFT formula, it is common to plot graphs of G(W) and f(
7、W) over the frequency range -p to p rather than 0 to p. As G(W) = G(-W) for a real filter the gain-response will always be symmetric about W=04.2. Design of an FIR low-pass digital filterWG(W)p/3-p/30p-p1Assume we require a low-pass filter whose gain-response approximates the ideal brick-wall gain-r
8、esponse in Figure 4.2.Fig. 4.2If we take the phase-response (W) to be zero for all W, the required frequency-response is:- and by the inverse DTFT, = (1/3)sinc(n/3) for all n. where A graph of sinc(x) against x is shown below:x12-1-2-331sinc(x)-4Main lobeZero-crossings at x =1, 2, 3, etc.RipplesFig
9、4.3aThe ideal impulse-response hn with each sample hn = (1/3)sinc(n/3) is therefore as follows:Fig. 4.3bIdeal impulse response for low-pass filter cut-off p/3n36-3-6-991/3hn-12Reading from the graph, or evaluating the formula, we get:hn = ., -0.055, -0.07, 0, 0.14, 0.28, 0.33, 0.28, 0.14, 0, -0.07,
10、-0.055, . A digital filter with this impulse-response would have exactly the ideal frequency-response we applied to the inverse-DTFT i.e. a brick-wall low-pass gain response & phase = 0 for all W. But hn has non-zero samples extending from n = - to , It is not a finite impulse-response. It is also n
11、ot causal since hn is not zero for all n 0. It is therefore not realisable in practice.To produce a realisable impulse-response of even order M:(2) Delay resulting sequence by M/2 samples to ensure that the first non-zero sample occurs at n = 0.The resulting causal impulse response may be realised b
12、y setting an = hn for n=0,1,2,.,M. Taking M=4, for example, the finite impulse response obtained for the /3 cut-off low-pass specification is : .,0,.,0, 0.14, 0.28, 0.33 , 0.28 , 0.14 , 0 ,.,0,.The resulting FIR filter is as shown in Figure 4.1 with a0=0.14, a1=0.28, a2=0.33, a3=0.28, a4=0.14. ( Note: a 4th order FIR filter has 4 delays & 5 multiplier coefficients ). The gain & phase responses of this FIR filter are sketched below. G(WdB0-10-20-30p/3-6 dBpW-f(W)-pp/3pW Fig. 4.4Clearly, the effect of the truncation of hn to M/2 and the M/2 samples delay is to produce gain and phase respon
