IIR digital filter design for low pass filter based on impulse invariance and bilinear transformation methods using butterworth analog filter Nasser M. Abbasi May 5, 0 compiled on hursday January, 07 at 08:4 PM Contents Introduction. Filter specifications.......................................... he design steps. Impulse invariance method...................................... bilinear transformation method................................... 5.3 Summary of analytical derivation method............................. 9 3 Numerical design examples 9 3. Example............................................... 9 3.. using impulse invariance method using =....................... 3.. bilinear method........................................ 3. Example............................................... 3.. using impulse invariance method.............................. 3.. Using bilinear......................................... 3 4 IIR design for minmum order filter 3 4. Impulse invariance method..................................... 3 4. bilinear transformation method................................... 5 5 References 7 A low pass digital filter IIR is designed based on given specifications. Butterworth analog filter Hs is designed first, then it is converted to digital filter Hz Analytical procedure is illustrated below and simplified to allow one to more easily program the algorithm. 4 numerical examples are used to illustrate the procedure. Introduction his report derives a symbolic procedure to design a low pass IIR digital filter from an analog Butterworth filter using methods: impulse invariance and bilinear transformation. wo numerical examples are then used to illustrate using the symbolic procedure. here are a total of 3 steps. A Mathematica program with GUI is written to enable one to use this design for different parameters. A Matlab script is being written as well.
. Filter specifications Filter specifications are 5 parameters. he frequency specifications are analog frequencies, while the attenuation for the passband and the stopband are given in db f c f s δ p δ s he sampling frequency in Hz he passband cutoff frequency in Hz he stopband corner frequency in Hz he attenuation in db at f c he attenuation in db at f s his diagram below illustrates these specifications he frequency specifications are in Hz and they must be first converted to digital frequencies ω where 0 ω π before using the attenuation specifications, he sampling frequency is used to do this conversion since corresponds to π on the digital frequency scale. he design steps. Convert specifications from analog to digital frequencies. Based on design method impulse invariance of bilinear apply the attenuation criteria to determine and N the filter order 3. Using and N find the locations of the poles of the butterworth analog filter Hs. 4. Find Hz from Hs. he method of doing this depends if we are using impulse invariance or bilinear. his step is much simpler for the bilinear method as it does not require performing partial fractions decomposition on Hs Now we begin the analytical design procedure.. Impulse invariance method F We first convert analog specifications to digital specifications: s π = fp ω p, hence ω p = π fp Convert the criteria relative to the digital normalized scale: herefore 0 log H e jωp δ p 0 log H e jωs δ s and ω s = π fs
H e jωp δp 0 A H e jω s δs 0 B Notice that we added the scaling factor. Now, Butterworth analog filter squared magnitude Fourier transform is given by H a jω = + jω j hence equations A and B above are now written in terms of the analog Butterworth amplitude frequency response and become herefore the above becomes Ωp + + Ω s δp 0 δs 0 Ωp + + Ω s δp δs Now, for impulse invariance, Ω p = ωp and Ω s = ωs. Change inequalities to equalities and simplify Divide the above equations ωp / + ωs / + ωp / ωs / δp δs = δp = δs δp ωp = ω s δs [log ω p log ω s ] = log δp log δs [ ] [ ] N = log δp log δs log ω p log ω s 3
We need to round the above to then nearest integer using the Ceiling function N = N Now for impulse invariance method, use equation above to solve for ωp / log ω p / = δp log ω p / = = log δp ω p / = = log log δp ω p / δp log δp Now that we found N and, next find the poles of H s Since the butterworth magnitude square of the transfer function is H a s = + s j Hence H s poles are found by setting the denominator of the above to zero s + = 0 j s = j s j = e j = e jπ+πk k = 0,,, π+πk s = j e j π+πk = e j π e j π+πk = e j π+k+n We only need to find the LHS poles, which are located at i = 0 N, because these are the stable poles. Hence the i th pole is s i = e j π+i+n For example for i = 0, N = 6 we get s 0 = e j π+n = e j π7 Now we can write the analog filter generated based on the above selected poles, which is, for impulse invariance H a s = K s s i 3 K is found by solving H a 0 = hence k = s i 4
Now we need to write poles in non-polar form and plug them into 3 s i = e j π+i+n π + i + N π + i + N = cos + j sin i = 0 N Hence, Where and H a s = N = K s cos π+i+n = ω p / log δp + j sin π+i+n [ ] [ log δp log δs log ω p log ω s ] 4 Now that we have found H s we need to convert it to H z We need to make sure that we multiply poles of complex conjugates with each others to make the result simple to see. Now that we have H a s, we do the A D conversion. I.e. obtain H z from the above H s. When using impulse invariance, we need to perform partial fraction decomposition on 4 above in order to write H s in this form For example, to obtain A j, we write H s = A j = lim s sj H a s = Once we find all the A s, we now write H z as follows H z = s s i i j k s s i exp s i z his completes the design. We can try to convert the above form of H z to a rational expression as H z = Nz Dz. bilinear transformation method F We first convert analog specifications to digital specifications: s π = fp ω p, hence ω p = π fp Convert the criteria relative to the digital normalized scale: 0 log H e jωp δ p 0 log H e jω s δs and ω s = π fs Hence H e jωp δp 0 A H e jωs δs 0 B 5
Butterworth analog filter squared magnitude Fourier transform is given by H a jω = + jω j hence equations A and B above are now written in terms of the analog Butterworth amplitude frequency response and become Ωp + Ω + s δp δp 0 = δs 0 = δs Now we assign values for Ω p and Ω s as follows,ω p = tan ω p, Ωs = tan ω s + + tan ω p tan ω s Change inequalities to equalities and simplify Divide the above equations [ log tan ωp log tan ω p tan ω s ωp tan tan ω s tan ωs δp δs = δp = δs δp = δs ] = log δp log δs N = log δp log δs log tan ω p log tan ωs We need to round the above to then nearest integer using the Ceiling function. i.e. N = N 6
Now for bilinear transformation we used equation above to find Hence we now solve for + tan ω s log tan ω s = δs log tan ω s = tan ω s = = = log δs log log δs δs tan ω s log δs Now that we found N and we find the poles of H s. Since for bilinear the magnitude square of the transfer function is H a s = + s j Hence H s poles are found by setting the denominator of the above to zero s + = 0 j s = j s j = e j = e jπ+πk k = 0,,, π+πk s = j e j π+πk = e j π e j π+πk = e j π+k+n We only need to find the LHS poles, which are located at i = 0 N, because these are the stable poles. Hence the i th pole is s i = e j π+i+n For example for i = 0, N = 6 we get s 0 = e j π+n = e j π7 For bilinear, Hs is given by H a s = K is found by solving H a 0 =, hence we obtain K s s i 3 k = 7 s i
We see that the same expression results for k for both cases. Now we need to write poles in non-polar form and plug them into 3 s i = e j π+i+n π + i + N = cos + j sin π + i + N i = 0 N then Where and H a s = N = K s cos π+i+n = log δp log tan ωp tan ω s log δs log + j sin π+i+n δs log tan ωs 4 Now that we have found H s we need to convert it to H z. After finding Hs as shown above, we simply replace s by z +z. his is much simpler than the impulse invariance method. Before doing this substitution, make sure to multiply poles which are complex conjugate of each others in the denominator of Hs. After this, then do the above substitution 8
.3 Summary of analytical derivation method We will now make a table with the derivation equations to follow to design in either bilinear or impulse invariance. Note that the same steps are used in both designs except for step 5, 6, 8, 3. his table make it easy to develop a program. step Impulse invariance common equation bilinear ω p = π fp ω s = π fs 3 α p = δp 4 α s = 5 Ω p = ωp 6 Ω s = ωs 7 N = 8 = Ω p δs log[α p ] log[α s ] logω p logω s log [ αp ] = 9 poles of HS s i = e j π+i+n k = K H a s = s s i do partial fractions: H a s = 3 H z = s s i s i tan ω p tan ω s H a s = Ω s log [αs ] K s s i exp s iz H z = H a s s= z +z 3 Numerical design examples 3. Example Sampling frequency = 0khz, passband frequency f p = khz, stopband frequency f s = 3khz, with δ p db and δ stop 5db 9
3.. using impulse invariance method using = step Impulse invariance ω p = π fp ω s = π fs 3 α p = δp 4 α s = 5 Ω p = ωp 6 Ω s = ωs 7 N = 8 = π000 0000 0.π π3000 0000 0.3π. 58 9 3. 63 δs 5 ωp 0.π 0.3π 0.3π log[α p ] log[α s ] logω p logω s Ω p [ ] 0.π log αp log. 58 9 log 3. 63 log 0.π log 0.3π 5. 885 9 6 0.703 6 log. 58 9 9 poles of HS s i = e j π+i+n s i = 0.703 e j π7+i i = 0 5 s 0 = 0.8 + j0.679 5, s = 0.497 4 + j0.497 4, s = 0.679 5 + j0.8 s 3 = 0.679 5 j0.8, s 4 = 0.497 4 j0.497 4, s 5 = 0.8 j0.679 5 k = s i 0.8 j0.679 5 0.497 4 j0.497 4 0.679 5 j0.8 0.679 5 + j0.8 0.497 4 + j0.497 4 0.8 + j0.679 5 0. 9 K H a s = s s i 0. 9 s+0.8 j0.679 5s+0.497 4 j0.497 4s+0.679 5 j0.8 s+0.679 5+j0.8s+0.497 4+j0.497 4s+0.8 +j0.679 5 0. 9 multiply complex conjugates s +0.364 s+0.494 5s +0.994 48s+0.494 50s +. 358 5s+0.494 5 0. 9 s 6 +. 77 0s 5 +3. 69 0s 4 +3. 78 9s 3 +. 85 s +0.664 38s+0. 9 A partial fraction H a s = i 0.43 54+j0.48 6. 07 4 j. 66 8 5 0.97 85 j. 607 s s i s+0.8 j0.679 5 s+0.497 4+j0.497 4 + s+0.679 5 j0.8 0.97 85+j.607.07 4+j.66 8 5 0.43 54 j0.48 6 s+0.679 5+j0.8 s+0.497 4 j0.497 4 + s+0.8 +j0.679 5 0.43 54+j0.48 6 exp s iz exp 0.8 +j0.679 5z.07 4 j.66 8 5 0.97 85 j.607 exp 0.497 4 j0.497 4z + exp 0.679 5+j0.8 + 3 H z = + 0.97 85+j.607.07 4+j.66 8 exp 0.679 5 j0.8 z 5 0.43 54 j0.48 6 exp 0.497 4+j0.497 4z + exp 0.8 j0.679 5z 0.43 54+j0.48 6.07 4 j.66 8 5 0.97 85 j.607 0.648 58+j0.53 68z 0.534 55 j0.90 z + 0.498 6+j9.76 5 z 0.97 85+j.607.07 4+j.66 8 5 0.43 54 j0.48 6 0.498 6 j9.76 5 z 0.534 55+j0.90z + 0.648 58 j0.5368z H z = +
3.. bilinear method Using the above design table, these are the numerical values: = = 0000 step Bilinear ω p = π fp ω s = π fs 3 α p = δp π000 0000 0.π π3000 0000 0.3π. 58 9 4 α s = 3. 63 δs 5 5 Ω p = tan ω p 0000 tan 0.π 997 6 Ω s = tan ω s 0000 tan 0.3π 038 log[α 7 N = p ] log[α s ] logω p logω s log. 58 9 log 3. 63 log 997 log 038 5. 304 8 6 Ω 8 = s 038 535.6 6 log 3. 63 log [αs ] 9 poles of HS s i = e j π+i+n s i = 535 e j π7+i i = 0 5 s 0 = 3966. 4 + j4803, s = 836 + j836., s = 4803 + j3966.4 s 3 = 4803 j3966.4, s 4 = 836 j836., s 5 = 3966.4 j4803 k = s i 3966. 4 j4803 836 j836 4803 j3966.4 4803 + j3966. 4 836 + j836 3966.4 + j4803 multiply complex conjugate terms. 348 6 8. 348 4 8. 348 6 8. 95 4 5 K H a s = s s i. 95 4 5 s+3966. 4 j4803s+836 j836s+4803 j3966.4s+4803+j3966. 4s+836+j836s+3966.4+j4803 multiply complex conjugate. 95 4 5 s +793. 8s+. 348 6 8 s + 67s+34 837 79s +9606.s+. 348 6 8 657. 4s. 355 8 8 4670.s+5. 060 4 8 40445.s+8. 765 4 8 = s +793. 8s+. 348 6 8 s +67.s+. 348 4 + 8 s +9606.s+. 348 6 8. 95 4 5 s 6 +59.s 5 +. 753 0 9 s 4 +3. 90 3 s 3 +4. 6 9 7 s +3. 65 8 s+. 95 3 5 3 H z = H a s s= z ODO +z 3. Example Sampling frequency = khz, passband corner frequency f p = khz, stopband corner frequency f s = khz, with criteria δ p 3db and δ stop db
3.. using impulse invariance method = step Impulse invariance ω p = π fp ω s = π fs 3 α p = δp 4 α s = 5 Ω p = ωp 6 Ω s = ωs 7 N = 8 = π00 000 0.π π000 3 000 0.4π. 995 3.0 δs ωp 0.π 0.4π 0.4π log[α p ] log[α s ] logω p logω s Ω p [ ] 0.π log αp log. 995 3 log.0 log 0.π log 0.4π. 588 4 0.69 06 log. 995 3 9 poles of HS s i = e j π+i+n s i = 0.69 06 e j π3+i 4 s 0 = 0.444 8 + j0.444 8, s = 0.444 8 j0.444 8 k = K H a s = i = 0 s i 0.444 8 j0.444 8 0.444 8 + j0.444 8 0.395 7 s s i partial fraction H a s = 0.45 35z z. 57 z+0.4 8 3 H z = 0.395 7 s+0.444 8 j0.444 8s+0.444 8+j0.444 8 0.395 7 s +0.889 6s+0.395 7 s s i 0.444 8j s+0.444 8+0.444 8j 0.444 8j s+0.444 8 0.444 8j exp s iz 0.444 8i exp 0.444 8 j0.444 8z 0.444 8i exp 0.444 8+j0.444 8z
3.. Using bilinear = 000 step Impulse invariance ω p = π fp ω s = π fs 3 α p = δp π00 000 0.π π000 3 000 0.4π. 995 3 4 α s =.0 δs 5 Ω p = tan ω p 000 tan 0.π 6498. 4 6 Ω s = tan ω s 000 tan 0.4π 453. log[α 7 N = p ] log[α s ] logω p logω s log. 995 3 log.0 log 6498. 4 log 453.. 368 Ω 8 = s 453 8389. 5 log.0 log [αs ] 9 poles of HS s i = e j π+i+n s i = 8389. 5 e j π3+i 4 s 0 = 593. 3 + j593. 3, s = 593. 3 j593. 3 k = K H a s = i = 0 s i 593. 3 j593. 3 593. 3 + j593. 3 7. 038 4 7 s s i 7. 038 4 7 s +865.s+7. 038 4 7 3 H z = H a s s= z +z 7. 038 4 7 s+593. 3 i593. 3s+593. 3+i593. 3 0.09945 9z +0.98 9z+0.09945 9 z 0.93 56z+0.39 38 Mathematica GUI program is written to implement the above. It can be downloaded from here Also a Matlab script nma_filter.m.txt was written no GUI to implement this design. his script written on Matlab 007a. his script does not handle the conversion from Hs to Hz well yet, need to work more on this... ofcourse, one can just use Matlab butter function for this. Example output of the above Matlab script is matlab_output.txt 4 IIR design for minmum order filter his is another small note on IIR design for minmum order filter. his document describes how to design an IIR digital filter given a specification in which the filter order is specified. Given the following digram, the specifications for the design will be. Digital filter order N. f pass, the passband corner frequency or the cutoff frequency at 3db. his means A pass = 3db 4. Impulse invariance method We first convert analog specifications to digital specifications: hen the cutoff frequency = ωp for impulse invariance. π = fpass ω p, hence ω p = π fpass 3
Next find the poles of H s. Since the butterworth magnitude square of the transfer function is H a s = + s j Hence H s poles are found by setting the denominator of the above to zero s + = 0 j And as I did in the earlier document, the poles of H s are found at s = e j π+k+n We only need to find the LHS poles, which are located at i = 0 N, because these are the stable poles. Hence the i th pole is s i = e j π+i+n Now we can write the analog filter generated based on the above selected poles, which is, for impulse invariance K H a s = 3 K is found by solving H a 0 = hence k = s s i s i Now we need to write poles in non-polar form and plug them into 3 s i = e j π+i+n π + i + N π + i + N = cos + j sin i = 0 N 4
Hence, H a s = K s cos π+i+n + j sin π+i+n Now that we have found H s we need to convert it to H z We need to make sure that we multiply poles of complex conjugates with each others to make the result simple to see. Now that we have H a s, we do the A D conversion. I.e. obtain H z from the above H s. When using impulse invariance, we need to perform partial fraction decomposition on 4 above in order to write H s in this form For example, to obtain A j, we write H s = A j = lim s sj H a s = Once we find all the A s, we now write H z as follows H z = s s i i j k s s i exp s i z his completes the design. We can try to convert the above form of H z to a rational expression as H z = Nz Dz 4. bilinear transformation method We first convert analog specifications to digital specifications: π = fpass ω p, hence ω p = π fpass Now = tan ω p Now that we have N and we find the poles of H s. Since for bilinear the magnitude square of the transfer function is H a s = + s j Hence H s poles are found by setting the denominator of the above to zero s + = 0 j Which leads to poles at s i = e j π+i+n. We only need to find the LHS poles, which are located at i = 0 N, because these are the stable poles. For bilinear, Hs is given by H a s = K is found by solving H a 0 =, hence we obtain k = 5 K s s i s i 4 3
We see that the same expression results for k for both cases. Now we need to write poles in non-polar form and plug them into 3 s i = e j π+i+n π + i + N π + i + N = cos + j sin i = 0 N then H a s = K s cos π+i+n + j sin π+i+n Now that we have found H s we need to convert it to H z. After finding Hs as shown above, we simply replace s by z +z. his is much simpler than the impulse invariance method. Before doing this substitution, make sure to multiply poles which are complex conjugate of each others in the denominator of Hs. After this, then do the above substitution 4 6
5 References. Alan Oppenheim, Ronald Schafer, Digital Signal Processing, Prentice-Hall, inc. 975, Chapter 5.. Mostafa Shiva, Electrical engineering department, California state university, Fullerton, Lecture notes, handout H. 3. John Proakis, Dimitris Manolakis, digital signal processing, 3rd edition, section 8.3. 7