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# SIGNALS & SYSTEMS LAB EE 422G

UK

GPA 3.81

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This 13 page Class Notes was uploaded by Adaline Pollich on Friday October 23, 2015. The Class Notes belongs to EE 422G at University of Kentucky taught by Staff in Fall. Since its upload, it has received 49 views. For similar materials see /class/228327/ee-422g-university-of-kentucky in Electrical Engineering at University of Kentucky.

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Date Created: 10/23/15

EE 422G Notes Instructor Cheung Analysis and Design of Digital Filter I What is lter design H 6139 l lt gt N E L I Z 39 gt i 0 gt I I gt 1 E K I Z 139 Filter speci cation in terms of amplitude response and TranSfer functlon error to lerance x Dlrec Form reallzallon II Two Inai or categories 1 IIR typically based on transforming a continuous time analog filter such as Butterworth into discrete time Advantages Decades of experiences in designing analog filters Typical less complex fewer registers arithmetic units than FIR in realizing the design filter spectrum 2 FIR entirely discrete time domain method Advantages Advantages of any FIR filter always stable linear phase Optimal CAD method not covered in this class III IIR Filter Design General procedure 1 Design an analog filter Has that satisfies the specification 2 Map the analog filter Has into a discrete time filter Hdz Focus on step 2 There are two approaches 1 IHVariaIzce Design 33 Simple to understand 33 Requires high sampling rate to mitigate aliasing Page 9 1 EE 422G Notes Instructor Cheung 2 leznear Mapping Design 7 Frequency warping 9 No aliasin 49 Frequency distortion III 1 Invariant Design yt gt yltnTgt xaamm Innpl oal for a GIVEN xt we want identical to the sampled v the output of o exsion ysnT of Ifxt is the delta function we effective 1 n ur digital filter ynT to be the continu time lter output 1y have ha nT This is called IWULSEeWVARIANCE design Let s see an example to illustrate all the steps 0 5s 4 Example Given 5745 m Find digital filter Hz by impulsecinvariance Page 972 EE 422G Notes Instructor Cheung Step 1 Compute the response of the analog filter yat Since Xat 5t gt Xasl thus yaU L71 HaSXaSl L 1Hs 51 05s4 sls2 L 1 71395 1 1 5672 8722 sl s2 Step 2 Sample yat to get yanT yanT 15 6439 Step 3 Perform Ztransform on yanT l 5 l YZZya m m Step 4 Perform Ztransform on Xa1 1T Xz ZxanT 1 Step 5 Find Hdz YzXz l l H 7 Z l 677271 1 6727271 Step 6 Finally choose a sampling period T and obtain the numerical representation of the digital filter gtgt Impulse Invariance Design gtgt num 05 l 4 gtgt den convl l l 2 gtgt Fs 10 Sampling freq lOHz gtgt bzaz impinvarnumdenFs gtgt gtgt Compare Analog and Digital Filter gtgt haw freqsnumden gtgt f w2pi Convert to Normal Freq gtgt hd freqzbzazfFs gtgt plotwabsha b wabshd r x BluezAnalog RedzDigital Page 93 EE 422G Notes Instructor Cheung Amplitude Response 0 10 20 30 40 50 60 70 80 90 100 Frequency rads What happen around 63 rads Ans The FT of the discrete lter is periodic with period 21Ef521l3102 63 rads If your client cannot stomach the divergence starting at 30Hz what should you do Ans Increase the sampling rate 3 S What about if we want to design a highpass filter H HPs 2 s 2s 2s 4 18 1 6 7 7 Important We cannot use 39 invariance method to design 14 any non bandlimited filter g 12 7 7 such as highpass or band 1 stop To design these filters w we need to understand the E 08 concept of frequency EL 06 transformation lt 04 7 r 02 r 0 r r l r 0 2 4 6 8 10 Frequency Hz Page 94 EE 422G Notes Instructor Cheung 1112 Bilinear Transformation The basic idea of frequency transformation is straightf01ward 1 Start with a lowpass prototype filter Hpj 03 2 Design a frequency transform c0 f03 such that 3 Hpj 03 is your desired filter Frequency transformation is useful in both analog filter design and analog to discrete conversion Example Convert a lowpass analog lter to highpass analog filter Let s say you have a lowpass analog filter with cutoff frequency 3dB frequency at l rads as follows 1 SDUZSD To design a HIGHPASS filter with cutoff frequency at up we use the following transformation s cup cap a or equivalently s a s l 33 Let 2 H s 2 mp 2 22 2 02x 2s 71 7 71 S S S 1 08 gt 2 27067 7 DC 039439 03 oo gtco 0 o330 gtoa ltgtltgt E quot 02 xxquot I 0 v 0 2 4 6 8 10 Frequency rads Page 95 EE 422G Notes Instructor Cheung Question How do we use frequency transformation to map analog filter to digital CTFT extends from negative in nity to positive infinity while DTFT is periodic with period equal to the sampling frequency ZnT To ELIMINATE ALIASING we need to come up with a frequency transform f such that f maps analog frequency 006 oooo to DdE nT nT One such example of f is the arctan 0d 2 arctan 0 In the following example T 2 39 139 05a 1 05a5 When 00 is small 13 00 1 the distOItion is relatively small and we have 2 cod z WC Beyond that the transformation compresses 00 causing a fair amount of distortion If we are designing a lowpass filter with passband CS 00 SC we can scale 00 first by C and use the transform 0d 2 arctan In most applications the analog filter is specified in terms of the Laplace variable sj be and the target discrete filter needs to be expressed in Ztransform variable zede Thus it will be nice if we can derive a transform directly from s to z 1 Z 1 Bilmear Transformation S C 1 l z Page 96 EE 422G Notes Instructor Cheung Proof l 87de 1 06 1 eijwzeijwz l e jdeAe JWA e Z 371de 2139 Sin T C C t d 2cosmd 1 an lt2 05 Ctan d 2 0d 2 arctan To determine C we typically assume there is a fixedpoint or anchor frequency or which maps to itself 2 a a wrz arctan 3C2 r T C 0 tan 7 2 2 xame Hsz c orer ueiwo 1erw1 E p197 a 0 2 d d Btt rthflt th s2 1260 062 035100015 as the anchor frequency and 035400015 Applying bilinear transformation 2 a Hd Z 71 2 C 71 2 0 2722 26051 271a 2 l Z l Z a 1z 12 C2l z 12 2acl z 2 a 1z 12 c Solving C C L 22916876 a T tan 5 0292893 05857862 1 02928932 2 Hd Z 2 9 l0l71573z Page 97 EE 422G Notes Instructor Cheung IV FIR Digital Filter IV 1 Design Procedure 1 Start With desired amplitude response With zero phase I A r3 39 392 T r I f L L 2 2 Apply inverse DTFT to recover acausal symmetric and possibility infinitely long impulse response ha nT DTFT Formula xnT L XeJquot TeJquot 17da 27139 or in normalized frequency xnT IfOSSXeJ2 eJ 2 dr 3 Truncate ha nT to nite extent by multiplying it by an appropriate Window function and form hm nT 2 39r T 139 T TuliTNV r J T r T r u u u 4 Delay hmho to make it casual 039 g 0 Page 9 8 EE 422G Notes Instructor Cheung 1 Example Des1gn a FIR filter that appr0X1mates H r 31 cos27r r Solution 1 Inverse DTFT explicit calculation 1 w 7 39n In HO 2 51 0052717 2 Zhd Te I 2 Notice n ranges from ltgtltgt to co n Compute hd nT COSzmeJn2mdr h0ll21 2 d112d 112 2d1 Forn0 d EL2 cos 72 r E rEL2cos 727 r E 712 For 11 i0 1 12 n r 1 12 e12 e 1392 r hdnT J e 2 dr J e27m dr 2 712 2 712 2 1 12 1 12 1 J e mmmdr 1ze2 quot 1rdr Z for n i1 and 0 otherwise 4 712 Hr ihdnTe jzmr ler r llerj2m 4 4 2 4 9 1 1 Ila07 min hdltnro M122 The impulse response is nite duration Thus there is no need to truncate 3 Make it acasual by delaying one sample Hz llz 1 1272 4 2 4 IV2 Effect of windowing 2M1 terms M 39 w Truncation gt Hltz mepz quot wanwdmnz quot l n S M 0 n gt M Truncation time multiplication with rectangular window function Cigt ConVOlVing in frequency domain with a sine function where wr n 2 Page 99 EE 422G Notes Instructor Cheung w ejan39r iej27rnr Sinc function in frequency domain PM smrt r Suppose the original analog filter Hr is an ideal lowpass filter windowing in time domain corresponds to convolution in frequency with a sinc function Sliding sinc function in convolution 39 A Ideal LP l I l I A 39 l A I A quot w v 397 v 2 quot v 6 39 39I l l I I Vquot 39v v v 2 39v Frequency Spectrum of the resulting truncated sequence Spectrum of the rectangular window that retains 15 samples M7 sin wlM 12 Sin ca2 M 7 8 EEA NA 21 Zn I 1r 2 M 1 M 1 739 w Mainlobe Aw width Figure 728 Magnitude of the Fourier transform of a rectangular window M 7 The frequency spectrum of a good truncation window should resemble as much as possible to a delta function Typically it is measured based on two criteria 1 Narrow Main Lobe for rectangular window main lobe width 27tMl a narrow main lobe produce better frequency transition Page 910 EE 422G Notes Instructor Cheung 2 Small Side Lobes measured by the highest peaks caused by the sharp cutoff of the rectangular window By tapering the window smoothly to zero at each end the height of the sidelobes can be diminished however this is achieved at the expense of a wider mainlobe and thus a wider transition at the discontinuity Some commonly used windows include 211 M O S n S M 2 Bartlett triangular wn 2 2n M M 2 s n s M 0 otherwise 05 05cos27mM OSnSM Hanning wn 0 otherwise 054 046cos OSnSM Hamming who 2 M 0 otherwise 042 05 2 M 008 47mM OS SM Blackman wn COS 7m COS n 0 otherwise w n Rectangular O n 39 39 08 39 I Hummin I I g 06 I Hanning 39 Blackman 04 I I Bartet t 39 fll 02 39 n M 2 Hanning vs Rectangular Hamming vs Rectangular Blackman vs Rectangular Page 911 EE 422G Notes EX ample approxima Instructor Cheung Design a l7 tap low pass FIR digital filter with Hamming window to te 1 r s 015 Hr o o15ltrso5 Step 1 Inverse Fourier Transform 0 5 hdnTJ 6J2 quot dr e h ee m 1 sin037rn 70 15 jzm hdlt0Tgt hm dsan37z39 nd7z39 1 lim 03 cosO37z39n 03 7tan d7z nd7z 1 7tan 1 110 03 sgt haltnTgt san37zn n 0 7171 Step 2 amp 3 Multiply by l7 tap Hamming window 1I 8 and make it aeausal 3 Hm z ZhaO1TgtWhltngtz q HC 7 Z SHNcZgt YABLE 92 Flller Weights for FIR LowPass Finer I g 0151 unn l uls Response e nmmn g indnw Unit Pulse Response with Rectangular Funcuon with Hamming n wnndow mun Window wnux morn s o 003r 7 00006 e o n uobgb es en 7 I 4 er a 2 l 4 2 0 IJ97 I n 4x50 0 n J 0 14850 3 0 1309 z 3 0 021477 4 002 s 5 ea 0 7 o ed mew e 7 0 00101 x 0 037x41 01301027 Page 97 l 2 EE 422G Notes Instructor Cheung Lo mm m we Fmquot won a can unau cu 0 g u o 2 FIlr mum m ummm Mn 0 z D n I n 039 V u o c s in z Nammxm mums rr FIGURE 9 31 Ampmude response of dlglla lowpass ner The negative portions are shown negauva for convenience Page 9 1 3

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