Homework 3 Solution

Homework 3 Solution -...

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Unformatted text preview: !!"#$#%"&'(')*+",'(-*+"./01233'-("""""""""""""""""""""""""""""""""""""""""""""""""4*++"5667" ./08+29",2)":"""""""""""""""""""""""""""""""""""""""""""""""""""""""""&;2",2<)2982/"57="5667"""" ! 1. Consider!the!filter!with!transfer!function!given!by:! 1 " z !1 H ( z) # ! 1 ! 0.9 z !1 i. ii. iii. Determine!the!impulse!response!for!a!causal!implementation.!Is!the!filter!stable!or! unstable?! Sketch!the!magnitude!response!of!the!filter.!Is!the!filter!Lowpass,!Highpass,! Bandpass,!Bandstop,!or!allpass?! n Sketch!the!magnitude!response!for!the!filter g[n] # $! 1% h[ n] .!Comment!on!your! result.! ! Solution:! i. ! h[n] # Z !1 &H ( z )' # 0.9 n ([n] " 0.9 n !1 ([n ! 1] ! Since!the!pole!is!inside!the!unit!circle,!the!system!is!stable!for!a!causal!impulse! response.!In!addition,!the!impulse!response!is!absolutely!summable.! ! ii. ! H (e j 0 ) # 20, H (e j( ) # 0, H e j0 / 2 # iii. $ % !!Lowpass!filter! 1" j 2 .0 + # 1.0512 / M, ) # 1 ! 0.9 j 1.81 -2* ! G $ z % # H ! z / G e j1 # H e j $1 !0 % ! $ % $ % $ % G (e j 0 ) # 0, H e j0 / 2 # ! H (0 ) # 20, $ % !Highpass!filter! 1! j 2 .0 + / M, ) # # 1.0512 1 " 0.9 j 1.81 -2* The!magnitude!response!curves!could!be!sketched!by!hand!based!on!calculation!of!! M(1)!at!a!few!points.!For!convenience,!however,!the!magnitude!and!phase!response! of!H(z)!and!G(z)!are!shown!in!the!figure!below:! !!"#$#%"&'(')*+",'(-*+"./01233'-("""""""""""""""""""""""""""""""""""""""""""""""""4*++"5667" ./08+29",2)":"""""""""""""""""""""""""""""""""""""""""""""""""""""""""&;2",2<)2982/"57="5667"""" ! Frequency Response of H(z) 25 20 M agnitude 15 10 5 0 0 1 2 3 4 M agnitude 20 15 10 5 0 0 1 2 3 4 Frequency Response of G(z) 0 Phase, R ad Phase, R ad 0 1 2 3 Frequency, rad 4 -0.5 -1 -1.5 -2 2 1.5 1 0.5 0 0 1 2 3 Frequency, rad 4 ! ! 2. Let! x[n] !be!a!purely!real!sequence.!You!are!given!the!following!information!about! x[n] !and! must!determine!what!it!is.!Even!if!you!are!unable!to!specify! x[n] !fully,!you!may!receive! partial!credit!by!describing!which!features!of! x[n] !are!determined!by!each!clue.! i. x[n] !is!a!causal!sequence.! ii. iii. iv. v. vi. Solution:! i. x[ n] # 0 n 6 0 ! ii. This!condition!implies!that! x[n] !is!symmetric!and,!together!with!condition!i,!that x[ n] # 0 7 n 5 4 .! iii. x[0] # 2 .! iv. x[1] # 0 ! Let! v[ n] # x[n " 2] .!Its!DTFT, V e j1 ,!is!purely!real.! lim X ( z ) # 2 ! z 32 $ % 1 20 1 20 0 ! 1 1 40 X $e % e d1 # 0 ! j j x[ 2] 5 0 ! 0 !0 4 $ % X e j1 2 d1 # 9 ! !!"#$#%"&'(')*+",'(-*+"./01233'-("""""""""""""""""""""""""""""""""""""""""""""""""4*++"5667" ./08+29",2)":"""""""""""""""""""""""""""""""""""""""""""""""""""""""""&;2",2<)2982/"57="5667"""" ! v. x[n] !is!even!symmetric!!! x[ 4] # x[0] # 2 &! x[3] # x[1] # 0 !.! 1 20 0 / x[n] # &2,0,1,0,2' ! 3. The!magnitude!responses!of!4!LTI!systems!are!given!in!the!figure:! ! (a) 30 40 30 20 10 10 0 -4 5 4 3 2 1 -2 0 2 Frequency, radians 4 0 -4 -2 0 2 Frequency, radians 4 (b) !0 4 $ % X e j1 2 d1 # 9 / 8 x[n] n #0 4 2 # 2 2 " 0 " x[2]^ 2 " 0 " 2 2 / x[2] # 1 ! 20 0 -4 8 6 4 2 0 -4 -2 0 2 Frequency, radians (c) 4 -2 0 2 Frequency, radians (d) 4 9 9 i. ii. iii. iv. State whether these filters act as LPF, BPF, or HPF. Also, which filter has the lowest cutoff frequency and which one has the highest. Which of the 4 magnitude responses matches: Filter 1 represented by the difference equation: y[n] # x[n] " x[ n ! 1] " x[n ! 2] " x[n ! 3] " x[ n ! 4] Filter 2 represented by the difference equation: y[ n] # 1.6 y[n ! 1] ! 0.64 y[ n ! 2] " x[ n] Filter 3 represented by the difference equation: y[ n] # x[ n] " x[ n ! 1] " x[ n ! 2] " x[ n ! 3] " x[ n ! 4] " x[ n ! 5] " x[n ! 6] " x[n ! 7] Filter 4 represented by the difference equation: y[ n] # 1.4 y[n ! 1] ! 0.49 y[ n ! 2] " x[ n] " x[ n ! 1] " x[n ! 2] In each case, explain clearly the basis of your choice, i.e., no credit for mere guessing. Solution: !!"#$#%"&'(')*+",'(-*+"./01233'-("""""""""""""""""""""""""""""""""""""""""""""""""4*++"5667" ./08+29",2)":"""""""""""""""""""""""""""""""""""""""""""""""""""""""""&;2",2<)2982/"57="5667"""" ! All the systems shown in the figure are LP filters. System (a) has the smallest cutof frequency while system (d) has the largest one (based on 3-dB cutoff criterion). Simplist way to identify which frequency response belongs to which difference equation is to find H(z) for each equation and find the dc gain, i.e., H(1). Another simple solution is to identify the number of zeros of the transfer function. 1 ! z !5 . H(1)=5 also four zeros on the UC (with pole-zero 1 ! z !1 cancellation at z=1). This matches frequency response shown in (d). a) H ( z ) # 1 " z !1 " z ! 2 " z !3 " z ! 4 # b) H ( z ) # 1 1 # . H(1)=25 (also system has no zeros on the UC). !2 1 ! 1.6 z " 0.64 z $1 ! 0.8z !1 %2 !1 This matches the frequency response shown in (a). c) H ( z ) # 1 " z !1 " z !2 " z !3 " z !4 " z !5 " z !6 " z !7 # 1 ! z !8 . H(1)=8 also seven zeros on the UC 1 ! z !1 (with pole-zero cancellation at z=1). This matches frequency response shown in (c). 1 " z !1 " z !2 1 " z !1 " z !2 # . H(1)=3/0.09~33 (also system has two nontrivial d) H ( z ) # 1 ! 1.4 z !1 " 0.49 z !2 1 ! 0.7 z !1 2 $ % zeros at w~=2.0944 rad on the UC). This matches the frequency response shown in (b). 4. Problem!3.41!P&M.!! ! !!"#$#%"&'(')*+",'(-*+"./01233'-("""""""""""""""""""""""""""""""""""""""""""""""""4*++"5667" ./08+29",2)":"""""""""""""""""""""""""""""""""""""""""""""""""""""""""&;2",2<)2982/"57="5667"""" ! a1-a2<1 and a1+a2>-1 ! 5. Problem!4.17!P&M.! !!"#$#%"&'(')*+",'(-*+"./01233'-("""""""""""""""""""""""""""""""""""""""""""""""""4*++"5667" ./08+29",2)":"""""""""""""""""""""""""""""""""""""""""""""""""""""""""&;2",2<)2982/"57="5667"""" ! ! Note:!The!solution!of!Part!(!e)!is!somewhat!simplistic!and!is!missing!some!terms,!but!it!is!acceptable!at! this!point.!!We!will!have!an!opportunity!to!derive!this!result!properly!at!a!later!point!(after!we!derive! the!sampling!theorem).! ! 6. Problem!4.18!P&M!(Grad!Students!only).! !!"#$#%"&'(')*+",'(-*+"./01233'-("""""""""""""""""""""""""""""""""""""""""""""""""4*++"5667" ./08+29",2)":"""""""""""""""""""""""""""""""""""""""""""""""""""""""""&;2",2<)2982/"57="5667"""" ! ! 7. Problem!4.22!P&M.! !!"#$#%"&'(')*+",'(-*+"./01233'-("""""""""""""""""""""""""""""""""""""""""""""""""4*++"5667" ./08+29",2)":"""""""""""""""""""""""""""""""""""""""""""""""""""""""""&;2",2<)2982/"57="5667"""" ! ! 8. Use!Assignment3.m!to!generate!the!figures!below!and!answer!the!following!questions:! i. What!is!the!90%!bandwidth!(in!Hz)!of!the!aortic!pressure!waveform?!See!Section!4.2.9! P&M!for!definition.!You!may!have!to!use!the!zoom!and!the!Datatip.!You!can!also!use!the! MATLAB!command!find,!but!this!will!require!a!little!work.! 2.01 hz ii. What!is!the!90%!bandwidth!(in!Hz)!of!the!filtered!waveform?!1.94hz iii. What!is!the!6"dB!bandwidth!of!the!MA!filter!used?! 8 hz !!"#$#%"&'(')*+",'(-*+"./01233'-("""""""""""""""""""""""""""""""""""""""""""""""""4*++"5667" ./08+29",2)":"""""""""""""""""""""""""""""""""""""""""""""""""""""""""&;2",2<)2982/"57="5667"""" ! iv: components outside the bandwith of the filter are surpressed iv. v. vi. Describe!the!effects!of!the!filter!on!the!PSD!of!the!signal.!In!particular,!what!frequency! components!are!affected!(if!any)?!! Can!you!estimate!the!SNR!for!both!signals?! Repeat!(i.!!v.)!!the!above!for!the!ecg!waveform.! ! ! v. The SNR is evaluated approximately from the results of pwelch by identifying the maximum signal level within the signal band and a a flat region in the noise band. From the figures in the assignment, the signal level for unfiltered aortic is ~27 dB and the noise level is approximately -30 dB. There fore, the SNR for unfiltered aortic is approximately 57 dB ...
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This homework help was uploaded on 04/17/2008 for the course EE 4541 taught by Professor Ebbini during the Fall '08 term at Minnesota.

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