3015_Homework_13_Solutions

3015_Homework_13_Solutions - 7 7.21. Given the...

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Unformatted text preview: 7 7.21. Given the z—transferm pair sr[r:-.] % fl with R00 c“: 4. use the z—transforrn properties :0 determine the :-transform of the following signals: :mmm=xm—m {h} yln] = [HEP-TM N me=érwfl {UMM=IM*fln—fl N 11g=xop4xup= fin] = It'll * ml“ - 3] 7.24. Use the method of partial fractions to obtain the time—domain signals corresponding to the fol— lowing z—transforms: A B X; = —+— [2' 1— %;‘1 1+ %;‘1 l = A+B 7 1 l — = —A——B E 3 2 2 —1 Kb?) = —_+—_ —%;1 l+§31 1 l I[n] = [2(ajn—[—§jn]un] 1+?!—1 1 1 {Ci Xifi} = —H—1——1][1+57_13 i '5 |3| “‘1 E [l—EI same as [3,), but fin] is two—sided 7.29. A causal system has input .r[n] and output y[n]. Use the transfer function to determine the impulse response of this system. {3.) I[J‘1]= fin] —|— ifln — 1] — fififn — 2]. = [fin] — %r§[n — l] 1 1 X(:) = 1+Es—1—Es—1 3 _ Ye) = l—Esl Y: He) = —%1 +3: = + l—%:‘1 l+%:‘1 7.31. Determine transfer function and [ii]: impulse response representations for the systems de— scribed the following difference equations: [11: _r,r[ra] —_Tll_t‘r[:n — 1: — E—{Eyh —2: — Erlri] +.r[H — 1] 11:10 =I:* 3) = [34 '].t[;] U .1} He} — ’ 73‘ 2— H 1'0" _ ____'_______—-——-— —v — + 3 JEN ' 7.37. Use the graphical method to sketch the magnitude response of the systems having the ihilowing transfer functions: .1 {b1 H153] = 3 2 1 . 3 +2 +1 H (z) = —322 Q. 6J2fl + (in +1 H (‘J J = 3.32:: poles at: z = . {double} zeros at: z = 6:25: I111 TI. 3 ' I \& /1 Re Figure P737. {b} Graphical method. P737 (b1 Magnitude Response 1 I I I I I I [19- - 0.8 — — 0.1— — Ln I m | M I a M m .5. 7.42. Use the unilateral z-trsnsform to determine the forced response the natural response, and the complete response of the systems described by the following difference equations with the given inputs and initial conditions. (a) Mal—1mm — 1] = Erin], y[—1]=1.. Ilnl=[‘71)“u[n1 1 2“ l Y[:j—§[:‘1Y|::}+l} = 2X(:) 1 1 Y: 1——:-1 = — 2X: {}( 3 ) 3+ {1 1 1 1 Y{:] = +2 X3) 31—13—1 1—31.53—1‘: \—~.—l \—~,—l YWT'LI} Yul?) Natural Response 1 1 ymlml = _ 31—%:—1 1 1 '1 ylfllifll = §(§) “[11] Forced Response wing] = m+fig yl'fl[fl] = (—%)n+ g ulni Complete Response slfll = [ ...
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This note was uploaded on 12/20/2009 for the course EE 3015 taught by Professor Staff during the Fall '08 term at Minnesota.

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