ECE 4213/5213 Test 2

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$ECE 4213/5213 Test 2 Fall 2018 Dr. Havlicek Wednesday, November 14, 2018 4:30 PM - 5:45 PM Name: SOLU1 \ 0 N Student Num: Directions: This test is open book.$

1 ECE 4213/5213 Test 2 Fall 2018 Dr. Havlicek Wednesday, November 14, :30 PM - 5:45 PM Name: SOLU1 \ 0 N Student Num: Directions: This test is open book. You may also use a calculator, a clean copy of the course notes, and a clean copy of the formula sheet from the course web site. Other materials are not allowed. You have 75 minutes to complete the test. All work must be your own. Students enrolled for undergraduate credit: work any four problems. Each problem counts 25 points. Below, circle the numbers of the four problems you wish to have graded. Students enrolled for graduate credit: work all five problems. Each problem counts 20 points. SHOW ALL OF YOUR WORK for maximum partial credit! GOOD LUCK! SCORE: 1. (25/20) 2. (25/20) 3. (25/20) 4. (25/20) 5. (25/20) TOTAL (100): On my honor, I affirm that I have neither given nor received inappropriate aid in the completion of this test. Name: Date: 1

$Use the DPT to find the 3-point circular convolution y[n] = x[n]@h[n]. 2. xc1._ L "X[nJ W3fl :;. 1 W3o'(:. + 2w;t:. + 3 w;k =.i +-2W k T 311\12 1< 0 "2 n ;o 3 '.3 ),,?. 2 Hct<1 c"j w"'\=. = -2.$ $wJ* + 4wJ-2 w 3 1' -2 + 4 w}:: 1-2w3 1<.) " " 2 n=o 3 YEt<] = XT:k)HC\!1 = (1t2W3k+3W/ 1 (.)(-2+4W3\4.-2.Wl\<.) : -2 -t- 4 'Nf - 2IJJ32k.. - 4w.{ --+ ew'l.tc. - 4-W}J<. 3 - b w}k + t 2 vj;arr-.$

2 1. 25/20 pts. Let x[n] and h[n] be 3-point discrete-time signals given by x[n] = [1 2 3] = 8[n] + 28[n - 1] + 38[n - 2],_ 0 n 2, and h[n] = [-2 4-2] = -28[n] + 48[n - 1] - 2o[n - 2], 0 n 2. Use the DPT to find the 3-point circular convolution y[n] = x[n]@h[n]. 2. xc1._ L "X[nJ W3fl :;. 1 W3o'(:. + 2w;t:. + 3 w;k =.i +-2W k T 311\12 1< 0 "2 n ;o 3 '.3 ),,?. 2 Hct<1 c"j w"'\=. = -2.wJ* + 4wJ-2 w 3 1' w}:: 1-2w3 1<.) " " 2 n=o 3 YEt<] = XT:k)HC\!1 = (1t2W3k+3W/ 1 (.)(-2+4W3\4.-2.Wl\<.) : -2 -t- 4 'Nf - 2IJJ32k.. - 4w.{ --+ ew'l.tc. - 4-W}J<. 3 - b w}k + t 2 vj;arr-.. - b Wa 4k "2. {'.)_ 3 k 1 1 A 1 4 i'- "( (\<1 = -2-\- o wk. + C> \J.)3 + ow3 - (0 vva -= -2 t e.> w J -+ D w]l<-+ B - b w;<- = 0 - b W 3 "' +- o w f ; o \<. 1 2 ( ) By de.t\ Vl\-\ \M 1 \ 2 "<n,.1 z!-o'f 'CYlj w ;I<- -::: f[o1-r.11)1 wt + ':1 t-z.j W3 NoTE:.' Wi'f..=i W 4-w\(.. l..,,.,a l nt1+s po s-.11) y rv-.1 :: [ - o J - b 8( n J - b o ( Ill \J J O V\ 2-2

$g., IH(eiw)I = IG(eiw)I V w ER s ov-. "'--\-(?,<t.,;::,<.\_e, 7,'2-fb) we "'-v-e...j \ 4 -, - 4 6('!o) =- (1- ',.$

3 A 2. 25/20 pts. The causal, stable IIR digital filter G has transfer function 1 lzl > 4. Note that G(z) does not have minimum phase because there is a zero outside the unit circle at z = 4. Give the transfer function H(z) for a new causal, stable IIR digital filter H such that H(z) has minimum phase and G and H have the same-magnitude response, e.g., IH(eiw)I = IG(eiw)I V w ER s ov-. "'--\-(?,<t.,;::,<.\_e, 7,'2-fb) we "'-v-e...j \ 4 -, - 4 6('!o) =- (1- ',.?: )(1+}:r ) (1-42- ) -1_4 ( l-\-,(r') l I - -Jfe j rv+ l"' )( \- -.¾: e--j Tr, i-') kea.j e. f o a "' \.i =---- Mlc) t -4:;;,-t?_,_ + a..ll f'1-s5 J lt- 1 "iy 3

$H(z) has a real zero at z = - Find the transfer function H(z) and the impulse response h[n]. No+.. r 7..-tt; Bec\lSe. -\-'1.,,lf"'e. \s e 01\-e.. t i: -= ½ Note f -,,,o A -+ype It -Q; \er VV\.'J.$ $\ h\j-e.. The e-'l'o5 t.:tre,, = 3; 1-,... : 1; d'!::. i. H(): K!, l l- """" -t) J wfare l'\.,s -= (\-i-e ')(l-\1!- 1 )[1- i-') -=- k(1-1t-i ' +2:--a.$

4 3. 25/20 pts. H is a causal, stable type IV linear phase FIR digital filter with order N = 3. The impulse response h[n] is real and h[o] = 1. H(z) has a real zero at z = - Find the transfer function H(z) and the impulse response h[n]. No+.. r 7..-tt; Bec\lSe. -\-'1.,,lf"'e. \s e 01\-e.. t i: -= ½ Note f -,,,o A -+ype It -Q; \er VV\.'J.\ h\j-e.. The e-'l'o5 t.:tre,, = 3; 1-,... : 1; d'!::. i. H(): K!, l l- """" -t) J wfare l'\.,s -= (\-i-e ')(l-\1!- 1 )[1- i-') -=- k(1-1t-i ' +2:--a.)(\-,-') =- k ( \ - Ii f-\ + t:"''2 - i-t + i 1:- - i-3 ). ero Sc.::. J" +e.re W\\4+\io "' te-<o + = 1... = k ( I - lf _, + i - - l- 3 ) ::: K - f k i_, + i l<- "2.. _ k 2-3 J R..D c : t l "' a., letfo\e: [V\1 = k.srl\j-1k8t-l)-t!¾j'(- f)- kj'[vl-1 k [DJ :: i :: V, l-l ("l) = l - :t.-l -t z:.---a J le J > D \ \., (11 l :::: n1 - ii("'-,; + 1l-bt"' 21 - Cc-3) I

$) uvv'lbe,,i/) bo1c-e, h '6:-e. lcil axe ca..\c1.t\tt"tbr re 1{k-el/. 4. 25/20 pts. Design an analog Butterworth low pass filter to meet the following analog specification: \. A.:: o.$ 90 Give the analog filter transfer function Ha(s). Hint: The design formulas for the analog Butterworth filter are given on pages A-4 and A-5 of the notes file ECE5213NotesAnalogFil terdesign.

90 Give the analog filter transfer function Ha(s). Hint: The design formulas for the analog Butterworth filter are given on pages A-4 and A-5 of the notes file ECE5213NotesAnalogFil terdesign.

5 ) uvv'lbe,,i/) bo1c-e, h '6:-e. lcil axe ca..\c1.t\tt"tbr re 1{k-el/ /20 pts. Design an analog Butterworth low pass filter to meet the following analog specification: \. A.:: o.o, passband edge freq. Dp = l0001r rad/ sec A-= \DO stopband edge freq. Ds = 65001r_rad/sec min. stopband attenuation l/a = 0.01 max. passband attenuation 1; 1 + c 2 = 0.90 Give the analog filter transfer function Ha(s). Hint: The design formulas for the analog Butterworth filter are given on pages A-4 and A-5 of the notes file ECE5213NotesAnalogFil terdesign. pdf and in Appendix A.2 on pages of the text. ::. J Q" t> 42.i, l = f Hint: Make sure that your calculator is set for radians and not degrees! l r,.l "Le> ' (I..$./.. 2 bsooi'l O f\1,.=- \.!ld CfiJ 2, ; If> l'z. 13 -"'eij to :n..p t() \ j_ - f\'2. - \ => (A. I\).. r 1 ::. n.c. J 1\ l 3 t1-1) 1, j1$'3+ Ac.. $/' pa: Jl.c. e \1\ l 3+i,-t)/G ra:..flc e

$( s - z. Rei r I l s + l f I \ a ) (.s t=2.) - _ [ s2.- lf,ft*)s + Ir, \2.J(s-r.2.) t ( 2.-2 JLc.cos1r s _-t-.$ $0s.,_f J s - -'l:-1 "1,...n.l.:: s 3 - (_2.n.c. CO;; - ftg) s 2 -+ (_.al=- 2.fl. t ws ) -r Jl @ ::. 53 4,- 2..{)_c. s-a + 211.; s +.n.. (..n.(.. =- 4, 3 9 I,os) 6$

6 More Workspace for Poblem 4... (A,to) µt:tl) -::. -3..n. :: n- (s::-pj.),t.-: I.Jlt - = (.s-r,)(.s-rj(s-p,.) -..12:f ::. ( s - z. Rei r I l s + l f I \ a ) (.s t=2.) - _ [ s2.- lf,ft*)s + Ir, \2.J(s-r.2.) t ( 2.-2 JLc.cos1r s _-t-.ai)( S-fz) c;,j-2flc,s,2. -r..qs-ris't2..a.c.. Pz. C(!f/f s - Alfa..fl! - s 3 - ( 212c v-5 f + f z..) s Lill -t G, fz. 0s.,_f J s - -'l:-1 "1,...n.l.:: s 3 - (_2.n.c. CO;; - ftg) s 2 -+ (_.al=- 2.fl. t ws ) -r ::. 53 4,- 2..{)_c. s-a ; s +.n.. (..n.(.. =- 4, 3 9 I,os) 6

$Nu Vv\elfs '"' boi<e \; \<..e. @II e.t If' c G\.. \cv...l't or v-e'j, sfev-} ¼-=-oD3\" (. \1) 5. 25/20 pts. Use the bilinear transform with T = 2 to design a Type 1 Chebyshev -::.$ 7661r rad/sample v'l 2 = 0.9500 min. stopband atten. l/a = 0.0316 Give the digital filter transfer function H(z).

7661r rad/sample v'l 2 = 0.9500 min. stopband atten. l/a = 0.0316 Give the digital filter transfer function H(z).

7 Nu Vv\elfs '"' boi<e \; e.t If' c G\.. \cv...l't or v-e'j, sfev-} ¼-=-oD3\" (. \1) 5. 25/20 pts. Use the bilinear transform with T = 2 to design a Type 1 Chebyshev -::.il,g-+551 tm f 9 digital lowpass filter that meets the following specifications: passband edge freq. stopband edge freq. max. passband ripple Wp = r rad/sample w 8 = r rad/sample v'l 2 = min. stopband atten. l/a = Give the digital filter transfer function H(z). Hint: The design formulas for the analog Type 1 Chebyshev filter are given on pages A-7 and A-8 of the notes file ECE5213NotesAnalogFil terdesign. pdf and in Appendix A.3 of the text on pages Hint: Make sure that your calcor is set for radians and not degrees! \Q!r.l... 7:.11r::. t(;tv\ ( ) :: a, '32-tt,-I,,. 2. m JJ.5,-,,_I\() :::.. 2.'S'JG9f1 fil r ::: f:os' ( / )l : fc.,«1 ( % ') l 5",2.5'9811 ::: / Cos l ( n.s;,n.f) \ c>-s\.."'\ (.,.,, 2 t3) 1 ::. -0,24124 ' \ (A.\B) =- -./1.1 7 No-tes p, A-8 : ' - o,. 95

$More Workspace for Problem 5... \\lo-\ec; r A-s; l-l"-l) = C.o (-f,)(-r7.$ $- Lr IT p rl') s + r It,* c= \ri\1.. - --------, rflj \_\Gl ( $) = 0./ bo 72.$

8 More Workspace for Problem 5... \\lo-\ec; r A-s; l-l"-l) = C.o (-f,)(-r7..) ( s -f1 ) l s -1 1.) P1 F1--ll : C.o s '2. - Lr IT p rl') s + r It,* c= \ri\ , rflj \_\Gl ( $) = 0./ bo 72.. s'2. + o.4s'2.4ss + No \-es t -7 H ("! ) = \,,_ls) Is= -- (\-rr-l)' (_ l +1t 1 ) 2 0 \<le... 8 '

$More Workspace for Problem 5... c.-o - (r.r'\ 2 -till) l \ + 2 i:""\ i-a.) -=ca === 11 la' u lb4.a- WLA.www...um _,$ $r-) + (-2.. +l (0-1,.. r,o.l )) 2- I + (1-r 2<rj + Oi1."'A) z-'2- @ f.e -@ @ pll.lj',, +ta.e 1\\l""'be"'s. tm /w tfi) e>. U, a 7 2 + O, 3 2.$

9 More Workspace for Problem 5... c.-o - (r.r'\ 2 -till) l \ + 2 i:""\ i-a.) -=ca === 11 la' u lb4.a- WLA. _, mw-...,.w - r: 1-2i-' t a; +20i i-, +(011-+-r)-r2(0i,fJ2r) r- 1 +(o-?+.n.l)i--'l. ta) ---, - c. ( tr, -i.. +A.t) ( \ -l- 2 i-\ ct- r."'a.) - (\- 2c:r; -t- o-.2 +.ll.r-) + (-2.. +l (0-1,.. r,o.l )) 2- I + (1-r 2<rj + Oi1."'A) f.e pll.lj',, +ta.e 1\\l""'be"'s. tm /w tfi) e>. U, a O, 3 2. I f 4 c -, + o, lie,() 12 i-"" l-\ Le) = L. b \ 6 b - l., 6 1 ' o, s b 7 o ----._...,.._-= --=.-..i-..=z"""":n:-.nri_m,,,,_iiw -4;--...,,n t).jj4,2. i:-\ + D-0' l-\(:c),... \ - \ C.: ,415"7-6 e iz,v - rrtitw 9

ECE 4213/5213 Test 1. SHOW ALL OF YOUR WORK for maximum partial credit! GOOD LUCK!

ECE 4213/5213 Test 1. SHOW ALL OF YOUR WORK for maximum partial credit! GOOD LUCK! ECE 4213/5213 Test 1 Fall 2018 Dr. Havlicek Monday, October 15, 2018 4:30 PM - 5:45 PM Name: S_O_L_U_'T_I Q_t\l Student Num: Directions: This test is open book. You may also use a calculator, a clean copy