SoLUT10N [Name: [ Final Exam ECE301 Signals and Systems Wednesday, May 6, 2015 Cover Sheet Write your name on this page and every page to be safe. Test Duration: 120 minutes. Coverage: Comprehensive Open Book but Closed Notes. Three two-sided handwritten sheets. Calculators NOT allowed. This test contains four problems, each with multiple parts. You have to draw your own plots. You must show all work for each problem to receive full credit. Good luck! It was great having you in class this semester! Have a great Summer!
Problem 1. Using the Z-Transform, determine the output of the systern y[n] = ay[n - I]+ x[n] for the input x[n] =,enu[n]. You must use the Z-Transform to solve this problem, i.e., use basic Z-Transform properties and a basic Z-Transform pair plus partial fraction expansion to ultimately determine the output y[n]. You are given that lal < 1 and l,81 < 1. ""'"l_\ af J:),'ft-e,,-e~CQ l::: c"' ="J ~~th. <;.i.~e5 Yc~1:. ol -z:-1 Yc=r-) ~ 't< c.:c) '-{Cr-) \ ~ n - --- c'.61-=xc~) l - ol.1:--1 - -o<. Y. (?::)::: t ~-;& -,~ \--\-- c =E") ~ ( ~) Jc~1 ~ 'L (t -e>z)c "b-~) -=r-- A\ AL -1... -z: Ye :z.) - + c"b- ~ '1 c e- -Pj c -t?z. ~-f> Cc-~) ~ \ (~-olj Ct:--t>) c--13 buth C:;\. ~es> of- @ (3 - d.
Problem 2. The rectangular pulse Xin(t) = {u(t) - u(t - 1)} of duration 1 sec is input to the following integrator The output xa(t) is sampled every Ts= 1/3 seconds to form x[n] = Xa(nTs) The sampling rate is ls = 3 samples/sec. Show work. Clearly label and write your final answer in the space provided on the next few pages. (a) Do a stem plot of the DT signal x[n], or you can simply write the numbers that comprise x[n] (indicate with an arrow where then= 0 value is.) (b) The Discrete-Time (DT) signal x[n], created as described above, is input to the DT system described by the difference equation below: y[n] = x[n] - x[n - 1] (i) First, determine the impulse response h[n] for this system. (ii) Determine and plot the output y[n] by convolving the input x[n] defined above with the impulse response h[n]. Show all work in the space provided. Do the :'ltem plot fur y[n] on the grnph prnvided on the pttge ttfter next. S ''f 11-.e.,,.cp 1 5 s \' "'""p2f- i t ")( M ( t) :: L.""' ('r:-) Jco ~ I S. V\,..... t"' t v-~p\.::r...ce-.\ "' 'L ct J:: h I ( t -3 J h~ C -t I ~1----""T ~ V'I' p la. \ s.~ v ~ ~ pon.sq \ s. 3, ~ \ c -l) - s h I ( t - 3) -t-t. h Ct)~ ()._Ct) -uct-i) ~ ~r 2 \ (. L\ N \.,_ ( *: '\ \ h \ _>- ~ <;. CC 'A V 0 ) v e I::\ taj \.\ ~ f" v' ) + \) f QI/ 'IV' ~ ~ ~ t) ~ y: ~h ( +-) h.~ ct) i..
, Show your work and plots for Problem 2 here. fn>m CV\b,. 5\., e-t) 1..Ve f::""vlow ~ (~) j_b_ ~~ 1- b 0.1.. 2- Thv-5) 1' r. v o~'.""-8 l\ \111\~ - I.<r.VtA. V\.O\V\ ce CA.."'~?( ' (*-J-:.. 3 'd b (-'<:) - s 'db ( t - 3) 0!)( r~1 -= 0<v. c 'A-'~) '2:7 ':2_:, :> 0 t - _3:. -~ -3 =3 (,. { ~ > l ) 2)?, > ~ > 3 :>~) 2 >I) O)-\ :> -1.) -3-,-3.:>-~-> -3. )-<. J-l _j yv:. <) ( b J :r"" r \A \se R t':~ po "SQ ct ~ [ V' 1 ~ ')( r i,.._"j - <")<.r Y\ - ""D, :.., h r r.. J -= d' C f\.) - J Ch - \) 7 V\ -) '(!J thl-=!)( r~> ~ hrh1-= -Y rn.l - ~ ch-d / { o t 2 3 3 3 -s 2 ~ o " - ' - 1. )-3 - ~ - 3-3 -2 :> - \ ) or ) ) >) ::>:)) -:,:')../ ') ) ).;) > 1 :_ { O, o ) I, 7. > 3 > _> :> ~ > '3, > '< :>I, o ) - J.,, - C ) - 3, :> - 3. :> - 3., -:3> - l :>- ) ::: { o ") J > l > l ) o.,, 0 > o, - ) )- I,-J >-J) - \,-\ ) o > o, o) I -:;. I, I J A W'>
Problem 3 (a). Consider an analog signal xa(t) with maximum frequency (bandwidth) WJvI = 40 rads/sec. That is, the Fourier Transform of the analog signal Xa(t) is exactly zero for lwl > 40 rads/sec. This signal is sampled at a rate W8 = 1 rads/sec., where Ws = 27r /Ts such the time between samples is Ts = {~ sec. This yields the discrete-time sequence 1f sin ( ~ n) sin ( ~~ n) }.. ( 27f ) x[n] = Xa(nTs) = { - 2J sm -n 5 7rnTs 7rnTs 5 where: T 2 7f s - 1 A reconstructed signal is formed from the samples above according to the formula below. Determine a simple, closed-form expression for the reconstructed signal Xr(t). Show work. Xr(t) = L x[n]h (t - nts) n=~oo where: T - 27r and h(t) = Ts!!_ sin(lot) sin(50t) s - 1 10 7rt 7rt Problem 3 (b). Consider the SAME analog signal Xa ( t) with maximum frequency (bandwidth) WM = 40 rads/sec. This signal is sampled at the same rate W 8 = 1 rads/sec., but is reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does xr(t) = Xa(t)? For this part, you do not need to determine xr(t), just need to explain whether xr(t) = xa(t) or not. Xr(t) = L x[n]h (t - nts) n=~oo where: T = 27r s 1 s2 7rt 7rt and h(t) = T ~ {sin(40t) + sin(60t)} Problem 3 (c). Consider the SAME analog signal xa(t) with maximum frequency (bandwidth) WJvI = 40 rads/sec. This signal is sampled at the same rate ws = 1 rads/sec., but is reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does xr(t) = xa(t)? For this part, you do not need to determine Xr(t), just need to explain whether Xr(t) = xa(t) or not. Xr(t) = L x[n]h (t - nts) n=-oo where: T = 27r s 1 s2 7rt 7rt and h(t) = T ~ {sin( 45t) + sin(55t)} Show all your work for Prob. 3, parts (a)-(b)-(c) on next page.
Show your work for Prob. 3, parts (a)-(b)-(c) below. c v ''t ""'. ta.. -\-o c- \... ~c. \;(' : W~"'? "'\ 2.Wt.A. \ ~ : > \ > \I\ e. ce...s.c;.... [J ~S. h_(-t,) S\Aok +-'~ 4f H (Lo') "7- ~ +o.,, j \A.I \ < (.(.) ) M \h: '!:P \"""' S\A..lle.S..,..~A+ 0..,., 't \ f\.&j \ fou II, \v \VA~ s-+e1 II'... c -e 1"' -i- ~ II~ J.c.. t- u... r:::.:{) i <. r c;.. s So u.. \\ ~ I t- ~ II Q ~ Afvy (..(;:,')-:. ~o.. (-\:.) - ")((~1~ '1<.o,. (-<:.)\ t::. ~-s -4\:> \ t' - - sot ~ { \'"'\ ( l s -\:.) I J.~,Y~~~ 3z, CV\,\-Q1q'~ O.v~ ~Q~) +"hq.i"- n-+- 'Z-) s ~ '(\_ l 2 () t)
'3 CbJ Show your work and plots for Problem 2 her f-\- ( )) r e. w 'l fv I?.L I-\ C U-J ') ~ Ts ~ 0.1 \ ~ \ <. 4o 1-r lu...\ ~...,,,,,..,. I- 1.... - 7'. Ao Jo 0 0 f Uv l U.,,, \ ~ \).; S - (.u M -:.. l 60-4 0 ::.. b 0 7 Z. (4o) ~Cc) SA""e. pl~:r_,..~ "bov ) -'>)Ii cerj" 40-745 cah J 6 (J -::> $"$' H- (VJ"\-:.. Ts. -fo.,, \ ~\ ~ 4o -:::... 0 +o..j l 1...V \ "/ b 0
Pro bl em 3 ( d). Consider an analog signal Xa ( t) with maximum frequency (bandwidth) WM = 40 rads/sec. That is, the Fourier Transform of the analog signal xa(t) is exactly zero for JwJ > 40 rads/sec. This signal is sampled at a rate W 8 = 80 rads/sec., where W 8 = 2Jr /Ts such the time between samples is Ts = ~~ sec. This yields the discrete-time sequence Jr sm ( 8 n sm 7r ) ( 8 n 37r ) } Jr x[n] = Xa(nTs) = { - 2j sin (-n) 5 JrnTs JrnT 8 2 2Jr where: Ts = 80 A reconstructed signal is formed from the samples above according to the formula below. Determine a simple, closed-form expression for the reconstructed signal Xr ( t). Show work. Xr(t) = L x[n]h (t - nts) where: T = 2 Jr and h(t) = Ts sin( 4 0t) 8 80 Jrt n=-cxj Problem 3 (e). Consider the SAME analog signal xa(t) with maximum frequency (bandwidth) WM = 40 rads/sec. This signal is sampled at the same rate W 8 = 80 rads/sec., but reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does Xr(t) = xa(t)? For this part, you do not need to determine Xr(t), just need to explain whether Xr(t) = Xa(t) or not. Xr(t) = L x[n]h (t - nts) n=-oo where: T = 2Jr 8 s 80 5 Jrt Jrt and h(t) = T {~ sin(5t) sin(40t)} Problem 3 (f). Consider the SAME analog signal xa(t) with maximum frequency (bandwidth) WM = 40 rads/sec. This signal is sampled at the same rate W 8 = 80 rads/sec., but reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does Xr(t) = xa(t)? For this part, you do not need to determine Xr(t), just need to explain whether Xr(t) = xa(t) or not. Xr(t) = L x[n]h (t - nts) n=-oo where: T = 2Jr s 80 s 2 Jrt Jrt and h(t) = T! {sin( 40t) + sin(lot)} Show all your work for Prob. 3, parts (d)-(e)-(f) on next page.
Show your work for Prob. 3, parts (d)-(e)-(f) below. H C "--'}-=- TS +o,,, } vv\<+o v J-\ ( u..i )-::: (j -f o.1 l u...} > 4 O V v op l,' c~ 5 -f, \ t-e t.-e J. ~...,,,-;--?C \'" ( -\-: )-=. t)( A Ct) -:. x c~) -::: h~ t;l& U,,;th -::.4o ) ~ V~f> ): C~ Or-<:_ } 1c.J- +u1ach:~ ~ '"'- o t he v- )-+ ( W)-=- T; 1-t\U)}-=-u Y cu~ ~... vc.\\ ~ off of J-\ \ -\-9 y } s \~C\_ h,.c;..j.;:(- \ u.,., \ <'... 3 5.S ; " c! u.,; t.a-= 40 -=-"') w; 11 '~ "'l t-ev t.~ I ~\ )' 4 5 1-- I/'~ F) L CAs. 5+<Av* ~:+ UJ ~ - WM-:.. gu--4.0 ) -::...4~ ~ ( f) ~s~ : n) \-\ ( u,, 1 ~ T<> ' 0 ~ \ u.. \.c. 4lJ x
Problem 3 (g). Consider an analog signal xa(t) with maximum frequency (bandwidth) WM = 40 rads/sec. This signal is sampled at a rate Ws = 50 rads/sec., where Ws = 27f /Ts such the time between samples is Ts = ;~ sec, yielding the following discrete-time sequence: 27f where: Ts = 50 A reconstructed signal is formed from the samples above according to the formula below. Determine a closed-form expression for the reconstructed signal Xr(t). Show all work. Xr(t) = L x[n]h (t - nts) where: Ts= ~~ and h(t) =Ts sin~~ 5 t) n=-oo Problem 3 (h). Consider the SAME analog signal xa(t) with maximum frequency (bandwidth) WM = 40 rads/sec. This signal is sampled at the same rate W 8 = 50 rads/sec., where W 8 = 27f /Ts and the time between samples is Ts = ;~ sec, but at a different starting point. This yields the Discrete-Time x[n] signal below:. [ ]- ( )-{~} {sin(~(n+0.5))sin( -(n+0.5))}.. (47f( )) XE n - Xa nts + 0.5Ts -. ( ) ( ) 2J sm n + 0.5 5 7f n + 0.5 Ts 7f n + 0.5 ~' 5 A reconstructed signal is formed from the samples above according to the formula below. Determine a simple, closed-form expression for the reconstructed signal Xr(t). Hint: before you do a lot of work, look at the interpolating lowpass filter being used below. Xr(t) = L xe[n]h (t - (n + 0.5)Ts) n=-oo 27f where: Ts= - 50 sin(lot) and h(t) = Ts-7f_t_ Show your work for Prob. 3, parts (g)-(h) on next page.
Show all your work for Prob. 3, parts (g)-(h) on this page. w - 1o M- (JJ ~-:. So <:. ZC-40 -;> ~ \,'o...s~ \'\~ )-\- (w )=. ~ -t OV"' \ 1..U\ < 25 H c.u.; I ~ o for \ u.., \ '125 - 'Tr - - s - ~ 5 S) r... C-e_ H ( u:>}-=- C fov \ ~) > 2.S. X r cu...)
Show all your work for Prob. 3, parts (g)-(h) on this page.,1 --:-- I')( o-, ( t") pct--c) ~+7 p (~ )"".:S ~ct-..,15) _A'O I ' ' \ / '-,,' ) - - -...-r...::-=j,/'-----1: sr~c+vo.\ v Q p' : c.e._ c~ "'-+.e ve..j '- i LA.> -::: - OJ_$. -:::.-S'O -11 T.s / " I I I S<S(, \ '{>O\SSE' S O \I\~ c\ cv:+- c\:)~ \V'\~ Xv- Cw} :1 -lo Jb
Problem 4. For each part: show all work, state which Fourier Transform pairs and/or properties you are using, and clearly indicate your final answer. Use the result below: 1 -(x-m)2 _ e~ dx = y"i;- O' 0\ - \ - - - rrr er~ Problem 4 (b). The Gaussian pulse x( t) = 0 (-l) -:. 'X (-\; - ~) form y(t) = x(t - 3). Find the numerical value :f; = ~ 1 \ YCtL)j =. }YCUJ)\-i. - ~e -2o1, with O'i = 3 2, is time-shifted to 27r -oo IY(w)l2dw. --:-:> Yew)::. e-:> 3 w 'l<c~> I e- ) lw\ } X (u.;)) \)((v.-.)\l. \ 4 CO\)
to form y(t) = tx(t). Find the numerical value of A = l: y(t)dt. Show all work, state Problem 4 (c). The Gaussian pulse x(t) = form y(t) = fkx(t). Find the numerical value of A= l: y(t)dt. \ 2 c e- 2 ""r, with O"i = 3 2, is differentiated to i71 v L,7f M (-t)-: ~')< (t) -=-/ u di. '(cw) -= ~w X Cw) A~_f;ctJr!-\:: - YCo)-=- J (1>) )(Cc:.) - -z -t '2 0""1' -e t 1- z. i l: 0 - o~j. X eve-ii\ - f V\. r-"" Problem 4 (d). The Gaussian pulse x(t) = 1,,, 2 ;-::- e-~, with O"i = 3 2, is multiplied by t i71 v L,7f which Fourier Transform pairs and/ or properties you are using, and clearly indicate your final answer. '/(. ( t). I '\ Ot'\ Ave~ G
Problem 4 (e). The Gaussian pulse x(t) = 1 ;n::- 2 e- 2 ""i, with cri = 3 2, is compressed in time 0-1 v.<?r by a factor of 2 to form y(t) = x(2t). Find the numerical value of the energy E = j'xj y 2 (t)dt. J... '2. ~ _ 2 '1. t 1...)"2- \ -: Z ~ t L Al). c -t.) ~ :x C2 t.).:: ~ e 2 a \t -=- 2r~)'1.. e z.c:r,"" U, '1 err er\ \4'ttf 0--;.. ()'\ t &. 0\'2rr e -2 "[- c-~-.) \ z Problem 4 (f). The Gaussian pulse x(t) = - 1 -e-~ with CJ 2 = 3 2 is input to an a1v'2?r ' 1 ' LTI system with a Gaussian shaped impulse response, h(t) = 1~2 e - 2 0-~, with er~ = 4 2. a2v.<?r Determine a simple, closed-form expression for the output y(t) = x(t) * h(t). t;z. a-3>.: ~
Problem 4 (g). The Gaussian pulse x(t) = - 1 -e- 2 " I with CJ 2 = 3 2 is input to an 0-1 v12i ' 1 ' LTI system with a Gaussian shaped impulse response, h(t) = \ 2,= e -~, with CJ~ = 4 2. 0-2 v L.7' Determine the numerical value of the area, A=; = y(t)dt, under the output y(t) = x(t) * -DO ~ h(t). - t_ ~ t t..) - "/.. ( t) * h ( l) -=- \ e ~er~'?- cr3 -:: ~ u 0-~~ 60 Av... -;:: A=-f 0 (t)j-'c."' 1 -C>o Problem 4 (h). The Gaussian pulse x(t) = - 1 -e -~ with CJ 2 = 3 2 is input to an <71 v12i ' 1 ' LTI system with a Gaussian shaped impulse response, h(t) = 1 ;;;-::: 2 e-~, with CJ~ = 4 2. <72V L.7' Determine the numerical value of the energy E = j_: y 2 (l)dt of the output y(t) = x(t)*h(l). t'l.. M ( t) -:.. I)( ( -t') ~ h C-l).: t e - :z 5 'L o-a -:.. S 0 CT~~...:> E ~ j~'<c~j<h- - - a-'" 1 2tr j~- 0--z. 2 i\ 3' -oo \ O'"~ \ - \c~ ~ '2.~..fi er~ 2 {n- ANS
_(t-2j2 Pro bl em 4 ( i). The Gaussian pulse x ( t) = 1= 2 e 2 " 1, with CTI = 3 2, is input to an 0"1 V L/7r (t-3)2 LTI system with a Gaussian shaped impulse response, h(t) = 1= 2 e-~, with CTi = 4 2. <72V L/7r Determine the numerical value of the energy E = 1: y 2 (t)dt of the output y(t) = x(t)*h(t). \ ~ Problem 4 (j). The signal z(t) = x(t)y(t) is the PRODUCT of the Gaussian pulse x(t) = - 1 -e-2;[ with CT2 = 3 2 and the Gaussian pulse y(t) = -1 -e-~ with CT2 = 4 2. 0"1 v'2k ' 1 ' 0"2..;z:;r ' 2 Determine a simple expression for the Fourier Transform, Z(w), of z(t) = x(t)y(t)~ 1:.1..(' \ \ - - + c-c t; -=- -x ct.) ~ \~) = e -:z tr'\,.. o-'l.z.. '2 rr o;- O" i.. t1.. \ J"-zl\'o-~ \ - -i. rr O"' 1 a t. --f ti. rr 1 er 3. \ (J~ '2. d ') e 0-: i.. ~Zl\ "\ 0\ O" l.. L z er: 1.,. "> 2. \ wt.. 0-3. = \ \ -}- oc... o:"t.. ~ 3 \ i:.