NAME: EE301 Signals and Systems Exam 3. NAME In-Class Exam Thursday, Apr. 20, Cover Sheet

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1 NAME: EE31 Signals and Systems Exam 3 NAME n-class Exam Thursday, Apr. 2, 217 Cover Sheet Test Duration: 75 minutes. Coverage: Chaps. 5,7 Open Book but Closed Notes. One 8.5 in. x 11 in. crib sheet Calculators NOT allowed. All work should be done on the sheets provided. You must show all work for each problem to receive full credit. For Problem 4, plot your answers on the graphs provided. VP Note Regarding DTFT Plots: The abscissa in each plot is the frequency axis. For each plot, the abscissa goes from -27f to 27f with tic marks every 7f /8. There is a dashed vertical line at w = -Jf and another dashed vertical line at w = +K. You only have to plot any DTFT over -Jf < w < Jf. 1

2 Problem 1: True False Questions. Circle the True (T) or False (F) for each part The Nyquist rate for the square of a signal is twice the Nyquist rate of the The highest frequency for the DTFT is effectively JT, corresponding to half of the sampling rate, whereas the highest frequency for the CTFT is oo. When you multiply a signal by a real-valued sinewave, the Nyquist rate of the resulting modulated signal is greater than the Nyquist rate for the original signal. (T) When you pass an analog signal through an LT filter, the Nyquist rate for the output signal is greater than the Nyquist rate for the input signal. (T)@)The inverse DTFT is a summation over frequency whereas the inverse CTFT is an integration over The DTFT is always periodic with period 2JT whereas the CTFT is generally not periodic except under special conditions. The time-domain variable for the DTFT is a discrete (integer-valued) variable for the DTFT whereas the time-domain variable for the CTFT is a continuous-valued variable. When you form the product of two analog signals, the Nyquist rate for the product signal is the sum of the respective Nyquist rates for the two individual (F) The DTFT is a summation over time whereas the CTFT is an integration over time. (T)@The Nyquist rate for the derivative of a signal is greater than the Nyquist rate for the original signal. 2

3 Problem 2. Short answer questions. (a) Briefly write and explain ONE of the main advantages of digital over analog, in terms of storage, transmission, and/ or processing, in a coherent sentence. L. t:_ '-" <." o ~ ~ \'\..$' S: "1. ""- p) e A""' f> \: h, J@:s c.\.s. -1 '_s. f Ov' ~ f''/" v" Co "'r (' \ Cc \{ e..c+, (.) \.'\,. Cc. J. ~ \\ g. CoV"vec.T f-oi1 b:+ Ji?V"{'OVS: J~Q to V\ '\.$-1( Q\ "".l 6 ~ o.. \ ) (,\ w5 i-o o. ~ s- o V\<'\ o=t f.,: c.~~ o.."'j ). r'v\ pq v-fec:..1 t't~... ~ ~' R~t 'V'-~V c~,.. ~ O'\. F \ ~ 1' ib~ \, \J. ~J. tov ~L t< +""~ Co ntfv\u..vl{ C.e...,. ~c.i V.). p V'~ C. \' S~ G <\. COV\V ~V',:>~6\t\ C\.~+e" A- /D.. t ~\~, t~\ s-\g\'\."\ f'vo(t"'.ss;~ (b) f one samples at a rate W 8 Ws A 2 + u.w aliased to? Assume O < ~w < ~s in radians/sec, what analog frequency is the frequency - ~ ~ ~ ~~ S \ C V\. <::\ \ 5 V e-~j - 'v "'- \ la...e cl ) {>, ~ "'> \o~ ( (-~ ~bv...)-t) C<>s (C~ -A~)-t) Th.\..\s.: p.., \ \,~so J -t~ : Lu s. - ( ~ -\- e.v.../

4 Additional CoS (({+A~ tj \ ± '-' ~cs(:~:_!~ ~ - ~-o~c~v.. - ~'" ~ ~,~-- '. COS + + cus-( \'\ 's 4' -~~ _\,_ ~J ~\r\+?,w:t,. \~~( 'J1h::- - AWis- ~w~ (~-n- ~ H~ -~ - ll.w Ts.. Cu~(-~) ~- c (.) ~ (~) V\) - *"... \!\=- ~.. -~v-l~)- -_ '>< f h)

5 Problem 3. Consider the input signal x (t) below. This signal is first input to an analog filter with impulse response ( ) 7r {sin (5t) sin (lot)}.. ( ) hlp t = - 2J sm 15t 5 Jrt Jrt to form x(t) = x(t) * hlp(t), and then x(t) is sampled at a rate of W 8 = 6 to form x[n], so 27r that the time between samples is T 8 = 6. The DT signal x[n] thus obtained is then input to a DT LT system with impulse response sm 4n 7r {. (7r ) } h[n] = Jrn 2cos ( 2 n) (1) Show all work. Write your expression for the output y[n] = x[n] * h[n] in the space below. Plot both the Fourier Transform of hlp(t) and the DTFT of h[n] to help solve the problem, but since the input is a sum of sinewaves, it is not necessary to plot the Fourier Transform of x (t) or the DTFT of the sampled signal x[n]. \-\ \w}. i LP -\ \o 3 - i. \ - z_ \~~ Ts 'St- - ~ \o-1::. -j\1i-\: 2 (; t- z,s:t ) -J -j e e - -e - e 'Z. e f- v' ~Eu..~""- C\. S> ± ~() \ V\. c:.(..rs<> Q. Ve'" 7.e- v t1ej CH.A :.~ w~-=- UJa.. \s lu ~s 'Z...'\t - ~ ~~ bo ~ bc - ) 4 \ ~(.:( 5.,..R_ -;\:- ~

6 Problem 3. You can continue your work for 3 here. $T'. 1t \'\ - ~ V\ s..!f"'. 41'.\ -- \.\. ')( r "'l:: -::i:: e -r ej +- e +c + z:: \._, \ j " ~ j 6 n-j 6 "' \.o'-' ~,r... 4> fi.i-1. ca - "ti L l _;\\ _1\' - - -\ \\... _A ~..._ 7 ~. :T: h - '"tr" ' tr V\. e j ~ ej~\o\. ~CY')::. +- -'re).,. 'lt ' '11" -j~i'\ -_; :;:~. 1:n- - -k... e e - e-> b 5

7 Workout Problem 4. Consider the continuous-time signal xa(t) below. Note that the multiplication by the scalar j is included to make the Fourier Transform Xa(w) be purely real-valued, and the multiplication by the scalar Ts is intended to offset the amplitude-scaling by the sampling rate Fs = A that inherently occurs in the process of sampling. x (t) = _ y _]!},_ { sin(2t} a J s 1 dt Jrt (a) A discrete-tirn~e signal is created by sampling xa(t) according to x[n] = Xa(nTs) for Ts = 2 7r. Plot the DTFT of x[n], X(w), over -Jr < w < Jr. Show your work on this 5 page and the next page, and do your plot in the space provided on the next page. 27r (b) Repeat part (a) for Ts = 25. Plot the new DTFT of x[n], X(w), over -Jr < w < 7r. Show your work and do your plot in the space provided on the sheets attached. \ w \ ( t7\) fo11 XJcw) X~( rswi... - '><s ( ~) t:> T\-::. T z.,,... -c -:.. ='> l.ns. ::: 5 / s ='""'> \""\. i)?d" ' \ ~ o...s.)v'\.~ -"") O'Y'\ ~ V... ' 2 u..\"' (.)..." ~ ve,r.:::..u +~v VV\ - T' L la.j.::::ct" Ll_) -=- 2-C tv\ (Of' T /, ht... ia.5> \ l () r ~ <. {;. (.::!.) :::. _.!_ \>s. 4o \ e> 7 8n -- s

8 Plot your answer to Problem 4 (a) here. Show work. Plot your answer to Problem 4 (b) on next page. 1--~~~~~~...--it-'""-.,-~~~--ft----+~;--+---'T... ~~,._~~~--1-2n -t 41\ 1t 2n

9 Plot of X_s (ω) analog frequency (rads/sec) aliasing starts at ω_s ω_μ = 25 2 =5 2

10 Plot of DTFT digital frequency (rads/sec) aliasing starts at 5 T_s = 5 *2 π / 25 = 2 π / 5 6

11 =-~s o -5 o 5 1 o Analog Frequency: omega (rads/sec) i i i. 1--~~~----~--~~-1-~~~~-1-~~~--~ -2n. -n 2n -.t> i. -1 i i r.. i i Digital Frequency: omega (rads/sec) 9