ECE 45 Homework 3 Solutions

1 UC San DiegoJ. ConnellyECE 45 Homework 3 SolutionsProblem the Fourier transform of the function (t) = 1 2|t| |t| 1 the statement of Problem to verify your answer. Note: the function (t) is sometimes called the unit triangle function, as it a triangular pulse with height 1, width 1, and is centered at : Recall the trig identity 1 cos(2x) = 2 sin2(x). Also, the time-reversal property can be :Letz(t)= 1 2t0 t 1/20otherwiseThen (t) =z(t) +z( t), so by the linearity and time reversal properties of the Fourier Transform,F( ) =Z( ) +Z( ).Z( ) =Z z(t)e j tdt=Z1/20(1 2t)e j tdt=j (2t 1) + 2(j )2e j t 1/2t=0=2 j 2e j /2 2 ThereforeD( ) =2 j 2e j /2 2+2+j 2ej /2 2=4 2 ej /2+e j /2 2=4 4cos( /2) 2=8sin2( /4) 2=sinc2( /4)2wheresinc (x) = corresponds to the function in Problem withA=W= 1andt0= report any typos/errors to , W,andt0be real numbers such thatA, W >0, and suppose thatg(t)is given byg(t)At0t0 W2t0+W2 Show the Fourier transform ofg(t)is equal toAW2sinc2(W /4)e j t0 Wusing the results of Problem and the properties of the Fourier : You do NOT have to re-integrate, this should only take a few :g(t) is a triangular pulse of height A, width W , and is centered at t0.

2 (t), from Problem , is atriangular pulse of height 1, width 2, and is centered at 0. T th uts g(t) is an amplitude-scaled, time-0scaled, time-shifted version of (t). In particular, g(t) = A .By the amplitude scaling, time scaling, and time shift properties:G( ) =A1|1W|D 1W e j t0=AW D(W )e j t0=AW2sinc2(W /4)e j t0 Problem the Fourier transform of the functionx(t) = 1if1 |t| 3 1if|t|< : Recall rectangle functions to reduce amount of :Recall the unit rectangle functionrect (t) = 1|t| 1 denote the Fourier transform ofrect (t)byR( ). ThenR( ) =Z rect (t)e j tdt=Z1/2 1/2e j tdt=1 j (e j /2 ej /2)=2 sin( /2) = sinc ( /2).We havex(t) = rect t6 2rect t2 ,so by the time-scaling property, we haveX( ) = 6R(6 ) 4R(2 ) = 6 sinc (3 ) 4 sinc ( ).

3 Alternatively,x(t) = rect t+22 rect t2 +rect t 22 ,so by the time-scaling and time-shift properties,we haveX( ) = 2ej2 R(2 ) 4R(2 ) + 2e j2 R(2 )= 2R(2 ) (ej2 +e j2 2)= 4R(2 ) (cos(2 ) 1) = 4 sinc ( /2) (cos(2 ) 1).It can be shown with trig manipulations that these two functions are the inverse Fourier transform of the functionF( ) =12 + 7j 2( 2 2j 1)( 2+j 6)Hint: Use Partial :By factoring each of the quadratic polynomials, we haveF( ) =(3 +j )(4 +j )(1 +j )2(2 j )(3 +j )=(4 +j )(1 +j )2(2 j )=A(1 +j )2+B1 +j +C2 j where(4 +j ) =A(2 j ) +B(1 +j )(2 j ) +C(1 +j )2. Therefore4 = 2A+ 2B+C1 = A+B+ 2C0 =B Cwhich yields,A= 1andB=C= Discussion Notes 6, for anya >0, we haveF(e atu(t)) =1a+j F(eatu( t)) =1a j F(te atu(t)) =1(a+j )2whereu(t)is theunit step by the linearity of the Fourier transform,f(t) =te tu(t) +23e tu(t) +23e2tu( t) =((t+23)e tt 023e2tt < Suppose a function f(t) has Fourier transformF ( ) = 2 j e | |.)

4 Is f(t) purely real? Is f(t) purely imaginary? Is f(t) even? Is f(t) odd? What is f(0)? Calculate f(t) and verify these : For finding the inverse Fourier transform, use this property: Let G(w) = 2 e^( | |) and F( ) = j G( ) and use the derivative property. Solution:Recall the Fourier transform is a one-to-one mapping, so y(t) = z(t) if and only if Y ( ) = Z( ). For any function x(t),x (t) =12 Z X (W)e jW tdW=12 Z X ( )ej t( d ) =12 Z X ( )ej td =F 1(X ( ))Therefore, the Fourier transform ofx (t)isX ( ).Ifx(t)is real, thenx(t) =x (t), soX( ) =X ( ).Ifx(t)is imaginary, thenx(t) = x (t), soX( ) = X ( ).x( t) =12 Z X(W)e jW tdW=12 Z X( )ej t( d ) =12 Z X( )ej td =F 1(X( ))Therefore, the Fourier transform ofx( t)isX( ).

5 If a functionx(t)is even, thenx(t) =x( t), soX( ) =X( ).If a functionx(t)is odd, thenx(t) = x(t), soX( ) = X( ).For the functionf(t), we haveF ( ) = 2 j e | | = 2 j e | |=F( )F( ) = 2 j e | |= 2 j e | |= F( )Therefore,f(t)is real and odd. For any odd function,f(0) = f( 0) = f(0) = ( ) = 2 e | |. ThenF( ) =j G( ), so by the time-derivative propertyf(t) =ddtg(t).g(t)=12 Z ]G( )ej td =Z e| |ej td =Z0 e ej td +Z 0e ej td =Z0 e (1+jt)d +Z 0e (jt 1)d =11 +jt+11 jt=(1 jt) + (1 +jt)(1 +jt)(1 jt)=21 +t2 Hencef(t) =ddt11 +t2= 2t(1 +t2)2We also havef(0) = 0,f(t) = f( t), andf(t)is real, thus verifying our (t)is the input to an LTI system with transfer functionH( ), andy(t)is theoutput of this system, wherex(t) =e |t|cos(At)andH( ) = 1 +e j +e 3j.

6 Find a real numberA >0such thaty(0) = 1. Is your answer unique?Solution:We haveY( ) =X( )H( ) =X( ) +e j X( ) +e j3 X( ).Therefore by the linearity and time-shift properties of theFourier transform,y(t) =x(t) +x(t 1) +x(t 3).Theny(0) =x(0) +x( 1) +x( 3).x(0) =e0cos(0) = 1x( 1) =e 1cos( A)x( 3) =e 3cos( 3A).IfA= (n+ 1/2), for any integern, we havex( 1) =x( 3) = 0, soy(0) =x(0) = (t)is the input to an LTI system with transfer functionH( ), andG( )is theFourier transform ofg(t). Find the output of the systemy(t).111 12555 5 5 56 68 8G( )|H( )| H( )Solution:LetX( ) = 1 | | | | 10otherwiseIn the interval[ 5,5],|H( )|= 1and H( ) = , andH( ) = ,Y( ) =H( )G( ) =X( )e j , which, by the time-shifting property of the Fourier transform,impliesy(t) =x(t 1).

7 X(t) =12 Z X( )ej td =12 Z0 1(1 + )ej td +12 Z10(1 )ej td =jt ejt+ 12 t2 jt e jt 12 t2=2 2 cos(t)2 t2=4 sin2(t/2)2 t2=sinc2(t/2)2 Thus we havey(t) =sinc2 t 12 2 Alternatively, we can use theDuality Propertyand our results from Problem Property:If the Fourier transform off(t)isF( ), then the Fourier transform ofF(t)is2 f( ).Letf(t)be a triangular pulse of height12 , width2, centered at0. ThenF( ) =12 sinc2( /2).We haveX( ) = 2 f( ), and sincefis an even function,X( ) = 2 f( ).Therefore, by the Duality Property, the Fourier transform ofF(t)isX( ), sox(t) =F(t) =12 sinc2(t/2).Problem theunit step functionu(t)is given byu(t) = 0ift <01ift we have a system for which the outputy(t)isy(t) =Zt x( )d when the input isx(t).

8 Findy(t)and its Fourier transformY( )when the input isx(t) =u(t+ 1) 2u(t 1) +u(t 3).Solution:y(t) =Zt x( )d = Rt 11d 1 t <1R1 11d Rt11d 1 t <30otherwise= 1 +t 1 t <13 t1 t <30otherwise= 2 t 14 By Problem withA= 2,W= 4, andt0= 1, we haveY( ) = 4 sinc2( )e j Problem LTI system has impulse responseh(t) =e 3tu(t). What was the inputx(t), whenthe output ise 3tu(t) e 4tu(t)?Solution:For alla >0, we haveF(e atu(t)) =Z e atu(t)e j tdt=Z 0e t(a+j )dt=1a+j ThereforeH( ) =13 +j andY( ) =13 +j 14 +j =1(3 +j )1(4 +j ).We also haveY( ) =X( )H( ) =X( )13 +j ThusX( ) =14 +j , sox(t) =u(t)e (t) =u(t)e 3t. Findy(t)whend2y(t)dt2+dy(t)dt 2y(t)=x(t).Solution:For alla>0, we haveF(e atu(t)) =1a+j andF(eatu( t)) =1a j By taking the Fourier transform of both sides of the differential equation, we have(j )2Y( ) +j Y( ) 2Y( ) =X( ) =13 +j ThereforeY( ) =X( )(j )2+j 2=1(3 +j )(j 1)(j + 2)=A3 +j +Bj 1+C2 +j Solving forA,B, andCgives usA(j 1)(2 +j ) +B(3 +j )(2 +j ) +C(3 +j )(j 1) = 1 A( 2+j 2) +B( 2+ 5j + 6) +C( 2+ 2j 3) = 1which impliesA+B+C= 0A+ 5B+ 2C= 0 2A+ 6B 3C= 1and soA=14,B=112,C= 13 Thusy(t) =14u(t)e 3t 112u( t)et 13u(t)e : Use Partial Problem 4In this problem, I am giving you a data file consisting of several amplitude modulated audio signals (similar to AM radio).

9 Let x1(t), .. , x9(t) denote the 9 audio signals. I have provided the signalr(t) =9Xk=1cos(2 fkt)xk(t)wherefk= 25000 (k 1).The signalsx1(t), .. , x9(t)can be viewed as the audio content of a radio station, thecos(2 fkt)xk(t)signals can be viewed as what the radio stations are transmitting, andr(t)can be viewed as the signalyour antenna is receiving (since by Maxwell s equations, EM waves are additive). Your goal will be to tune into each of the 9 stations and decipher the modulated audio messages. Place the files and into your MATLAB directory and load them into yourworkspace using load ; and load ; (warning, these files are quite large atroughly 27MB) is the signal9Xk=1xk(t).In other words, is what happens when radio stations DO NOT perform is akin to multiple people talking to you at once.

10 It is very difficult to pick out any one ofthe messages, since they are all communicating in the same frequency 1:try running modplay(noisy); and describe what you that you will need to use this custom function to play the audio files in this problem (asopposed to the usual sound function). We will make use of the fact thatx(t) cos( 0t) X( + 0) +X( 0)2 Each of the signalsx1(t), .. , x9(t)was multiplied by a differentcarrier frequency, and thebandwidth of each signal is smaller than the differences in the carrier frequencies, soR( ) =9Xk=1 Xk( 2 fk) +Xk( 2 fk) This allows us to spread out the signals in the frequency domain. Plot the modulated signal inthe frequency domain using: Len=length(mod); Fs= 811025; f=Fs ( Len/2 :Len/2 1)/Len; ModFreq = fft(mod); plot(f,abs(fftshift(ModFreq))); Output 2:include well-labeled plots of both the modulated (mod) and unmodulated (noisy)signals in the frequency domain, and explain the differences you notice.