M. J. Roberts - 7/12/03Solutions 2-1 Chapter 2 - Mathematical Description of SignalsSolutions1. If g t e t( ) 72 3 write out and simplify(a) g 3 79( ) e(b) g 2 7 72 2 3 7 2 ( ) ( ) +t e et t(c) g tet104 7511+j(, \,( (d) g jt e j t( ) 72 3(e)g gcosjt jte e ee tj t j t( ) + ( )

+ ( )2727 232 23(f)g gcosjt jte etjt jt j(, \,( + j(, \,(

+ ( )323227272. If g x x x ( ) +24 4 write out and simplify(a) g z z z ( ) +24 4(b) g u v u v u v u v uv u v + ( ) + ( ) + ( ) + + + +22 24 4 2 4 4 4(c) g e e e e e ejt jt jt j t jt jt( ) ( ) + + ( )2224 4 4 4 2(d) g g g t t t t t t t ( ) ( ) + ( ) + ( ) + ( ) +2 2224 4 4 4 4 4 4 4g g t t t t t ( ) ( ) + +4 3 28 20 16 4(e) g 2 4 8 4 0 ( ) + 3. What would be the numerical value of g in each of the following MATLABinstructions?(a) t = 3 ; g = sin(t) ; 0.1411(b) x = 1:5 ; g = cos(pi*x) ; [-1,1,-1,1,-1](c) f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w) ;M. J. Roberts - 7/12/03Solutions 2-20 0247 0 1550 0920 0 28910 0920 0 2890 0247 0 155. .. .. .. .++,,,,,,,]]]]]]]]jjjj4. Let two functions be defined byx, sin, sin11 20 01 20 0t tt( ) ( ) ( ) < and x, sin, sin22 02 0t t tt t( ) ( ) ( ) < .Graph the product of these two functions versus time over the time range, < < 2 2 t .t-2 2x(t)-225. For each function, g t ( ), sketch g ( ) t , ( ) g t , g t ( ) 1 , and g 2t ( ).(a) (b)tg(t)24tg(t)1-13-3tg(-t)-24tg(-t)1-13-3t-g(t)24t-g(t)1-13-3tg(t-1)3 14tg(t-1)1 23-3tg(2t)14tg(2t)13-3212-6. A function, G f ( ), is defined byM. J. Roberts - 7/12/03Solutions 2-3G rect f e fj f( ) j(, \,( 22 .Graph the magnitude and phase of G G f f ( ) + + ( ) 10 10 over the range, < < 20 20 f .G G rect rect f f e fe fj f j f ( ) + + ( ) j(, \,( + + j(, \,( ( ) + ( )10 101021022 10 2 10 f -20 20|G( f )|1f -20 20Phase of G( f )-7. Sketch the derivatives of these functions.(All sketches at end.)(a) g sinc t t ( ) ( ) ( ) ( ) ( )( ) ( ) ( )gcos sin cos sint t t ttt t tt 22 2(b) g u t e tt( ) ( ) ( )1 ( ) < ( )g,,u t e tt e ttt00 0t-4 4x(t)-11t-4 4dx/dt-11 t-1 4x(t)-11t-1 4dx/dt-11 (a) (b)8. Sketch the integral from negative infinity to time, t, of these functions which are zero forall time before time, t 0.M. J. Roberts - 7/12/03Solutions 2-4g(t)t11 2 312g(t)t11 2 3 g(t) dt g(t) dtt11 2 312t11 2 39. Find the even and odd parts of these functions.(a) g t t t ( ) + 2 3 62ge t t t t t tt ( ) + + ( ) ( ) +

+ +2 3 6 2 3 624 1222 62222go t t t t t tt ( ) + ( ) + ( )

2 3 6 2 3 6262322(b) g cos t t ( ) j(, \,( 20 404 gcos cose tt t( ) j(, \,( + j(, \,( 20 40420 4042 Using cos cos cos sin sin z z z z z z1 2 1 2 1 2+ ( ) ( ) ( ) ( ) ( )gcos cos sin sincos cos sin sine tt tt t( ) ( ) j(, \,( ( ) j(, \,(,, ]]]+ ( ) j(, \,( ( ) j(, \,(,, ]]]20 40440420 4044042 gcos cos sin sincos cos sin sine tt tt t( ) ( ) j(, \,( + ( ) j(, \,(,, ]]]+ ( ) j(, \,( ( ) j(, \,(,, ]]]20 40440420 4044042 M. J. Roberts - 7/12/03Solutions 2-5g cos cos cose t t t ( ) j(, \,( ( ) ( ) 2044020240 gcos coso tt t( ) j(, \,( j(, \,( 20 40420 4042 Using cos cos cos sin sin z z z z z z1 2 1 2 1 2+ ( ) ( ) ( ) ( ) ( )gcos cos sin sincos cos sin sino tt tt t( ) ( ) j(, \,( ( ) j(, \,(,, ]]] ( ) j(, \,( ( ) j(, \,(,, ]]]20 40440420 4044042 gcos cos sin sincos cos sin sino tt tt t( ) ( ) j(, \,( + ( ) j(, \,(,, ]]] ( ) j(, \,( ( ) j(, \,(,, ]]]20 40440420 4044042 g sin sin sino t t t ( ) j(, \,( ( ) ( ) 2044020240 (c) g t t tt( ) ++2 3 612ge tt ttt tt( ) ++ + + +2 3 612 3 6122 2ge tt t t t t tt t( ) + ( ) ( ) + + + ( ) + ( )+ ( ) ( )2 3 6 1 2 3 6 11 122 2ge t t tttt( ) + +( ) +4 12 62 16 512 2222go tt ttt tt( ) ++ + +2 3 612 3 6122 2M. J. Roberts - 7/12/03Solutions 2-6go tt t t t t tt t( ) + ( ) ( ) + + ( ) + ( )+ ( ) ( )2 3 6 1 2 3 6 11 122 2go t t t tt t tt( ) ( ) +6 4 122 12 913222(d) g sinc t t ( ) ( ) gsin sinsine ttttt tt( ) ( )+ ( )

( ) 2go t ( ) 0(e) g t t t t ( ) ( ) + ( ) 2 1 42 2 g t t t t ( ) ( ) + ( )oddeven even{12312 4 3 42 1 42 2Therefore g t ( ) is odd, g ge ot t t t t ( ) ( ) ( ) + ( ) 0 2 1 42 2and(f) g t t t t ( ) ( ) + ( ) 2 1 4ge t t t t t t t( ) ( ) + ( ) + ( ) + ( ) ( ) 2 1 4 2 1 42ge t t ( ) 72go t t t t t t t( ) ( ) + ( ) ( ) + ( ) ( ) 2 1 4 2 1 42go t t t ( ) ( ) 2 4210. Sketch the even and odd parts of these functions.M. J. Roberts - 7/12/03Solutions 2-7tg(t)11tg(t)2 11-1tg (t)11tg (t)2 11-1tg (t)11tg (t)2 11-1e eo o (a) (b)11. Sketch the indicated product or quotient, g t ( ), of these functions.t1-11-1t1 -11g(t)Multiplicationt1-11-1t1-1-11g(t)g(t) g(t)Multiplication(a) (b)t1-11-1t1 -11-1M. J. Roberts - 7/12/03Solutions 2-8t11g(t)Multiplication(c)t-11t11g(t)g(t)Multiplication(d)t11t-1-1 1g(t)t-1 11 (e) (f)t1 -11-1t1 -11g(t)Multiplication... ... t11-1t1-11g(t)Multiplication g(t)t1 -11-1... ... g(t)t11-1t11g(t)Division Division(g)t-1-1 -1 11 1g(t)g(t)(h)t1 t-1 11tg(t)t-112. Use the properties of integrals of even and odd functions to evaluate these integrals inthe quickest way.M. J. Roberts - 7/12/03Solutions 2-9(a) 2 2 2 2 411111101+ ( ) + t dt dt t dt dteven odd{ {(b)4 10 8 5 4 10 8 5 8 108101201201201201201200120cos sin cos sin cos t t dt t dt t dt t dt ( ) + ( ) [ ] ( ) + ( ) ( ) even odd1 2 4 3 4 1 2 4 3 4(c) 4 10 0120120t t dtoddevenodd{12 4 3 41 2 4 3 4cos ( ) (d) t t dt t t dt t t tdtoddoddeven{12 4 3 41 2 4 3 4sin sincos cos10 2 10 210101010110110011001100110 ( ) ( ) ( )+ ( ),,,,]]]]] t t dt toddoddeven{12 4 3 41 2 4 3 4sinsin10 21100101015011011020110 ( ) + ( )( ),,,,]]]]]

(e) e dt e dt e dt e et t t t [ ] ( ) even{1101010112 2 2 2 1 1 264 .(f) t e dttodd evenodd{ {123 11013. Find the fundamental period and fundamental frequency of each of these functions.(a) g cos t t ( ) ( ) 10 50 f T0 025125 Hz s ,(b) g cos t t ( ) +j(, \,( 10 504 f T0 025125 Hz s ,(c) g cos sin t t t ( ) ( ) + ( ) 50 15 f T0 0251522 512 50 4 j(, \,( GCD , . ,.. Hz s(d) g cos sin cos t t t t ( ) ( ) + ( ) + j(, \,( 2 3 534 M. J. Roberts - 7/12/03Solutions 2-10f T0 013252121122 j(, \,( GCD , , Hz , s14. Find the fundamental period and fundamental frequency of g t ( ).g(t)t... ...1t... ...1g(t)t... ...1+(a) (b)t... ...1g(t)t... ...1+(c)(a) f T0 0313 Hz and s(b) f T0 06 4 212 ( ) GCD , Hz and s(c) f T0 06 5 1 1 ( ) GCD , Hz and s15. Plot these DT functions.(a) x cos sin n n n[ ] j(, \,( ( ) j(, \,(421232 28 , < 24 24 nn-24 24x[n]-77(b) x n nen[ ] 35 , < 20 20 nn-20 20x[n]-66(c) x n nn [ ] j(, \,( + 2121423 , < 5 5 nM. J. Roberts - 7/12/03Solutions 2-11n-5 5x[n]-2000200016. Let x cos1528n n[ ] j(, \,( and x2682n en[ ] j(, \,(. Plot the following combinations of thosetwo signals over the DT range, < 20 20 n . If a signal has some defined and someundefined values, just plot the defined values.(a) x x x n n n [ ] [ ] [ ]1 2n-20 20x[n]-4040(b) x x x n n n [ ] [ ] + [ ] 4 21 2n-20 20x[n]-4020(c) x x x n n n [ ] [ ] [ ]1 22 3n-20 20x[n]-4020(d) xxxn nn[ ] [ ] [ ]122n-20 20x[n]-5000010000(e) x x x n n n[ ] ,, ]]] + ,, ]]]22431 2n-20 20x[n]-405M. J. Roberts - 7/12/03Solutions 2-1217. A function, g n [ ] is defined byg,,,nnn nn n[ ] < 0,Strength ( ) ( ) da a da a1 1 1and for a < 0,Strength ( ) ( ) ( ) da a da da a1 1 1 1Therefore for a > 0 and a < 0,Strength 1a and a t ta t t ( )[ ] ( )0 01 .40. Using the results of Exercise 39, show that(a) comb axa x nan( ) j(\,1From the comb definition, comb ax ax n ( ) ( ) .Then, using the property from Exercise 39, comb ax a x na a x na( ) j(, \,(,, ]]] j(, \,( 1.(b) the average value of comb ax ( ) is one, independent of the value of aThe period is 1/a. Thereforecomb comb comb axaax dx a ax dx a ax dxttaaaaa( ) j(\,( ) ( ) ( )+ 1100112121212comb ax x dxaa( ) ( ) 12121M. J. Roberts - 7/12/03Solutions 2-37(c) a comb function of the form, 1atacombj(\, is a sequence of unit impulses spaced a units apart.1 1ata a a t an t ann ncombj(\, ( ) ( ) and (d) even though ata t ( ) ( )1, comb comb axa x ( ) ( )1 ax na x nn n ( ) ( ) 11 1a x na a x nn n j(, \,( ( ) QED41. Sketch the generalized derivative of g sin rect t tt ( ) j(\, ( ) 32.Except at the discontinuities at t 12, the derivative is either zero, for t >12, or itis the derivative of 32sin t j(, \,( , 32 2 cos t j(, \,( , for t >>Better figures needed(a)comb cos cos cos t t dt t n t dt t n t dtn n( ) ( ) ( ) ( ) ( ) ( ) 48 48 48 comb cos cos t t dt nn n( ) ( ) ( ) 48 48 1 M. J. Roberts - 7/12/03Solutions 2-40(b)comb sin t t dt ( ) ( )2comb sin sin sin t t dt t n t dt nn n( ) ( ) ( ) ( ) ( ) 2 2 2 0 (c)comb recttt dtj(\, ( )240204comb rect rect recttt dt t n t dt nn nj(\, ( ) ( ) ( ) + ( ) 244 2 4 4 2 4 0020020(d)comb sinc sinc sinc t t dt t n t dt nn n( ) ( ) ( ) ( ) ( ) 22221 45. Sketch the derivatives of these functions.(a) g sin sgn t t t ( ) ( ) ( ) 2 ( ) ( ) u ,F120 , f ft t ( ) ( )22 2 , x xx u X t e t fjft( ) ( ) ( ) +211020 Ft-0.016667 0.066667x(t)8(e) cos 2120 0 0 f t f f f f ( ) ( ) + + ( ) [ ]FM. J. Roberts - 7/12/035-25x cos X t t f f f( ) ( ) ( ) ( ) + + ( ) 1363 36 Ft-1 1x(t)-0.50.5(f) x X t f f f ( ) ( ) ( ) ( )858 585F t-1 1x(t)2(g) sgn tj f( ) F1x sgn X t t fj f( ) ( ) ( ) 33Ft-1 1x(t)-3320. Sketch the inverse CTFTs of these functions. (a) e et 224F , ( ) j(, \,( 414 4, x x t tx X t e j e e et( ) ( ) ( )14222216416444 FM. J. Roberts - 7/12/035-26t-8 8x(t)0.2 (b) tri sinc t ( ) j(, \,(F 22 , ( ) j(, \,( 212 2, x x t tx tri X sinc t tj ( ) j(, \,( ( ) j(, \,(72 272 F t-4 4x(t)4(c) sin 0 0 0t j ( ) + ( ) ( ) [ ]Fx sin X t t j j ( ) ( ) ( ) + ( ) ( ) [ ] 10 10 10 Ft-0.4 0.4x(t)-11 (d) comb comb t ( ) j(, \,(F 2 , ( ) j(, \,( 818 8, x x t tx comb Xcombt tj ( ) j(, \,( ( ) j(, \,(140 845Ft-40 40x(t)0.2M. J. Roberts - 7/12/035-27(e) sgn tj( ) F2 , 1 2F ( ) x sgn X t t jj( ) ( ) + ( ) + ( )525510 Ft-1 1x(t)-418 (f) e ta j aat ( ) + > u ,F10x u X t e t jjt( ) ( ) ( ) +6633 Ft-0.2 1.5x(t)6 (g) sinc tri22t ( ) j(, \,(F , ( ) j(, \,( 16116 16, x x t tx sinc X tri t tj ( ) j(, \,( ( ) ( )54 1620 82 Ft-200 200x(t)0.521. Find the CTFTs of these signals in either the f or form, whichever is moreconvenient.M. J. Roberts - 7/12/035-28(a) x cos sin t t t ( ) ( ) + ( ) 3 10 4 10X f f f j f f ( ) j(, \,( + +j(, \,(,, ]]]+ +j(, \,( j(, \,(,, ]]]325 525 5 X f jf jf ( ) j(, \,( j(, \,( + + j(, \,( +j(, \,(3 425 3 425 X. .f e f e fj j( ) j(, \,( j(, \,( + j(, \,( +j(, \,(525 5250 927 0 927 X. .j e ej j ( ) ( ) ( ) + ( ) + ( )5 10 5 100 927 0 927(b) x comb comb t t t( ) j(, \,( j(, \,(212X comb comb comb f f f e f ej f j f( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 12 2 X comb comb sin f e f e e j e f fj f j f j f j f( ) ( ) ( ) ( ) ( ) 2 2 4 2X comb sin j j e j ( ) j(, \,( j(, \,(422t1x(t)-66(a)-20 20|X(j)|20-20 20Phase of X(j)- t-4 4x(t)-22(b)f -4 4|X( f )|2f-4 4Phase of X( f )- (c) x sinc sinc sinc t t t t ( ) ( ) j(, \,(j(, \,( +j(, \,(j(, \,(4 4 2 4142 414M. J. Roberts - 7/12/035-29X rect rect rect f f fe fej fj f( ) j(, \,( j(, \,( j(, \,(412 412 42 2 X rect rect rect rect cos f f fe e f f f j fj f( ) j(, \,( j(, \,( +,, ]]]

j(, \,( j(, \,( j(, \,(412 4 4 4 22 2 X rect rect cos j ( ) j(, \,( j(, \,( j(, \,(8 8 4(d) x u t e e tj t j t( ) +[ ] ( ) + ( ) ( )2 21 2 1 2 Using e ta j aat ( ) + > u ,F10X jj j j jjj j j j ( ) + ++ + + + ( ) + + ( )21 221 24 41 2 1 2X j jj ( ) ++ ( ) + ( )411 22 2X f j fj f( ) ++ ( ) + ( )42 12 1 22 2 Alternate Solution:x u cos u t e e t e t tj t j t t( ) +[ ] ( ) ( ) ( ) + ( ) ( ) 2 2 4 21 2 1 2 Using e t t a jj aat ( ) ( ) ++ ( ) +cos u 0 202FX j jj ( ) ++ ( ) + ( )411 22 2M. J. Roberts - 7/12/035-30t-2 2x(t)-44(c)f-4 4|X( f )|2f-4 4Phase of X( f )- t-1 3x(t)-44(d)f -3 3|X( f )|2f-3 3Phase of X( f )- (e) x t et( ) 416Using e aa aa t + ( ) >F202 2, Re ,et j(, \,( +162218116F4121161281 25616222et j(, \,( +

+F t-50 50x(t)4(e)-0.25 0.25|X(j)|128-0.25 0.25Phase of X(j)- 22. Sketch the magnitudes and phases of these functions. Sketch the inverse CTFTs ofthe functions also.(a) x u X t e e t jj jt t( ) ( ) ( ) ( ) + + 10 4103453 5 F M. J. Roberts - 7/12/035-31t-1 2x(t)6-20 20|X(j)|4-20 20Phase of X(j)- (b) x rect cos X sinc sinc t t t f f f( ) ( ) ( ) ( ) j(, \,( + +j(, \,(,, ]]]16 2 2 41212 Ft-1 1x(t)-1616f -10 10|X( f )|5f-10 10Phase of X( f )(c) x . sinc sin X tri tri t t t f j f f( ) ( ) ( ) ( ) +j(, \,( j(, \,(,, ]]]1 6 8 41028282 Ft-0.5 0.5x(t)-0.50.5f -15 15|X( f )|0.1f -15 15Phase of X( f )- (d)x cos cos X t t t f f ff f( ) ( ) + ( )[ ] ( ) + ( ) + + ( )+ ( ) + ( )2 2100 19001050 950950 1050 Forx cos cos X t t t f f ff f( ) ( ) ( ) ( ) + ( ) + + ( )+ ( ) + ( )4 100 20001050 950950 1050 FM. J. Roberts - 7/12/035-32t-0.04 0.04x(t)-44f -1200 1200|X( f )|1f -1200 1200Phase of X( f )-(e) xcoscoscosX ttttff ff ff f( ) ( )+ ( )+ ( ),,,,]]]]] ( ) + ( ) + + ( )+ + ( ) + ( )+ ( ) + ( ),,,,]]]]]2210019002 20001050 2 1000950 9502 1000 1050 Forxcos coscosX t t tt ff ff ff f( ) ( ) ( )+ ( ),, ]]] ( ) + ( ) + + ( )+ + ( ) + ( )+ ( ) + ( ),,,,]]]]]42000 10020001050 2 1000950 9502 1000 1050 Ft-0.04 0.04x(t)-88f -1000 1000|X( f )|2f -1000 1000Phase of X( f )- 23. Sketch these signals versus time. Sketch the magnitudes and phase of their CTFTs ineither the f or form, whichever is more convenient.(a) x rect comb rect comb t t t t t ( ) ( ) ( ) ( ) j(, \,( 2 212X sinc comb f ff e j f( ) j(, \,( ( ) ( )12 21 X sinc comb sin f je ff f j f( ) j(, \,( ( ) j(, \,( 22 2X sinc sin f e k kf kj kk( ) j(, \,( j(, \,( ( ) ( ) 212 2M. J. Roberts - 7/12/035-33Non-zero only for odd values of k. At those odd values, e k j k ( ) j(, \,(212sin ,always evaluates to +1. ThereforeX sinc f kf kkk( ) j(, \,( ( )20t-3 3x(t)-11f -8 8|X( f )|1f -8 8Phase of X( f )-(b) x rect comb t t t ( ) + ( ) ( ) 1 2 2X sinc comb f ff f ( ) j(, \,( ( ) ( )2 X sinc sinc f f kf k kf kk kk( ) ( ) + j(, \,( ( ) j(, \,( ( ) 2 20Same as answer in part (a).t-3 3x(t)-11f -8 8|X( f )|1f -8 8Phase of X( f )- (c) x u sin t e t tt( ) ( ) ( )42X jjj ( ) ++ ( ) 2 ( ) [ ]1142X j jj j ( ) + ( ) 2 ( )+,,,,]]]]]2142142M. J. Roberts - 7/12/035-34Rationalizing the denominator,X jj j ( ) + ( )+ ( ) ++ ( ) 2 ( )42116224211622222X jj ( ) + ( )+ ( ) 2 ( )[ ]+ ( )+ ( ) + 2 ( )[ ]411622211622222x sin cos t t t ( ) + ( )( ) + ( )( )14116222116222 2 xsin cost t t( ) ( ) ( )+4 2 32 21 642 t-2 2x(t)-0.20.2 -8 8|X(j)|1 -8 8Phase of X(j)-(d) x rect comb t e t tt( ) ( ) ( ) [ ]22X sinc comb f e fff( ) j(, \,( ( )2 12 2X sinc sinc f e ff k e kf kfkkk( ) j(, \,( ( ) j(, \,( ( ) 12 212 22 2 x sinc sinc cos t e ke e kktk j ktkkk( ) j(, \,( + j(, \,( ( )

12 212 222 221 M. J. Roberts - 7/12/035-35t-2 2x(t)1f -8 8|X( f )|1f -8 8Phase of X( f )-(e) x rect tri comb t t t t ( ) ( ) ( ) ( )[ ]2X sinc sinc comb f f ff ( ) ( ) j(, \,( ( )12 22All the comb impulses have zero weight except the one at zero.X f f ( ) ( )12x t ( ) 12t-2 2x(t)1f -8 8|X( f ) |1f -8 8Phase of X( f )-(f) x sinc . comb t t t ( ) ( ) ( ) 2 01X.rect.comb.f ff f f f ( ) j(, \,( ( ) + ( ) + ( ) + ( ) [ ]12 01 2 0112 011 1 xcos.t t( ) + ( ) 1 2 22 01M. J. Roberts - 7/12/035-36t-2 2x(t)-12f -1 1|X( f )|1f -1 1Phase of X( f )-(g) x sinc . comb t t t ( ) ( ) ( ) 1 99X.rect.comb.f ff f ( ) j(, \,( ( ) ( )11 99 1 9911 99x.. t ( ) 11 990 5025t-2 2x(t)1f -2 2|X( f ) |1f -2 2Phase of X( f )-(h) x t e et t( ) 2 2X f e e ef f f( ) 2 2 2 2 2 22x t e et t( ) j(, \,( 2 22222t-4 4x(t)2f -1 1|X( f )|4f -1 1Phase of X( f )-24. Sketch the magnitudes and phases of these functions. Sketch the inverse CTFTs ofthe functions also.M. J. Roberts - 7/12/035-37(a) X sinc f ff f ( ) j(, \,( ( ) + + ( ) [ ]1001000 1000 x rect rect cos t t e e t tj t j t( ) ( ) + ( ) ( ) ( )100 100 200 100 20002000 2000 f -1000 1000|X( f )|1f -1000 1000Phase of X( f )- t-0.01 0.01x(t)-200200(b) X sinc comb f f f ( ) ( ) ( ) 10x rect comb rect t tt nt nn( ) j(, \,( ( ) j(, \,( ( )110 10110 10 f -2 2|X( f )|1f -2 2Phase of X( f )- t-10 10x(t)0.125. Sketch these signals versus time. Sketch the magnitudes and phases of the CTFTs ofthese signals in either the f or form, whichever is more convenient. In some casesthe time sketch may be conveniently done first. In other cases it may be moreconvenient to do the time sketch after the CTFT has been found, by finding the inverseCTFT.(a) x sin t e tt( ) ( )220X f e jf f j e eff f( ) + ( ) ( ) [ ]

+ ( ) ( ) 22 2210 10210 10M. J. Roberts - 7/12/035-38t-2 2x(t)-11f -12 12|X( f )|0.5f -12 12Phase of X( f )- (b) x cos comb cos t t t n t nn( ) ( ) ( ) ( ) j(, \,(400 10011004100 X comb f f f f( ) ( ) + + ( ) [ ] j(, \,(12200 2001100 100 X comb combcomb combf f ff f( ) j(, \,( + +j(, \,(,,,,,,]]]]]]]

j(, \,( j(, \,(1200200100200100100 1001 2 44 3 44 1 2 44 3 44X comb f f( ) j(, \,(1100 100t-0.1 0.1x(t)0.01f -1000 1000|X( f )|0.01f -1000 1000Phase of X( f )- (c) x cos cos t t t ( ) + ( )[ ] ( ) 1 400 4000 X f f f f f f ( ) ( ) + ( ) + + ( ) [ ] ( ) + + ( ) [ ] 12200 200122000 2000X ff f ff f ff f f( ) ( ) ( ) + + ( )[ ]+ ( ) ( ) + + ( )[ ]+ + ( ) ( ) + + ( )[ ]122000 200012200 2000 200012200 2000 2000 M. J. Roberts - 7/12/035-39X ff ff f f f( ) ( ) + + ( )+ ( ) + + ( ) + ( ) + + ( )[ ]122000 2000122200 1800 1800 2200 t-0.01 0.01x(t)-22f -2500 2500|X( f )|0.5f -2500 2500Phase of X( f )- (d) x rect comb cos t t t t ( ) + ( ) ( )[ ] ( ) 1 100 50 50 500X sinc comb f f f ff f ( ) ( ) + j(, \,( j(, \,(,, ]]] ( ) + + ( )[ ] 1100 100 5012250 250X sinc f f kf k f fk( ) ( ) + j(, \,( ( ),, ]]] ( ) + + ( )[ ] 12 25012250 250Xsincff fkf k f kk( ) ( ) + + ( ) [ ]+ j(, \,( ( ) + + ( ) [ ],,,,,]]]]]]12250 25014 2250 50 250 50 t-0.04 0.04x(t)-0.080.08f -500 500|X( f )|1f -500 500Phase of X( f )- (e) x rect comb t tt ( ) j(, \,( ( )7X sinc comb sinc f f f f f kk( ) ( ) ( ) ( ) ( )7 7 7 7 X sinc f f kk( ) ( ) ( )7 7M. J. Roberts - 7/12/035-40t-6 6x(t)1f -4 4|X( f )|7f -4 4Phase of X( f )- 26. Sketch the magnitudes and phases of these functions. Sketch the inverse CTFTs ofthe functions also.(a) X sinc comb f ff ( ) j(, \,( ( )4x rect comb rect t t t t t nn( ) ( ) ( ) ( ) ( )4 4 4 4 x rect t t nn( ) ( ) ( )4 4t-2 2x(t)4f -16 16|X( f )|1f -16 16Phase of X( f )- M. J. Roberts - 7/12/035-41(b) X sinc sinc comb f f ff ( ) j(, \,( + +j(, \,(,, ]]] ( )1414x rect comb rect cos t t e e t t t t nj t j tn( ) ( ) +( ) ( ) ( ) ( ) ( )4 4 8 4 22 2 x rect cos t t n t nn( ) ( ) ( ) ( ) ( )8 4 2t-2 2x(t)1f -8 8|X( f )|2f -8 8Phase of X( f )- (c) X sinc sinc f f f ( ) ( ) ( ) 2x rect rect rect rect t t tt t( ) ( ) j(, \,( ( ) j(, \,(12 212 2The result of the graphical convolution can be expressed in the form,x tri tri t tt ( ) j(, \,( ( ),, ]]]143232t-2 2x(t)0.5f -2 2|X( f )|1f -2 2Phase of X( f )- 27. Sketch these signals versus time and the magnitudes and phases of their CTFTs.(a) x sincsin cos sint ddt t ddtttt t tt( ) ( )[ ]

( ) ,, ]]]

( ) ( ) 2X rect f j f f ( ) ( ) 2M. J. Roberts - 7/12/035-42t-8 8x(t)-22(a)f -1 1|X( f )|f -1 1Phase of X( f )- (b) x rect t ddtt( ) j(, \,(,, ]]]46X sincsinsin f j f f j f ff j f ( ) ( ) ( ) ( ) 2 24 6 2 24668 6 Alternate Solution:ddttf f e ej f j f464 3 3 46 6rectj(, \,(,, ]]] + ( ) ( )[ ] ( ) F4 3 3 8 6 f f j f + ( ) ( ) [ ] ( )Fsint-33x(t)-44f -0.5 0.5|X( f )|8f -0.5 0.5Phase of X( f )- (c)x tri comb rect rect comb t ddt t t t t t ( ) ( ) ( ) [ ] +j(, \,(j(, \,( j(, \,(j(, \,(,, ]]] ( )2 2 214214X sinc comb sinc f j f ff j k kf kk( ) j(, \,( ( ) j(, \,( ( )212 2 22 2 M. J. Roberts - 7/12/035-43t-2 2x(t)-22f -8 8|X( f )|2f -8 8Phase of X( f )- 28. Sketch these signals versus time and the magnitudes and phases of their CTFTs.(a) x sin t dt( ) ( )2 X sin fj fjf f t f f fff( ) + ( ) ( )[ ]+ ( ) ( ) [ ] ( ) + ( ) ( )

12 21 11221 140 FX f f ff f ( ) + ( ) ( ) + ( ) + ( ) [ ] 141412121 1t-2 2x(t)-0.20.2f -1 1|X( f )|0.1f -11Phase of X( f )- (b) x rect,,,t dtt ttt( ) ( ) < + 0121212112Xsincrectsincf fj f t f fj f ff( ) ( )+ ( ) ( ) [ ] ( ) ( )+ ( )

212 2120 Ft-1 1x(t)1f -2 2|X( f )|1f -22Phase of X( f )- M. J. Roberts - 7/12/035-44 (c) x sinc t dt( ) ( )3 2 Let u 2. Then x sincsint udu uu dut t( ) j(, \,( ( ) 32322 2 For t 0:xsin sin sin sint uu du uu du uu du uu dutt( ) ( ) ( ),, ]]] ( )+ ( ),, ]]] 32320200 02 xsin sinSi SiSiSit d uu du t ttt( ) ( )+ ( ),,,,,,]]]]]]] + ( ),, ]]] ( ),, ]]] ( ) ( ) 3232 2232 2202022 1 2 4 3 4 1 2 4 3 4For t 0:xsin sin sin sint uu du uu du uu du uu dut t( ) ( )+ ( ),, ]]] ( )+ ( ),, ]]] 32320020 02 x Si t t ( ) + ( ),, ]]]32 22Therefore, for any t, x Si t t ( ) + ( ),, ]]]32 22 .X rect sinc rect fj fft fj ffff( ) j(, \,( + ( ) [ ] ( ) j(, \,( + ( )

1232 2123 234 2340 Ft-4 4x(t)-12f -2 2|X( f )|1f -2 2Phase of X( f )- 29. From the definition, find the DTFT ofx rect n n [ ] [ ] 104 .M. J. Roberts - 7/12/035-45and compare with the Fourier transform table in Appendix E.X x rect F n e n e ej Fnnj Fnnj Fnn( ) [ ] [ ] 242 24410 10 X F e e e e eej F mmj F j Fmmj Fj Fj F( ) ( )

10 10 10112 4088 2088182 Xsinsindrcl , F e eee ee eFF Fj Fj Fj Fj F j Fj F j F( ) ( )( ) ( ) 10 10990 989 9 9 From the table,rect drcl ,N w wwn N F N [ ] + ( ) + ( )F2 1 2 1rect drcl ,49 9 n F [ ] ( )F10 90 94rect drcl , n F [ ] ( )FCheck30. From the definition, derive a general expression for the F and forms of the DTFT offunctions of the form,x sin sin n A F n A n [ ] ( ) ( ) 20 0 .(It should remind you of the CTFT of x sin sin t A f t A t ( ) ( ) ( ) 20 0 .) Compare with theFourier transform table in Appendix E.X x sin F n e A F n e A e ej ej Fnnj Fnnj F n j F nj Fnn( ) [ ] ( ) 2022 22220 0 X F Aj e ej F F n j F F nn( ) [ ] ( ) + ( )22 20 0 Then, usinge xj xnn2 ( ) combM. J. Roberts - 7/12/035-46we getX comb comb comb comb F Aj F F F F A jF F F F ( ) ( ) ( ) [ ] ( ) + ( ) [ ]2 20 0 0 0X comb comb F A jF F F F ( ) + ( ) ( ) [ ]20 0The form can be found by the transformation, F 2.X comb comb j A j ( ) +j(, \,( j(, \,(,, ]]]2 2 2 2 20 0 X comb comb j j A ( ) + ( ) ( ) [ ]0 031. A DT signal is defined byx sinc n n[ ] j(, \,(8 .Sketch the magnitude and phase of the DTFT of x n [ ] 2 .From the table of transform pairs,sinc combnw w wF Fj(, \,( ( ) ( )Frectsinc combnF F88j(, \,( ( ) ( )F8rectsinc combnF F e j F j(, \,( ( ) ( ) [ ] 2884 F8rect sinc ne F kj Fkj(, \,( ( ) ( )2884 F8 rectM. J. Roberts - 7/12/035-47n-32 32x[n]1 F -1 1|X( F )|8F -1 1Phase of X( F )|- 32. A DT signal is defined byx sin n n[ ] j(, \,(6 .Sketch the magnitude and phase of the DTFT of x n [ ] 3 and x n + [ ] 12 .From the table,sin comb comb 220 0 0F n jF F F F ( ) + ( ) ( ) [ ]Fsin comb combn jF F6 2112112j(, \,( +j(, \,( j(, \,(,, ]]]Fsin comb comb n jF F e j F ( ) j(, \,( +j(, \,( j(, \,(,, ]]] 36 21121126 Fsin n je F k e F kj F j Fk ( ) j(, \,( + j(, \,( j(, \,(,, ]]] 36 21121126 6 Fsin n je F k e F kj kjj kj k ( ) j(, \,( + j(, \,( j(, \,(,,,]]]] j(, \,(

+j(, \,( 36 211211261126112F1 2 4 3 4 1 2 4 3 4sin nF k F kk ( ) j(, \,( + j(, \,( + j(, \,(,, ]]]3612112112Fsin comb comb nF F ( ) j(, \,( +j(, \,( + j(, \,(,, ]]]36121212FSimilarly,sin n je F k e F kj k j kk+ ( ) j(, \,( + j(, \,( j(, \,(,,,]]]]+ j(, \,( + +j(, \,(126 21121122411224112FM. J. Roberts - 7/12/035-48sin n je F k e F kj k j kk+ ( ) j(, \,( + j(, \,( j(, \,(,, ]]]+ ( ) + + ( )126 211211224 2 24 2 Fsin n jF k F kk+ ( ) j(, \,( + j(, \,( j(, \,(,, ]]]126 2112112Fsin comb comb n jF F+ ( ) j(, \,( +j(, \,( j(, \,(,, ]]]126 21212FThis is the same as the transform of the unshifted function because a shift of 12 is a shiftover exactly one period.n-12 12x[n]-11F -1 1|X( F )|0.5F -1 1Phase of X( F )|- Phase of X( F )|n-12 12x[n]-11 F -1 1|X( F )|0.5F -1 1- 33. The DTFT of a DT signal is defined byX rect rect comb j ( ) j(, \,(j(, \,( + +j(, \,(j(, \,(,, ]]] j(, \,( 42222 2 .Sketch x n [ ].From the table,sinc combnw w w j(, \,( j(, \,( j(, \,(Frect 2 2 sinc combn422j(, \,( j(, \,( j(, \,(F4rect M. J. Roberts - 7/12/035-49e n j n2422 2sinc combj(, \,( j(, \,(j(, \,( j(, \,(F4rect e n j n j(, \,( +j(, \,(j(, \,( j(, \,(2422 2sinc combF4rect sinc combne ej n j n42222 22 2j(, \,( +j(, \,( j(, \,(j(, \,( + +j(, \,(j(, \,(,, ]]] j(, \,( F4 rect rect 24 22222 2sinc cos combnnj(, \,( j(, \,( j(, \,(j(, \,( + +j(, \,(j(, \,(,, ]]] j(, \,(F4 rect rect Thereforex sinc cos n nn [ ] j(, \,( j(, \,( 24 2n-16 16x[n]-2234. Sketch the magnitude and phase of the DTFT ofx rect cos n n n[ ] [ ] j(, \,(426 .Then sketch x n [ ] .From the table,rect drcl ,N w wwn N F N [ ] + ( ) + ( )F2 1 2 1M. J. Roberts - 7/12/035-50rectsinsinNwwn F NF[ ] + ( ) ( )( )F 2 1andcos comb comb 2120 0 0F n F F F F ( ) ( ) + + ( ) [ ]FXsinsincomb comb F FF F F ( ) ( )( ) j(, \,( + +j(, \,(,, ]]]9 121616The function, sinsin9FF( )( ) is periodic with period, one. To prove that, let k be any integer.Thensinsinsinsinsin cos sin cossin cos sin cos9 9 9 9 9 9 9 F kF kF kF kF k k FF k k F+ ( ) ( )+ ( ) ( ) + ( )+ ( ) ( ) ( ) + ( ) ( )( ) ( ) + ( ) ( )sinsinsinsinsinsin,9 9 9 F kF kFFFF k+ ( ) ( )+ ( ) ( ) ( ) ( ) ( )( )oddsinsinsinsin,9 9 F kF kFF k+ ( ) ( )+ ( ) ( ) ( )( )evenXsinsinF FF F k F kk( ) ( )( ) j(, \,( + + j(, \,(,, ]]]129 1616 XsinsinsinsinFkkF kkkF k ( ) +j(, \,(j(, \,(+j(, \,(j(, \,( j(, \,( +j(, \,(j(, \,(j(, \,(j(, \,(+ j(, \,(,,,,,]]]1291616169161616 ]]]]kSince sinsin9FF( )( ) is periodic with period, one, so are sinsin91616kkj(, \,(j(, \,(j(, \,(j(, \,( and sinsin91616kk+j(, \,(j(, \,(+j(, \,(j(, \,(.Therefore, for any k,sinsinsinsin916163261122kkj(, \,(j(, \,(j(, \,(j(, \,(

j(, \,(j(, \,(

and sinsinsinsin916163261122kk+j(, \,(j(, \,(+j(, \,(j(, \,(

j(, \,(j(, \,(

M. J. Roberts - 7/12/035-51and thereforeX comb comb comb comb F F F F F ( ) j(, \,( +j(, \,( j(, \,( + +j(, \,(122162161616 .Then, usingcos comb comb 2120 0 0F n F F F F ( ) ( ) + + ( ) [ ]F j(, \,( j(, \,( + +j(, \,(,, ]]]2261616cos comb combnF FFand, therefore,x cos n n[ ] j(, \,( 226n-12 12x[n]-22 F -1 1|X( F )|1F -1 1Phase of X( F )|- 35. Sketch the inverse DTFT ofX rect comb comb F F F F ( ) ( ) ( )[ ] ( ) 4 2 .From the table,sinc combnw w wF Fj(, \,( ( ) ( )Frect and comb combN n N F00[ ] ( )FTherefore using multiplication-convolution duality,x sinc comb n nn [ ] j(, \,( [ ]14 42 .M. J. Roberts - 7/12/035-52n-16 16x[n]-0.10.2536. Using the differencing property of the DTFT and the transform pair,tri cosnF21 2j(, \,( + ( )F ,find the DTFT of 121 1 2 n n n n + [ ] + [ ] [ ] ( ) ( ) . Compare it with Fouriertransform found using the table in Appendix E.The first backward difference of tri n2j(, \,( is 121 1 2 n n n n + [ ] + [ ] [ ] ( ) ( ) .Applying the differencing property,tri tri cosn ne Fj F2121 1 22j(, \,( j(, \,( ( ) + ( ) ( ) F 121 1 2 1 1 22 n n n n e Fj F+ [ ] + [ ] [ ] [ ] ( ) ( ) + ( ) ( ) Fcos121 1 2 12 222 222 2 n n n n e e ee e ej Fj F j Fj Fj F j F+ [ ] + [ ] [ ] [ ] ( ) + + +F121 1 2 12 212 222 2 4 n n n n e e e ej Fj F j F j F+ [ ] + [ ] [ ] [ ] ( ) + + F121 1 21212 2 4 n n n n e e ej F j F j F+ [ ] + [ ] [ ] [ ] ( ) + ( ) FOther route to the DTFT:121 1 21212 2 4 n n n n e e ej F j F j F+ [ ] + [ ] [ ] [ ] ( ) + ( ) FCheck.37. Using Parsevals theorem, find the signal energy ofM. J. Roberts - 7/12/035-53x sinc sin n n n[ ] j(, \,( j(, \,(1024 .E n F dFxn [ ] ( ) x X2 21From the table,sinc combnw w wF Fj(, \,( ( ) ( )Frectandsin comb comb 220 0 0F n jF F F F ( ) + ( ) ( ) [ ]FUsing the multiplication-convolution duality of the DTFT,sinc sin comb comb combn nF F jF F10241021414j(, \,( j(, \,( ( ) ( ) +j(, \,( j(, \,(,, ]]] F10rectPeriodic convolution is the same as aperiodic convolution with one period.sinc sin combn nj F F F F1024101414j(, \,( j(, \,( ( ) ( ) +j(, \,( j(, \,(,, ]]] F5rectsinc sin comb combn nj F F F1024101414j(, \,( j(, \,( ( ) +j(, \,( j(, \,(,, ]]] F5rectsinc sin comb combn nj F F F F102410141014j(, \,( j(, \,( ( ) +j(, \,( ( ) j(, \,(,, ]]] F5 rect rectE j F F F F dFx ( ) +j(, \,( ( ) j(, \,(,, ]]] 5 rect rect 1014101421comb combSince we are integrating only over a range of one, only one impulse in each comb issignificant.E F F F F dFx ( ) +j(, \,( ( ) j(, \,(,, ]]] 25 1014101421rect rect M. J. Roberts - 7/12/035-54E F F dFx +j(, \,(j(, \,( j(, \,(j(, \,(,, ]]]25 1014101421rect rectThe square of the sum equals the sum of the squares because there is no cross product; thetwo rectangles do not overlap.E F dF F dFx +j(, \,(j(, \,( + j(, \,(j(, \,( 25 101410141 1rect rectE dF dFx + +j(, \,( ++ 25 2511011051412014120141201412038. Sketch the magnitude and phase of the CTFT ofx rect1 t t ( ) ( )and of the CTFS ofx rect comb218 8t t t( ) ( ) j(, \,( .For comparison purposes, sketch X1 f ( ) versus f and T k0 2X [ ] versus kf0 on the same setof axes. ( T0 is the period of x2 t ( ) and Tf001 .)X sinc1 f f ( ) ( )X sinc218 8k k[ ] j(, \,(M. J. Roberts - 7/12/035-55f -4 4|X ( f )|1f -4 4Phase of X ( f )- kf0 -4 4|T0X [ k]|1kf0 -44Phase of T0X [ k]|- 112239. Sketch the magnitude and phase of the CTFT ofx cos14 4 t t ( ) ( ) and of the DTFT ofx x2 1n nTs[ ] ( )where Ts 116. For comparison purposes sketch X1 f ( ) and T T fs sX2( ) versus f on thesame set of axes.X12 2 2 f f f ( ) ( ) + + ( ) [ ] x cos cos24 4 44n nT ns[ ] ( ) j(, \,( X comb comb221818F F F ( ) j(, \,( + +j(, \,(,, ]]]X221818F F k F kk( ) j(, \,( + + j(, \,(,, ]]] X221618 1618T f fk fksk( ) j(, \,( + + j(, \,(,, ]]] X232 2 16 2 16 T f f k f ksk( ) ( ) + + ( )[ ] T T f f k f ks skX22 2 16 2 16 ( ) ( ) + + ( )[ ] M. J. Roberts - 7/12/035-56f -16 16|X1( f )|2 f -16 16Phase of X1( f )- Tsf -16 16|TsX2(Tsf )|2 Tsf -16 16Phase of TsX2(Tsf )- ... ...... ...40. Sketch the magnitude and phase of the DTFT ofxsinc1164nn[ ] j(, \,(and of the DTFS ofxsinccomb2 32164nnn [ ] j(, \,( [ ] .For comparison purposes sketch X1 F ( ) versus F and N k0 2X [ ] versus kF0 on the same setof axes.From the table,sinc combnw w wF Fj(, \,( ( ) ( )FrectsinccombnF F16416j(, \,( ( ) ( )F4rectM. J. Roberts - 7/12/035-57thereforeX F F F F qq1= 4rect = 4 rect ( ) ( ) ( ) ( ) ( )16 16 comb .The fundamental frequency of the periodic signal is the reciprocal of its period,FN001 132 .Using the results of the analysis of periodic extensions of aperiodic DT signals,X X201 01kN kF [ ] ( )X X2 1132 3218163218 216 k k kq kqq q[ ] j(, \,( j(, \,(j(, \,( j(, \,( rect rectF -1 1|X1(F )|4F -1 1Phase of X1(F )- kF0 -1 1N0|X2[k]|4kF0 -32 32Phase of X2[k]- M. J. Roberts - 7/12/035-5841. A system is excited by a signal,x rect t t( ) j(, \,( 42and its response isy u u t e t e tt t( ) ( ) + ( ) ( ) ( )[ ] + ( ) ( )10 1 1 1 11 1 .What is its impulse response?Y jj j ej j ej j ( ) + ( ) +j(, \,( + ( ) +j(, \,(,, ]]]101 111 11Y sin jj j e e jj jj j ( ) +j(, \,( ( ) +j(, \,( ( )101 11201 11Hsinsincsinsinjjj jj j jjj j ( ) +j(, \,( ( )j(, \,(

+j(, \,( ( )( ) +j(, \,(201 118521 11 521 11H j jj j j ( ) + ( ) +521521h u t e tt( ) ( )5242. Sketch the magnitudes and phases of the CTFTs of the following functions.(a) g t t ( ) ( ) 5 4 5 454 t ( ) F(b) g comb comb t t t( ) + j(, \,( j(, \,(1434y u u t e t e tt t( ) ( ) + ( ) ( ) ( ) + ( ) ( )4 1 1 4 1 11 1M. J. Roberts - 7/12/035-59comb comb comb combt tf e f ej f j f+ j(, \,( j(, \,( ( ) ( ) 14344 4 4 42 6 F comb comb combt tf e e ej f j f j f+ j(, \,( j(, \,( ( ) ( ) 14344 42 4 4 F comb comb sin combt tj e f fj f+ j(, \,( j(, \,( ( ) ( ) 14348 4 4 02 F Also, g comb comb t t t( ) + j(, \,( j(, \,( 14340f -1 1|G( f )|2(a)f -1 1Phase of G( f )- f -1 1|G( f )|2(b)f -1 1Phase of G( f )- (c) g u u t t t ( ) ( ) + ( ) 2 1u u 2 1121221212122t tj f fj f e fj f( ) + ( ) j(, \,(+ ( ) + + ( ) F u u 2 1112t tj f e fj f( ) + ( ) + ( ) + ( ) F (d) g sgn sgn t t t ( ) ( ) ( )sgn sgn t tj f j f j f( ) ( ) ( ) F1 111 2 M. J. Roberts - 7/12/035-60f -2 2|G( f )|4(c)f -2 2Phase of G( f )- f -2 2|G( f )|8(d)f -1 1Phase of G( f )- (e) g rect rect t t t( ) + j(, \,( + j(, \,(1212rect rect sinc sinct tf e f ej f j f+ j(, \,( + j(, \,( ( ) + ( ) 12122 2 2 22 2 F rect rect sinc cost tf f+ j(, \,( + j(, \,( ( ) ( )12124 2 2F(f) g rect t t( ) j(, \,(4rect sinctf44 4j(, \,( ( )Ff -1 1|G( f )|4(e)f -1 1Phase of G( f )- f -1 1|G( f )|4(f)f -1 1Phase of G( f )- M. J. Roberts - 7/12/035-61(g) g tri tri t t t( ) j(, \,( j(, \,( 5522552225 5 4 22 2tri tri sinc sinct tf fj(, \,( j(, \,( ( ) ( )F(h) g rect rect t t t( ) j(, \,( j(, \,(32 8 232 8 224 8 2 rect rect sinc sinct tf fj(, \,( j(, \,( ( ) ( )Ff -0.4 0.4|G( f )|25(g)f -0.4 0.4Phase of G( f )- f -0.5 0.5|G( f )|24(h)f -0.5 0.5Phase of G( f )- 43. Sketch the magnitudes and phases of the CTFTs of the following functions.(a) rect 4t ( )rect sinc 414 4t f( ) j(, \,(Frect sinc 414 8t ( ) j(, \,(F M. J. Roberts - 7/12/035-62t11818-rect(4t) f sinc f 414( )| |sinc f 414( ) 4 -4 f 144 -4(b) rect( ) 4 4 t t ( ) rect( ) sinc sinc 4 414 444t t f f ( ) j(, \,( j(, \,( Frect( ) sinc 4 48t t ( ) j(, \,( Ft41818-rect(4t)4(t) f sinc f 4( )| |sinc f 4( ) 4 -41 f 4 -4(c) rect( ) 4 4 2 t t ( ) rect( ) sinc sinc 4 4 214 4444 4t t fe fej f j f ( ) j(, \,( j(, \,( Frect( ) sinc 4 4 282t t e j ( ) j(, \,( FM. J. Roberts - 7/12/035-63t4rect(4t)4(t-2)2182-182+ f 4 -4 f 4-41sinc f 4( )| |e-j4fsinc f 4( ) e-j4f16(d) rect 4 4 2 t t ( ) ( ) rect sinc sinc 4 4 214 4212 4t t f f( ) ( ) j(, \,( j(, \,( Frect sinc 4 4 212 8t t ( ) ( ) j(, \,( Ft21818-rect(4t)4(2t) f sinc f 4( )| |4 -4 f 4 -41212sinc f 4( ) 12(e) rect comb 4t t ( ) ( )rect comb sinc comb sinc 414 414 4t t ff ff kk( ) ( ) j(, \,( ( ) j(, \,( ( )Frect comb sinc 414 4t t kf kk( ) ( ) j(, \,( ( )Frect comb sinc sinc 414 4 2 2 42 t t kk kkk k( ) ( ) j(, \,( j(, \,( j(, \,( ( ) F M. J. Roberts - 7/12/035-64t11818-rect(4t)comb(t)1 -1 -2... ... f 4 -4 f 144 -4sinc f 414( )| |comb( f )sinc f 414( ) comb( f )(f) rect comb 4 1 t t ( ) ( )rect comb sinc comb sinc 4 114 414 42 2t t ff e fe f kj f j fk( ) ( ) j(, \,( ( ) j(, \,( ( ) F rect comb sinc sinc 4 114 414 421t t ke f k kf kj kk k( ) ( ) j(, \,( ( ) j(, \,( ( ) F 123rect comb sinc 42 42 t t kkk( ) ( ) j(, \,( ( )F Same as part (e).t11818-rect(4t)comb(t-1)1 -1 -2... ... f 4 -4 f 144 -4sinc f 414( )| |comb( f )sinc f 414( ) comb( f )(g) rect comb 4 2 t t ( ) ( )rect comb sinc comb sinc 4 214 412 218 4 2t t f f f fkk( ) ( ) j(, \,( j(, \,( j(, \,( j(, \,(Frect comb sinc sinc 4 214 4214 22 t t ff k kf kk k( ) ( ) j(, \,( ( ) j(, \,( ( ) F M. J. Roberts - 7/12/035-65rect comb sinc sinc 4 214 2 222 24 t t kk kkk k( ) ( ) j(, \,( j(, \,( j(, \,( ( ) F t211818-rect(4t)comb(2t)1 -1 -2... ... f 4 -4 f 144 -4sinc f 418( )| |f 2( )combsinc f 418( ) f 2( )comb(h) rect comb t t ( ) ( ) 2rect comb sinc comb sinc t t f ff fkk( ) ( ) ( ) j(, \,( ( ) j(, \,(212 212 2Frect comb sinc sinc t t f f k k f k fk k( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2F rect comb t t ( ) ( ) j(, \,( ( ) 222F trect(t)comb(2t)1 -1 -2... ...1 f f 1|( f )|( f )44. Plot these signals over two periods centered at t 0.(a) x cos sin cos sin t t t t t ( ) ( ) + ( ) + ( ) ( ) 2 20 4 10 3 20 3 10 (b) x cos sin t t t ( ) ( ) + ( ) 5 20 7 10 M. J. Roberts - 7/12/035-66t1x(t)-1010(a)t1x(t)-1010(b)Compare the results of parts (a) and (b).M. J. Roberts - 7/12/035-6745. A periodic signal has a period of four seconds.(a) What is the lowest positive frequency at which its CTFT could be non-zero?14Hz(b) What is the next-lowest positive frequency at which its CTFT could benon-zero?12Hz46. Sketch the magnitude and phase of the CTFT of each of the following signals (form):(a) x . rect t t( ) j(, \,( 0 140X sinc f f ( ) ( ) 4 40X sinc j ( ) j(, \,( 420tx(t)0.120 20 4sinc(40 f )| | -440 402 22 240 404sinc( )20(b) x rect t t( ) + j(, \,( 3510X sinc f f e j f( ) ( ) +30 1010M. J. Roberts - 7/12/035-68X sinc j e j ( ) j(, \,( +3055tx(t)310 3030sinc( )| |ej5 10 102 22530sinc( ) ej5 5(c) x comb t t( ) j(, \,(75 5X comb f f ( ) ( ) 7 5X comb j k kk k ( ) j(, \,( j(, \,( j(, \,( 75275214525tx(t)7... ...5 5 10 10 |X(j)|... ...1455 5 54 2 2 44 2 2 45 X(j)... ...5 5 5 5(d) x comb comb t t t( ) j(, \,( + j(, \,(75257535X comb comb f f e f ej f j f( ) ( ) ( ) +7 5 7 54 6 X comb j e k e kej jkj kk ( ) j(, \,( j(, \,( j(, \,( 752752145252 245M. J. Roberts - 7/12/035-69X comb j e k e kej jkj kk ( ) j(, \,( j(, \,( j(, \,(+ ++ 752752145253 365tx(t)7... ...2 3 7 8 13 |X(j)|... ...5 5 54 2 2 45 X(j)65 45 ............14547. Sketch the inverse CTFTs of the following functions:(a) X rect f f( ) j(, \,( 208x sinc t t ( ) ( ) 160 820-4 4-4 4f |X( f )|f X( f )t160sinc(8t)1 2 3 4 51608 8 8 8 858 48 38 28 18(b) X rect f fej f( ) j(, \,( 2088x sinc t t ( ) +j(, \,(j(, \,(160 811620-4 4-4 4f |X( f )|f X( f )22t1 2 3 4 51608 8 8 8 858 48 38 28 18116160sinc[8(t + )]116M. J. Roberts - 7/12/035-70(c) X f f f ej f( ) + ( ) + ( ) [ ]2 5 55 X f f e f e f fj j( ) + ( ) + ( ) [ ] + ( ) + ( ) [ ]2 5 5 2 5 5 x cos t t ( ) ( ) 4 10f | X( f ) |5 5f X( f )5 52-t-4cos(10t)1 2 3 45 5 5 545 35 25 154(d) X f f f f ( ) ( ) + ( ) + + ( ) 8 5 5 5 5 x cos t t ( ) + ( ) 8 10 10f | X( f ) |5 5f X( f )5 558t8+10cos(10t)1 25 525 1518-248. Find the inverse CTFT of this real, frequency-domain function (Figure E48) andsketch it. (Let A 1, f195 kHz and f2105 kHz.)X rect rect f A f fff ff( ) j(, \,( + + j(, \,(,, ]]]0 0 x sinc sinc t A f f t e f f t ej f t j f t( ) ( ) + ( ) [ ]+ 2 20 0 M. J. Roberts - 7/12/035-71x sinc sinc cos t A f f t e e A f f t f tj f t j f t( ) ( ) +[ ] ( ) ( )+ 2 200 02 2 x , sinc , cos t t t ( ) ( ) ( ) 20 000 10 000 2 105f1-f1 f2-f2X( f )f Figure E48 A real frequency-domain function-5 -4 -3 -2 -1 0 1 2 3 4 5x 10-4-2-1012x 104x(t)t49. Find the CTFT (either form) of this signal (Figure E49) and sketch its magnitude andphase versus frequency on separate graphs. (Let A B 1 and let t11 and t22 .)Hint: Express this signal as the sum of two functions and use the linearity property.x rect rect t A tt A B tt( ) j(, \,( ( ) j(, \,(2 22 1X sinc sinc f t A t f t A B t f ( ) ( ) ( ) ( ) 2 2 2 22 2 1 1X sinc sinc f f f ( ) ( ) ( ) 4 4 4 2t x(t)ABt 1- t 1- t 2 t 2Figure E49 A CT functionM. J. Roberts - 7/12/035-72f -4 4|X( f )|3f -4 4Phase of X( f )- 50. In many communication systems a device called a mixer is used. In its simplestform a mixer is simply an analog multiplier. That is, its response signal, y(t), is theproduct of its two excitation signals. If the two excitation signals arex sinc110 20 t t ( ) ( ) and x cos25 2000 t t ( ) ( ) plot the magnitude of the CTFT of y(t), Y f ( ), and compare it to the magnitude of theCTFT of x1 t ( ). In simple terms what does a mixer do?y x x sinc cos t t t t t ( ) ( ) ( ) ( ) ( )1 250 20 2000Y X X rect f f f ff f ( ) ( ) ( ) j(, \,( ( ) + + ( ) [ ]1 254 201000 1000 Y rect rect f f f( ) j(, \,( + +j(, \,(,, ]]]5410002010002010 -1012f|X ( f )|154f |Y( f )|990 -990 1010 -1010M. J. Roberts - 7/12/035-7351. Sketch a graph of the convolution of the two functions in each case:(a) rect rect t t ( ) ( )t11 1rect(t) rect(t)*(b) rect rect t t j(, \,( +j(, \,(1212t11 1rect(t- ) rect(t+ )*1212(c) tri tri t t ( ) ( ) 1ttri(t)tri(t-1)12 1 -1tri(t-) tri(-1)12 1 t t+1 t-1 -1For t < -1, the non-zero portions of the two functions do not overlap and the convolution is zero.For t > 3, the non-zero portions of the two functions do not overlap and the convolution is zero.For -1 < t < 0:The non-zero portions overlap for 0 < < t+1 and, in that range of ,M. J. Roberts - 7/12/035-74tri t t ( ) + 1 and tri ( ) 1Therefore, for -1 < t < 0,tri tri t t t d t dt t( ) ( ) + ( ) + ( ) [ ]+ + 1 1 101201 tri tri t t t t t tt( ) ( ) + ( ) ,, ]]] + ( ) + ( )

+ ( )+1 12 31213162 3013 3 3 For 0 < t < 1:tri(t-) tri(-1)12 1 t t+1 t-1 -1tri tri tri tri t t t d ( ) ( ) ( ) ( )1 The non-zero portions overlap for 0 < < t+2 and, in that range of , there are three cases toconsider, 0 < < t, t < < 1 and 1 < < t+1. Thereforetri tri tri tri tri tri tri tri t t t d t d t dttt( ) ( ) ( ) ( ) + ( ) ( ) + ( ) ( ) +10111 Case 1:0 < < ttri and tri t t ( ) + ( ) 1 1Case 2:t < < 1tri and tri t t ( ) + ( ) 1 1Case 3:1 < < t+1tri and tri t t ( ) + ( ) 1 1 2ThereforeM. J. Roberts - 7/12/035-75tri tri t t t d t d t dttt( ) ( ) + ( ) + + ( ) + + ( ) ( ) +1 1 1 1 20111 tri tri t t t d t d t t dttt( ) ( ) ( ) +[ ] + + ( ) [ ] + + ( ) + ( ) +[ ] +1 1 1 2 1 2 12021211 tri tri t t t t t tttt( ) ( ) ( ) +,, ]]] + + ( ) ,, ]]] + + ( ) + ( ) +,, ]]]+1 12 312 32 1 12 32 302 3122 311 tri tri t t t t tt t t tt t t t tt t( ) ( ) ( ) +,, ]]]+ + ( ) + ( ) +,, ]]]+ + ( ) + ( ) + ( ) + ( )+ + ( ) + ( ) + + + ( ) ,1 12 31121312 32 1 1 112132 1 1 112132 3 2 32 22 3, ]]]tri tri t t t t tt tt t( ) ( ) ( ) + + ( ) +j(, \,( + + + ( ) + ( )1 12231 12131162 3 223tri tri t t t t t( ) ( ) + + + 12 2 2163 2For the remaining regions of t, the convolution simply repeats with even symmetry about thepoint, t = 1. The analytical solutions can be found by the following successive changes ofvariable:t t t t t t + 1 1 , ,These three successive changes of variable can be condensed into one,t t + 2Then, for 1 < t < 2,tri tri t t t t t t t tt t( ) ( ) + + +,, ]]] ( )+ ( )+ ( )+,, ]]] +12 2 216222222163 223 2and, for 2 < t < 3,M. J. Roberts - 7/12/035-76tri tri t t t tt t( ) ( ) + ( ),, ]]] + ( ) +[ ] +1162 16323-5 50.8tri(t) tri(t -1)t*(d) 3 10 t t ( ) ( ) cos t3(t) 10cos(t)302(e) 10comb rect t t ( ) ( ) The constant, 10.(f) 5comb tri t t ( ) ( ) The constant, 10.52. In electronics, one of the first circuits studied is the rectifier. There are two forms, thehalf-wave rectifier and the full-wave rectifier. The half-wave rectifier cuts off half ofan excitation sinusoid and leaves the other half intact. The full-wave rectifier reversesthe polarity of half of the excitation sinusoid and leaves the other half intact. Let theexcitation sinusoid be a typical household voltage, 120 Vrms at 60 Hz, and let bothtypes of rectifiers alter the negative half of the sinusoid while leaving the positive halfunchanged. Find and plot the magnitudes of the CTFTs of the responses of bothtypes of rectifiers (either form).Half-Wave Case:x cos rect comb t t t t ( ) ( ) ( ) ( ) [ ] 120 2 120 120 60 60 X sinc comb f f f f f( ) ( ) + + ( )[ ] j(, \,( j(, \,(,, ]]]60 2 60 601120 120 60 X sinc comb f f f f f( ) ( ) + + ( )[ ] j(, \,( j(, \,(,, ]]]2260 60120 60 M. J. Roberts - 7/12/035-77X sinc f f f k fkk( ) ( ) + + ( )[ ] j(, \,( j(, \,(,, ]]]2260 602 60 X sinc f kf f f kk( ) j(, \,( ( ) + + ( )[ ] ( )30 2260 60 60 X sinc f kf k f kk( ) j(, \,( ( ) + + ( )[ ]30 2260 60 60 60 Full-Wave Case:x cos rect comb t t t t ( ) ( ) ( ) ( ) [ ] 120 2 120 2 120 60 60 1 X sinc comb f f f f ff ( ) ( ) + + ( )[ ] j(, \,( j(, \,( ( ),, ]]]60 2 60 60160 120 60 X sinc comb f f f f ff ( ) ( ) + + ( )[ ] j(, \,( j(, \,( ( ),, ]]]2260 60120 60 X sinc f f f kf k f kk( ) ( ) + ( ) + j(, \,( ( ) + + ( )[ ],, ]]]60 2 60 60260 60 60 60 f |X( f )|30 260f |X( f )|60 212053. Find the DTFT of each of these signals:(a) x u n nn[ ] j(, \,( [ ]131X x u j n e n e ej nnnj nnnj nn ( ) [ ] j(, \,( [ ] j(, \,(

131131M. J. Roberts - 7/12/035-78X j e emj mmj mm ( ) j(, \,( j(, \,(+ + ( )

+

13 311010X j e e eeeej j mmjjjj ( ) j(, \,(

3 3 311330Alternate Solution:x u u n n n nn n[ ] j(, \,( [ ] j(, \,( [ ] [ ]13113 Usingnjneu[ ] F11 and n [ ] F1x u n n nenj[ ] j(, \,( [ ] [ ] 13131131 x neeeeeejjjjjj[ ] j(, \,(

1 13133133Second Alternate Solution:x u n nn[ ] j(, \,( [ ]131311X jee ee jjjj( )

1311133(b) x sin u n n nn[ ] j(, \,(j(, \,( [ ]4142x u u n e ej nje enj n j n n jnjn[ ] j(, \,( [ ] j(,,\,(( j(,,\,((j(,,,\,((( [ ] 4 4 4 4214212 4 42M. J. Roberts - 7/12/035-79x u nje e e enj jnj jn[ ] j(,,\,((j(,,\,(( j(,,\,((j(,,\,((j(,,,\,((( [ ] 12 4 4 4 4242424242 eneejnjj444114j(,,\,(( [ ] u Fen eeejnjjj42244214j(,,\,(( [ ] u F X jje eeee eeejjjjjjjj( ) j(,,\,((j(,,\,((j(,,,,,\,(((((12 41441442244224X j eje ee e eeee eejj jjj jjjjjj ( ) j(,,\,(( j(,,\,(( j(,,\,(( j(,,\,((j(,,\,(( j(,,2424 4244 42414 4141414 \\,((j(,,,,,,,\,(((((((X j eje e ee e e eee eejj j jjj j jjj j ( ) j(,,\,(( j(,,\,(( j(,,\,(( +j(,,\,(( +j(,,\,(( 242424 424244 424 4 4 4 4 414 4 jj je + 2M. J. Roberts - 7/12/035-80Xcosj eje e ee eee ejj j jjjjj j ( ) + j(, \,( + 22 2 4 42216 16 64 64112 4 Xcosj ejj e eee ejj jjj j ( ) j(,,\,(( j(, \,( + 24 4228 64112 4 Xsincosj e ee ejjj j ( ) j(, \,( j(, \,( + 2216114 4112 4(c) x sinc sinc n n n[ ] j(, \,( ( ) j(, \,(282 48 Usingsinc combnw w wF Fj(, \,( ( ) ( )FrectX comb comb F F F F F e j F( ) j(, \,( ( ),, ]]] j(, \,( ( ),, ]]] rect rect82828 X comb F F F e j F( ) j(, \,( ( ),, ]]] rect828x sinc n n[ ] ( ) j(, \,(2 48(d) x sinc n n[ ] j(, \,(228Usingsinc combnw w wF Fj(, \,( ( ) ( )FrectM. J. Roberts - 7/12/035-81X comb comb F F F F F ( ) j(, \,( ( ),, ]]] j(, \,( ( ),, ]]]82rect8282rect82 X comb comb F F F F F ( ) j(, \,( j(, \,( ( ),, ]]] j(, \,( ( ),, ]]]82rect82rect82 2X comb F F F F ( ) j(, \,( j(, \,( j(, \,( ( )82rect82rect82 2X tri comb tri comb F F F F F ( ) j(, \,( j(, \,( ( ) j(, \,( ( )8228828282 254. Sketch the magnitudes and phases of the DTFTs of the following functions:(a) rect2 n [ ]Using rect drcl ,N w wwn N F N [ ] + ( ) + ( )F2 1 2 1 ,rect drcl ,25 5 n F [ ] ( )FF -1 1|X( F )|5F -1 1Phase of X( F )- (b) rect25 n n [ ] [ ] ( ) Using n [ ] F1 and x y X Y n n F F [ ] [ ] ( ) ( )Frect drcl , drcl ,25 5 5 5 25 5 n n F F [ ] [ ] ( ) ( ) ( ) ( ) FM. J. Roberts - 7/12/035-82F -1 1|X( F )|25F -1 1Phase of X( F )- (c) rect23 3 n n [ ] + [ ] Using x X n n e Fj Fn [ ] ( )020F rect drcl ,263 3 15 5 n n F ej F[ ] + [ ] ( ) FF -1 1|X( F )|15F -1 1Phase of X( F )- (d) rect rect2 25 4 5 n n n n [ ] [ ] ( ) [ ] [ ] ( ) rect drcl ,25 4 25 5 n n F [ ] [ ] ( ) ( ) FM. J. Roberts - 7/12/035-83F -1 1|X( F )|25F -1 1Phase of X( F )- (e) rect comb2 8n n [ ] [ ]Usingcomb combN n N F00[ ] ( )F,rect comb drcl , comb2 85 5 8 n n F F [ ] [ ] ( ) ( )Frect comb drcl ,2 85 5 8 n n F F mm[ ] [ ] ( ) ( )Frect comb drcl ,2 85858n n F F mm[ ] [ ] ( ) j(, \,(Frect comb drcl ,2 858 858n n mF mm[ ] [ ] j(, \,( j(, \,(FF -1 1|X( F )|0.625F -1 1Phase of X( F )- M. J. Roberts - 7/12/035-84(f) rect comb2 83 n n [ ] [ ]Using the result of (e) and x X n n e Fj Fn [ ] ( )020F ,rect comb drcl ,2 86358 858n n e mF mj Fm[ ] [ ] j(, \,( j(, \,(F F -1 1|X( F )|0.625F -1 1Phase of X( F )- (g) rect comb rect rect2 8 2 22 2 8 2 4 n n n n m n n mm m[ ] [ ] [ ] [ ] [ ] ( )[ ] rect comb rect rect comb2 8 2 2 42 4 n n n n m n nm[ ] [ ] [ ] [ ] [ ] [ ] Thereforerect comb drcl , comb2 82 5 5 4 n n F F [ ] [ ] ( ) ( )Frect comb drcl ,2 82 5 5 4 n n F F mm[ ] [ ] ( ) ( )Frect comb drcl ,2 825454n n F F mm[ ] [ ] ( ) j(, \,(Frect comb drcl ,2 8254 454n n mF mm[ ] [ ] j(, \,( j(, \,(FM. J. Roberts - 7/12/035-85F -1 1|X( F )|1.25F -1 1Phase of X( F )- (h) rect comb2 5n n [ ] [ ]rect comb drcl , comb2 55 5 5 n n F F [ ] [ ] ( ) ( )Frect comb drcl ,2 55 5 5 n n F F mm[ ] [ ] ( ) ( )Frect comb drcl ,2 55555n n F F mm[ ] [ ] ( ) j(, \,(Frect comb drcl ,2 5555n n mF mm[ ] [ ] j(, \,( j(, \,(FThen, using drcl , combnm m nm2 12 12 1+ +j(, \,( [ ]+rect comb comb2 5 55n n m F mm[ ] [ ] [ ] j(, \,(Frect comb comb2 5n n F [ ] [ ] ( )Fand, since1 F ( ) comb F , that implies that rect comb2 51 n n [ ] [ ] .M. J. Roberts - 7/12/035-86F -1 1|X( F )|1F -1 1Phase of X( F )- 55. Sketch the inverse DTFTs of these functions.(a) X comb comb F F F ( ) ( ) j(, \,(12Using 1 F ( ) comb F and e n F Fj F n 200x X [ ] ( )F112 ( ) j(, \,( e F Fj n Fcomb combe e e F Fj nj nj n 2 2 212 +j(, \,( ( ) j(, \,(Fcomb comb j(, \,( ( ) j(, \,( j e nF Fj n22122sin comb combF j(, \,( ( ) j(, \,(+ ( )221221e nF Fj nsin comb combF + ( )j(, \,( + + ( )j(, \,(,, ]]] j(, \,( ( ) j(, \,( 22121212cos sin sin comb comb n j n nF FF + ( )j(, \,( j(, \,( + + ( )j(, \,( j(, \,(,,,,,,]]]]]]] ( ) j(, \, j(, \,( 2212 2121220cos sin sin sin comb combsin n nj n nF Fn1 2 44 3 44 1 2 4444 3 4444F((M. J. Roberts - 7/12/035-8722122sin comb combnF Fj(, \,( ( ) j(, \,(Fn-12 12x[n]2(b) X comb comb F j F j F ( ) +j(, \,( j(, \,(1818Using 1 F ( ) comb F and e n F Fj F n 200x X [ ] ( )Fje je j F j Fj nj n +j(, \,( j(, \,( 4 41818Fcomb comb j(, \,(j(, \,( +j(, \,( j(, \,( j j nj F j F 241818sin comb comb F241818sin comb combnj F j Fj(, \,( +j(, \,( j(, \,(Fn-12 12x[n]-22(c) X sinc sinc comb F F F F ( ) j(, \,(j(, \,( + +j(, \,(j(, \,(,, ]]] ( ) 10141014Usingsinc comb cos drcl ,tw f f t wf twTw f t Twj(, \,( ( ) j(, \,( + j(, \,( j(, \,(,, ]]] 0 0 00001 1M. J. Roberts - 7/12/035-88from Appendix A (because Tw02 is an integer),X cos drcl , cos drcl , F F F F F ( ) j(, \,(j(, \,( + j(, \,( + +j(, \,(j(, \,( + +j(, \,(11010149149 10149149 cos cos 10 1021 F e dF Fj Fn( ) ( ) F121010 10 21e e e dF Fj F j F j Fn +( ) ( ) Fcos12102 512 51e dF e dF Fj F n j F n + ( ) ( ) +[ ] ( )Fcos122 5 2 52 5 2 51011cos sincos sincos F n j F n dFF n j F n dFF+ ( ) ( ) + + ( ) ( ) [ ]+ ( ) ( ) + ( ) ( ) [ ],,,,]]]]] ( )FThese integrals are zero unless n 5. Therefore125 5 10 n n F + [ ] + [ ] ( ) ( )Fcos .Then using rect drcl ,N w wwn N F N [ ] + ( ) + ( )F2 1 2 1 ,rect drcl ,49 9 n F [ ] ( )FCombining inverse transforms,125 5 10 9 94 n n n F F + [ ] + [ ] ( ) + [ ] ( ) + ( ) rect cos drcl ,F .Then, using e n F Fj F n 200x X [ ] ( )Fe n n n F Fj n 24125 5 10149149 + [ ] + [ ] ( ) + [ ] j(, \,(j(, \,( + j(, \,( rect cos drcl ,Fande n n n F Fj n+ [ ] + [ ] ( ) + [ ] +j(, \,(j(, \,( + +j(, \,( 24125 5 10149149 rect cos drcl ,F .M. J. Roberts - 7/12/035-89Then, finally110125 5125 511010149142424e n n ne n n nF Fj nj n + [ ] + [ ] ( ) + [ ]+ + [ ] + [ ] ( ) + [ ]j(,,,,\,(((( j(, \,(j(, \,( + rectrectcos drclF,,cos drcl ,910149149j(, \,(+ +j(, \,(j(, \,( + +j(, \,( F FThe impulses on the left side cancel and we getrectcoscos drcl ,cos drcl ,45 21101014914910149149n n F FF F[ ] j(, \,( j(, \,(j(, \,( + j(, \,(+ +j(, \,(j(, \,( + +j(, \,( Fn-12 12x[n]-0.20.2(d) X comb F F F F F ( ) j(, \,( + j(, \,( + j(, \,(,, ]]] ( ) 143165162X comb comb F F F F F F ( ) j(, \,( + j(, \,( + j(, \,(,, ]]] ( ) + j(, \,(,, ]]] 143165161212XcombcombFF F F FF F F F( ) j(, \,( + j(, \,( + j(, \,(,, ]]] ( )+ j(, \,( + j(, \,( + j(, \,(,, ]]] j(, \,(12143165161431651612 Xcomb comb combcomb comb combFF F FF F F( ) j(, \,( + j(, \,( + j(, \,(,, ]]]+ j(, \,( + j(, \,( + j(, \,(,, ]]]121431651614123161251612M. J. Roberts - 7/12/035-90Then, using 1 F ( ) comb F and e n F Fj F n 200x X [ ] ( )F we get121214316516141231612385832118138e e ee e eF F FF Fj nj nj nj nj nj n + ++ + +j(,,,\,((( j(, \,( + j(, \,( + j(, \,(+ j(, \,( + Fcomb comb combcomb comb2251612j(, \,( + j(, \,(comb For1212143165163411162385825838e e ee e eF F FF Fj nj nj nj nj nj n + ++ + +j(,,,\,((( j(, \,( + j(, \,( + j(, \,(+ j(, \,( + j(, Fcomb comb combcomb comb \\,( + j(, \,(comb F1316orcos cos coscomb comb combcomb comb comb n n n F F FF F F2385812143165163411161316j(, \,( + j(, \,( + j(, \,( j(, \,( + j(, \,( + j(, \,(+ j(, \,( + j(, \,( + j(, \,(Fn-16 16x[n]-3356. Using the relationship between the CTFT of a signal and the CTFS of a periodicextension of that signal, find the CTFS ofx rect comb t tw TtT( ) j(, \,( j(, \,(10 0and compare it with the table entry.X sinc f w wf ( ) ( )X X sinc sinc k f kf f w wkf wTwT k [ ] ( ) ( ) j(, \,( 0 0 0 00 057. Using the relationship between the DTFT of a signal and the DTFS of a periodicextension of that signal, find the DTFS ofM. J. Roberts - 7/12/035-91rect combN Nwn n [ ] [ ]0and compare it with the table entry.X drcl , F N F Nw w( ) + ( ) + ( ) 2 1 2 1From the text,X XpppkN kF [ ] ( )1 .Therefore,X drcl ,pwwk NNkN N [ ] ++j(, \,(2 12 10 0 .M. J. Roberts - 7/12/03Solutions 6-1Chapter 6 - Fourier Transform Analysis ofSignals and SystemsSolutions(In this solution manual, the symbol, , is used for periodic convolution because thepreferred symbol which appears in the text is not in the font selection of the word processorused to create this manual.)1. A system has an impulse response,h uLPtt e t ( ) ( )310 ,and another system has an impulse response,h uHPtt t e t ( ) ( ) ( ) 310 .(a) Sketch the magnitude and phase of the transfer function of these two systems in aparallel connection.H , HLP HPjj jj ( ) + ( ) +3101310HP jj j ( ) + + + 31013101-40 40|HP(j)|1-40 40Phase of HP(j)- (b) Sketch the magnitude and phase of the transfer function of these two systems in acascade connection.HC jj j j jjj ( ) + +,, ]]] + + ( ) + ( ) + ( )31013103109103 10 9102 2M. J. Roberts - 7/12/03Solutions 6-2HC j jjjj ( ) ++ ( ) ++ ( )3 211037102 2-40 40|HC(j)|0.25-40 40Phase of HC(j)- 2. Below are some pairs of signals, x t ( ) and y t ( ). In each case decide whether or not y t ( )is a distorted version of x t ( ).t2x(t)-22(a)t2y(t)-22 t2x(t)-22(b)t2y(t)-22 (a) Inverted and amplified, undistorted(b) Time shifted, undistortedM. J. Roberts - 7/12/03Solutions 6-3t2x(t)-22(c)t2y(t)-22 t2x(t)-11(d)t2y(t)-11 (c) Clipped at a negative value, distorted(d) Time compressed, distortedt2x(t)-22(e)t2y(t)-22 t2x(t)-22(f)t2y(t)-22(e) Constant added, distorted(f) Log-amplified, distorted3. Classify each of these transfer functions as having a lowpass, highpass, bandpass orbandstop frequency response.M. J. Roberts - 7/12/03Solutions 6-4f -10 10|H( f )|1(a)F -2 2|H(F)|1(b) -4 4|H(j)|1(c)-100 100|H(j)|1(d)(a) Lowpass (b) Bandpass (c) Lowpass (d) BandpassF -2 2|H(F)|1(e) -4 4|H(j)|1(f)(e) Highpass (f) Bandstop4. Classify each of these transfer functions as having a lowpass, highpass, bandpass orbandstop frequency response.(a) H rect f f( ) j(, \,(110010Bandstop(b) H rect comb F F F ( ) ( ) ( ) 10 Lowpass (c) H rect rect comb j ( ) j(, \,(j(, \,( + +j(, \,(j(, \,(,, ]]] j(, \,( 204204 2 Bandpass5. A system has an impulse response,h rect..t t( ) j(, \,( 100 010 02 .What is its null bandwidth?The transfer function isH . sinc ..f f e j f( ) ( ) 0 2 0 020 02.M. J. Roberts - 7/12/03Solutions 6-5Its first positive-frequency null occurs at f 50. The null bandwidth is therefore 50.6. A system has an impulse response,h u n nn[ ] j(, \,( [ ]78 .What is its half-power DT-frequency bandwidth?Using njneu[ ] F11 the transfer function isH je e j j ( )

117888 7.The low-frequency gain isH 0 8 ( ) The -3 dB point occurs whereH j dB( ) 3228232 .Solving,H je e e ej jj j ( )

+ ( ) + j(, \,(2211781178117878Hcosj dBdB( ) ( ) + j(, \,(

323211747832cos . ( ) + j(, \,(

+

32178132746464496426474111 464 70 991dBhp n 0 1337 2 . So the -3 dB DT-frequency bandwidth in radians is 0.1337. In cycles it is 0.0213.7. Determine whether or not the CT systems with these transfer functions are causal.(a) H sinc f f ( ) ( ) h rect t t ( ) ( ) Not CausalM. J. Roberts - 7/12/03Solutions 6-6(b) H sinc f f e j f( ) ( ) h rect t t ( ) j(, \,(12Causal(c) H rect j ( ) ( ) h sinc t t( ) j(, \,(12 2 Not Causal(d) H rect j e j ( ) ( ) h sinc t t( ) j(, \,(1212 Not Causal(e) H f A ( ) h t A t ( ) ( ) Causal(f) H f Aej f( ) 2h t A t ( ) + ( ) 1 Not Causal8. Determine whether or not the DT systems with these transfer functions are causal.(a) HsinsinF FF( ) ( )( )7h rect n n [ ] [ ]3Not Causal(b) HsinsinF FF e j F( ) ( )( ) 72h rect n n [ ] [ ]31 Not Causal(c) HsinsinF FF e j F( ) ( )( ) 32h rect n n [ ] [ ]21 Causal(d) H rect comb F F F ( ) ( ) ( ) 10 h sinc n n[ ] j(, \,(110 10Not Causal9. Find and sketch the frequency response of each of these circuits given the indicatedexcitation and response.(a) Excitation, vi t ( ) - Response, vL t ( )R = 10 C = 1 F L = 1 mH v (t)i+-v (t)L+-HVVj jjj Lj Lj C RLCLC j RCLi ( ) ( )( ) + +

+1122M. J. Roberts - 7/12/03Solutions 6-7-150000 150000|H( j )|3-150000 150000Phase of H( j )- (b) Excitation, vi t ( ) - Response, iC t ( )R = 1 k C = 1 F v (t)i+-i (t)CHIVj jjRj Cj Cj RCCi ( ) ( )( ) +

+111-1500 1500|H( j )|0.001-15001500Phase of H( j )- (c) Excitation, vi t ( ) - Response, vR t ( )M. J. Roberts - 7/12/03Solutions 6-8R = 1 k C = 1 F L = 1 mH v (t)i+-v (t)R+ -HVVj jjRRj Lj Cj Lj CRR j LLCR LCR LC j LRi ( ) ( )( ) ++

+

( ) ( ) +1111222-50000 50000|H( j )|1-50000 50000Phase of H( j )- (d) Excitation, ii t ( ) - Response, vR t ( )R = 100 C = 1 F L = 1 mH i (t)iv (t)R+-HVj jI j R I jI j R j Lj L Rj CR LCLC j RCRiRCi ( ) ( )( ) ( )( ) + + +1122M. J. Roberts - 7/12/03Solutions 6-9-1000000 1000000|H( j )|100-1000000 1000000Phase of H( j )- 10. Classify each of these transfer functions as having a lowpass, highpass, bandpass orbandstop frequency response.(a) H fjf( ) +11Lowpass(b) H f jfjf( ) + 1Highpass(c) H j jj ( ) +10100 102Bandpass(d) HsinsinF FF( ) ( )( )3Lowpass(e) H sin sin j j ( ) ( ) + ( ) [ ] 2 Bandpass11. Plot the magnitude frequency responses, both on a linear-magnitude and on a log-magnitude scale, of the systems with these transfer functions, over the frequency rangespecified.(a) H ff j f( ) +2020 4 422 2 , < < 100 100 f(b) H jj j ( ) + ( ) + ( )2 10100 1700 2052 , < < 500 500 M. J. Roberts - 7/12/03Solutions 6-10f -100 100|H( f )|1(a)f -100 100ln(|H( f )|)-10-500 500|H( j )|2(b) -500 500ln(|H( j )|)-1012. Draw asymptotic and exact magnitude and phase Bode diagrams for the frequencyresponses of the following circuits and systems.(a) An RC lowpass filter with R 1 M and C 0 1 . F.H.j j Cj C R j RC j j ( ) +

+ + +1111110 10 110 1 16 710-1100101102103-50-40-30-20-100 |H(j)|dB10-1100101102103-2-1.5-1-0.50 Phase of H(j)(b)M. J. Roberts - 7/12/03Solutions 6-11R = 10 C = 1 F L = 1 mH v (t)i+-v (t)L+-From Exercise 9 (a)HVVj jjj Lj Lj C RLCLC j RCLi ( ) ( )( ) + +

+1122H jj ( ) + 101 10 109 29 2 5102103104105106107-100-80-60-40-20020 |H(j)|dB10210310410510610700.511.522.533.5 Phase of H(j)13. Find the transfer functions, HVVf ffoi( ) ( )( ), of these active filters and identify them aslowpass, highpass, bandpass or bandstop.(a)M. J. Roberts - 7/12/03Solutions 6-12v (t)iv (t)x+-v (t)o+-KR2R1C2C1V V VV Vx i ox of j fCj fC RG f j fC f Gf Rj fC RK f( ) +++j(,,,\,((( ( ) ( ) ( )+ ( )21122 0121221 1 1222V V VV Vx i ox of j fC j fCj fR C G f j fC f Gf j fR Cj fR C K f( ) ++ +j(, \,( ( ) ( ) ( )+ ( )221 22 021 2122 21 1 12 22 2 j fC j fCj fR C G Gj fKR C j fR Cffj fCfxoi221 22 1 220122 21 12 2 2 21 ++ +j(, \,( + ( ),,,,]]]]]( )( ),, ]]] ,, ]]] ( )VVV ++ +j(, \,( + ( ) + j fC j fCj fR C G j fR C j fKGR C 221 21 2 2122 21 2 2 1 2 2 ( ) + + ( ) [ ] 2 2 122 1 2 1 2 1 2 2 1 f R CC j f C C GR C K GV Vo if j fC j fCj fR C G j fCj fKR Cf ( ) ++ +j(, \,(( )1 221 222 0122 21 12 2 HVVf fff KR CCf R CC j f C C GR C K Goi( ) ( )( ) ( )( ) + + ( ) [ ] 22 2 122 1 222 1 2 1 2 1 2 2 1 M. J. Roberts - 7/12/03Solutions 6-13H f f KRR CCf RR CC j f R C C R C K( ) ( )( ) + ( ) + ( ) ( ) 22 2 1 121 2 1 221 2 1 2 1 1 2 2 2 Highpass(b)v (t)iv (t)x+-v (t)o+-KR2R1C2C1V V VV Vx i ox of GRj fCj fC f G f j fCf j fCj fC RK f( ) +++j(,,,\,((( ( ) ( ) ( )+ ( )1221 1 12221122 2 01212 V V VV Vx i ox of G j fCj fR C j fC f G f j fCf K j fC R f( ) ++ +j(, \,( ( ) ( ) ( ) + ( ) ( ) 122 21 1 12 222 12 2 01 2 0 G j fCj fR C j fC j fCK j fC RffGfxoi122 21 12 2122 12 21 20++ +j(, \,( + ( ),,,,]]]]]( )( ),, ]]] ,, ]]] ( ) VVV + ( ) ++ +j(, \,( + 1 222 12 22 2 122 21 1j fC R G j fCj fR C j fC jK fC ( ) ( ) + ( ) 2 2 122 1 2 1 2 2 1 2 1 f R CC j f GR C C K C GV Vo if G j fCj fR C j fC GKf ( ) ++ +j(, \,(( )122 120122 21 1 M. J. Roberts - 7/12/03Solutions 6-14H f KGf R CC j f GR C C K C G( ) ( ) ( ) + ( ) 122 1 2 1 2 2 1 2 12 2 1 H f Kf RR CC j f RC R C RC K( ) ( ) + + ( ) ( ) 2 2 1 121 2 1 2 1 2 2 2 1 1 Lowpass14. Show that this system has a highpass frequency response.x(t)y(t)Writing the differential equation,ddt t t t t t ( ) ( ) ( ) [ ] ( ) ( ) ( ) ( ) [ ]y x x y xor ( ) + ( ) ( ) + ( ) ( )ddt t ddt t t t t y x x y xorddt t t ddt t y y x ( ) + ( ) ( )Fourier transforming both sides,j j j j j Y Y X ( ) + ( ) ( )orHYXj jjjj ( ) ( )( ) +1Highpass15. Draw the block diagram of a system with a bandpass frequency response using twointegrators as functional blocks. Then find its transfer function and verify that it has abandpass frequency response.x(t) y(t)From Exercise 14, the first stage is characterized by the transfer function,M. J. Roberts - 7/12/03Solutions 6-15H11j jj ( ) + .The second stage differential equation is( ) ( ) ( ) y x y t t t .Then, Fourier transforming both sides,j j j j Y X Y ( ) ( ) ( )andH211jj( ) +Therefore the overall system transfer function isH j jj jjj ( ) + + + ( )11112 .Bandpass16. Find the transfer function, HYXj jj ( ) ( )( ), and sketch the frequency response of eachof these DT filters over the range, < < 4 4 .(a)Dx[n] y[n]y x x n n n [ ] [ ] [ ] 1Y X X j j e jj ( ) ( ) ( )HYXj jj e j ( ) ( )( ) 1M. J. Roberts - 7/12/03Solutions 6-16-2 2|H(j)|2-2 2Phase of H(j)- (b)Dx[n] y[n]y x y n n n [ ] [ ] [ ] 1Y X j e jj ( ) + ( ) ( )1HYXj jj e j ( ) ( )( ) + 11-2 2|H(j)|5-2 2Phase of H(j)|- (c)M. J. Roberts - 7/12/03Solutions 6-17Dx[n]Dy[n]y x x y n n n n [ ] [ ] [ ] + [ ] ( ) 1 1H j eejj( ) +11-2 2|H(j)|5-2 2Phase of H(j)- (d)DDx[n] y[n]y x x zz x zn n n nn n n[ ] [ ] [ ] + [ ][ ] [ ] [ ]11Y X X ZZ Z Xj j e j jj e j jjj ( ) ( ) ( ) + ( )( ) + ( ) ( )Y X ZZXj e j jj jejj ( ) ( ) ( ) + ( )( ) ( )+11M. J. Roberts - 7/12/03Solutions 6-18Y XXj e j jejj ( ) ( ) ( ) + ( )+11HYXj jjeejj ( ) ( )( ) +212-2 2|H(j)|5-2 2Phase of H(j)|- 17. Find the minimum stop band attenuation of a moving-average filter with N 3. Definethe stop band as the frequency region, F Fc < F202 2, Re ,h t e t( ) 49147Not Causal25. Determine whether or not the DT systems with these transfer functions are causal.(a) H rect comb F F F e j F( ) ( ) ( ) [ ] 1020h sinc n n[ ] j(, \,(1101010Not Causal(b) H sin F j F e ej F j F( ) ( ) 222 2 h n n n [ ] + [ ] [ ] ( )121 1 Not Causal(c) H F e j F( ) 14h n n n [ ] [ ] [ ] 2 Causal(d) H j eeeejjjj( ) 88 5158h u n nn[ ] j(, \,( + [ ]+5811Not Causal26. Find and sketch the frequency response of each of these circuits given the indicatedexcitation and response.(a) Excitation, vi t ( ) - Response, vC t2( )M. J. Roberts - 7/12/03Solutions 6-31R = 1 k C = 1 F v (t)i+-+-R = 10 k C = 0.1 F v (t)C211 22HVVj jjZ jR Z jj CRj CZ jR Z j j R CCi ( ) ( )( ) ( )+ ( )+

( )+ ( ) +212221 2 21111Z j j C Rj Cj C Rj CRj Cj C R CCj R Cj C C R CC ( ) +j(, \,(+ +

++ +

++ ( ) 1 11 1111122122221 2122 21 222 1 2H jj R Cj C C R CCR j R Cj C C R CCj R C ( ) ++ ( ) + ++ ( ) +2 21 222 1 212 21 222 1 22 21111H jj C C R CC R j R C ( ) + ( ) ( ) + +111 222 1 2 1 2 2H jRR CC j C C R R C ( ) + + ( ) +[ ]1121 2 1 2 1 2 1 2 2-15000 15000|H( j )|1-15000 15000Phase of H( j )- (b) Excitation, vi t ( ) - Response, iC t1( )M. J. Roberts - 7/12/03Solutions 6-32R = 1 k C = 1 F v (t)ii (t)C1+-R = 10 k C = 0.1 F 11 22HIVIVj jjjjRj Cj C Rj CZ jj R Cj R C CCCiRi i ( ) ( )( ) ( )( )++ +

( )++ +1 1221222 22 22111 11 11Z j R Z j R j R Cj C C R CCi ( ) + ( ) + ++ ( ) 1 12 21 222 1 21Z j R j C C R CC j R Cj C C R CCi ( ) + ( ) [ ]+ ++ ( ) 1 1 222 1 2 2 21 222 1 21H j j C C R CCR j C C R CC j R Cj R Cj R C CC ( ) + ( ) + ( ) [ ]+ +++ +1 222 1 21 1 222 1 2 2 22 22 221111H jj C CC j R CR j C C R CC j R Cj R Cj R C CC ( ) +j(, \,( +,, ]]]+ ( ) [ ]+ +++ +1212 21 1 222 1 2 2 22 22 2211111H j j C j R CR j C C R CC j R C ( ) + ( )+ ( ) [ ]+ +1 2 21 1 222 1 2 2 211H j j C j R CRR CC j R C C R C ( ) + ( ) + + ( ) +[ ]1 2 221 2 1 2 1 1 2 2 211M. J. Roberts - 7/12/03Solutions 6-33-15000 15000|H( j )|0.001-15000 15000Phase of H( j )- (c) Excitation, vi t ( ) - Response, vR t2( )C = 1 F v (t)i+-v (t)R2+-R = 10 k 2R = 10 k 11 C = 1 F 2HVVj jjZ jj C Z jRj C Rj C Z jj C Z jj R Cj R CRi ( ) ( )( ) ( )+ ( ) +

( )+ ( ) +21222112 22 21 11 1Z jRj C RRj C RR j R Cj C R R ( ) +j(, \,(+ +

++ ( ) +12212212 22 1 21111H jj C R j R Cj C R Rj C R j R Cj C R Rj R Cj R C ( ) ++ ( ) ++ ++ ( ) ++1 12 22 1 21 12 22 1 22 22 2111111H jRR CCj C R Rj C R j R Cj C R R ( ) + ( ) ++ ++ ( ) +21 2 1 22 1 21 12 22 1 21111M. J. Roberts - 7/12/03Solutions 6-34H j RR CCj C R R j C R j R C ( ) + ( ) + + + ( )21 2 1 22 1 2 1 1 2 21 1H j RR CCRR CC j C R R C R ( ) + + ( ) + ( )21 2 1 221 2 1 2 2 1 2 1 11-2000 2000|H( j )|1-2000 2000Phase of H( j )- (d) Excitation, ii t ( ) - Response, vR t1( )C = 1 F i (t)iv (t)R1+-R = 10 k 2R = 10 k 11C = 1 F 2HVIIIj jj R jj RRj CR Rj CRiRi ( ) ( )( ) ( )( ) ++ +1111221 2211H j R j R Cj R R C ( ) ++ ( ) +12 21 2 211M. J. Roberts - 7/12/03Solutions 6-35-500 500|H( j )|10000-500 500Phase of H( j )- (e) Excitation, vi t ( ) - Response, vRL t ( )v (t)iv (t)RL+-+-R = 10 k 1 R = 1 k LR = 10 k 2C = 1 F 1 C = 1 F 2V V VV V VR i RLRL L i Rj j C j C G j j C j j Cj j C G G j G j j C111 2 1 1 22 2 2 200 ( ) + + [ ] ( ) ( ) ( ) + + [ ] ( ) ( ) j C C G j Cj C j C G Gjjj CG jLRRLi 1 2 1 22 2 2121+ ( ) + + + ( ),, ]]] ( )( ),, ]]] ,, ]]] ( )VVV + ( ) +[ ] + + ( ) [ ]+ j C C G j C G G CL 1 2 1 2 2222 + + ( ) + ( ) +[ ]+ + ( ) 21 2 1 2 2 1 2 1 2CC j C C G G GC G G GL LV VRL ij j C C G j Cj C G j ( ) + ( ) + ( )11 2 1 12 2M. J. Roberts - 7/12/03Solutions 6-36HVVj jjCC j C C G GGCC j C C G G GC G G GRLi L L ( ) ( )( ) + + ( ) + + + ( ) + ( ) +[ ]+ + ( )21 2 1 2 2 1 221 2 1 2 2 1 2 1 2H jCC j C CR RRCC j C CR RCR R R RL L ( ) + ++ + + ( ) +j(, \,( +,, ]]] + +j(, \,(21 21 22 1 221 2 1 2221 1 211 1 1 1 1H j RR CC j R C CRR CC j C C RR R R C R RRLLL ( ) + + ( ) + + + ( ) +j(, \,( +,, ]]] + +21 2 1 2 1 1 221 2 1 2 1 221 2 2211-25000 25000|H( j )|1-25000 25000Phase of H( j )- 27. Find and sketch versus frequency the magnitude and phase of the input impedance,ZVIiniif ff( ) ( )( ) and transfer function, HVVf ffoi( ) ( )( ), for each of these filters.(a)1 Fv (t)ii (t)iv (t)o+-+-1 kZins ssC R ( ) +1 , Zin fj fC Rj f( ) + +1212 1010006 Hs s RRsCsCRsCR( ) +

+11 , H f j fCRj fCRj fj f( ) + +22 12 102 10 133M. J. Roberts - 7/12/03Solutions 6-37f -800 800|Zin( f )|50000f -800 800Phase of Zin( f )- f -800 800|H( f )|1f -800 800Phase of H( f )- (b)10 nFv (t)iv (t)o+-+-100 50 mH i (t)iZins s R sLsC( ) + +1Z .in f R j fLj fC j fj f( ) + + + + 212100 0 512 108 Hs s sCR sLsCs LC sCR LCs sRL LC( ) + +

+ + + +11111 1122H fLCj f j f RL LCf j f( ) ( ) + + + +1 12 212 1014 2 10 4000292 2 9 f -40000 40000|Zin( f )|100000f -40000 40000Phase of Zin( f )- f -40000 40000|H( f )|25f -40000 40000Phase of H( f )- M. J. Roberts - 7/12/03Solutions 6-3828. The signal, x(t), in Exercise 23 is the excitation of an RC lowpass filter with R 1kand C 0 3 . F. Sketch the excitation and response voltages versus time on the samescale.From Exercise 23,x rect comb t t t ( ) ( ) ( ) 500 1000 500X sinc f nf nn( ) j(, \,( ( )12 2500 The transfer function isH fj fRC( ) +12 1 Therefore the output isY sinc fj fRCnf nn( ) +j(, \,( ( )1212 1 2500 Y sinc f nj nRC f nn( ) j(, \,(+ ( )12 211000 1500 Converting to the time domain,ysinctnj nRC ej ntn( ) j(, \,(+1221000 11000ory sinc t n ej nRCej nRCj nt j ntn( ) + j(, \,(+ + +,, ]]]

1212 1000 1 1000 11000 10001 y sinc t n j nRC e e e enRCj nt j nt j nt j ntn( ) + j(, \,( ( ) + +( ) +,,,]]]]

121210001000 11000 1000 1000 100021 y sincsin cost n nRC nt ntnRCn( ) + j(, \,( ( ) + ( )( ) +,, ]]] 12122000 1000 2 10001000 121 -2 ms 2 ms1t29. Draw asymptotic and exact magnitude and phase Bode diagrams for the frequencyresponses of the following circuits and systems.(a)M. J. Roberts - 7/12/03Solutions 6-39R = 1 k C = 1 F v (t)i+-+-R = 10 k C = 0.1 F v (t)C211 22From Exercise 26(b)H jRR CC j C C R R C ( ) + + ( ) +[ ]1121 2 1 2 1 2 1 2 2H.jj ( ) + 11 10 2 1 106 2 3101102103104105106-140-120-100-80-60-40-200 |H(j)|dB101102103104105106-3.5-3-2.5-2-1.5-1-0.50 Phase of H(j)(b)10j+10jj+10X(j) Y(j) H jjjjjjjj jj ( ) + + + ( ) +j(, \,( +j(, \,(

1010 10101011010110120frequencyindependentgainzero atpole at j 10pole at j 10{}1 2 4 3 4 1 2 4 3 4M. J. Roberts - 7/12/03Solutions 6-4010-1100101102103-50-40-30-20-100 |H(j)|dB10-1100101102103-2-1012 Phase of H(j)(c) A system whose transfer function is H,j jj ( ) +2010 000 202H. .j jj j j jjj ( ) + ( ) + + ( ) + ,, ]]]2010 99 5 10 99 52010000 1500 100002H j jj ( ) + 15001500 100002100101102103104-60-50-40-30-20-100 |H(j)|dB100101102103104-2-1012 Phase of H(j)30. Find the transfer function for the following circuit. What function does it perform?M. J. Roberts - 7/12/03Solutions 6-41v (t)iv (t)ov (t)x+-+-RCi (t)ii (t)ffiSumming currents at the virtual ground,j fCV f V fRio2 ( ) ( )orV f j fRCV f RC j fV fo i iV fi( ) ( ) ( )( )2 2 derivative of1 2 4 3 4This is a differentiator.31. Design an active highpass filter using an ideal operational amplifier, two resistors andone capacitor and derive its transfer function to verify that it is high pass.v (t)iv (t)ov (t)x+-+-RR Ci (t)ii (t)ffi iSumming currents at the virtual ground,V fRj fCV fRiiiof( )+ ( )12orV fV fj fC Rj fC Roii fi i( )( ) +22 132. Find the transfer functions, HVVf ffoi( ) ( )( ), of these active filters and identify them aslowpass, highpass, bandpass or bandstop.(a)M. J. Roberts - 7/12/03Solutions 6-42v (t)iv (t)x+-v (t)o+-C3C4R1R2R5V V VV Vx i ox of G G j fC j fC f G f j fCf j fC f G( ) + + + ( ) ( ) ( ) ( ) ( ) 1 2 3 4 1 43 52 2 2 02 0 G G j fC j fC j fCj fC GffGfxoi1 2 3 4 43 512 2 22 0+ + + ( ) ,, ]]] ( )( ),, ]]] ,, ]]] ( ) VVV + + + ( ) ( ) G G j fC j fC G f C C1 2 3 4 523 42 2 2 ( ) + + ( ) + + ( ) 2 223 4 5 4 3 1 2 5 f C C j fG C C G G GV Vo If G G j fC j fC Gj fC f ( ) + + + ( )( )1 2 22 01 2 4 3 13 H f j fGCf C C j fG C C G G G( ) ( ) + + ( ) + + ( )22 21 323 4 5 4 3 1 2 5 H f j fR Cf RR C C j fR C C RR( ) ( ) + ( ) +j(, \,(22 2 15 321 5 3 4 1 4 312 Bandpass(b)v (t)iv (t)x+-v (t)o+-C3C4C1R2R5M. J. Roberts - 7/12/03Solutions 6-43V V VV Vx o ix of j f C C C G j fC f j fC fj fC f G f( ) + + ( ) +[ ] ( ) ( ) ( ) ( ) 2 2 2 02 01 4 3 2 4 13 5 j f C C C G j fCj fC Gffj fCfxoi2 22201 4 3 2 43 51 + + ( ) + ,, ]]] ( )( ),, ]]] ,, ]]] ( )VVV ( ) + + + ( ) + 2 223 4 5 1 4 3 2 5 f C C j fG C C C G GV Vo if j f C C C G j fCj fC f ( ) + + ( ) +( )1 2 22 01 4 3 2 13 HVVf fff CCf C C j fG C C C G Goi( ) ( )( ) ( )( ) + + + ( ) +22 221 323 4 5 1 4 3 2 5 H f f R R CCf R R C C j fR C C C( ) ( )( ) + + + ( ) +22 2 122 5 1 322 5 3 4 2 1 4 3 Highpass(c)v (t)iv (t)x+-v (t)o+-R3R4R1C2C5V V VV Vx i ox of G G G j fC G f G fG f j fC f( ) + + + ( ) ( ) ( ) ( ) ( ) 1 3 4 2 1 43 52 02 0G G G j fC GG j fCffGfxoi1 3 4 2 43 5122 0+ + + ,, ]]] ( )( ),, ]]] ,, ]]] ( )VVV ( ) + + + ( ) + 2 222 5 1 3 4 5 3 4 f C C j f G G G C G GV Vo if G G G j fC GG f ( ) + + +( )1201 3 4 2 13M. J. Roberts - 7/12/03Solutions 6-44HVVf ffGGf C C j f G G G C G Goi( ) ( )( ) ( ) + + + ( ) +1 322 5 1 3 4 5 3 42 2 H f Rf RR R C C j f R R RR RR C R( ) ( ) + + + ( ) +421 3 4 2 5 3 4 1 4 1 3 5 12 2 Lowpass33. When music is recorded on analog magnetic tape and later played back, a high-frequencynoise comp