Solutions Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 3rd Edition by Roberts IBSN 0078028124

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Solutions Manual

for Signals and Systems

for Signals and Systems Analysis Using Transform Methods

Analysis Using Transform Methods

and MATLAB 3rd Edition by Roberts IBSN 0078028124

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Chapter 2 -

Chapter 2 - Mathematical De

Mathematical Description o

scription of

f

Continuous-T

Continuous-Time

ime Signals

Signals

Solutions

Exercises With Answers in Text

Signal Functions Signal Functions

1. If

1. If ggt t 77ee22tt 33 write out and simplifywrite out and simplify (a) (a) gg 33 77ee99 8.63878.6387 101044(b)(b) g g22 t t 77ee2222t t 33 77ee7722tt(c)(c) ggtt // 1010 44 7 7eet t /511/511 (d) (d) gg jt jt 77ee j j22t t 33 (e) (e) gg jt jt gg jt jt 2 2 77ee 3 3 ee j j22tt ee j j22 tt 2 2 7

7ee33coscos22tt 0.3485 cos20.3485 cos2tt

g g jt jt33 jt jt33 g g _jt_jt _jt_jt (f) (f) 2 2 22 77e e ee 7 cos7 costt 2 2 22 2. If

2. If gg x x xx22 44 x x 44 write out and simplifywrite out and simplify (a)

(a) gg z z z z22 44 z z 44

(b)

(b) ggu u vv uu vv22 44uu vv 4 u4 u22 vv22 22uvuv 44uu 44vv 44

(c)

(2)

(d) (d) ggggt t ggt t 22 44tt 44 t t 22 44tt 4422 44 t t 22 44tt 44 44 g gggt t t t 44 88t t 33 2020t t 22 1616tt 44 (e) (e) gg22 4 8 4 0 4 8 4 0 3.

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(a)

(a) ee

3 3 j j 2.32.3

ee33 j j 2.32.3 ee33 ee j j2.32.3 ee33 coscos2.32.3 j jsin2.3sin2.3

ee3 j3 j 2.32.3 ee33 coscos222.32.3 sinsin222.32.3 ee33 0.04980.0498

(b) (b) ee22 j j 66 ee22 j j 66 ee22 ee j j66 ee22 7.38917.3891 (c) (c) 100 100 8 8 j j1313 100 100 100100 100100 6.5512 6.5512 8 8 j j1313 88 j j1313 ₈₈22 13 1322 4. Let 4. Let GG f f j j44 f f .. 2 2 j j77 f f / / 1111 (a)

(a) What What value value does does the the magnitude magnitude of of this this function function approach approach asas f f approachesapproaches positive infinity? positive infinity? lim lim j j44 f f jj44 f f 4 4 4444 6.2856.285 f f ₂₂_j_j₇₇_f_f _{/ 11}_{/ 11} _j_j₇₇_f_f _/_{/ 11}₁₁ ₇_{7 /}_{/ 11}₁₁ ₇₇

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0 0

//22

(b)

(b) What What value value (in (in radians) radians) does does the the phase phase of of this this function function approach approach asas f f

approaches zero from the positive side? approaches zero from the positive side? li limm f f 00 j j44 f f 22 j j77 f f // 1111 limlim j j44 f f 22 // 22 0 0 / / 22 5. Let 5. Let XX f f jf jf jf jf 1010 (a)

(a) Find Find the the magnitudemagnitude XX 44 and and the the angle angle in radiansin radians

X X f f jf jf XX 44 j4 j4 44ee 0.37140.3714 ee j j1.191.19 jf jf 1010 j j4 4 1010 10.7710.77ee j j0.38050.3805 X X 44 0.3714 0.3714 (b)

(b) What What value value (in (in radians) radians) does does approach approach asas f fapproaches zero fromapproaches zero from the positive side?

the positive side?

Shifting and Scaling Shifting and Scaling

6.

6. For For each each functionfunction ggtt graphgraph ggtt,, gt gt ,, ggtt 11, and, and gg22_{tt .. (a)}_(a)

(b) (b) g( g(t t )) g(g(t t )) 4 4 33 2 2 tt 11 t t -3 -3 g(-g(-t t )) g(-g(-t t )) -g(-g(t t )) -g(-g(t t )) 4 4 33 -2 -2 tt 11 t t -3 -3 3 3 2 2 tt _-1_-1 t t 4 4 -3-3

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(a) (a) (a) (a) 2 2 22 1 1 11 0 0 00 -1 -1 -1-1 -2 -2 -2-2 -4 -2 -4 -2 t t 00 22 44 -4 -4 -2-2 t t 00 22 44 (b) (b) (b) (b) 2 2 22 1 1 11 0 0 00 -1 -1 -1-1 -2 -2 -2-2 -4 -2 -4 -2 t t 00 22 44 -4 -4 -2-2 t t 00 22 44 (c) (c) (c) (c) 2 2 22 1 1 11 0 0 00 -1 -1 -1-1 -2 -2 -2-2 -4 -4 -2-2 00 22 44 -4-4 -2-2 00 22 44 t t t t g g ( ( t t ) ) g g 1 1 ( ( t t ) ) g g 1 1 ( ( t t ) ) 1 1 g g ( ( t t ) ) g g ( ( t t ) ) g g ( ( t t ) ) 2 2 2 2 2 2 g( g(t t -1)-1) g(g(t t -1)-1) g(2g(2t t )) g(2g(2t t )) 4 4 1 1 33 t t 3 3 1 1 22 t t -3 -3 4 4 33 -- 11 1 1 tt 11 t t 2 2 -3 -3 7.

7. Find Find the the values values of of the the following following signals signals at at the the indicated indicated times.times. (a)

(a) xxt t 22 r r ectecttt // 44 ,, xx 11 22 r r ectect11 // 44 22

(b)

(b) xxt t 5r 5r ectecttt // 2sgn2sgn22t t ,, xx 0.50.5 5r 5r ectect 11 // 44sgnsgn₁₁ 55

(c)

(c) xxt t 99 rectrecttt //10sgn10sgn33t t 2 ,, x2 x 11 99 rectrect 11 //10sgn10sgn 33 99

8.

8. For For each each pair pair of of functions functions in in Figure Figure E-8 E-8 provide provide the the values values of of the the constantsconstants A A,,

t

t ₀₀ andandwwin the functional transformationin the functional transformation gg₂₂t t A Agg₁₁t t t t ₀₀ // ww ..

Figure E-8 Figure E-8 Answers: (a)

Answers: (a) A A 2 2 ,,t t ₀₀ 1,1,ww 11 , (b), (b) A A 2 2 ,,t t ₀₀ 0 0 ,,ww 1 / 21 / 2 , (c), (c) A A

1 / 2 ,

1 / 2 ,t t ₀₀ 1,1,ww 22 9.

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-10 -10 -5-5 00 5 5 10 10 -10 -10 -5-5 00 55 1010 t t t t -10 -10 -5-5 00 5 5 110 0 -1-10 0 -5-5 00 55 1010 t t t t 0 0 0 0 g g ( ( t t ) ) g g ( ( t t ) ) g g ( ( t t ) ) 1 1 1 1 1 1 g g ( ( t t ) ) g g ( ( t t ) ) g g ( ( t t ) ) 2 2 2 2 2 2 t

t ₀₀ andandaa in the functional transformationin the functional transformation gg₂₂t At Agg₁₁wwtt t t ₀₀..

A A= 2,= 2,t t ₀₀ = 2,= 2,ww== -2-2 8 8 88 4 4 44 (a) (a) 0 0 00 -4 -4 -4 -4 -8 -8 -8 -8

Amplitude comparison yields

Amplitude comparison yields A A 22. . Time Time scale cscale comparison yieldsomparison yields ww 22.. g g₂₂22 22 gg₁₁2222 t t ₀₀ 22 gg₁₁00 44 22t t ₀₀ 00 t t ₀₀ 22 A A= 3,= 3,t t = 2, = 2,ww= 2= 2 (b) (b) 8 8 88 4 4 44 0 0 00 -4 -4 -4 -4 -8 -8 -8 -8

Amplitude comparison yields A A 33. . Time scale Time scale comparison yielcomparison yieldsds ww 22.. g g₂₂22 3g3g₁₁2222 t t ₀₀ 3g3g₁₁00 4 4 22t t ₀₀ 00 t t ₀₀ 22 A A= -3,= -3, tt= -6,= -6, ww= 1/3= 1/3 8 8 88 4 4 44 (c) (c) 0 0 00 -4 -4 -4 -4 -8 -8 -10 -10 -5-5 0 0 5 5 1010 -8-8-10 -10 -5-5 00 55 1010 t t t t

Amplitude comparison yields A A 33. . Time scale Time scale comparison ycomparison yieldsields ww 11 // 33.. g

g₂₂00 3g3g₁₁11 // 3030 t t ₀₀ 3g3g₁₁22 t t ₀₀ // 33 22 t t ₀₀ 66 OR

OR

Amplitude comparison yields A A 33. . Time scale Time scale comparison ycomparison yieldsields ww 11 // 33.. g

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g g 1 1 ( ( t t ) ) g g 1 1 ( ( t t ) ) g g 2 2 ( ( t t ) ) g g 2 2 ( ( t t ) ) A A= -2,= -2,t t ₀₀= -2,= -2,ww = 1/3= 1/3 8 8 88 4 4 44 (d) (d) 0 0 00 -4 -4 -4 -4 ---10 -10 -5-5 0 0 5 5 1010 ---10 -10 -5-5 00 55 1010 t t t t

Amplitude comparison yields A A 22. . Time scale Time scale comparison ycomparison yieldsields ww 11 // 33.. g g₂₂44 22gg₁₁11// 3344 t t ₀₀ 3g3g₁₁22 t t ₀₀ // 33 4 / 4 / 33 22 t t ₀₀ 22 A A= 3,= 3,t t ₀₀= -2,= -2, ww= 1/2= 1/2 8 8 88 4 4 44 (e) (e) 0 0 00 -4 -4 -4 -4 ---10 -10 -5-5 00 5 5 1010 ---10 -10 -5-5 00 55 1010 t t t t

Amplitude comparison yields A A 33. . Time scale Time scale comparison yielcomparison yieldsds ww 11 // 22 .. g

g₂₂00 3g3g₁₁11// 2200 t t ₀₀ 3g3g₁₁11 t t ₀₀ / / 22 11 t t ₀₀ 22 Figure E-9

Figure E-9 10.

10. In In Figure Figure E-10 E-10 is is plotted plotted a a functionfunction gg₁₁tt which is zero for all time outside thewhich is zero for all time outside the range plotted.

range plotted. Let some other Let some other functions be defined byfunctions be defined by

tt 33 g

g₂₂t t 3g3g₁₁ 22 t t ,, gg₃₃t t 22gg₁₁tt // 44 ,, gg₄₄tt gg₁₁

2 2

Find these values. Find these values. (a) (a) gg₂₂11 33 (b)(b) gg₃₃11 3.53.5 3 3 33 11 (c) (c) gg44ttgg33tt_t_t₂₂ 2 2 11 22 (d)(d) ₃₃gg44t dt t dt The function

The function gg₄₄tt is linear between the integration limits and the area under it isis linear between the integration limits and the area under it is aa triangle.

triangle. The base width is The base width is 2 and the heig2 and the height is -ht is -2. 2. Therefore the Therefore the area is -2.area is -2.

1 1 g g₄₄t dtt dt 22 3 3

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1 1 g ( g (t t )) -4 -3 -4 -3 4 4 3 3 2 2 1 1 -2 -1 -2 -1 -1 -1 -2 -2 -3 -3 -4 -4 1 2 3 4 1 2 3 4 t t 11.

11. A A functionfunction GG f f is defined byis defined by

Figure E-10 Figure E-10

G

G f f ee j j 22 f f rectrect f f // 22 .. Graph the magnitude and phase of

Graph the magnitude and phase of GG f f 1010 GG f f 1010 over the range,over the range, 20

20 f f 2020 ..

First imagine what

First imagine what GG f f looks like. looks like. It It consists of a consists of a rectangle centered rectangle centered atat f f 00 ofof width, 2, mul

width, 2, multiplied by tiplied by a complex exponential. a complex exponential. Therefore for Therefore for frequencies greatfrequencies greaterer than one i

than one in magnitude it is n magnitude it is zero. zero. Its magIts magnitude is simply nitude is simply the magnitude of tthe magnitude of thehe rectangle function because the magnitude of the complex exponential is one for rectangle function because the magnitude of the complex exponential is one for any

any f f ..

ee j j22 f f coscos 22 f f j jsinsin22 f f coscos22 f f j jsinsin22 f f ee j j22 f f coscos222 f2 f sinsin2222 f f 11 The phase (angle) of

The phase (angle) of GG f f is simply the phase of the complex exponentialis simply the phase of the complex exponential between

between f f 11 andand f f 11 and undefined outside that range because the phase of theand undefined outside that range because the phase of the rectangle function is zero between

rectangle function is zero between f f 11 andand f f 11and undefined outside thatand undefined outside that rangerange and the phase

and the phase of a product is of a product is the sum of the sum of the phases. the phases. The phase of the The phase of the complexcomplex exponential is

exponential is

ee j j 22 f f ccosos22 f f j jssiinn 22 f f tantan11 ssiinn22 f f tantan11 ssiinn22 f f

ccosos22 f f ccosos22 f f

ee j j 22 f f tantan11tan2tan2 f f The inver

The inverse tangent se tangent function is function is multiple-valued. multiple-valued. Therefore therTherefore there are e are multiplemultiple correct answers

correct answers for this phase. for this phase. The simplest of them iThe simplest of them is found by choosings found by choosing ee j j 22 f f 22 f f

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1

1 22

which is simply the coefficient of

which is simply the coefficient of j j in the original complex exponential expression.in the original complex exponential expression. A more general solution would be

A more general solution would be ee j j22 f f 22 f f 22nn ,, nn anan iintnteeger ger . The. The solutionsolution of the original problem is simply this solution except shifted up and down by 10 in of the original problem is simply this solution except shifted up and down by 10 in

f

fand added.and added.

G

G f f 1010 GG f f 1010 ee j j22 f f1010 rectrect f f 1010 ee j j2 f2 f1010 rectrect f f 1010 2

2 22

12. Let

12. Let xx t t 3r 3r ectect t t 11 // 66 and xand x t t r r aampmpt t uu t t uu t t 44 ..

1

1 22

(a)

(a) Graph Graph them them in in the the time time rangerange 1010 tt 1010. . Put scale Put scale numbers onnumbers on thethe vertical axis so that actual numerical values could be

vertical axis so that actual numerical values could be read from theread from the graph. graph.

x

x ((t

t

))

1 1

x

x ((t

22

t

))

6 6 66 -10 10 -10 10 -6 -6 -10 10 -10 10 -6 -6 (b) Graph

(b) Graph xx t t xx 22t t xx t t // 22 in the time rangein the time range 1010 tt 1010. Put. Put

scale numbers on the vertical axis so that actual numerical values scale numbers on the vertical axis so that actual numerical values could be read from the graph.

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5 5

11 11

x

x 22t t 3r 3r ectect 22t t 11 // 66 and xand x t t // 22 r r aampmpt t // 22 uu t t // 22 uu t t / 2/ 2 44

1

1 22

x

x 22t t 33r r ectectt t 11 // 22 // 33 andand xx t t // 22 11 // 2r 2r aampmpt t uu t t uu t t 88

1 1 22

x

x (2

(2t

t

)-x

1 1 22

((t

t

/2)

8 8 -10 10 -10 10 -8 -8 13.

13. Write Write an an expression expression consisting consisting of of a a summation summation of of unit unit step step functions functions to to represent represent aa signal which consists of rectangular pulses of width 6 ms and height 3 which occur signal which consists of rectangular pulses of width 6 ms and height 3 which occur at a uniform rate of 100 pulses per second with the leading edge of the first pulse at a uniform rate of 100 pulses per second with the leading edge of the first pulse occurring at time occurring at time tt 00.. x xt t 33 n n00 u utt 0.010.01nn uutt 0.010.01nn 0.0060.006 14.

14. Find Find the the strengths strengths of of the the following following impulses.impulses. (a) (a) 3344t t 3 3 44t t 33 11 t t Strength Strength is is 3 3 / / 44 4 4 (b) (b) 5533t t 11 5 533tt 11 55tt 11 Strength isStrength is 55 3 3 33 15.

15. Find Find the the strengths strengths and and spacing spacing between between the the impulses impulses in in the the periodic periodic impulseimpulse 9 9 55t t .. 9 9 55t t 99 55t t 1111k k 99 t t 1111kk // 55 11 11 k k k k

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The strengths are all -9/5 and the spacing between

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1 1 1 1 t t Derivatives

Derivatives and Integrals and Integrals of of FunctionsFunctions

16.

16. Graph Graph the the derivative derivative ofof xxt t 11 eettuut t ..

This function is constant zero for all time before time,

This function is constant zero for all time before time, tt 00, therefore its, therefore its derivative during that

derivative during that time is zero. time is zero. This function is This function is a constant minus a decaya constant minus a decayinging exponential after time,

exponential after time, tt 00, and its derivative in th, and its derivative in that time is therefore aat time is therefore a positive positive decaying exponential. decaying exponential. x x t t ee ,, tt 00 0 0 ,, tt 00

Strictly speaking, its derivative is not defined at exactly

Strictly speaking, its derivative is not defined at exactly tt 00. . Since tSince the value he value of of aa physical signal at a single

physical signal at a single point has no point has no impact on any phimpact on any ph ysical system (as long as itysical system (as long as it is finite) we can choose any finite value at time,

is finite) we can choose any finite value at time, tt 00, without changing the effect, without changing the effect ofof this sig

this signal on any nal on any physical system. physical system. If If we choose 1/2, we choose 1/2, then we can wrthen we can write theite the derivative as derivative as x x t t eettuut t .. x( x(t t )) t t -1 -1 -1 -1 dx/d dx/d t t t t -1 -1 -1 -1

Alternate Solution using the chain rule of differentiation and the fact that the Alternate Solution using the chain rule of differentiation and the fact that the impulse occurs at time

impulse occurs at time tt 00 ..

d d x xtt 11 eett t t eettuut t eettuutt dt dt 1144224433 0 for 0 for t t 00 17.

17. Find Find the the numerical numerical value value of of each each integral.integral.

11 11 44 (a) (a) u u 44 t t dt dt dtdt 66 2 2 22 8 8 (b) (b) tt 33 22 44t dt t dt 1 1

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3 3 2 2 4 4 8 8 8 8 88 tt 33 22 44t dtt dt tt 33dtdt 22 44t dt t dt 1 1 1 1 11 8 8 88 tt 33 22 44t dtt dt 0 0 22 11 // 44 t dtt dt 11 / / 22 (c) (c) 1 1 5/2 5/2 2 2 33t dt t dt 1/2 1/2 5/2 5/2 _5/2_5/2 1 1 1 15/25/2 2 2 33t dtt dt 33tt 22nn dtdt 3 3 tt 22nn//33dt dt 1/2 1/2 1/21/2 nn 5/2 5/2 1/2 1/2 nn 1 1 2 2 33t dtt dt 11 11 11 11 1/2 1/2 (d)

(d) tt 44 r r aampmp 22t dtt dt r r aampmp 2244 r r aampmp88 88

10 10 10 10 10 10 (e) (e) r r aampmp 22tt 44dtdt 22tt 44dt tdt t22 44t t 100 40 4 8 64100 40 4 8 64 3 3 22 (f) (f) 82 82 3s

3siinn200200t t t t 77 dtdt 00 (Impulse does not occur between 11 and 82.)(Impulse does not occur between 11 and 82.)

11 11 (g) (g) 5 5 sin

sintt/ 20/ 20dtdt = = 0 0 Odd function Odd function integrated integrated over over

5 5 symmetrical limits. symmetrical limits. 10 10 10 10 (h) (h) 3939t t 22 t t 11dtdt 3939 t t 22 t t 11 44k k dt dt 4 4 2 2 22 k k 10 10 39 39t t 22 10 10 t t 11dtdt 3939 t t 22 t t 11 t t 55 t t 99 dt dt 4 4 2 2 22 10 10 39 39t t 22 2 2 t t 11dtdt 39391122 5522 9922 41734173

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(i) (i)

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prior written consent of McGraw-Hill Education.ill Education.

t t

ee1818ttuu t t 1010t t 22 dt dt ee1818ttuu t t 1010t t 11 // 55dt dt

Using the scaling property of the impulse, Using the scaling property of the impulse,

ee1818ttuu t t 1010t t 22 dtdt 11 ee1818ttuu t t t t 11 // 55 dt dt 10

10

Using the sampling property of the impulse, Using the sampling property of the impulse,

ee1818ttuu t t 1010t t 22 dtdt 11 ee18/518/5 0.0027320.002732 10 10 9 9 99 (j) (j) 99 t t 44 // 55d d t t 4545 t t 44 dtdt 4545 2 2 22 3 3 33 (k) (k) 5 5 33t t 44 dtdt 55 t t 44 dtdt 00 6 6 3366 (l) (l) r r aamp3mp3t t t t 44 dtdt r r aampmp33 44 1212 (m) (m) 17 17 3 3ttccosos22tt // 33dt dt 1 1

The impulses in the periodic impulse occur at ...-9,-6,-3,0,3,6,9,12,15,18,... The impulses in the periodic impulse occur at ...-9,-6,-3,0,3,6,9,12,15,18,... At each of these points the cosine value

At each of these points the cosine value is the same, one, because itsis the same, one, because its period is 3 seconds also.

period is 3 seconds also. So the integral value is simple the sum of theSo the integral value is simple the sum of the strengths of the impulse that occur in the time range, 1 to 17 seconds strengths of the impulse that occur in the time range, 1 to 17 seconds

17 17 3 3ttccosos22tt // 33dtdt 55 1 1 18.

18. Graph Graph the the integral integral from from negative negative infinity infinity to to timetime tt of the functions in Figure E-18of the functions in Figure E-18 which are zero for all time

which are zero for all time tt 00.. This is the integral

This is the integral gg d d which, in geometrical terms, is the accumulated which, in geometrical terms, is the accumulated area under the function

area under the function ggtt from from time time to to timetime t t . . For For the case the case of the of the twotwo back-to-back rectangular pulses, there is no accum

back-to-back rectangular pulses, there is no accum ulated area until after timeulated area until after time tt 00

and then in the time interval

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prior written consent of McGraw-Hill Education.ill Education.

dt dt

a maximum area of one at time

a maximum area of one at time tt 11. . In In the second the second time intertime intervalval 11 tt 22 the areathe area isis linearly declining at half the rate at which it increased in the first time interval linearly declining at half the rate at which it increased in the first time interval

0

0 tt 11down to a value of 1/2 down to a value of 1/2 where it stays because there is no accumulation of areawhere it stays because there is no accumulation of area for

for tt 2.2.

In the second case of the triangular-shaped function, the area does not accumulate In the second case of the triangular-shaped function, the area does not accumulate linearly, but rather non-linearly because the integral of a linear function is a linearly, but rather non-linearly because the integral of a linear function is a second-degree polynomial.

second-degree polynomial. The rate of accumulatiThe rate of accumulation of area is on of area is increasing up toincreasing up totimetime

tt 11 and then decreasing (but still positive) until timeand then decreasing (but still positive) until time tt 22 at which time itat which time it stopsstops completely.

completely. The final value The final value of the accumulated area of the accumulated area must be the total must be the total area of tarea of thehe triangle, which, in this case, is one.

triangle, which, in this case, is one.

g( g(t t )) g(g(t t )) 1 1 1 1 2 2 33 _t_t 1 1 2 2 1 1 t t 1 1 2 2 33 Figure E-18 Figure E-18 g( g(t t )) d d t t g(g(t t ) d ) d t t 1 1 1 1 2 2 t t 1 1 22 33 1 1 t t 1 1 2 2 33 19.

19. If If 4 4 uutt 55 dd xxtt, what is the function, what is the function xxtt ?? xx t t 44 ramprampt t 55

Generalized Derivative Generalized Derivative

20.

20. The The generalized generalized derivative derivative ofof 18 rect18 rect tt 22 consists of two impulses.consists of two impulses. 3

3 Find their numerical locations and strengths. Find their numerical locations and strengths. Impulse

Impulse #1: #1: Location Location isistt = 0.5 and strength is 18.= 0.5 and strength is 18. Impulse

Impulse #2: #2: Location Location isistt = 3.5 and strength is -18.= 3.5 and strength is -18.

Even

Even and and Odd Odd FunctionsFunctions

21.

21. Classify Classify the the following following functions functions as as even, even, odd odd or or neither neither .. (a)

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15 15

Cosine is even but the shifted triangle is neither even

Cosine is even but the shifted triangle is neither even nor odd whichnor odd which means it has a non-zero even

means it has a non-zero even part and a non-zero odd part. part and a non-zero odd part. So the productSo the product also has a non-zero even part and a non-zero odd part.

also has a non-zero even part and a non-zero odd part. (b)

(b) ssiinn 22t t r r ectecttt // 55 Odd Odd

Sine is odd and

Sine is odd and rectangle is rectangle is even. even. Therefore the Therefore the product is odd.product is odd. 22.

22. An An even even functionfunction ggtt is described over the time rangeis described over the time range 00 tt 1010 by by 2 2tt , , 00 tt 33 g gtt 33tt , , 33 tt 7 7 .. 2 2 , , 77 tt 1010 (a)

(a) What What is is the the value value ofof ggtt at timeat time tt 55?? Since

Since ggtt is even,is even, ggt t ggt t gg55 gg55 15 15 33 5 5 00 .. (b) (b) What What is is thethe value of the first derivative of g(

value of the first derivative of g(t t ) at time) at time tt 66?? Since

Since ggtt is even,is even,

d d g gt t dd ggt t dd ggt t dd ggt t 33 33 .. dt dt dt dt dtdt _t_t 6₆ dtdt _t_t 6₆ 23.

23. Find Find the the even even and and odd odd parts parts of of these these functions.functions. (a) (a) ggt t 22t t 22 33tt 66 g g_eet t 22tt22 33tt 6 6 22 tt 22 3 3tt 66 2 2 2 2tt22 33tt 6 6 22tt2233tt 66 4 4tt22 1212 2 2 6 6t t 2 2t t 22 66 g g_o_ot t 33t t 2 2 22 (b) (b) ggt t 2020 coscos 4040 tt // 44 20 20 ccos 40os 40tt / 4 20/ 4 20 ccos 40os 40tt / / 44 g g_eet t 2 2 Using

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2 2

20

20 coscos 4040t t cos // 4cos 4 sinsin 4040t t sinsin // 44

g

g_eet t 20 20 cos cos 4040ttcos / cos / 4 si4 sin 40n 40ttsin sin / / 44

2 2

20

20 coscos 4040t t coscos // 44 sinsin 4040t t sinsin // 44

g

g_eet t 20 20 cos cos 4040ttcos / cos / 4 sin 4 sin 4040ttsin / 4sin / 4

2 2

g

g_eet t 2020ccosos / / 44ccosos 4040 t t 2020 / / 22ccosos 4040 t t 20 20 ccos 40os 40tt / 4 20/ 4 20 ccos 40os 40tt / / 44 g g_o_ot t 2 2 Using

Using coscos z z₁₁ z z₂₂ coscos z z₁₁cos zcos z₂₂ sinsin z z₁₁sinsin z z₂₂,, 20

g

g_o_ot t 20 20 cos cos 4040ttcos / cos / 4 sin 4 sin 4040ttsin / 4sin / 4

2 2

20

g

g_o_ot t 20 20 cos cos 4040ttcos / cos / 4 sin 4 sin 4040ttsin / 4sin / 4

2 2

g

g_o_ot t 2020 ssiinn // 44ssiinn 4040 t t 2020 / / 22 ssiinn 4040 t t

(c) (c) ggtt 22tt 33tt66 1 1 t t 2 2tt22 33tt 66 22tt22 33tt 66

(19)

g g_eett 11tt 11tt 2 2 2 2tt22 33tt 6611 tt 22tt22 33tt 6611tt 11 tt11 tt g g_eett 2 2

(20)

2 2 2 2 2 2 g g_eet t 44t t 22 12 12 66t t 22 2 211 t t 22 6 6 55t t 22 1 1 t t 22 2 2tt22 33tt 66 22tt22 33tt 66 g g_o_ott 11tt 11tt 2 2 2 2tt22 33tt 6161 tt 22tt22 33tt 6161 tt 11 tt11 tt g g_o_ott 2 2 g g_o_ot t 66tt 44t t 33 1212t t 2 211 t t 22 t t 2 2t t 22 99 1 1 t t 22 (d) (d) ggt t t t 22 t t 2211 44t t 22 g gt t tt 22 tt221 1 44tt22 odd odd1212331144224433 even

even eveneven

Therefore

Therefore ggtt is odd,is odd, gg_eet t 0 and g0 and g_o_ot t t t 22 t t 11 44t t 22

(e) (e) ggt t t t 22 t t 11 44t t t t 22 t t 1144t t t t 22 t t 1144t t g g_eet t 2 2 t t 22 t t 1144t t t t 22 t t 1144t t g g_eet t 77t t 2 2 g g_o_ot t 2 2 ggoot t t t 22 44tt (f) g (f) gtt 20 420 4tt 77t t 1 1 t t g g_eett 20 20 44tt22 77t t 1 1 t t 20 20 44tt22 77t t 1 1 t t 2 2 40 40 88tt22 1 1 t t 2 2 20 20 44tt22 1 1 t t

(21)

1 1 1 1 1 1 1 1 1 1 ,, g g_o_ott 20 20 44tt22 77t t 1 1 t t 20 20 44tt22 77t t 1 1 t t 2 2 14 14t t 1 1 t t 2 2 7 7t t 1 1 t t 24.

24. Graph Graph the the even even and and odd odd parts parts of of the the functions functions in in Figure Figure E-24.E-24.

To graph the even part of graphically-defined functions like these, first graph To graph the even part of graphically-defined functions like these, first graph

g

gtt.. Then add it (graphically, point by point) toThen add it (graphically, point by point) to ggtt and (graphically) divide theand (graphically) divide the sum

sum by by two. two. Then, to graph Then, to graph the odd the odd part, spart, subtractubtract ggtt fromfrom ggtt (graphically) and divide the difference b

(graphically) and divide the difference b y two.y two.

(a) (b) (a) (b) g( g(t t )) g(g(t t )) 1 1 t t ₁₁ ₂₂ t t -1 -1 Figure E-24 Figure E-24 g g_ee((t t )) gg_ee((t t )) 1 1 tt ₁₁ ₂₂ t t -1 -1 g ( g (oot t )) ggoo((t t )) 1 1 1 1 tt 1 1 22 t t -1 -1 (a) (b) (a) (b) 25.

(22)

1 1 -1 -1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 -1 1 -1 -1 11 -1 1 -1 1 1 1 (a) (b) (a) (b) 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 11 t t g( g(t t )) Multiplication Multiplication t t 1 1 -1 -1 t t 1 1 -1 -1 1 1 -1 -1 t t 1 1 -1 -1 g( g(t t )) Multiplication Multiplication g( g(t t )) g(g(t t )) 1 1 -1 -1 t t 1 1 -1 -1 1 1 t t -1 -1 11 -1 -1 (c) (c) (d)(d) tt t t g( g(t t )) g(g(t t )) Multiplication Multiplication t t Multiplication Multiplication t t g( g(t t )) g(g(t t )) t t t t (e) (f) (e) (f) 1 1 -1 -1 11 -1 -1 1 1 -1 -1 11 t t t t g( g(t t )) Multiplication Multiplication 1 1 1 1 t t -1 -1 g( g(t t )) Multiplication Multiplication 1 1 t t -1 -1 g( g(t t )) 1 1 -1 -1 11 -1 -1 t t g( g(t t )) 1 1 1 1 t t -1 -1

(23)

1 1 -1 -1 1 1 1 1 0 0 (g) (g) (h)(h) t t g( g(t t )) 1 1 -1 -1 -1-1 11 t t g( g(t t )) Division Division Division Division tt 1 1 t t g( g(t t )) -1 -1 t t Figure E-25 Figure E-25 g( g(t t )) 1 1 -1 1 -1 1 t t 26.

26. Use Use the the properties properties of of integrals integrals of of even even and and odd odd functions functions to to evaluate evaluate these these integralsintegrals in the quickest way.

in the quickest way.

1

1 1 1 1 1 11

(a)

(a) 22 t dtt dt 22_{_{ dt dt _{{{ dt}t t dt 22 22dtdt 44

1

1 11 eveneven 11 oddodd 00

(b) (b)

1/20 1/20

4

4 ccosos1010tt 88 ssiinn 55t t dtdt

1/20 1/20 4 4 ccos10os10t dtt dt 4 4224433 1/20 1/20 8 8 ssiinn 55t dt t dt 1 144224433 1/20 1/20 1/20 1/20 1/20 1/20 1 1 1/20 1/20 even

even 1/201/20 odd odd

1/20 1/20

4

4 ccosos1010tt 88 ssiinn55tt _dt_dt ₈₈

0 0 ccosos1010tt dt dt 88 10 10 (c) (c) 1/10 1/10 1/20 1/20 1/20 1/20 4 4 tt cos 10cos 10t dtt dt 00 odd odd1144224433 14 144422evev₄₄enen ₄₄₃₃ odd odd 1/10 1/10 cos 10

cos 10tt 1/101/10 1/101/10 cos 10cos 10tt

(d)

(d) t t _{{ sin 10sin 10t dtt dt 22 ttsin 10sin 10t dtt dt 22t t dtdt

1/10 1/10 oddodd 1 144224433 0 0 1010 00 1010 14 1422od od ₄₄d d ₃₃ even even 1/10 1/10 1 1 sin 10sin 10tt 1/10 1/10 1 1

(24)

1 1 t t {{ sinsin1010t t dt dt 22 22 1/10 1/10 oddodd1144224433 100100 1010 5050 1 14422od od ₄₄d d ₃₃ 00 even even 1 1 1 1 11 (e) (e) _{_{eett dtdt 22 eett dtdt 22 eettdtdt 22eett 2211 ee11 1.2641.264 1 1 eveneven 00 0 0 0 0

(25)

0 0 5 5 (f) (f) 1 1 t t _{{{ ee}tt dtdt 00 1 1 oo{{dd dd eveneven odd odd Periodic Signals Periodic Signals 27.

27. Find Find the the fundamental fundamental period period and and fundamental fundamental frequency frequency of of each each of of these these functions.functions. (a)

(a) ggt t 1010 coscos5050 t t f f ₀₀ 25 25 Hz Hz ,, T T ₀₀ 1 1 / 25 / 25 ss (b)

(b) ggt t 1010 coscos5050tt // 44 f f ₀₀ 25 25 Hz Hz ,, T T ₀₀ 1 1 / 25 / 25 ss (c)

(c) ggt t ccos50os50t t ssiinn1515 t t

The fundamental period of the sum of two periodic signals is the least The fundamental period of the sum of two periodic signals is the least common multiple

common multiple (LCM) of (LCM) of their two individual ftheir two individual fundamental periods. undamental periods. TheThe fundamental frequency of the sum of two periodic signals is the greatest fundamental frequency of the sum of two periodic signals is the greatest common divisor (GCD) of their two individual fundamental frequencies. common divisor (GCD) of their two individual fundamental frequencies. f

f ₀₀ GCDGCD25,1525,15 // 22 2.5 Hz , 2.5 Hz , T T ₀₀ 11 / 2/ 2.5 .5 0.4 s0.4 s (d)

(d) ggt t ccosos22 t t ssiinn 33t t ccosos55 tt 3 /3 / 44

f

f ₀₀ GCD 1GCD 1, 3 / , 3 / 2, 5 2, 5 // 22 1 1 / / 2 Hz 2 Hz ,, T T ₀₀

(e)

(e) ggt t 3s3siinn2020t t 88 ccosos 44t t

1 1 1 / 2 1 / 2 2 s2 s 10 10 2 2 22 f f ₀₀ GCDGCD , , Hz Hz ,,TT / 2 s/ 2 s (f)

(f) ggt t 1010 ssiinn2020t t 77 ccos10os10t t

f

f GCDGCD 1010,, 0 Hz ,0 Hz ,T T ,, ggtt is not periodicis not periodic

0

0 00

(g)

(g) ggt t 33ccosos20002000t t 88 ssiinn25002500t t f

(26)

(h) (h)

g

g₁₁t t isis periodic periodic withwith fundamental periodfundamental period T T ₀₁₀₁ 1515ss g

g₂₂t t isis periodic periodic withwith fundamental periodfundamental period T T ₀₂₀₂ 4040ss

T T LCMLCM 1515ss,40,40ss 120120ss ,, f f 11 8333833311 HzHz 0 0 00 120 120ss 33 28.

28. Find Find a a function function of of continuous continuous timetimett for which the two successive transformationsfor which the two successive transformations

t

t tt andand t t tt 11leave the function unchanged.leave the function unchanged. cos

cos22 nt nt ,, one. one.

1/

1/nnt t ,, nn an integer. an integer. Any even periodic Any even periodic function with function with a period a period of of

29.

29. One One period period of of a a periodic periodic signalsignal xxtt with periodwith period T T ₀₀ is graphed in Figure E-29.is graphed in Figure E-29. Assuming

Assuming xxtt has a periodhas a period T T ₀₀, what is , what is the value the value ofof xxtt at time,at time, tt 220ms220ms??

x(

t

))

4 4 3 3 2 2 1 1 -1 -1 -2 -2 -3 -3 -4 -4

t

5ms 10ms 15 5ms 10ms 15 ms 20msms 20ms T T ₀₀ Figure E-29 Figure E-29 Since the function is periodic with period 15 Since the function is periodic with period 15 ms,ms,

x

x220ms220ms xx220ms220ms nn 15ms15ms wherewherenn is any is any integer. integer. If If we choosewe choose nn 1414 wewe getget x

x220ms220ms xx220ms220ms 14 15ms14 15ms xx 220ms220ms 210ms210ms xx10ms10ms 22 ..

Signal Energy

Signal Energy and Pand Power of ower of SignalsSignals

30.

30. Find Find the the signal signal energy energy of of each each of of these these signals.signals.

1/2 1/2 2

2

(a)

(a) xxt t 22 r r ectectt t EE_x_x 22 r r ectecttt dtdt 44

1/2 1/2

dt dt 44

(b)

(27)

2 2

E

E _x_x A Auutt uutt 1010 dt dt AA22

10 10 dt dt 1010 A A22 0 0 (c) (c) xxt t uut t uu1010 t t 0 0 2 2 E E _x_x uutt uu1010 tt dt dt dt dt dtdt 10 10 (d)

(d) xxt t rectrect t t coscos22t t

2 2 1/21/2 2 2 11 1/2 1/2 E

E _x_x r r ectectttccosos 22tt dtdt

1/2 1/2 cos cos 22t dtt dt 2 2_1/2_1/2 11 c cosos 44t dt t dt (e)

(e) xxt t rectrect t t coscos 44 t t

E

E _x_x r r ectectttccosos 44tt 22 dtdt

1/2 1/2 1/2 1/2 cos cos22 44t dtt dt 11 2 2 1/2 1/2 1/2 1/2 1 1 c cosos 88t dt t dt 1 1 1/21/2 1/21/2 11 E E_x_x 2 2 dtdt cos 8cos 8t dtt dt 22 1/2 1/2 ₁₁1/21/2₄₄₄₄₂₂₄₄₄₄₃₃ 0 0 (f)

(f) xxt t rectrect t t sinsin22t t

E

E _x_x r r ectectttssiinn22tt 22 dtdt

1/2 1/2 1/2 1/2 sin sin22 22t dtt dt 11 2 2 1/2 1/2 1/2 1/2 1 1 c cosos 44t dt t dt

(28)

E E 2 2 2 2 (g) x (g) xtt 1 1 tt 1 1 ,, tt 11 0 0 , , otherwiseotherwise 1 1 11 2 2 x x tt 11 1 1 3 3 dt dt 22 0 0 1 1 t t 1122dtdt 22 0 0 t t 22 22tt 11dt dt 2 2 tt E E 22 tt22 t t 22 1 1 1 1 11 x x ₃₃ 0 0 3 3 33 31.

31. A A signal signal is is described described byby xxt t A Ar r ectect t t B Br r ectecttt 0.50.5 . . What What is is its its signalsignal energy?_energy?

E

E _x_x A Ar r ectecttt B Br r ectecttt 0.50.5 22dt dt

Since these are purely real functions, Since these are purely real functions,

E

E _x_x A Ar r ectecttt B Br r ectecttt 0.50.5 _dt_dt

E

E _x_x A A rectrect22 tt B B22 rectrect22tt 0.50.5 22 AB ABr r ectectttr r ectecttt 0.50.5dt dt

1/2 1/2 11 2 2 22 1/2 1/2 2 2 22 E

E_x_x A A dt dt BB dtdt 22 AB AB dt dt AA B B ABAB

1/2

1/2 0 0 00

32.

32. Find Find the the average average signal signal power power of of the the periodic periodic signalsignal xxtt in Figure E-32.in Figure E-32.

x( x(t t )) 3 3 2 2 1 1 -4 -3 -2 -4 -3 -2 -1-1 -1 -1 -2 -2 -3 -3 1 2 3 4 1 2 3 4 Figure E-32 Figure E-32

(29)

0 0 t t ₀₀T T ₀₀ 22 1 1 11 33 11 P P 11 xxtt 22dtdt 11 xxtt 22dtdt 11 22tt 22dtdt 44 tt22dtdt 44 tt 88 T T ₀₀ _t_t 33₁₁ 33₁₁ 33₁₁ 3 3 33₁₁ 99

(30)

0 0 T T 4 4 44 1 1 2 2 2 2 1 1 2 2 0 0 5 5 33.

33. Find Find the the average average signal signal power power of of each each of of these these signals.signals. (a)

(a) xxt t AA PP_x_x limlim

T T /2/2

A

A22dtdt limlim A A

T T /2/2

dt

dt limlim A A T T AA22

T T TT

T

T/2/2 T T T T TT /2/2 T T T T

(b)

(b) xxt t uut t PP_x_x limlim

T T /2/2 u utt 22 dtdt limlim 11 T T /2/2 1 1 TT 11 dt dt limlim T T _T_T T T /2/2 (c)

(c) xxt t A Acoscos22 f f ₀₀tt

T T _T_T 0 0 T T TT 2 2 22 P P_x_x 11 T T T T 00/2/2 A

Accosos22 f f ₀₀tt _dt_dt A A22 T T T T 00/2/2 cos cos2222 f f 00t t dt dt 0 0 T T ₀₀/2/2 00 T T ₀₀/2/2 A A22 T T 00/2/2 A A22 ssiinn44 f f tt22T T 00/2/2 P P_x_x 2 2T T 11 c cosos 44 f tt f 22dtdt ₂₂_T_T tt 0 0 0 0 T T ₀₀/2/2 00 44 f f 00 _T_{T /2}_/2 A

A22 sin4sin4 f T f T / / 2 2 22 sinsin44 f T f T / / 2 2 22 A A22

0 0 0 0 0 0 00 P P_x_x 2 2T T T T 00 44 f f 44 f f 22 0 0 ₁₄₄₄₄₁₄₄₄₄00₄₄₄₄₄₄₄₄₂₂₄₄₄₄₄₄₄₄₄₄00₄₄₄₄₄₄₃₃ 0 0

The average signal power of a periodic power signal is unaffected if it is The average signal power of a periodic power signal is unaffected if it is shifted in t

shifted in time. ime. Therefore we cTherefore we could have found the averould have found the average signal powerage signal power of

of A Acoscos22 f f ₀₀t t instead, which is somewhat instead, which is somewhat easier algebraically.easier algebraically. (d)

(d) xxt t is periodic with fundamental period four and one fundamental period isis periodic with fundamental period four and one fundamental period is described by described by xx t t t t 11 t t ,, 11 tt 55 .. P P 11 xxt t 22dtdt 11 t t t t 1122dtdt 11 5 5 t t 44 22t t 33 t t 22 dt dt x x 0 0 00 1 1 11

(31)

3 3 5 4 3 5 4 3 55 P P 11 t t tt t t 11 3125 625 125 1 1 13125 625 125 1 1 1 x x 4 4 5 5 22 4 4 55 1 1 2 2 3 3 5 5 22 33 88.5333 88.5333 (e)

(e) xxt t is periodic with is periodic with fundamental period sifundamental period six. x. This signal This signal is described is described over one fundamental period by

(32)

x

xt t rectrect tt 22 4 rect4rect tt 44 , 0, 0 tt 66.. 3

3 22

The signal can be described in the time period

The signal can be described in the time period 00 tt 66 by by 0 , 0 0 , 0 tt 1 1 / / 22 1 1 , , 1 1 / / 22 tt 33 x xtt 3 3 , , 33 tt 77 2 2 4 , 4 , 7 7 / / 22 tt 55 0 , 5 0 , 5 tt 66 1 1 22 11 22 55 22 11 22 33 22 2.5 4.5 242.5 4.5 24 P P_x_x 6 6 00 22 11 2233 2244 22 00 11 66 5.1675.167

Exercises Without Answers in Text

Signal Functions Signal Functions

34.

34. Let Let the the unit unit impulse impulse function function be be represented represented by by the the limit,limit, xx limlim 11 // aarectrect x/x/aa ,, aa 00 ..

a a00

The function

The function 11 // aarectrect x x//aa has an area of one regardless of the value ofhas an area of one regardless of the value ofaa. (a). (a)

What is the area of the function

What is the area of the function 44 x x limlim11 // aarectrect 44 x x//aa ??

a a00

This is a rectangle with the same height as

This is a rectangle with the same height as 11 // aarectrectx/x/aa but but 1/4 1/4 times times the basethe base

width.

width. Therefore its area is Therefore its area is 1/4 times as gr1/4 times as great or 1/4.eat or 1/4. (b)

(b) What What is is the the area area of of the the functionfunction 66 x x lim1lim1 // aarect 6rect6 x x//aa ??

a a00

This is a rectangle with the same height as

This is a rectangle with the same height as 11 // aarectrect x/x/aa but but 1/6 1/6 timestimes

the base width. (The fact that the factor is

the base width. (The fact that the factor is ““_--6”6”_{instead of}_{instead of} ““₆₆””_{just means}_{just means}

that the rectangle is reversed in time which

that the rectangle is reversed in time which does not change its shape odoes not change its shape or area.)r area.) Therefore its area is 1/6 times as great or 1/6.

Therefore its area is 1/6 times as great or 1/6. (c)

(c) What What is is the the area area of of the the functionfunction bxbx limlim 11 // aarectrect bxbx //aa forforbb

a a00

positive and for

(33)

It is simply

(34)

a a

35.

35. Using Using a a change change of of variable variable and and the the definition definition of of the the unit unit impulse, impulse, prove prove thatthat a att t t ₀₀ 11 // aa tt t t ₀₀ .. x x 0 0 ,, x x 00 ,, x x dxdx 11 a att t t ₀₀ 0 , where a0 , whereatt t t ₀₀ 0 or0 or t t t t ₀₀ Let Let Strength Strength aatt t t ₀₀ dt dt Then, for Then, for aa > 0,> 0, a

att t t ₀₀ andand adt d adt d

Strength Strength d d 11 d d 1 1 11 and for and for aa< 0,< 0, a a a a aa aa Strength Strength d d 11 d d 11 d d 1 11 1 a a aa Therefore for

Therefore for aa> 0 and> 0 and aa< 0,< 0,

a

a a a aa

Strength

Strength 11 andand aa t t t t 11 t t tt ..

a

a 00 aa 00

36.

36. Using Using the the results results of of Exercise Exercise 35,35, (a)

(a) Show Show thatthat ₁₁axax 11 // a a x x nn// aa

n n

From the definition of the periodic impulse

From the definition of the periodic impulse ₁₁axax axax nn .. Then, using the property from Exercise 35

Then, using the property from Exercise 35 1

1

1aa x x aa x x nn//aa x x nn//aa..

(b)

(35)

The period is

The period is 11// aa . Therefore. Therefore

1 1axax 11 1 1 //aa t t 001/1/aa t t ₀₀ 1 1aa x x dx dx aa 1/2 1/2 aa 1/2 1/2 aa 1 1axax dx dx aa 1/2 1/2 aa 1/2 1/2 aa ax ax dxdx

(36)