Chapter 18 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006

1. (a) 5, 10, 15, 20, 25 (all rad/s) (b) 5, 10, 15, 20, 25 (all rad/s) (c) 90, 180, 270, 360, 450 (all rad/s)

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2. (a) ωo = 2π rad/s, f = 1 Hz, therefore T = 1 s. (b) ωo = 5.95 rad/s = 2π f rad/s, f = 0.947 Hz, therefore T = 1.056 s. (c) ) ωo = 1 rad/s = 2πf rad/s, f = 1/2π Hz, therefore T = 2π s.



3.

( ) 3 3cos(100 40 ) 4sin(200 10 ) 2.5cos300 Vv t t t tπ π π= − − ° + − ° + (a) V 3 0 0 0 3.000 Vav = − + + =

(b) 2 2 2 21V 3 (3 4 2.5 ) 4.962 V2eff = + + + =

(c) 2 2T 0100o

π .02 sπω π

= =

v ms = − − ° + ° + °

=

(d) (18 ) 3 3cos( 33.52 ) 4sin(2.960 ) 2.5cos(19.440 ) 2.459 V= −



4. (a)

t v t v

0 2 0.55 -0.844

0.05 2.96 0.6 0.094

0.1 3.33 0.65 0.536

0.15 2.89 0.7 0.440

0.2 1.676 0.75 0

0.25 0 0.8 -0.440

0.3 -1.676 0.85 -0.536

0.35 -2.89 0.9 -0.094

0.4 -3.33 0.95 0.844

0.45 -2.96 1 2

0.5 -2 (b) (c) min 3.330v =

2 2

2

max

4 sin 2 7.2 cos 4 04sin 2 7.2(cos 2 sin 2 )

4 16 414.724sin 2 7.2(1 2sin 2 ) 0.5817, 0.8595 sin 228.8

0.09881,0.83539 0.5593 for smaller max)

tt t t

t t x t

t v

v tπ π π π

π π π

3.330(

π π π

′ = − + =

∴ = −

− ± +∴ = − ∴ = = − =

∴ = ∴ =



5. (a) a0 = 0 (b) a0 = 0 (c) a0 = 5 (d) a0 = 5



6. (a) a0 = 0 (b) a0 = 0 (c) a0 = 100 (d) a0 = 100



7. (a) a0 = 3, a1 = 0, a2 = 0, b1 = 0, b2 = 0 (b) a0 = 3, a1 = 3, a2 = 0, b1 = 0, b2 = 0 (c) a0 = 0, a1 = 0, a2 = 0, b1 = 3, b2 = 3 (d) 3 o o ocos(3 10 ) 3cos3 cos10 3sin 3 sin10t t t− = + a0 = 0, a1 = 3cos10o = 2.954, a2 = 0, b1 = 3sin10o = 0.521, b2 = 0



8. ao = 0

1 ( ) 2.5T

f t dtT

=∫ . a1 = a2 = 0 since function has odd symmetry

2

2

1 00 11

2 2 5b ( )sin 5sin cos 102 0 2

Tf t tdt tdt t

Tω π π

π π= = = − = −

−∫ ∫

2

2

2 00 11

2 2 5b ( )sin 2 5sin 2 cos 2 02 0 2

Tf t tdt tdt t

Tω π π

π= = = − =

−∫ ∫



9. ao = 2

2

0 00

1 1 2( ) 2 43 3 3

Tf t dt dt t

T= =∫ ∫ = .

a1 =2

2

00 00

2 2∫ ∫ 2 4 3( ) cos 2cos sin

3 3 3 2 3T

f t tdt t dt tT

π πωπ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 0.551=

a2 =2

2

00 00

2 2∫ ∫ 4 4 3( ) cos 2 2cos sin

3 3 3 4 3T

f t tdt t dt tT

π πωπ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

4 0.276=

a3 =2

2

00 00

2 2∫ ∫ 6 4 3( )cos3 2cos sin

3 3 3 6 3T

f t tdt t dt tT

π πωπ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

6 0

2

2

1 00 00

2 2 2 4 3 2( )sin 2sin cos3 3 3 2 3

Tf t tdt t dt t

Tπ πω

π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∫ ∫ 0.955= b

2

2

2 00 00

2 2 4 4 3 4( )sin 2 2sin cos3 3 3 4 3

Tf t tdt t dt t

Tπ πω

π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∫ ∫ 0.477 b



10. h(t) = –3 + 8 sin πt + f(t) Use linearity and superposition. T = 2 s.

ao = 0

1 13 ( ) 3 2.52

Tf t dt

T= − + = −∫− + .

a2 = 0

b1 = 2

00 0

2 2 2( )sin 8 (1)sin 8 7.362

Tf t tdt tdt

Tω π8

π+ = + = − =∫ ∫

b2 = 2

00 1

2∫ ∫ 2( )sin 2 (1)sin 2 0

2T

f t tdt tdtT

ω π= =



11. (a) T 10 , F 0.1(2 4 2 2) 1.200av os a= = = × + × = (b)

2 22 2

0 0

222 2 3

000

1

(c)

F = −f (4 ) 0.2 (16 8 )5

1 80.2 16 4 0.2 32 163 3

= − +

⎡ ⎤ ⎛ ⎞= − + = − + =⎢ ⎥ ⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

∫ ∫ef t dt t t dt

t t t 1.9322

2 2 2

30 0 0

2 2

200

2

2 22 (4 )cos3 0.4 4cos 0.6 0.4 cos 0.610 10

1 11.6 sin 0.6 0.4 cos 0.6 sin 0.60.6 0.36 0.6

8 10 4sin1.2 (cos1.2 1) sin 1.2 0.045813 9 3

= × − × = −

⎛ ⎞= − +⎜ ⎟⎝ ⎠

= − − − = −

∫ ∫ ∫ta t dt t dt t t dt

tt t t

π π π

π π ππ π π

π ππ π π

π



12. (a) T = 8 − 2 = 6 s

(b) 1 Hz6of =

(c) 2 rad/so of 3π ω π= =

(d) 1 (10 1 5 1) 2.5= 6oa = × + ×

(e) 3 4

22 3

43

2 3

2

2 2 210sin 5sin6 3 3

1 30 2 15 2cos cos3 2 3 2 3

1 15 4 7.5 8 1 15 7.5cos 2 cos cos cos 2 (1.5) ( 1.5) 1.19373 3 3 3

⎡ ⎤= +⎢ ⎥

⎣ ⎦⎡ ⎤

= − −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤∴ = − − − − = − − − = −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦⎣ ⎦

∫ ∫t tb dt dt

t t

b

π π

π ππ π

π ππ ππ π π π



13.

433 4

3322 3

433 4

32 32 3

2 6 6 1 10 510cos 5cos sin sin6 6 6 3

10 1 1sin 3 sin 2 sin 4 sin 3 03 2 2

1 1 1010sin 5sin cos cos3 3

10 1cos3 cos 2 cos3 2

5

⎡ ⎤⎡ ⎤= + = −⎢ ⎥⎢ ⎥

⎢ ⎥⎣ ⎦ ⎣ ⎦⎛ ⎞= − + − =⎜ ⎟⎝ ⎠

⎡ ⎤⎡ ⎤= + = − −⎢ ⎥⎢ ⎥

⎢ ⎥⎣ ⎦ ⎣ ⎦

= − − +

∫ ∫

∫ ∫

t ta dt dt t

b tdt t dt t t

π π π ππ π

π π π

t

π

π π π ππ π

π ππ

π

2 23 3

1 104 cos3 ( 1)2 3

⎛ ⎞− = − − =⎜ ⎟⎝ ⎠

+ =a b

π ππ

1.0610

1.0610



14.

(a) 2 23.8cos 80 1.9 1.9cos160 , T 12.5 ms, ave value 1.9160

ππ = + π = = =

πt t

(b)

(c) 23.8cos70 3.8sin80 ; , , T 2 ; ave value 0ππ − π ω = π ω = π = = =

πo ot t t t s

33.8cos 80 (3.8cos80 )(0.5 0.5cos 160 )1.9cos80 0.95cos 240 0.95cos80 2.85cos80 0.95cos 240

2T 25ms, ave value 80

π = π + π= π + π + π = π + π

π= =

π

t t tt t t t t

0=



15. T = 2 s (a)

( )

11

400

4 1

1 1

2 4 2 1sin cos 42 2 4

1 1 cos 44

max when 4 , 0.1252

× π= = −

ππ

∴ = − ππ

ππ = =

∫tt tb dt t

b t

t t s

(b) 41

4=

πb



16.

( ) 5 8cos10 5cos15 3cos 20 8sin10 4sin15 2sin 20= + − + − − +g t t t t t t t

(a) 25 T 1.25665π

ω = ∴ = =o s

(b) 5 104 3.183 Hz2

= β = = =π πo of f

(c) G 5= −av

(d) 2 2 2 2 2 2 21G ( 5) (8 5 3 8 4 2 ) 116 10.7702

= − + + + + + + = =eff

(e)



17.

[ ]0.1 0.1

0.1 0.10.1

0.1

T 0.2, ( ) V cos5 , 0.1 0.1

2 V cos5 cos10 5V cos(5 10 ) cos(10 5 )0.2

1 15V sin(10 5 ) sin(10 5 )10 5 10 5

V 2 2sin(10 5 )0.1 sin(102 1 2 1

− −

−

= = π − < <

= π π = π + π + π −

⎡ ⎤= π + π + π − π⎢ ⎥π + π π − π⎣ ⎦

= π + π +π + −

∫ ∫

m

n m m

m

m

f t t t

a t n t dt n t n t dt

n t n tn n

nn n

π

2 2

5 )0.1

V 2 2sin( 0.5 ) sin( 0.5 )2 1 2 1

V 2V2 2 1 1cos ( cos ) cos2 1 2 1 2 1 2 1

2V 4V2 1 2 1 coscos4 1 4 1

1 1V cos5 5V sin sin0.2 5 2 2

⎡ ⎤π − π⎢ ⎥⎣ ⎦⎡ ⎤= π + π + π − π⎢ ⎥π + −⎣ ⎦⎡ ⎤ ⎛= π + − π = π −⎜ ⎟⎢ ⎥π + − π + −⎣ ⎦ ⎝

− − − π= π = −

π − π −π π⎛= π = − −⎜π ⎝

m

m m

m m

o m m

n

n nn n

n n nn n n n

n n nnn n

a t dt

⎞⎠

0.1

0.1

2V

−

⎡ ⎤⎞ =⎟⎢ ⎥ π⎠⎣ ⎦

+

∫ m

2V 4V 4V 4V 4V( ) cos10 cos 20 cos30 cos 40 ...3 15 35 63

∴ = + π − π + π − ππ π π π π

m m m m mv t t t t t



18.

(a) 1even, wave2

−

(b) 0 for all ; 0; 0= = =n even ob n a a (c) 1 2 3 2

22

11

1 3

0, 0

8 10 6 205cos sin sin sin12 6 3 6 3 6

20 20 20sin sin 2.330, sin sin 2.1223 6 3 2 3

= = = =

π π π π⎛ ⎞= = = −⎜ ⎟π π ⎝ ⎠π π π⎛ ⎞ ⎛ ⎞∴ = − = = π − = − = −⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠

∫n

b b

π

n t n t n na dtn n

b a

a a



19. (a) 0

( ) 0.2sin1000 0.6sin 2000 0.4sin 3000= =

∴ = π + π +o na a

πy t t t t (b) 2 2 2Y 0.5(0.2 0.6 0.4 ) 0.5(0.56) 0.5292= + + = =eff (c) (2ms) 0.2sin 0.2 0.6sin 0.4 0.4sin 0.6 1.0686= π + π + π =y



20. (a) (b) (c) (d) (e)

33

5 522

3

5 52

5 5

4 2 5 32 6 5 3.2 15 10[ ] 0, 8cos sin sin sin6 6 6 10 3 3

4 2 5 32 6 15 10 3.2[ ] 0, 8sin cos cos ( 0.5)6 6 6 10 3 3

8 2 5 64 12 15[ ] 0, 8cos sin12 12 12 10

π π π π⎛ ⎞= = = = − =⎜ ⎟π π ⎝ ⎠

π − π π⎛ ⎞⎛ ⎞= = = − = − −⎜ ⎟⎜ ⎟π π⎝ ⎠⎝ ⎠

π= = =

π

∫

∫

t ta b a dt

tb a b dt

tc b a dt

0.88213

0.5093=

3

23

5 52

10sin6 6

8 10 64 12 15 10[ ] 0, 8sin cos cos12 12 12 10 6 6

π π⎛ ⎞− =⎜ ⎟⎝ ⎠

π π⎛ ⎞⎛= = = − −⎜ ⎟⎜π⎝ ⎠⎝

∫

∫td a b dt

3.801

1.0186π ⎞ =⎟⎠



21. T = 4 ms (a)

0.0040.004

00

1000 250 88sin125 cos1254 125

16 16cos 1 5.0932

×= π =

− π

π⎛ ⎞= − − = =⎜ ⎟π π⎝ ⎠

∫o πdt ta t (b)

0.004

10

0.004 0.004

10 0

0.004

0

1

24000 sin125 cos0.004

4000 sin125 cos500 2000 (sin 625 sin 375 )

cos625 cos375 3.2 5.3332000 (1 cos 2.5 ) (1 cos1.5 ) 0.6791625 375

4000 sin125

π= π

∴ = π π = π − π

π π⎛ ⎞= − + = − π − − π = −⎜ ⎟π π π π⎝ ⎠

= π

∫

∫ ∫

ta t dt

a t t dt t t dt

t t

b

0.004 0.004

0 0

sin 500 2000 (cos375 cos 625 )

1 1 12000 (sin1.5 ) (sin 2.5 ) 2000375 625 375 625

π = π − π

−⎡ ⎤ ⎛ ⎞= π − π = = −⎜ ⎟⎢ ⎥π π π⎣ ⎦ ⎝ ⎠

∫ ∫t t dt t t dt

1 2.716−π

(c) 4 0 : 8sin125− < < πt t (d)

[ ]

( )

0.004

1 10

0.0040.004

100

40000, 8sin125 cos 2508

cos375 cos1252000 sin 375 sin125 2000375 125

5.333 161 cos1.5 cos 1 3.3952

+

= = π π

π π⎡ ⎤∴ = π − π = − +⎢ ⎥π π⎣ ⎦

π⎛ ⎞= − π + − = −⎜ ⎟π π ⎝ ⎠

∫

∫

b a t t dt

t ta t t dt



22.

0.0010.001

0 0

1 3 5 7 9

1odd and wave 0, 0, 02

T 10 0.01

8 110sin 200 8000 cos 2000.01 200

40 40(cos 0.2 1) (1 cos 0.2 )

2.432, 5.556, 5.093, 2.381, 0.27

− ∴ = = =

= =

⎡ ⎤ −⎛ ⎞= π =⎢ ⎥ ⎜ ⎟π⎝ ⎠⎣ ⎦π

∴ = − π − = − ππ π

∴

∫

o n even

odd

odd

a a b

ms s

b n t dt n tn

b n nn n

= = = = =b b b b b 02



23.

( )

( )

/ 4

00.001

0

2

60.0013

2 2 2 0

2 2

1 8odd and wave, T 8 ( )sin2 T

2 250 1000 1000 sin 250T

1Now, sin sin cos , 250

10( ) 10 sin 250 250 cos 250250

16 sin 04

− = ∴ = ω

πω = = π ∴ = π

= − = π

= ∴ = π − π ππ

π∴ = −

π

∫

∫

∫

T

n o

o n

x

n

n

ms b f t n t dt

b t nt dt

x ax dx a ax ax a na

f t t b n t n t n tn

nbn 1 2

3 52 2

16cos 0 sin cos 0.24604 4 4 4 4

16 3 3 3 16 5 5 5sin cos 0.4275 ; sin cos 0.134219 4 4 4 25 4 4 4

0

−

π π π π π⎛ ⎞ ⎛ ⎞− + ∴ = − =⎜ ⎟ ⎜ ⎟π⎝ ⎠ ⎝ ⎠π π π π π π⎛ ⎞ ⎛ ⎞= − = = − =⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠

=even

n n b

b b

b



24. (a) odd, T = 4 (b) even, T = 4:

(c) 1odd, wave: T 82

− =

(d) 1even, wave, T 8 :2

− =



25. (a)

1,

2 2

2 2

20 1 2 20 205 sin sin 5 , V ( 1)0.4

V 20 5Z 4 5 2 4 10 , IZ (4 10 ) 1 2.5

5 1 2.5 12.5 5I1 6.25 (1 6.25 )

12.5 1 5 1cos5 sin 51 6.25 1 6.25

∞ π= + ∴ = = −

π π π π

−= + = + = = = −

π + +− +

∴ = − = −π + π +

∴ = − +π + π +

∑s sn snodd

snn fn

n

fn

fn

ntv v ntn n n

j jj n j nn j n j

j j n jn n n n

i ntn n n

j

n

nt

21,

1 12.5 51.25 cos5 sin 51 6.25

∞ ⎡ ⎤∴ = + − +⎢ ⎥+ π π⎣ ⎦∑fodd

i ntn n

(b) 2

21,

21,

22

1,

1 12.5Ae , , (0) 0, (0) 1.251 6.25

2 1 2(0) 1.25 1.25 tanh 0.2 0.553880.16 4 0.4

1 12.5 5A 0.55388, 0.55388 1.25 cos5 sin 51 6.25

∞−

∞

∞−

⎛ ⎞= = + = = + −⎜ ⎟+ π⎝ ⎠π

∴ = − = − π =π + π ×

⎡ ⎤∴ = − = − + + − +⎢ ⎥+ π π⎣ ⎦

∑

∑

∑

tn f n f

odd

fodd

t

odd

i i i i i in

in

i e nt nt

n n

nt



26. (a) 2 0.40 0.2 : 2.5(1 ) (0.2 ) 2.5(1 ) 1.78848 A− − π< < π = − ∴ π = − =tt i e i e (b) 2( 0.2 )0.2 0.4 : 1.78848 (0.4 ) 0.50902 A− − ππ < < π = ∴ π =tt i e i (c) 2( 0.4 )0.4 0.6 : 2.5 (2.5 0.50902) , (0.6 ) 1.9335− − π −π < < π = − − π =tt i e i



27. (a)

( )

1,

2 2

2

20 15 sin 5

20 sin 5

20V

1 1 20 / 1 20 / 1 20Z 2 2 V5 2 10 2 1/ 10 10 1 20 1 20

20 1 20 20 1V , 20 cos5 sin 51 400 1 400

20 1 15 sin 51 400

∞

= +π

=π

= −π

− π − π −= + = + ∴ = × = ×

+ +− −

−

∴ = × = − ++ π π +

∴ = +π +

∑sodd

sn

sn

n cn

cn cn

cf

v ntn

v ntn

jn

j n j n j n j n j n j n j n j n j n

n j v n nt ntn n n n

vn n1,

20cos5∞ ⎛ ⎞−⎜ ⎟

⎝ ⎠∑odd

nt nt (b) / 4Ae−= t

nv (c)

2 2 21, 1,

2 21,

/ 42

1,

20 20 1 1(0) A 5 A 51 400 (1/ 20)

1 tanh 5 tanh 1.23117(1/ 20) 4(1/ 20) 20 2 40

1A 0 5 1.23117 4.60811

∞ ∞

∞

−= + + = + −

π + π +

π π π= = π =

+ ×

∴ 20 1 1( ) 4.60811 5 sin 5 20cos5

1 400

∞−

= − + × = −π

⎛ ⎞∴ = − + + −⎜ ⎟π + ⎝ ⎠∑t

codd

∑ ∑

∑

codd odd

odd

n n

n

v t e nt ntn n

v



28. At the frequency ω = 10nπ

( )

( )3

3

10 10 10 5 10

20 10 5 10n

j n

j n

π

π

−

−

⎡ ⎤+ ×⎣ ⎦= Ω+ ×

Z and ( )8Sn j

nπ= −I

Therefore ( )80 10 0.0520 0.05n

j njn j n

ππ π

⎡ ⎤+= − ⎢ ⎥+⎣ ⎦

V .

In the time domain, this becomes

( )2

o 1 11 2

1 ( )

1 (0.005 )40( ) cos 10 90 tan 0.005 tan 0.00251 (0.0025 )n odd

nv t n n n

n n

πππ π π

π ππ

∞− −

=

+⎛ ⎞= − +⎜ ⎟⎝ ⎠ +

∑ −



29. At the frequency ω = nπ

( ) ( )

( )1

223

10 32 and 120 5 10

n

Ln Sn Sn jjn nπ π

−

−= = −

+ ×I I I −

Thus, in the time domain, we can write

( )( )

( )( )

1o 12

2 21 (odd)

320 1( ) 1 cos 90 tan 0.00025n 20 1 0.00025

n

Ln

i t n t nn

π ππ π

∞ −−

=

⎡ ⎤= − − −⎢ ⎥

⎢ ⎥ +⎣ ⎦∑



30.

( )

30.001 0.0053

3 2 / 6 10 1003

0 0.003

0.001 0.00551000 1000

0 0.003

5 3

3 3

3

10 100 1006

10 1 16 1000 1000

100 1001 (1 1 1 1) 10.6106 6

10.610; 10.610

2

−− × π × − π

− π − π

− π − π − π

−

⎡ ⎤= −⎢ ⎥

⎣ ⎦⎡ ⎤−

= +⎢ ⎥π π⎢ ⎥⎣ ⎦

= + + − = + − + = −π π

∴ = =

=

∫ ∫j t j t

j t j t

j j j

c e e

e ej j

e e e jj j

c j c

a

( )

0.001 0.0053

0 0.003

5

2 23 3 3 3 3 3 3

10 100cos100 100cos10006

2 10 1 sin 0 sin 5 sin 3 06 1000

1 1( ) 21.22 and 21.222 2

⎡ ⎤×π − π⎢ ⎥

⎣ ⎦×

= π − − π + π =π

= − = − ∴ = + =

∫ ∫t dt t dt

c a jb j b b a b



31. 0.001 0.002

5 400 400

0 0.001

0.001 0.002400 400

0 0.001

0.004000.001 40002 2

0.001

1T 5 10 1000.005

20,000 1000

120,000 ( 400 1)160 400

j nt j ntm

j nt j ntn

j ntj nt

n

ms c te dt e dt

c t e dt e dt

ec j nt en j n

− π − π

− π − π

− π− π

⎡ ⎤= = +⎢ ⎥

⎣ ⎦⎡ ⎤

∴ = +⎢ ⎥⎣ ⎦

∴ = π + +π − π

∫ ∫

∫ ∫

( )

2

3 3

0.4 0.8 0.41 2 2

2

1(50 10 100 10 ) 0.15 200 300.005

1 1 120,000 (1 0.4 )160 160 400

125 (1 72 ) (1.60597 51.488 ) 12.66515 15.91548 90 (1 144 1 72 )

12.665(1 72

o o

j j j

c a

c e j e ej

− −

− π − π − π

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

∴ = = × + × = × =

⎡ ⎤= + π − − −⎢ ⎥π π π⎣ ⎦

= ∠ − ° ∠ ° − + ∠ ° ∠ − ° − ∠ − °π

= ∠ −

2

) (1 1.2566) 12.665 15.915(1 144 1 72 )20.339 20.513 12.665 18.709 108 24.93 88.613.16625 144 (1 2.5133) 3.16625 7.9575(1 288 1 144 )8.5645 75.697 3.16625 15.1361 144 13.309 1

j j

c j j

° + − + ∠ − ° − ∠ − °= ∠ − ° − + ∠ − ° = ∠ − °= ∠ − ° + − + ∠ − ° − ∠ − °= ∠ − ° − + ∠ ° = ∠ 77.43°

(a)

(b)



32. Fig. 17-8a: V 8 V, 0.2 , 6000o os f ppsτ μ= = =

(a) 1 1T , 6000, 0.2 5 MHz6000 of s fτ μ

τ= = = ∴ = =

(b) 6000 Hzof = (c)

6 6

3

3

8 0.2 10 sin(1/ 2 3 12,000 0.2 10

(d) 6 6 6

7.270 mV× ×=

× ×3333 6

2 10 8 0.2 10 sin(1/ 2 333 12,000 0.2 10333.36 10 1/ 6000 1/ 2 333 12,000 0.2 10

c ππ

− −

−

× × × × ×= ∴ =

× × ×

(e) 1/ 5 MHzβ τ= = (f) (g)

6000 3 18,000 (closest) 1/ 6000 0.0036

c

c

ππ

− −× × × × × ×× = ∴ =

9.5998 mV∴ =

2000 22002 < < 2.2 Mrad/s kHz or 318.3 350.1 kHz2 2

6 kHz 6 53 318; 324,330,336,342,348kHz o

f f

f f n

ωπ π

∴ < < < <

= ∴ = × = ∴ = 5

6 6

2278 0.2 10 sin(1/ 2 227 12,000 0.2 10 8.470 mV

1/ 6000 ( )227 6 1362 kHz

c

f

π− −× × × × × ×= =

′′

= × =



33.

1 2 3T 5 ; 1, 0.2 0.2, 0.5 0.25, 1 2, 0, 4= = = − = + = − − =o nms c c j c j c j c n ≥ (a) (b)

1 1 1 2 2 3 32 1, 0.4 0.4 0.4, 1 0.5, 2( )

n n n o ojb c a c a jb jb j a jb j a jbv t t t t t t t

a1 0.4cos 400 cos800 2cos1200 0.4sin 400 0.5sin800 4sin1200π π π π π π

= − = ∴ = = − = − = − − = + − = −∴ = + + − + − +

(1 ) 1 0.4cos 72 cos144 2cos 216 0.4sin 72 0.5sin144 4sin 216s = + ° + ° − ° + ° − ° + °v m ∴

(1 ) 0.332Vv ms∴ = −



34. (a)

( )

( )

6

6

0.6 106

60.4 10

65

10T 5 2 1cos 25 5 10

5 104 10 sin 43.2 sin 28.82

n

n

ts c n dt

c n nn

n

μ π

π

−

−

×

−×

−

= ∴ = ××

×

(b) 41 (sin172.8 sin115.2 ) 0.06203

4c

π= ° − ° = −

(c) 6 6

6

0.2 10 0.2 10 0.085 10o oc a

− −

−

× + ×= = =

×

(d) maxa little testing shows is max 0.08oc c∴ = (e)

(f) 6740 10740 148 MHz

5×

= = =ofβ

1 sin 43.2 sin 28.8nc nnπ

∴ = × ° − °

∴ = ° − °

∫

( )

( )

3 310.01 0.08 0.8 10 sin 43.2 sin 28.8 0.8 10

125 sin 43.2 sin 28.8 1

− −× = × ∴ ° − ° ≤ ×

ok for 740

° − ° ≤

n nn

n nn

π

π>n

∴



35. T 1/16, 32oω π= = (a)

1/961/9696 96

30 0

3

16 4016 4096

20 40( 1) 4.244 V3 3

j t j

j

c e dt ej

c j e j

π π

π

π

j (b)

π π

− −

−

×= −

−

= − = −

∫

= −∴

3 3 3 3 3

23 3

642

Near harmonics are 2 32 Hz, 3 48 HzOnly 32 and 48 Hz pass filter 2

2 8.488 0, 8.488 V8.488 1I 1.4536 31.10 A; P 1.4536 5 5.283 W

5 0.01 96 2

1 640401/16 6

o o

n n n

j t

f fa jb c

a jb c j a b

j

c e dtj

π

π

−

= =− =

− = = − ∴ = =

= = ∠ − ° = × × =+ ×

= =−

1/9664 /96

0

2 2 2

2

22

( 1) 2.7566 4.7746 V4

2 5.5132 9.5492 11.026 6011.026 60I 2.046 65.39 A5 0.01 641P 2.046 5 10.465 W P2

j

tot

e j

a b c j

j

π

π

π

− − = −

− = = − = ∠ − °∠ − °

15.748 W

∴ = = ∠ − °+ ×

= × × = ∴

∫

=∴



36. ( ) 5[ ( 3) ( 2) ( 2) ( 3)]f t u t u t u t u t= + + + − − − − (a)

-3 -2 -1 0 1 2 3

f(t)

t

∞(b)

2 2 3

3 2 2

2 3 2 2 3 2

3 3 2 2 2 2

F( ) ( )

F( ) 5 10 5

5 10 5F( ) ( ) ( ) ( )

5 5 10( ) ( ) ( )

5 5( 2)sin 3 (

− ω

−∞

−− ω − ω − ω

− −

ω ω − ω ω − ω − ω

ω − ω ω − ω ω − ω

ω =

∴ ω = + +

∴ ω = − + − + −− ω − ω − ω

= − + + − + − +− ω − ω − ω

= − ω +− ω − ω

∫

∫ ∫ ∫

j t

j t j t j t

j j j j j j

j j j j j j

t e dt

j e dt e dt e dt

j e e e e e ej j j

e e e e e ej j j

jj j

j f

102)sin 2 ( 2)sin 2ω + − ω− ω

10 10 20 10F( ) sin 3 sin 2 sin 2 (sin 3 sin 2 )∴ ω = ω− ω+ ω = ω+ ωω ω ω ω

j

j jj



37. (a) (b) (c)

0

( )

0

( ) ( ), 0 F( ) ( )

F( )

∞ ∞− − − −

−∞

∞− +

= > ∴ = =

−1 1∴ = =

∫ ∫at j t at j t

a j t

+ +

t e u t a j f t e dt e e dt

j ea j a j

ω ω

ω

ω

ωω ω

f

6 ( )

( )( )

( ) ( ), 0 F( )

1 1F( )

∞− − +

∞− + −− +

= − > ∴ =

− − ⎡ ⎤ 1∴ = = − =⎣ ⎦+ + +

o j t

∫o

o

o o o

o

at atat a j to

t

at at a j ta j t

t

e e u t t a j e e dt

j e e e e ea j a j a j

ωf t

ω ωω

ω

ωω ω ω

[ ]

( )

0( )

2 20 2

( ), 0 F( )

F( ) ( ) 1 0 [ 1]( ) ( )

∞− − +

− +∞

= > ∴ =( )

1 1( )

∴ = − + − = −+ + +

∫at a j t

a j t

t te u t a j te dt

ej a j ta j a j a j

ω

ω

ω

ω ω − =ω ω

f

ω



38.

0 4

4 00

14

4 4

10 0

4 0 : ( ) 2.5( 4); 0 4 : ( ) 2.5(4 )

F( ) 2.5( 4) 2.5(4 5)

ln 1 , let I 2.5(4 ) ( )

I 2.5(4 ) F( ) 2.5 (4 )( )

F( ) 5 (4 )cos 20

− ω − ω

−

ωτ

ωτ ω − ω

− < < = + < < = −

∴ ω = + + −

= τ ∴ = − τ − τ

∴ = − τ τ ∴ ω = − +

∴ ω = − ω = ×

∫ ∫

∫

∫ ∫

j t j t

st j

j j t j

t f t t t f t t

j t e dt e dt

t e d

e d j t e e dt

j t t dt

t

44 4

0 00

402

2 2 2

2sin 210× ω⎛ ⎞= ⎜ ⎟ω ω⎝ ⎠2

2

1 sin 5 cos

20 5F( ) sin 4 (cos sin )

20 5 5 5sin 4 (cos 4 1) 4 sin 4 (1 cos 4 )

2 5or, F( ) sin 2

ω − ωω

∴ ω = ω− ω + ω ωω ω

= ω − ω − − ω ω = − ωω ω ω ω

ω = ω

∫ ∫t t dt

j t t t

j



39.

(1 ) (1 )

(1 ) (1 ) (1 ) (1 )

( ) 5sin F( ) 5sin

5F( ) ( )2

5 [ ]2

5 1 1F( ) ( ) ( )2 (1 ) (1 )2.5 2.5( ) (

1 1

π− ω

−π

π− − ω

−π

π−ω − +ω

−π

π −ω − π −ω − π +ω π +ω

− πω πω

= − π < < π ∴ ω =

∴ ω = −

= −

⎡ ⎤ω = − − −⎢ ⎥− ω − + ω⎣ ⎦

−= − + −

− ω + ω

∫

∫

∫

j t

jt jt j t

jt jt

j j j j

j j

f t t t j t e dt

j e e e dtj

e e dtj

j e e e ej j j

e e

2 2 2

)

2.5 2.5 1 1( 2sin ) ( 2sin ) 5sin1 1 1

,

1

1

− ⎛ ⎞−⎜ ⎟+ ω⎝ ⎠

j 1 1 10sin 10sin5sin ( 1)1 1

− πω πω− +

= πω − πω = πω −− ω + ω − ω

+ ω + − ω πω πω⎛ ⎞= πω − = − =⎜ ⎟− ω − ω ω −⎝ ⎠

j je e

j j j

j j



40.

/ 2/ 2

/ 2 / 2

/ 2 / 2/ 2

(1 ) (1 )

/ 2

/ 2

/ 2

/ 2

( ) 8cos [ ( 0.5 ) ( 0.5 )]

F( ) 8cos 4 ( )

4

1 14(1 ) (1 )

14 ( )(1 )

π− π

π π− ω − − ω

−π −π

π−ω − +ω

−π

π− ω − − ω

−π

− πω

= + π − − π

∴ ω = = +

⎡ ⎤= +⎣ ⎦

⎧ ⎫⎪ ⎪= −⎨ ⎬− ω + ω⎪ ⎪⎩ ⎭

= − −− ω

∫ ∫

∫

j t jt jt j t

jt jt

j tjt jt j t

j

f t t u t u t

j te dt e e e dt

e e dt

e e e ej j

je j ej

/ 2 / 2 / 2

2 2

1(1 )

1 1 14 2cos 2cos 8cos1 2 1 2 2 1 1

2 cos / 28cos 162 1 1

πω − πω πω⎧ ⎫⎡ ⎤ ⎡− − −⎨ ⎬⎣ ⎦ ⎣+ ω⎩ ⎭πω πω πω⎧ ⎫ ⎛= × + × = +⎨ ⎬ ⎜ ⎟− ω + ω − ω + ω⎩ ⎭ ⎝

πω πω= =

− ω − ω

j jje jej

1

⎤⎦

⎞⎠

j

ω = ∴ j

(a) 0 F( 0) 16=

(b) 16cos 720.8, F( 0.8)0.36

°ω = =j 13.734=

(c) 16cos(3.1 90 )3.1, F( 3.1) × 0.29071 3.12

°ω = =

−j = −



41.

[ ]

( )22

2 2

2 2

1(a) (b) (c)

F( ) 4 ( 2) ( 2) ( ) F( )2

4 2 1 2( )22 4 5( ) 2sin 2 sin 2 (0.8) sin1.6

2

∞ω

−∞

ω ω −

− −

ω = ω+ − ω ω− ∴ = ω ωπ

1.5909

∴ = ω = = −π π π

= = ∴ =π π π

∫

∫

j t

j t j t j t j t

rad

f t e j d

f t e d e e ejt j t

f t j t t ft t

=

j u

∴

2 2

0(2 ) ( 2 )

0

2

2

4F( j e) 4 ( )2

2 2( )

2 1 1 2 1 1 2 4(1 0) (0 1)2 2 2 2 48 8( ) (0.8)

(4 ) 4.64

∞− ω − ω + ω

−∞

∞+ ω − + ω

−∞

ω = ∴ = ωπ

∴ = ω + ωπ π

⎡ ⎤ ⎛ ⎞= − + − = + =⎜ ⎟⎢ ⎥π + − + π + − π +⎣ ⎦ ⎝ ⎠

∴ = ∴ =π + π×

∫

∫ ∫

j t

jt j t

f t e d

f t e d e d

0.5488=

jt jt jt jt t

f t ft

[ ]

( )

( )

0.5 0.5

0.5 0.50.5

( ) ( 0.5 0.5 )

0.5

0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5

F( j u) 4cos ( 0.5) ( 0.5)

4 2 1( ) cos2 2

1

1 1 1( ) ( )

ω πω − πω ω

− −

π+ ω − π− ω

−

π+ − π− − π+ π−

ω = πω ω+ − ω −

∴ = πω× ω = + ωπ π

⎡ ⎤= + ω⎣ ⎦π

= − + −π π + −π +

∫ ∫

∫

j t j j j t

j jt j j t

j j t j j t j j t j j

u

f t e d e e e d

e e d

e e e ej t j t ( )

( ) ( )0.5 0.5 0.5 0.5

2 2 2 2

1 1 1( ) ( )

1 1 1 2cos 0.5 1 12cos 0.5 2cos0.5

2 42cos 0.5 cos 0.5 (0.8)

− −

⎡ ⎤⎢ ⎥⎣ ⎦⎡ ⎤

= + + − −⎢ ⎥π π + −π +⎣ ⎦⎡ ⎤ ⎛ ⎞= − = −⎜ ⎟⎢ ⎥π π + −π + π π + −π +⎣ ⎦ ⎝ ⎠

−⎛ ⎞= = =⎜ ⎟− π π −⎝ ⎠

t

j t j t j t j tje je je jej t j t

tt tt t t t

t tt t

0.3992∴ f



42. 1.5( ) 20 ( 2) V= − −tv t e u t (a)

21.5 1.5

2(1.5 ) 3 2 3

F j e( ) 20 ( 2) 20

20 20 20F ( 0)1.5 1.5 1.5

∞ −− ω − ω

−∞ −∞

−− ω − + ω −

−∞

ω = − − =

= = ∴ =− ω − ω

∫ ∫t j t t j tv

j t jv

u t e dt e dt

e e j ej j

0.6638=

3 2

3 4

20(b) F ( ) A ( ) B ( )

1.520F ( 2) 0.39830 282.31 0.08494 0.38913

1.5 2A (2)

− ω

−

ω = ω + ω =− ω

∴ = = ∠ ° = −−

0.08494∴ =

jv v v

jv

v

j e ej

j e e jj

(c) B (2) 0.3891= −v

(d) F ( 2) 0.3983=v j (e) φv(j2) = 282.3o or -77.69o



43. [ ]I( ) 3cos10 ( 0.05 ) ( 0.05 )ω = ω ω + π − ω − πj u u 0.05

2 2

0.05/ 20/ 20

/ 20/ 20

1 2(a) W 4 I( ) 9cos 102

18 1 1 9 9 1cos 20 0.1 sin 202 2 20

∞ π

−∞ − π

ππ

−π−π

= × ω ω = ω ωπ π

⎛ ⎞= + ω ω = × π +⎜ ⎟π π π⎝ ⎠

∫ ∫

∫

j d d

d

0.9 Jω = (b)

9 9 1(1 cos 20 ) 0.45 2 2sin 2020

0.05 2 0.1sin 20

ω

−ω

⎡ ⎤+ ω ω = = ω + × ω⎢ ⎥π π

, 0.04159 rad/s⎣

π = ω + ω

∫x

x

ω =

x⎦

x

x x

d

x∴



44. 4( ) 10 ( )−= tf t te u t (a) 8

2 2 8 21

0 0 0

∞

(b)

(c) ( )22

2

16100 )(+

=ω

ωjF

mJ/Hz 90.63 )( 2

0=

=ωωjF , mJ/Hz 7.669 )( 2

4=

=ωωjF

W ( ) 100 100 (64 16 2)( 512)

100 2512

0.3906 J

∞ ∞ −−

Ω = = = × + +−

= ×

∫ ∫t

t ef t dt t e dt t t

=

(4 )4 (4 )

20 0

2 2

10F( )16

ω =ω +

j

10F( ) {10 ( )} 10 [ (4 ) 1(4 )

10(4 )

∞∞ − + ω− − + ωω = = = − + ω −

+ ω

= ∴+ ω

∫j t

t j t ej te u t t e dt j tj

j

F



45. 2( ) 8 V−= tv t e

(a) 2 41

0

W ( ) 2 64 32 J∞ ∞

−Ω

−∞

= = × =∫ ∫ tv t dt e dt

(b) 2

0(2 ) (2 )

00

(2 ) (2 )2

0

(c)

F ( ) ( ) 8

F ( ) 8 8

8 8 82 2 2

∞ ∞−− ω − ω

−∞ −∞

∞− ω − + ω

−∞

∞− ω − + ω

−∞

ω = =

8 32 F ( )2 4

∴ ω = +

= − = + ω− ω + ω − ω + ω

∫ ∫

∫ ∫

tj t j tv

j t j tv

j t j t

j e v t dt e e dt

j e dt e dt

e e jj j j

= =+ ω vj

1

1

2 21 1

2 2 21

11 1 1 12 21 1

11 112

1

1 32 32 10.9 32 tan2 ( 4) 2 8( 4) 16 2

16 1 2 20.9 2 tan8( 4) 16 2 4 2

20.45 tan s (by SOLVE)4 2

ω−

−ω

−

−

⎡ ⎤ω ω× = ω = +⎢ ⎥π ω + π ω +⎣ ⎦

⎡ ⎤ ⎡ ⎤ω ω ω ω

2.7174 rad/

∴ = × + = +⎢ ⎥ ⎢ ⎥π ω + π ω +⎣ ⎦ ⎣ ⎦ω ω

π = +ω +

∫ d

∴ω =∴



46. (a) Prove: { ( )} { ( )} ( ) Let

{ ( )} ( ) { ( )}

∞− ω − ω

−∞

∞− ω − ω− ωτ

−∞

− = = − − = τ

∴ − = τ =

∫

∫

o

o o

j t j to o

j t j tjo

f t t e f t f t t e dt t t

f t t f e e dt e f t

F F

F F

o (b)

(c) (d) Prove: { ( )} F( ) Let 1 in (c) above− = − ω =f t j kF (e)

Prove: { ( )} { ( )} Let , ,

, { ( )} ( ) ( )

We assume ( ) 0 { ( )} { ( )}

∞− ω − ω − ω

−∞

∞∞− ω − ω

−∞−∞

= ω = = = − ω

= = ∴ = + ω

±∞ = ∴ = ω

∫

∫

j t j t j t

j t j t

dff t j f t e dt u e du j edt

dv df v f f t f t e j f t e dt

f f t j f t

F F

F

F F

/

1Prove: { ( )} F ( ) Let , 0

1 1{ ( )} ( ) F

1If 0, limits are interchanged and we get: F

1{ ( )} F

∞− ω

−∞

∞− ωτ

−∞

ω⎛ ⎞= = τ = >⎜ ⎟⎝ ⎠

ω⎛ ⎞∴ = τ τ = ⎜ ⎟⎝ ⎠

ω⎛ ⎞< − ⎜ ⎟⎝ ⎠

ω⎛ ⎞∴ = ⎜ ⎟⎝ ⎠

∫

∫

j t

j k

jf kt f kt e dt kt kk k

jf kt f e dk k k

jkk k

jf ktk k

F

F

F

rove: { ( )} F( ) Now, F( ) ( )

dF( ) ( )( ) { ( )} { ( )} ( )}

∞− ω

−∞

∞− ω

−∞

= ω ω =ω

ω= − = − ∴ = ω

ω

∫

∫

j t

j t

dtf t j j j f t e dtd

j

P

f t jt e dt j tf t tf f j f td

F

F F F∴



47. (a) (b) ( ) 4[sgn( 1) ( )] {4sgn( 1) ( )} { 4 ( )} 4= − δ ∴ − δ = − δ =f t t t t tF F −

f ( ) 4[sgn( ) ( 1)] {4[sgn( ) ( 1)] {4sgn(1) ( 1)} {4 ( 1)} 4 − ω= δ − ∴ δ − = δ − = δ − = j

(c) (10 30 ) (10 30 )

30 10 30 10 / 6 / 6

/ 6 / 6

4( ) 4sin(10 30 ) {4sin(10 30 )2

{ 2 2 } 2 2 ( 10) 2 2 ( 10)4 [ ( 10) ( 10)]

− ° − − °

− ° ° − − π π

− π π

⎧ ⎫⎡ ⎤= − ° ∴ − ° = − =⎨ ⎬⎣ ⎦⎩ ⎭− + = − πδ ω − + πδ ω+

= − π δ ω− − δ ω+

j t j t

j j t j j t j j

j j

f t t t e ej

j e e j e e j e j e

F F

F

j e e

t t t t t t t eF F F



48. (a) ( ) A cos( ) F( ) {A cos cos A sin sin }

A cos { [ ( ) ( )]} Asin [ ( ) ( )]

A{cos [ ( ) ( )] sin [ ( ) ( )]}

F( ) A[ ( ) ( )]φ − φ

= ω + φ ∴ ω = φ ω − φ ω =

⎧ ⎫πφ π δ ω + ω + δ ω − ω − φ δ ω − ω − δ ω + ω =⎨ ⎬

⎩ ⎭π φ δ ω + ω + δ ω − ω + φ δ ω − ω − δ ω + ω

∴ ω = π δ ω − ω + δ ω + ω

o o

o o o o

o o o oj j

o o

ot j t t

jj

Ff t

j e e (b) 2

2

2 1( ) 3sgn( 2) 2 ( ) ( 1) F( ) 3 2 ( )

6 1F( ) 2 ( )

− ω − ω

− ω − ω

⎡ ⎤= − − δ − − ∴ ω = × × − − πδ ω +⎢ ⎥ω ω⎣ ⎦

⎡ ⎤∴ ω = − − − πδ ω −⎢ ⎥ω ω⎣ ⎦

j j

j j

f t t t u t j e ej

(c)

j

j j e e j

2 2 2 2

1( ) sinh ( ) F( ) [ ] ( )2

1 1 1 1F( )2 2 2( )

−⎧ ⎫= ∴ ω = −⎨ ⎬⎩ ⎭

+ ω+ − ω −∴ ω = − = =

− + ω + ω − − ω ω +

kt ktkt u t j e e u tFf t

k j k j kjk j k j k k



49.

113

3 3

1(a) F( ) 3 ( 3) 3 ( 1) ( ) [3 ( 3) 3 ( 1)]2

3 3 1 3( ) ( )2 2 2

3(5) (1 5 1 1510

∞ω

−∞

ω ω + −

− −

ω = ω+ − ω− ∴ = ω + − ω − ωπ

(b) so f(5) = 0.1039∠73.52o

(c)

) 0.10390 106.48

∴ = = = −π π π

= − ∠ − ∠ −π

∫

∫

j t

j t j t jt j t

rad rad = ∠ − °

j u u f t u u e d

f t e dt e e ejt j t

f j∴

3

F( ) 3 ( 3 ) 3 ( 1)F( ) 3 F ( )

3( ) 3 ( ) ( ) (5) 0 0.10390 106.482

−

ω = − − ω + ω− →∴ ω = − ω

= δ − − ∴ = − ∠ − °π

a

jt j t

uj j

f t t e e fj t

j u

3

2 1F( ) 2 ( ) 3 ( 3 ) 3 ( 1) Now, {2 ( )}2

1 3 1( ) ( ) (5) 0.10390 106.482

−

ω = δ ω + − − ω + ω − δ ω = =π π

⎡ ⎤0.3618 15.985+= + − − ∴ = − ∠ − °⎢ ⎥π π π⎣ ⎦

jt j t

j u u

f t e e fj t

F

= ∠ °∴



50. 3 3(a) jF( ) 3 3 ( 1)

1ω = + + + δ ω −

+ ω ω 1.5( ) 3 ( ) 1.5sgn( ) 3 ( )−∴ = + + δ +

πt jt

j j

f t e u t t t e 1 sin 8 / 2(b)

(c) 32 2 2

6(3 ) 6(3 )F( ) ( ) 3 cos 2 ( )(3 ) 4 (3 ) 2

−+ ω + ωω = = ∴ =

+ ω + + ω +tj jj f t t u t

j j

F( ) 5sin 4 8 2.58 / 2ω

ω = ω = ×ω ω

j

( ) 2.5[ ( 4) ( 4)]∴ = + − −f t u t u t



51.

1/ 2

11

/ 22 2

1

/ 2 / 22 2 2 2

/ 2 / 22 2

T 4, periodic; find exp l form

1 104

12.5/ 2 / 4

1 1 1 12.52 / 4 / 2 / 4

1 42.5 ( ) (/ 2

− π

−

− π

−

− π π

− π π

′=

∴ =

⎡ ⎤⎛ ⎞∴ = −⎢ ⎥⎜ ⎟− π − π⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞∴ = + − +⎢ ⎥⎜ ⎟ ⎜ ⎟− π π π π⎝ ⎠ ⎝ ⎠⎣ ⎦

= − − +π π

∫ jn tn

jn tn

jn jnn

jn jn

c te dt

tc ejn n

c e ejn n jn n

e ejn n

/ 2 / 2

2 2

/ 22 2

2 2

)

5 102cos 2sin2 2

10 20( ) cos sin2 2

10 20F( ) cos sin 22 2 2

− π π

∞π

−∞

∞

−∞

⎡ ⎤−⎢ ⎥

⎣ ⎦π π⎛ ⎞= × + −⎜ ⎟π π ⎝ ⎠

π π⎡ ⎤∴ = −⎢ ⎥π π⎣ ⎦π π π⎡ ⎤ ⎛ ⎞

⎝ ⎠

∑

jn jn

jn t

e e

j n njn n

j n nf t j en n

∴ ω = − πδ ω−⎜ ⎟⎢ ⎥π π⎣ ⎦∑ j n n nj j

n n



53. 1F( ) 20 ( 20 )! 1

1 1 1 1 120 ( ) ( 20) ( 20) ( 40) ( 40)1 1 1 1 1 1 2 1 3

1 1( 60) ( 60) ...7 7

20 2010 ( ) [ ( 20) ( 20)] [ ( 40) ( 40)]2 3

20 20[ ( 60) ( 60) [ (7 25

∞

−∞

ω = δ ω−+

⎡= δ ω + δ ω + + δ ω− + δ ω + + δ ω−⎢ + + + +⎣⎤+ δ ω+ + δ ω− + ⎥⎦

= δ ω + πδ ω+ + πδ ω− + πδ ω+ + πδ ω − +π π

πδ ω + + πδ ω− + πδπ π

∑j nn

80) ( 80)] ...

10 20 20 20 20( ) cos 20 cos 40 cos 60 cos80 ...2 2 3 7 2520 1 1 1 10.25 cos 20 cos 40 cos 60 cos80 ...

2 3 7 2520 1 1 1 1(0.05) 0.25 cos1 cos 2 cos3 cos 4 ... 1.3858

2 3 7 25

ω + + πδ ω − +

∴ = + + + + +π π π π π

⎡= + + + + +⎢π ⎣⎡ ⎤∴ = + + + + + =⎢ ⎥π ⎣ ⎦

rad

f t t t t t

t t t t

f



54. Input ( ) 5[ ( ) ( 1)] ( ) ( ) ( )t

x t u t u t y t x z h t z d−∞

= = − − = −∫ z

u t

(a) (b) h t( ) 2 ( )h t u t= ( ) 2 ( 1)− u t= − (c) h t ( ) 2 ( 2)= x(t – z)

zt-1 t h( z)

z

2

y(t)

t

10

32

t < 0: y(t) = 0 1: ( ) 0t y t< = 2 : ( ) 0t y t< = 0 t< <1: 2 : 1 t< < 2 3t :< <

0

( ) 10 10= =∫t

y t dz t ∫ ==t

tdzty1

1) - 10( 10 )( ∫ ==t

tdzty2

2) - 10( 10 )(

t > 1: t > 2: t > 3:

∫ ==t

t

dzty1-

10 10 )( ∫ ==t

t

dzty1-

10 10 )( ∫ ==t

t

dzty1-

10 10 )(

5

x(t – z)

z t-1 t

x(t – z)

zt-1 t h( z)

z

2

h( z)

z

2

1 2

y(t)

t

10

1

y(t)

t

10

1 2



55. ( ) 5[ ( ) ( 2)]; ( ) 2[ ( 1) ( 2)]x t u t u t h t u t u t= − − = − − −

1

0

2

2

( ) ( ) ( )

1: ( ) 0

1 2 : ( ) 10 10( 1)

2 3: ( ) 10

3 4 : ( ) 10 10(2 2) 10(4 )

4 : ( ) 0( 0.4) 0; (0.4) 0; (1.4) 4

(2.4) 10; (3.4) 6; (4.4) 0

−∞

−

−

= −

< =

< < = = −

< < =

< < = = − + = −

> =∴ − = = =

= = =

∫

∫

∫

t

t

t

y t x z h t z dz

t y t

t y t dz t

t y t

t y t dz t

t y ty y y

y y y

t

t

or…. same answers as above

0

1

2

2

( ) ( ) ( )

1: ( ) 0

1 2 : ( ) 10 10( 1)

2 3: ( ) 10

3 4 : ( ) 10 10(2 2) 10(4 )

4 : ( ) 0

∞

−

= −

< =

< < = = −

< < =

< < = = − + = −

> =

∫

∫

∫

t

t

y t x t z h z dz

t y t

t y t dz t

t y t

t y t dz t

t y t



56.

2

( ) 2( )

0

2 20

02 2

2 2

( ) 3[ ], ( ) ( )

( ) ( ) ( )

3[ ]

13 [ ] 32

3 ( 1) 1.5 ( 1)( ) 3(1 ) 1.5(1 ) 1.5 3 1.5 , 0

− −

−∞

− − − −

− −

− −

− − − −

= − =

= −

= −

⎡ ⎤= − ⎢ ⎥⎣ ⎦= − − −

∴ = − − − = − + >

∫

∫

t t

t

tt z t z

tt z t t Z

t t t t

t t t t

h t e e x t u t

y t x z h t z dz

e e dz

e e e e

e e e ey t e e e e t



57.

0

( ) ( 2) ( )

2( ) (5 ), 2 53

∞

= −

= − < <

∫y t x t h z dz

h t t t

(a)

5 5

2 2

2 20( ) 10 (5 ) (5 )3 3

Note: ( ) is in window for 4 6

= × − = −

< <

∫ ∫ (b)

y t z dz

h z t

z dz

52

2

20 1( ) (5 )3 210 (0 9) 30 at 53

z⎛ ⎞= − −⎜ ⎟⎝ ⎠

= − − t= =

y t



58. ( 2)

0

( ) 5 ( 2), ( ) (4 16) [ ( 4) ( 7)], ( ) ( ) ( )∞

− −= − = − − − − = −∫tx t e u t h t t u t u t y t x t z h z dz

=

(a) 6 : ( ) 0 (5) 0t y t y< = ∴ (b)

6(8 2)

46 6

6 6

4 46

6 6 6 4

46 6 4 2 2

2

8 : (8) 5 (4 16)

(8) 20 80

20 ( 1) 80 ( )1

20 (5 3 ) 80 80 20 80 6020(1 ) 22.71

− − −

− −

− −

− −

−

= = −

2− −

∴ = −

⎡ ⎤= − − −⎢ ⎥

⎣ ⎦= − − + = + −

= + =

∫

∫ ∫

z

z z

z

t y (c)

e z dz

y e z e dz e e dz

ee z e e e

e e e e e ee

7 4

7(10 2)

47

8

47 7

8 8 8 7 84

4 48 7 4 1 4 1 4

10 : (10) 5 (4 16)

(10) 20 ( 4)

(10) 20 80 20 [ ( 1)] 80 ( )

20 (6 3 ) 80( ) 40 20 15.081

− − −

−

− − − −

− − − − −

= = −

∴ = −

∴ = − = − −

= − − − = + =

∫

∫

∫ ∫

z

z

z z z −

e z dz

y e e z dz

t y

y e ze dz e e dz e e z e e e

e e e e e e e



59.

0

0 0

0

( ) sin , 0 ; 0 elsewhere, Let ( ) ( )

( ) ( ) ( )

0 : ( ) 0

0 : ( ) sin sin

1( ) (sin cos )2

1 [ (sin cos ) 1]21 (sin cos )2

−

∞

− + −

−

−

−

= < < π =

= −

< =

< < π = × =

⎡ ⎤∴ = −⎢ ⎥⎣ ⎦

= − +

= − +

∫

∫ ∫

t

t tt z t z

tt z

t t

t

h t t t x t e u t

y t x t z h z dz

t y t

t y t z e dz e e z d

y t e e z z

e e t t

t t e

z

y

y

(a) (1) 0.3345+= (b) (2.5) 0.7409= (c)

0

0

: ( ) sin

1 1: ( ) (sin cos ) ( 1) 12.0702 2

(4) 0.2211

π−

π− − π

> π =

⎡ ⎤> π = − = + =⎢ ⎥⎣ ⎦−

∴ =

∫t z

t z t t

t e e z dz

y y t e e z z e e e

y

y y



60.

0

1

21

2

2

2 3

( ) 0.8( 1)[ ( 1) ( 3)],( ) 0.2 ( 2)[ ( 2) ( 3)]

( ) ( ) ( ) ,

3 : ( ) 0

3 4 : ( ) 0.8( 1)0.2( 2)

( ) 0.16 ( 2 2 2)

1 10.16 [ ( 1) 2 2 ] 0.16 ( 1)3 2

∞

−

−

= − − − −= − − − −

= −

< =

< < = − − −

∴ = − − + − +

= − + + + − = − + +

∫

∫

∫

t

t

x t t u t u th t t u t u t

y t x t z h z dz

t y t

t y t t z z dz

y t tz t z z z dz

z t z t dz z t11

2

22

3 2

3 2 2 2

3 2

(2 2 )

1 8 1 10.16 ( 1) ( 1) ( 1) ( 1) 4 (2 2 ) ( 1 2)3 3 2 21 1 8 1( ) 0.16 ( 1) ( 1) 2 2 2 6 2 63 3 3 2

1 1 1 10.16 1 2 1 6 3 86 2 2 2

−− ⎡ ⎤+ −⎢ ⎥⎣ ⎦

⎡ ⎤= − − + + + − − + + − − −⎢ ⎥⎣ ⎦⎡ ⎤∴ = − + − + + + − − − − + − − +⎢ ⎥⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞= + − − + − − + + + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

∫tt

z t z

t t t t t t

y t t t t t t t t t t

t t t 3 21 3 9 90.166 2 2 2

⎛ ⎞= − +⎜ ⎟⎝ ⎠

−t

t t

(b)

(a)

3(4.8) 90.67 10y −∴ = ×

3

333 2

22

(3.8) 13.653 10

1 14 5 : ( ) 0.16( 1) ( 2) 0.16 ( 1) (2 2 )3 2

1 1( ) 0.16 (27 8) ( 1)5 (2 2 )13 219 110.16 2.5 2.5 2 2 0.16 0.53 6

−∴ = ×

⎡ ⎤< < = − − − = − + + + −⎢ ⎥⎣ ⎦

⎡ ⎤= − − + + + −⎢ ⎥⎣ ⎦⎡ ⎤ ⎛ ⎞= − + + + − = −⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

∫t y t t z z dz z t z t z

y t t t

t t t

y

∴



61. 2 2

0

2( ) 2

0

2 2

02

( ) 10 ( ), ( ) 10 ( )

( ) ( ) ( )

( ) 10 10

100 100

( ) 100 ( )

− −

∞

− − −

− −

−

= =

= −

∴ =

= = ×

∫

∫

∫

t t

tt z z

tt t

x

∴ = t

t e u t h t e u t

y t x t z h z dz

y t e e dz

e dz e t

y t t e u t



62. 4( ) 5 ( )th t e u t−= (a)

0.88 0.8 6.4

10.1

25W 25 ( ) 1.3990 J8

25% 1.3990 / 100%8

− − −Ω = = − =

⎛ ⎞

44.77%= ×⎜ ⎟

⎝ ⎠

∫ te dt e e

= (b)

∴

221

1 200

11

5 1 25 25 1H( ) W tan4 16 4 4

25 1 0.9224W tan 0.9224 J % 100%4 2 25 / 8

−Ω

−Ω

ωω = ∴ = ω =

ω + π ω + π

29.52%= = ∴ = ×π

∫j

=

j d

∴



63.

22 2 2F( ) ( ) (2 2 ) ( )(1 )(2 ) 1 2

− −ω = = − ∴ = −+ ω + ω + ω + ω

t tj f t e e u tj j j j

(a) 2 3 41

0

4 8 4 1W (4 8 4 )2 3 4

J∞

− − −Ω = − + = −∫ t t te e e dt

e e e e tf e e

− − −

− − ×

= − + = − + = = =

3+ =

(b) f t 2

0.69315 2 0.69315max

( ) 2 4 0, 2 4 0, 2, 0.693152( ) 0.5

t t t t

=

∴ = −



64.

2 3

1 1/ 6 1/ 2(a)

1/ 3)(2 )(3 ) 2 3

ω = = − +ω + ω + ω ω + ω + ω

jF(j

1 1 1( ) sgn( ) ( ) ( )12 2 3

− −∴ = − +t t

j j j j j

f t t e u t e u t

2 3

1 1/ 6 1/ 2(b) 2 / 3)(2 )(3 ) 2 3

+ ωω = = + −

ω + ω + ω ω + ω + ωjjF(

j 1 1 2( ) sgn( ) ( ) ( )

12 2 3− −∴ = + −t t

j j j j j

f t t e u t e u t 2

2 3

(1 ) 1/ 6 1/ 2 4 / 3(c) )(2 )(3 ) 2 3

+ ωω = = − +

ω + ω + ω ω + ω + ω (d)

F(

1 1 4( ) sgn( ) ( ) ( )12 2 3

− −∴ = − +t t

jj j j j j j

f t t e u t e u t

j

3

2 3

(1 ) 1/ 6 1/ 2 8 / 3) 1(2 )(3 ) 2 3

+ ωω = = + + −

ω + ω + ω ω + ω + ωF(

1 1 8( ) ( ) sgn( ) ( ) ( )12 2 3

− −∴ = δ + + −t t

jj j j j j j

f t t t e u t e u t

j



65. ( ) 2 ( )th t e u t−=

(a) 1 2H( ) 21 1

ω = × =+ ω + ω

jj j

(b) V1 1 1 1/V 1 1/

H( )2 1 2

ωω = =

+ ωo

i

=+ ω

j jj j

(c) Gain = 2



66.

2

2

2

2 2

1 1( ) 22V ( ) 1 1 ( ) 2( ) 21

2( ) 2( ) 2 2( ) 2( )V ( ) 1

( ) 2( ) 2 ( ) 2( ) 2

2 2 4 8Let V ( ) 1 ; 1 12 2 2

A B A BV ( ) 1 Let 0 01 1 1 1 1 1 1 1

Let

ω+ω +ωω = =

ω + ω ++ ω+ω

ω + ω + − ω − ω∴ ω = = +

ω + ω + ω + ω +

− ± −ω = ∴ = − = = − ±

+ +

∴ = + + = = ∴ + =+ + + − + −

o

o

o

o

jjjj

j jjj

j j j jjj j j j

xj x x x jx x

x xx j x j j j

( 1 1) ( 1 1)

A B B 2 B 1 2 A B 2, A B 21 1 1 1 1 1

B B 2 2 B B 0 B 1 1 A 1 11 1 1 1 1 1 1 1V ( ) 1 , V ( ) 11 1 1 1 ( ) 1 1 ( ) 1 1

( ) ( ) (1 1) ( ) (1 1) (− − − +

+= − ∴ + = ∴ − = = + ∴ + =

− +∴ − + + + + = ∴ = − − ∴ = − +

− + − − − +∴ = + + ω = − −

+ + + − ω + + ω + −

∴ = δ − − − +

o o

j t j to

jx j jj j j j

j j j j jj j j jx j

0−

x j x j j j j jv t t j e u t j e u

45 45

)

( ) 2 ( ) 2 ( )− °− − °+ −= δ − −j jt t j jt t

t

t e u t e u t

t( ) 2 2 cos( 45 ) ( )−= δ − + °tt e t u



67.

22

/ 6

5 / 10 /V ( ) 105 / 35 30( ) 1/ 7 6( )

10 10 / 6V ( ) 7 16( ) 7( ) 1 ( ) ( )6 6

49 24 1 10 / 6 2 27 / 6 / 2 , 1 V ( )36 36 6 ( 1/ 6)( 1) 1/ 6 1

( ) 2( ) ( )− −

ω ωω = =

ω + + ω ω + + ω

∴ ω = =ω + ω + ω + ω +

⎛ ⎞∴ ω = − ± − = − − ∴ ω = = −⎜ ⎟⎜ ⎟ ω + ω + ω+ ω⎝ ⎠ +

∴

c

c

c

j jjj j j j

jj j j j

j jj j j j

= −t tcv t e e u t



68. 2 3( ) 5 ( ), ( ) 4 ( )t tf t e u t g t e u t− −= = (a) f g

0

2 2 3 2

0 02

2 3

( ) ( )

5 4 20

20 ( 1) V( ) ( )

∞

− − − −

−

− −

∗ = −

= =

= − −

∴ ∗ = −

∫

∫ ∫t t

t z z t z

t t

t t

f t z g z dz

e e e dz e e dz

e e

f g e e u t

2 3

5 4 20(b)

F( ) , G( ) F( )G( )2 3 ( 2)( 3)

20 20F( )G( )2 3

ω = ω = ∴ ω ω =ω + ω + ω + ω +

20( 2 ) ( )− −ω ω = −ω + ω +

j jj j j j

j j f g e u tj j

∴ ∗ = −t t

j j

∴



69. 2( )4 2

jjjωω

ω=

+H

24( ) from Table 18.2i jj

ωω

=V

Therefore o2 24 24( )

4 2 2jj

j j jωω

ω ω ω⎡ ⎤ ⎛ ⎞

= =⎜ ⎟⎢ ⎥+ +⎣ ⎦ ⎝ ⎠V

In the time domain, then, we find 2( ) 24 ( ) Vt

ov t e u t−=



70. so fromTable 18.2, ( ) 2 cos 4th t e t−=

( )( )2

2 1( )

1 1

jj

j

ωω

ω

+=

+ +H

6. Define output function f(t).

(a) ( ) 4 ( )jω πδ ω=I

Therefore F(ω) = 2

8 (1 ) 8( ) ( )(1 ) 16 17

jj

π ω πδ ω δω

⎡ ⎤+=⎢ ⎥+ +⎣ ⎦

ω .

The time domain output is then given by f(t) = 4/17. (b) ( ) 2 jj e ωω −=I

Therefore F(ω) = 2

4(1 )(1 ) 16

jj ej

ωωω

−⎡ ⎤+⎢ ⎥+ +⎣ ⎦

.

( ) ( )14 cos 4 1 ( 1)te t u t− − ⎡ ⎤− −⎣ ⎦ The time domain output is then given by f(t) =

(c) We find the response due to a unit step u t and treat as two unit steps, each shifted appropriately.

( ) ( )i t

2

2(1 ) 1( ) ( )(1 ) 16

jjj j

ωω πδ ωω ω

⎡ ⎤+= +⎢ ⎥+ + ⎣ ⎦

R

[ ]1 1( ) sgn( ) 2 cos 4 4sin 4 ( )17 17 17

ter t t t t u t−

= + − −

Therefore the system response is

( ) { }

( ) { }

0.25

0.25

2 1 cos 4( 0.25) 4sin 4( 0.25) ( 0.25)17

2 1 cos 4( 0.25) 4sin 4( 0.25) ( 0.25)17

t

t

e t t u t

e t t u t

− +

− −

⎡ ⎤− + − + +⎣ ⎦

⎡ ⎤− − − − − −⎣ ⎦


Chapter 18 solutions_to_exercises(engineering circuit analysis 7th)

Business

Transcript of Chapter 18 solutions_to_exercises(engineering circuit analysis 7th)