Bài giải mạch P17

b yt is not periodic although sin t and 4 cos 2πt are independently periodic.. Since the sum of a periodic function and a constant is also periodic, ht is periodic... Calculate the Fouri

Chapter 17, Solution 1

(a) This is periodic with ω = π which leads to T = 2π/ω = 2

(b) y(t) is not periodic although sin t and 4 cos 2πt are independently

periodic

(c) Since sin A cos B = 0.5[sin(A + B) + sin(A – B)],

g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(–t)] = –0.5 sin t + 0.5 sin7t

which is harmonic or periodic with the fundamental frequency

ω = 1 or T = 2π/ω = 2π

(d) h(t) = cos 2 t = 0.5(1 + cos 2t) Since the sum of a periodic function and

a constant is also periodic, h(t) is periodic ω = 2 or T = 2π/ω = π

(e) The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic

(c) f3(t) = 4 sin2 600π t = (4/2)(1 – cos 1200π t)

ω = 1200π or T = 2π/ω = 2π/(1200π) = 1/600

(d) f4(t) = e j10t = cos 10t + jsin 10t ω = 10 or T = 2π/ω = 0.2π

ncos(

nsinn

nsinn

2

nsin(

ncosn

ncosn

10x

2 cosn tn

t

π = [–5/(n

2π2)](cos 2nπ – 1) = 0

2 sinn tn

5

ππ

t5

1 n

ππ

2

5.0]x2x[2

1dt)t(z

π

=π

−π

=π

−π

=ω

sinn

1ntdtcos2

1ntdtcos1

1dtncos)t(z

=π

+π

−

=π

−π

=ω

evenn

0,

oddn,n

6nt

cosn

2nt

cosn

1ntdtsin2

1ntdtsin1

1dtncos)t(z

65

1

n∑∞

=

= π+

−

=

Chapter 17, Solution 6

.0a,functionodd

anisthis

Since

32

6)1214(2

1dt)t(y2

1

a

2

2,

2

T

n

2 0 o

o

=

=

=+

ππ

+

=

=π

−π

=π

−π

−π

−π

=

π

−ππ

−

−ππ

−

=ππ

−ππ

−

=

π+

π

=ω

=

odd

n 1n

even n , 0

odd n , n 4

2 1

1 0

2 1

1 0

2

n

)tnsin(

n

14

3)

t

(

y

))ncos(

1(n

2))ncos(

1(n

2))ncos(

1(n4

))ncos()n(cos(

n

2)1)n(cos(

n

4)

tncos(

n

2)tncos(

n4

dt)tnsin(

2dt)tnsin(

4dt)tnsin(

)t(y2

2

b

Chapter 17, Solution 7

0a,6T/2,

/nsinn

106

/nsin

π

−π

2b

[cos n /3 cosn /3 2sin n /3]

n

106

/ntncosn

106

/ncosn

π+π

π

=

1 n

n

ncosn /6 b sinn /6a

)t(

where an and bn are defined above

Chapter 17, Solution 8

π

=π

=ω

tdt)1t(22

1dt

nsinn

1tnsinn

ttncosn

12tdtncos)1t(22

2dtncos

2 1

=π+

=ω

− π π

−

= π +

= ω

n cos n

1 t n cos n

t t n sin n

1 2 tdt n sin ) 1 t ( 2 2

2 dt n sin

2 1

1

T

0

o n

∑∞

=

π

−π

−

=

1 n

n

tncosn

)1(4

2,

T

183.3

104

/sin)

4(4

1004/cos108

2)

(

0 2

0 0

[ t n t n ]dt dt

t n t

dt n t

40cos

/sin

25]12/[cos

For n>1,

2

)1(sin)1(

202

)1(sin)1(

204

)1(sin)1(

204

)1(sin)

++

=

−

−+

++

n

n n

n n

t n n

π

ππ

ππ

ππ

0sin

102sin4

20,

3662.62/sin

20sin

ππ

3 2

=ω

0

t jn T

0

1 0

2 1

t jn t

jn t

o jn

e2jn

e42

1dte

)2(dte

42

1dte

)t(h

oddn,n

j6]

6ncos6[n

je

2e

24e

4n

−

=

−ππ

=+

−

−π

Thus,

t jn

odd n n

en

6j)

−

=

Chapter 17, Solution 11

2/T/2,

− ω

− T

0

0 1

1 0

2 / jn 2

/ jn t

o jn

4

1dte

)t(y

−

−π

−π

−

2 2

2 / jn

jn

2e

jn

2)12/jn(4/n

−π

+

−ππ

+π

−π

π

− π

π

jn

2e

jn

2e

jn

2)12/jn(en

4jn

2n

44

2 2 2

2

=

But

2/nsinj2/nsinj2/ncose

,2/nsinj2/nsinj2/ncos

e2/nsinn2/nsin)12/jn(j1n

1)

t(

−π+π

0

2 T

4)

3/tt(2

2 0 3

=

−π

π

=

−ππ

0 2

T 0

n

t2)ntcos(

n

21dt)ntcos(

)tt2(T

2

+

−π

0 2

4)n2cos(

nn

1)11(n

2(T

0

2

π π

−+

π

−

−π

0 2

2 3

0

n

1))ntcos(

nt)nt(sin(

n

1n

2

0n

4n

4

=

π+π

1

n 2

2

) nt cos(

n

4 3

0

an = (2/T) ∫T ω

0h(t)cos(n ot)dt = [2/(2π)] ∫0π + ∫ππ − π 

2

dt)ntcos(

)tsin(

20dt

)ntcos(

tsin10

Since sin A cos B = 0.5[sin(A + B) + sin(A – B)]

sin t cos nt = 0.5[sin((n + 1)t) + sin((1 – n))t]

sin(t – π) = sin t cos π – cost sin π = –sin t

sin(t – π)cos(nt) = –sin(t)cos(nt)

an = π ∫0π + + − − ∫ππ + + − 

2

dt)]t]n1sin([

)t]n1[sin([

20dt)]

t]n1sin([

)t]n1[sin([

+

++

2n

1

)t]n1cos([

2n

1

)t]n1cos([

n1

)t]n1cos([

π+

−

−

++

)]n1cos([

3n

1

)]n1cos([

3n1

3n1

35

But, sin A sin B = 0.5[cos(A–B) – cos(A+B)]

sin t sin nt = 0.5[cos([1–n]t) – cos([1+n]t)]

bn = (5/π){[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/(1+ n)]0π + [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ + π2π

)]

n1

n1

)]n1sin([

60 30

) 4 / n sin(

1 n

10 ) nt 2 cos(

) 4 / n cos(

1 n

10 2

Chapter 17, Solution 15

(a) Dcos ωt + Esin ωt = A cos(ωt - θ)

where A = D2 +E2 , θ = tan-1(E/D)

n

1)1n

1

2 1 6

2

1 n tan nt 10 cos n

1 ) 1 n ( 16

(b) Dcos ωt + Esin ωt = A sin(ωt + θ)

where A = D2 +E2 , θ = tan-1(D/E)

++

1

3 1 6

2

n 4 tan nt 10 sin n

1 ) 1 n ( 16

Comparing v2*(t) with v1(t) shows that

v2*(t) = 2v1((t + to)/2) where (t + to)/2 = 0 at t = -1 or to = 1

Hence v2*(t) = 2v1((t + 1)/2)

But v2*(t) = v2(t) + 1

v2(t) + 1 = 2v1((t+1)/2)

v2(t) = -1 + 2v1((t+1)/2)

= -1 + 1 − π82 cosπ +t 21 + 91cos3π +t 21 + 251 cos5π +t 21 +"

v2(t) = − π82 cosπ2t + π2 + 91cos32πt + 32π+ 251 cos52πt + 52π+"

1 2

t 3 sin 9

1 2

t sin

(c) cos nπ(-t) sin nπ(-t) = - cos nπt sin nπt, odd

(d) sin2 n(-t) = (-sin πt)2 = sin2 πt, even

(e) et, neither odd nor even

t41T

1 8

2 / T

6

2dt)t(T

2

= 13(2t − 4t) +4(3−2)

2 1

an = ∫T/4 π

0 (t)cos(n /3)dtT

tnsinn

36

163

tnsinn

33

tnsinn

t33

tncosn

96

n cos 3

n 2 cos n

1 24 2

At t = 2,

f(2) = 2 + (24/π2)[(cos(2π/3) − cos(π/3))cos(2π/3)

+ (1/4)(cos(4π/3) − cos(2π/3))cos(4π/3) + (1/9)(cos(2π) − cos(π))cos(2π) + -]

= 2 + 2.432(0.5 + 0 + 0.2222 + -) f(2) = 3.756

2

tncos)t1(24

4dt)tncos(

)t(T

4

1

t n cos 2

n cos 1 n

8 2

1

Chapter 17, Solution 22

Calculate the Fourier coefficients for the function in Fig 16.54

f(t) 4

t -5 -4 -3 -2 -1 0 1 2 3 4 5

Figure 17.61 For Prob 17.22 This is an even function, therefore bn = 0 In addition, T=4 and ωo = π/2

0 2 1

0

2 T

4

2dt)t(T

4

4dt)ntcos(

)t(T4

1

0 2

n

t2)2/tncos(

n

44

=

n

8 ) 1 ) 2 / n (cos(

−ππ

Chapter 17, Solution 23

f(t) is an odd function

f(t) = t, −1< t < 1

ao = 0 = an, T = 2, ωo = 2π/T = π

bn = ∫ ω = ∫1 π

0

2 / T

2

4dt)tnsin(

)t(T

4

0 2

2 sin(n t) n tcos(n t)n

π = −[2/(nπ)]cos(nπ) = 2(−1)n+1/(nπ)

) n sin(

n

) 1 ( 2

4

f(t) = 1 + t/π, 0 < t < π

bn = ∫π + π

π 0 (1 t/ )sin(nt)dt2

n

1)ntcos(

n

12

= [2/(nπ)][1 − 2cos(nπ)] = [2/(nπ)][1 + 2(−1)n+1]

a2 = 0, b2 = [2/(2π)][1 + 2(−1)] = −1/π = −0.3183 (b) ωn = nωo = 10 or n = 10

a10 = 0, b10 = [2/(10π)][1 − cos(10π)] = −1/(5π) Thus the magnitude is A10 = 2

(c) f(t) = ∑∞ π

=

π

−π

1 n

)ntsin(

)]

ncos(

21[n2

f(π/2) = ∑∞

=

ππ

−π

1 n

)2/nsin(

)]

ncos(

21[n

2

π

For n = 1, f1 = (2/π)(1 + 2) = 6/π For n = 2, f2 = 0

For n = 3, f3 = [2/(3π)][1 − 2cos(3π)]sin(3π/2) = −6/(3π) For n = 4, f4 = 0

For n = 5, f5 = 6/(5π), Thus, f(π/2) = 6/π − 6/(3π) + 6/(5π) − 6/(7π) -

= (6/π)[1 − 1/3 + 1/5 − 1/7 + -]

f(π/2) ≅ 1.3824 which is within 8% of the exact value of 1.5

3

4dt)tnsin(

)t(T

4

=

1

0 2

nt2cosn2

t33

nt2sinn4

93

33

n2sinn4

93

4

2 2

1

t 2 sin 3

n 2 cos n

2 3

n 2 sin n 3

1 0

T

4

1dt)t(T

1

= 1

an = (t)cos(n t)dtT

ππ+

ππ

22

tnsinn

42

tnsinn

22

nsinn4

bn = ∫T ω

0 (t)sin(n ot)dtT

2

2

tnsin1dt2

tnsin2dt2

tnsin142

−

ππ

22

tncosn

42

tncosn22

= [cos(n ) 1]

n

πHence

ππ

−ππ

+

1

n

) 2 / t n sin(

) 1 ) n (cos(

) 2 / t n cos(

)) 2 / n sin(

) 2 / n 3 (sin(

2 1

tncosn

t22

tnsinn

4dt

2

tnsint

ππ

=π

∫4

2

ncosn

22

nsinn

2

2

1a

Frms2 = 0.5Σbn2 = [1/(2π2)][(16/π2) + 1 + (16/(8π2)) + (1/4) + (16/(625π2))] = (1/19.729)(2.6211 + 0.27 + 0.00259)

Frms = 0.14659 = 0.3829

Compare this with the exact value of Frms = t dt 1/6

T

2 1 0

4dt)tncos(

)t(T

4

=

1

0 2

n

t)tncos(

n

1)tnsin(

n

14

−ππ

n

t)tnsin(

n

1)tncos(

1

k 2 2

) n sin(

n

4 ) n cos(

2

+π

− 2 cos(nt) ntsin(nt) 0n

2

= 4/(n2π)

−π

)ntsin(

)t(

1 ) nt cos(

2 / T 2 /

o t

o jn

T

1dte

)t(T

(b) The first term on the right hand side of (1) vanishes if f(t) is odd Hence,

Chapter 17, Solution 31

/T

2'T

2'/

T'T),t()

→

α

=α

0

o

'T

2tdt'ncos)t(h'T

2'a

Let αt=λ , dt =dλ/α, αT'=T

n T

0

o

n ( )cosn d / aT

2'

Similarly, bn'=bn

Chapter 17, Solution 32

When is = 1 (DC component)

i = 1/(1 + 2) = 1/3 For n 1, ≥ ωn = 3n, Is = 1/n2∠0°

I = [1/(1 + 2 + jωn2)]Is = Is/(3 + j6n)

n41n

1)

3/n(tann13

0n1

2 2

1 2

2

−

∠+

=

∠+

+

1 n

1 2

2 cos( 3 n tan ( 2 n ))

n 4 1 n 3

1 3

1

Chapter 17, Solution 33

For the DC case, the inductor acts like a short, Vo = 0

For the AC case, we obtain the following:

s o

o o

s o

VVn

5n.2j1

04

Vjnnj

V10

VV

=

π+π+

−

)5n

.2(jn

4

n

5n.2j1

1n

4A

n

5n.2j1

VV

2 2 n

n

s o

−π+

π

=Θ

−

=

Θ

−π+

π

n

5n

.2tan

;)5n

.2(n

4

2 2 2 2

2 n

A)

t(v

1 n

n n

=

Θ+π

tan)2/()4/n[(

4nn20

)n/)2n((

tan)2n(nn

)2/)4/n((

20

2 1 2

2

2 1 2

2 2 2

−

−π

−π

∠+

=

−

∠

−+

π

−π

2 1 2

2 n tan 2 4

n nt cos 4 n n 20

1

f 2 (t)

2

t -2 -1 0 1 2 3 4 5

Figure 17.57(b) For Prob 17.35 The signal is even, hence, bn = 0 In addition, T = 3, ωo = 2π/3

13

4

n

2)

3/n2sin(

n

6)

3/n2sin(

n

33

π

vs(t) = ∑∞

=

ππ

π

−

1 n

)3/tncos(

)3/n2sin(

n

1234

Now consider this circuit,

4(jn12

v9j

2 2

s

−π+

π

−

For the dc case, n = 0 and vs = ¾ V and vo = vs/2 = 3/8 V

We can now solve for vo(t)

3

t n 2 cos A 8

3

1 n

π

ππ

n 2

3 3

n tan 90

and 6

3

n 4 n

16

) 3 / n 2 sin(

n

6

n 2

2 2 2

2 n

where we can further simplify An to this,

81 n

4 n

) 3 / n 2 sin(

9

4 4 n

+ππ

12

nsinn

1n

2 2 2 2

π+

π

=

ππ

+π

ωn = n and 2 H becomes jωnL = j2n Let Z = 1||j2n = j2n/(1 + j2n)

If Vo is the voltage at the non-reference node or across the 2-H inductor

Vo = ZVs/(1 + Z) = [j2n/(1 + j2n)]Vs/{1 + [j2n/(1 + j2n)]}

= j2nVs/(1 + j4n) But Vs = An∠−θn

Vo = j2n An∠−θn/(1 + j4n)

nsin9n1

n4tan5.10010

−

io(t) = cos( nt 100 5 tan 4 n )

n 16 1

1 10

2

odd

n 1 n

1 2

+

1 k

),tnsin(

n

120

For the DC component, the capacitor acts like an open circuit

Vo = 5 For the nth harmonic,

Vs = [20/(nπ)]∠0°

10 mF becomes 1/(jωnC) = −j/(nπx10x10−3) = −j100/(nπ)

vo =

2 2

1 s

n25n

n

5tan90

100n

20100jn20

100j20

n

100j

Vn

100j

π+π

π+

°

−

∠

=π

−π

−

=+π

−π

π+π

n

5 tan 90

t n sin(

n 25 n

1

2 2

=

1 k

n

122

1)t(v

π

=ωω

+

ω

j1

j

n

n o

1 o

1 2

2

o o

n1

ntan2

90n

2ntann

1

90nV

π+

•π

∠π+

∠π

−

1kn),ntantncos(

n1

2)

t(

+

1 k

),tnsin(

n

110

T = 2, ωo = 2π//T = π, ωn = nωo = nπ For the DC component, io = 5/(20 + 40) = 1/12

For the kth harmonic, Vs = (10/(nπ))∠0°

Let Z = −j20/(nπ)||(40 + j0.1nπ) =

π+

+π

−

π+

π

−

n.0j40n

20j

)n.0j40(n

20j

=

)20n

.0(jn40

800jnn

.0jn4020j

n.0j40(20j

2 2 2

−π

=π+

π+

−

π+

−

Zin = 20 + Z =

)20n

.0(jn40

)1200n

(jn802

2 2

2 2

−π+

π

−π+

π

I =

)]

1200n

2(jn802[n

)200n

(jn400Z

V

2 2

2 2

in

s

−π+

ππ

−π+π

=

Io =

)20n

.0(jn40

I20j)

n.0j40(n

20j

In

20j

2

2π −+

π

−

=π+

+π

2(jn802[n

200j

2

2π −+

ππ

−

=

2 2

2 2

2 2 1

)1200n

()802(n

)}

n802/(

)1200n

{(

tan90

200

−π+π

π

−π

+

1 k

n

n sin( n t ) I

200 20

°

=

n802

1200n

2tan90

2 2 1 n

In =

)1200n

()n804(n

1

2 2

π

1

0

2 1

0

T

0v(t)cos(n t)dt 2(1 t)cos(n t)dtT

2

=

1

0 2

n

t)tncos(

n

1)tnsin(

n

12

−ππ

=

2 2

2 2 2

2

)1n(

4odd

n,n4

evenn

,0)ncos1(n

2

−π

=

=π

=

=π

−π

0

T

0v(t)sin(n t)dt 2 (1 t)sin(n t)dtT

−ππ

−

n

2)

tncos(

n

t)tnsin(

n

1)tncos(

n

12

1

0 2

2 1

n

)1n(

16n

4A

,n2

)1n(

−π

+π

=

−π

−

tan

=φ

For the DC component, vs = 1/2 As shown in Figure (a), the capacitor acts like an open circuit

Vo = 3i = 0.75 For the nth harmonic, we consider the circuit in Figure (b)

ωn = nπ, Vs = An∠–φ, 1/(jωnC) = –j4/(nπ)

At the supernode,

(Vs – Vx)/1 = –[nπ/(j4)]Vx + Vo/3

Vs = [1 + jnπ/4]Vx + Vo/3 (3) But –Vx – 2Vx + Vo = 0 or Vo = 3Vx

Substituting this into (3),

Vs = [1 + jnπ/4]Vx + Vx = [2 + jnπ/4]Vx = (1/3)[2 + jnπ/4]Vo = (1/12)[8 + jnπ]Vo

Vo = 12Vs/(8 + jnπ) =

)8/n(tann

64

A12

1 2

2

n

π

∠π+

16n

4n

64

4 4

2 2 2

−π

+ππ

+

−

−

Thus

vo(t) = ∑∞

=

θ+π+

1 n

n

n cos( n t ) V

4 3

2

2 ( 2 n 1 )

16 n

4 n

64

12

−π

+ππ

14

2

For the DC component,

Vin = 2/π The inductor acts like a short-circuit, while the capacitor acts like an open circuit

Vo = Vin = 2/π For the nth harmonic,

Vin = [–4/(π(4n2 – 1))]∠0°

2 H becomes jωnL = j4n 0.1 F becomes 1/(jωnC) = –j5/n

°

∠

−

−+

−

)1n4(

04)

10n(j4

10j

2

=

2 2

1

)10n(16)1n4(

)}

5.2n2(tan90{40

−+

−π

π n 1 A n cos( 2 nt n ) 2

where

An =

29 n 40 n 16 ) 1 n 4 (

+

=

1 k

s sinn , n 2k-1

n

1205

v

π

=ω

=ωω

=

→



−ω

=

−

nn

,VRC

jV

)V0(CjR

0

V

o n

s n o o

n s

For n = 0 (dc component), Vo=0

For the nth harmonic,

2 2

5 9

4 2 2 o

o o

n

1010

x40x10xn

2090

n

20RCn

901V

π

=π

=

−

∠ππ

10)t(v

Alternatively, we notice that this is an integrator so that

∑

=

=ππ

n

12

10dtvRC

1)t(v

Chapter 17, Solution 43

2

130)

ba(2

1

1 n

2 n

2 n

Z = R + jωnL + 1/(jωnC) = 10 + j2n – j25/n

I = V/Z For n = 1, V1 = 100, Z = 10 + j2 – j25 = 10 – j23

3

1 n

n n n

V21

0

T

2dt)t(fT1

f(t) =

2t1,0

1t0,22

0

2dt 2(t t t /3))

t1(44

6416642

14

1A

2

1

1 n

2 n

≅+ ∑∞

Z = 2||(–j/n) = 2(–j/n)/(2 – j/n) = –j2/(2n – j)

V = ZI = [–j2/(2n – j)]I For n = 1, V1 = [–j2/(2 – j)]16∠45° = 14.311∠–18.43° mV

1 n

n n n

V21

= 20x40 + 0.5x10x0.014311cos(45° + 18.43°) +0.5x12x0.005821cos(–60° + 135.96°)

= 800.1 mW Chapter 17, Solution 49

2

1dt4dt12

1dt)t(zT

1Z

0

2 T

0

2 rms

=

π

581.1

Zrms =(b)

9193.2

25

116

19

14

1118

14

1n

362

14

1)ba(2

1a

2 o

=π+

=+

rms =Z

Chapter 17, Solution 50

nt j

1

n,

dte)t(T

Using integration by parts,

1 1 t jn 1

1

t

jn2

1e

jn2

jn

)j(n2

1e

en

j

−

π

− π

π

−

−π++

+π

−

n

)1(j)nsin(

n2

j2n

)1

2 2

n

jThus

f(t) = ∑∞

−∞

=

ω n

t jn

n e n

j ) 1 (

Chapter 17, Solution 51

π

=π

=ω

0 T

0

2 2 2 2

t jn t

jn 2 t

o jn

)jn(

e2

1dte

t2

1dte

)t(T

2)n4jn

(n

−π

=

n

t jn 2

n

2)

t

(

Chapter 17, Solution 52

nt j

1

n,

dte)t(T

Using integration by parts,

1 1 t jn 1

1

t

jn2

1e

jn2

jn

)j(n2

1e

en

j

−

π

− π

π

−

−π++

+π

−

n

)1(j)nsin(

n2

j2n

)1

2 2

n

jThus

f(t) = ∑∞

−∞

=

ω n

t jn

n e n

j ) 1 (

0

t jn

nj1

1e

n2j1

0

t n j 1

π+

−

=π

n

t n 2 j

n 2 j 1

e 6321 0

2 / jn 1

0

2 /

e241

= [ − π − + − π − − π − − π + − π]

π

jn n

j 2 / jn jn

2 /

e2n2j

= [ − π − + − π]

π

jn 2

/

e3n2j

f(t) = ∑∞

−∞

=

ω n

t jn

ne oc

0

t0),tsin(

t

j2

12

1dte)tsin(

21

=

+ π

−

− π

π +

−

−

n1

1e

n1

1e

41

)n1(j

e)n1(j

ej41

) 1 n ( )

n 1 (

0

) n 1 ( jt )

n 1 ( jt

[e 1 ne n e 1 ne n]

)1n(4

−π

But ejπ = cos(π) + jsin(π) = –1 = e–jπ

)n1(2

e12ne

nee

e)1n(4

1

2

jn jn

jn jn

−

−

−

−π

π

− π

− π

− π

− π

−

−π

+

n

t jn 2

jn

e ) n 1 ( 2

e 1

0 n n

t jn

2 e ) 1 n ( 2

) jn 1 ( 10

nt 50 j

3 e 2 n

3 3

) n cos(

j n

2

1 ) n cos(

4

+

0 n n

t ) 1 n 2 (

) 1 n 2 (

e 5 j 2

= 6 + 4cos(t)cos(50°) – 4sin(t)sin(50°) + 2cos(2t)cos(35°)

– 2sin(2t)sin(35°) + cos(3t)cos(25°) – sin(3t)sin(25°) + 0.5cos(4t)cos(20°) – 0.5sin(4t)sin(20°)

= 6 + 2.571cos(t) – 3.73sin(t) + 1.635cos(2t)

– 1.147sin(2t) + 0.906cos(3t) – 0.423sin(3t) + 0.47cos(4t) – 0.171sin(4t)

(b) frms = ∑∞

=

+

1 n

2 n

2

2

1a

/220

o = = π → = π=

ω

(b) f(t) a A cos(n t ) 3 4cos(20t 90 ) 5.1cos(40t 90o)

1 n

o n

o n

1t0,1

2 / T

3

2dt)t(T

2 / T

3

4dt)tncos(

)t(T4

5 1

1

1

tn2sinn

63

tn2sinn

334

= [–2/(nπ)]sin(2nπ/3)

1

tn2cos3

nsinn

1234

=+

3

nsinn

2

b2 n

2 n

a

A1 = 0.5513, A2 = 0.2757, A3 = 0, A4 = 0.1375, A5 = 0.1103 The amplitude spectra are shown below

1.333

0.1103 0.1378

0 0.275 0.551

4 3 2

π

+π

=+

2 n

a

= 2 2

n

44.441n

φn = tan–1(bn/an) = tan–1{[–3/(nπ)][n2π2/20]} = tan–1(–nx0.4712)

The schematic is shown below The waveform is inputted using the attributes of

VPULSE In the Transient dialog box, we enter Print Step = 0.05, Final Time = 12, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier After simulation, the output plot is shown below The output file includes the following Fourier

components