Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
6th Edition
ISBN: 9780078028229
Author: Charles K Alexander, Matthew Sadiku
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 17, Problem 25P

Determine the Fourier series representation of the function in Fig. 17.63.

Figure 17.63

Chapter 17, Problem 25P, Determine the Fourier series representation of the function in Fig. 17.63. Figure 17.63

Expert Solution & Answer
Check Mark
To determine

Find the Fourier series representation of the function given in Figure 17.63.

Answer to Problem 25P

The Fourier series representation f(t) of the function given in Figure 17.63 is n=1n=odd([4nπsin(2nπ3)+6n2π2(cos(2nπ3)1)]cos(2nπ3)t+[6n2π2sin(2nπ3)4nπcos(2nπ3)]sin(2nπ3)t).

Explanation of Solution

Given data:

Refer to Figure 17.63 in the textbook.

Formula used:

Write the expression to calculate the fundamental angular frequency.

ω0=2πT (1)

Here,

T is the period of the function.

Write the general expression to calculate the Fourier series of f(t).

f(t)=a0+n=1(ancosnω0t+bnsinnω0t) (2)

Here,

a0 is the dc component of f(t),

an and bn are the Fourier coefficients,

n is an integer, and

ω0 is the angular frequency.

Calculation:

The given waveform is drawn as Figure 1.

Fundamentals of Electric Circuits, Chapter 17, Problem 25P

Refer to Figure 1, each half cycle is the mirror image of the next half cycle. Therefore, the waveform is called as half-wave symmetry and it is expressed as,

f(tT2)=f(t)

Write the expressions for Fourier coefficients for the half-wave odd symmetric function.

a0=0

an={4T0T2f(t)cosnω0tdt, n=odd0, n=even (3)

bn={4T0T2f(t)sinnω0tdt, n=odd0, n=even (4)

Refer to the Figure 1. The Fourier series function of the waveform is defined as,

f(t)={2t, 0<t<10, 1<t<1.52t+3, 1.5<t<2.50, 2.5<t<3

The time period of the function in Figure 1 is,

T=3

Substitute 3 for T in equation (1) to find ω0.

ω0=2π3

Substitute 2π3 for ω0 and 3 for T for odd n in equation (3) to find an.

an=43032f(t)cosn(2π3)tdt=4301.5f(t)cos(2nπ3)tdt=43(01f(t)cos(2nπ3)tdt+4311.5f(t)cos(2nπ3)tdt)=43(012tcos(2nπ3)tdt+4311.5(0)cos(2nπ3)tdt)

Simplify the above equation to find an.

an=43(012tcos(2nπ3)tdt+0)=43(012tcos(2nπ3)tdt)

an=83(01tcos(2nπ3)tdt) (5)

Assume the following to reduce the equation (5).

x=01tcos(2nπ3)tdt (6)

Substitute the equation (6) in equation (5) to find an.

an=83x (7)

Consider the following integration formula.

abudv=[uv]ababvdu (8)

Compare the equations (6) and (8) to simplify the equation (6).

u=t dv=cos(2nπ3)tdtdu=dt v=sin(2nπ3)t(2nπ3)

Using the equation (8), the equation (6) can be reduced as,

x=01tcos(2nπ3)tdt=[(t)sin(2nπ3)t(2nπ3)]0101sin(2nπ3)t(2nπ3)dt=[(1)sin(2nπ3)(1)(2nπ3)(0)sin(2nπ3)(0)(2nπ3)](1(2nπ3))[cos(2nπ3)t(2nπ3)]01=[sin(2nπ3)(2nπ3)0]+1(2nπ3)2[cos(2nπ3)(1)cos(2nπ3)(0)]

Simplify the above equation to find x.

x=sin(2nπ3)(2nπ3)+1(2nπ3)2[cos(2nπ3)cos0°]=32nπsin(2nπ3)+1(4n2π29)[cos(2nπ3)1] {cos(0°)=1}=32nπsin(2nπ3)+94n2π2(cos(2nπ3)1)

Substitute 32nπsin(2nπ3)+94n2π2(cos(2nπ3)1) for x in equation (7) to find an.

an=83(32nπsin(2nπ3)+94n2π2(cos(2nπ3)1))=246nπsin(2nπ3)+7212n2π2(cos(2nπ3)1)=4nπsin(2nπ3)+6n2π2(cos(2nπ3)1)

Substitute 2π3 for ω0 and 3 for T for odd n in equation (4) to find bn.

bn=43032f(t)sinn(2π3)tdt=4301.5f(t)sin(2nπ3)tdt=43(01f(t)sin(2nπ3)tdt+11.5f(t)sin(2nπ3)tdt)=43(012tsin(2nπ3)tdt+11.5(0)sin(2nπ3)tdt)

Simplify the above equation to find bn.

bn=43(012tsin(2nπ3)tdt+0)=43(012tsin(2nπ3)tdt)

bn=83(01tsin(2nπ3)tdt) (9)

Assume the following to reduce the equation (9).

y=01tsin(2nπ3)tdt (10)

Substitute the equation (10) in equation (9) to find bn.

bn=83y (11)

Compare the equations (8) and (10) to simplify the equation (10).

u=t dv=sin(2nπ3)tdtdu=dt v=cos(2nπ3)t(2nπ3)

Using the equation (8), the equation (10) can be reduced as,

y=01tsin(2nπ3)tdt=[(t)(cos(2nπ3)t(2nπ3))]0101(cos(2nπ3)t(2nπ3))dt=[(1)cos(2nπ3)(1)(2nπ3)(0)cos(2nπ3)(0)(2nπ3)]+1(2nπ3)[sin(2nπ3)t(2nπ3)]01=[cos(2nπ3)(2nπ3)0]+1(2nπ3)2[sin(2nπ3)(1)sin(2nπ3)(0)]

Simplify the above equation to find y.

y=cos(2nπ3)(2nπ3)+1(2nπ3)2[sin(2nπ3)sin(0°)]=32nπcos(2nπ3)+1(4n2π29)[sin(2nπ3)0] {sin(0°)=0}=32nπcos(2nπ3)+94n2π2sin(2nπ3)=94n2π2sin(2nπ3)32nπcos(2nπ3)

Substitute 94n2π2sin(2nπ3)32nπcos(2nπ3) for y in equation (11) to find bn.

bn=83(94n2π2sin(2nπ3)32nπcos(2nπ3))=7212n2π2sin(2nπ3)246nπcos(2nπ3)=6n2π2sin(2nπ3)4nπcos(2nπ3)

Substitute 2π3 for ω0, 0 for a0, 4nπsin(2nπ3)+6n2π2(cos(2nπ3)1) for odd n for an and 6n2π2sin(2nπ3)4nπcos(2nπ3) for odd n for bn in equation (2) to find f(t).

f(t)=0+n=1n=odd([4nπsin(2nπ3)+6n2π2(cos(2nπ3)1)]cosn(2π3)t+[6n2π2sin(2nπ3)4nπcos(2nπ3)]sinn(2π3)t)=n=1n=odd([4nπsin(2nπ3)+6n2π2(cos(2nπ3)1)]cos(2nπ3)t+[6n2π2sin(2nπ3)4nπcos(2nπ3)]sin(2nπ3)t)

Conclusion:

Thus, the Fourier series representation f(t) of the function given in Figure 17.63 is n=1n=odd([4nπsin(2nπ3)+6n2π2(cos(2nπ3)1)]cos(2nπ3)t+[6n2π2sin(2nπ3)4nπcos(2nπ3)]sin(2nπ3)t).

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