Chapter 17, Problem 8E

With respect to the periodic waveform sketched in Fig. 17.30, let g_n(t) represent the Fourier series representation of f(t) truncated at n. [For example, if n = 1, g₁(t) has three terms, defined through a₀, a₁ and b₁.] (a) Sketch g₂(t), g₃(t), and g₅(t), along with f(t). (b) Calculate f (2.5), g₂(2.5), g₃(2.5), and g₅(2.5).

■ FIGURE 17.30

To determine

Sketch $g_2\left(t\right)$, $g_3\left(t\right)$, and $g_5\left(t\right)$ for the periodic waveform $f\left(t\right)$ shown in Figure 17.30.

Answer to Problem 8E

The sketch for $g_2\left(t\right)$, $g_3\left(t\right)$, $g_5\left(t\right)$ is drawn as shown in Figure 1, and the sketch for $f\left(t\right)$ is drawn as shown in Figure 2, and the sketch for $g_2\left(t\right)$, $g_3\left(t\right)$, and $g_5\left(t\right)$ along with $f\left(t\right)$ is drawn as shown in Figure 3.

Explanation of Solution

Given data:

Refer to Figure 17.30 in the textbook.

Formula used:

Write the general expression for Fourier series expansion.

$f\left(t\right)=a_0+{\displaystyle \sum _{n=1}^{\infty }\left(a_n\cos n{\omega }_{0}t+b_n\sin n{\omega }_{0}t\right)}$        (1)

Write the general expression for Fourier series coefficient $a_0$.

$a_0=\frac{1}{T}{\displaystyle {\int }_0^{T}f\left(t\right)}dt$        (2)

Write the general expression for Fourier series coefficient $a_n$.

$a_n=\frac{2}{T}{\displaystyle {\int }_0^{T}f\left(t\right)\cos n{\omega }_{0}t}dt$        (3)

Write the general expression for Fourier series coefficient $b_n$.

$b_n=\frac{2}{T}{\displaystyle {\int }_0^{T}f\left(t\right)\sin n{\omega }_{0}t}dt$        (4)

Write the expression to calculate the fundamental angular frequency.

${\omega }_0=\frac{2\pi }{T}=2\pi f_0$        (5)

Here,

$T$ is the period of the function.

Calculation:

In the given Figure 17.29, the time period is $T=6$.

Substitute 6 for T in equation (5) to find ${\omega }_0$.

${\omega }_0=\frac{2\pi }{6}$

${\omega }_0=\frac{\pi }{3}$        (6)

Substitute 6 for T in equation (2) to find the value of coefficient $a_0$.

$\begin{array}{c}{a}_0=\frac{1}{6}{\displaystyle {\int }_0^{6}f\left(t\right)}dt\\ =\frac{1}{6}\left[{\displaystyle {\int }_0^2\left(0\right)}dt+{\displaystyle {\int }_2^{3}10}dt+{\displaystyle {\int }_3^{6}0}dt\right]\\ =\frac{1}{6}\left[0+10{\left[t\right]}_2^3+0\right]\\ =\frac{1}{6}\left[10\left[3-2\right]\right]\end{array}$

Simplify the above equation as follows,

$\begin{array}{c}{a}_0=\frac{10}{6}\\ =1.667\end{array}$

Substitute 6 for T in equation (3) to find the value of coefficient $a_n$.

$\begin{array}{c}{a}_n=\frac{2}{6}{\displaystyle {\int }_0^{6}f\left(t\right)\cos n{\omega }_{0}t}dt\\ =\frac{2}{6}\left[{\displaystyle {\int }_0^2\left(0\right)\cos n{\omega }_{0}t}dt+{\displaystyle {\int }_2^3\left(1\right)\cos n{\omega }_{0}t}dt+{\displaystyle {\int }_3^6\left(0\right)\cos n{\omega }_{0}t}dt\right]\\ =\frac{2}{6}\left[0+{\left[\frac{\sin n{\omega }_{0}t}{n{\omega }_0}\right]}_2^3+0\right]\\ =\frac{2}{6}\left[\frac{\sin 3n{\omega }_0}{n{\omega }_0}-\frac{\sin 2n{\omega }_0}{n{\omega }_0}\right]\end{array}$

The above equation as follows,

$a_n=\frac{2}{6n{\omega }_0}\left[\sin 3n{\omega }_0-\sin 2n{\omega }_0\right]$        (7)

Substitute equation (6) in equation (7) as follows,

$a_n=\frac{2}{6n\left(\frac{\pi }{3}\right)}\left[\sin 3n\left(\frac{\pi }{3}\right)-\sin 2n\left(\frac{\pi }{3}\right)\right]$        (8)

Now finding the Fourier coefficient $b_n$.

Substitute 6 for T in equation (4) to find the value of coefficient $b_n$.

$\begin{array}{c}{b}_n=\frac{2}{6}{\displaystyle {\int }_0^{6}f\left(t\right)\sin n{\omega }_{0}t}dt\\ =\frac{2}{6}\left[{\displaystyle {\int }_0^2\left(0\right)\sin n{\omega }_{0}t}dt+{\displaystyle {\int }_2^3\left(1\right)\sin n{\omega }_{0}t}dt+{\displaystyle {\int }_3^6\left(0\right)\sin n{\omega }_{0}t}dt\right]\\ =\frac{2}{6}\left[0+{\left[\frac{-\cos n{\omega }_{0}t}{n{\omega }_0}\right]}_2^3+0\right]\\ =\frac{2}{6}\left[-\frac{\cos 3n{\omega }_0}{n{\omega }_0}+\frac{\cos 2n{\omega }_0}{n{\omega }_0}\right]\end{array}$

The above equation as follows,

$b_n=\frac{2}{6n{\omega }_0}\left[\cos 2n{\omega }_0-\cos 3n{\omega }_0\right]$        (9)

Substitute equation (6) in equation (9) as follows,

$b_n=\frac{2}{6n\left(\frac{\pi }{3}\right)}\left[\cos 2n\left(\frac{\pi }{3}\right)-\cos 3n\left(\frac{\pi }{3}\right)\right]$        (10)

Substituting the value of $a_0$, $a_n$, and $b_n$ in equation (1) as follows,

$\begin{array}{l}f\left(t\right)=1.667+{\displaystyle \sum _{n=1}^{\infty }\left(\frac{2}{6n\left(\frac{\pi }{3}\right)}\left[\sin 3n\left(\frac{\pi }{3}\right)-\sin 2n\left(\frac{\pi }{3}\right)\right]\cos n\left(\frac{\pi }{3}\right)t\right)}+\\ {\displaystyle \sum _{n=1}^{\infty }\left(\frac{2}{6n\left(\frac{\pi }{3}\right)}\left[\cos 2n\left(\frac{\pi }{3}\right)-\cos 3n\left(\frac{\pi }{3}\right)\right]\sin n\left(\frac{\pi }{3}\right)t\right)}\left\{\because {\omega }_0=\frac{\pi }{3}\right\}\end{array}$        (11)

The function $g_n\left(t\right)$ represent the Fourier series representation of $f\left(t\right)$ truncated at n.

For $n=2,$

$g_2\left(t\right)=f\left(t\right)$ have the Fourier coefficients $a_o$, $a_1$, $a_2$ and $b_1$, $b_2$.

Therefore, equation (11) will be as follows,

$\begin{array}{l}{g}_2\left(t\right)=1.667+\left(\frac{2}{6\left(1\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(1\right)\left(\frac{\pi }{3}\right)-\sin 2\left(1\right)\left(\frac{\pi }{3}\right)\right]\cos \left(1\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(1\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(1\right)\left(\frac{\pi }{3}\right)-\cos 3\left(1\right)\left(\frac{\pi }{3}\right)\right]\sin \left(1\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(2\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(2\right)\left(\frac{\pi }{3}\right)-\sin 2\left(2\right)\left(\frac{\pi }{3}\right)\right]\cos \left(2\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(2\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(2\right)\left(\frac{\pi }{3}\right)-\cos 3\left(2\right)\left(\frac{\pi }{3}\right)\right]\sin \left(2\right)\left(\frac{\pi }{3}\right)t\right)\end{array}$

Simplify the above equation as follows,

$\begin{array}{c}{g}_2\left(t\right)=1.667+\left(\frac{2}{2\left(\pi \right)}\left[\sin \left(\pi \right)-\sin \left(\frac{2\pi }{3}\right)\right]\cos \left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{2\left(\pi \right)}\left[\cos \left(\frac{2\pi }{3}\right)-\cos \left(\pi \right)\right]\sin \left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{1}{2\pi }\left[\sin \left(2\pi \right)-\sin \left(\frac{4\pi }{3}\right)\right]\cos \left(\frac{2\pi }{3}\right)t\right)+\\ \left(\frac{1}{2\pi }\left[\cos \left(\frac{4\pi }{3}\right)-\cos \left(2\pi \right)\right]\sin \left(\frac{2\pi }{3}\right)t\right)\\ =1.667+\left(\frac{1}{\pi }\left[0-0.866\right]\cos \left(\frac{\pi }{3}\right)t\right)+\left\{\because \sin \pi =0,\cos \pi =-1\right\}\\ \left(\frac{1}{\pi }\left[-0.5+1\right]\sin \left(\frac{\pi }{3}\right)t\right)+\left(\frac{1}{2\pi }\left[0-\left(-0.866\right)\right]\cos \left(\frac{2\pi }{3}\right)t\right)+\\ \left(\frac{1}{2\pi }\left[-0.5-1\right]\sin \left(\frac{2\pi }{3}\right)t\right)\end{array}$

$\begin{array}{l}{g}_2\left(t\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t\\ +0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t\end{array}$        (12)

Similarly, for $n=3$,

$g_3\left(t\right)=f\left(t\right)$ have the Fourier coefficients $a_o$, $a_1$, $a_2$, $a_3$ and $b_1$, $b_2$, $b_3$.

Therefore, equation (11) will be as follows,

$\begin{array}{l}{g}_3\left(t\right)=1.667+\left(\frac{2}{6\left(1\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(1\right)\left(\frac{\pi }{3}\right)-\sin 2\left(1\right)\left(\frac{\pi }{3}\right)\right]\cos \left(1\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(1\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(1\right)\left(\frac{\pi }{3}\right)-\cos 3\left(1\right)\left(\frac{\pi }{3}\right)\right]\sin \left(1\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(2\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(2\right)\left(\frac{\pi }{3}\right)-\sin 2\left(2\right)\left(\frac{\pi }{3}\right)\right]\cos \left(2\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(2\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(2\right)\left(\frac{\pi }{3}\right)-\cos 3\left(2\right)\left(\frac{\pi }{3}\right)\right]\sin \left(2\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(3\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(3\right)\left(\frac{\pi }{3}\right)-\sin 2\left(3\right)\left(\frac{\pi }{3}\right)\right]\cos \left(3\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(3\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(3\right)\left(\frac{\pi }{3}\right)-\cos 3\left(3\right)\left(\frac{\pi }{3}\right)\right]\sin \left(3\right)\left(\frac{\pi }{3}\right)t\right)\end{array}$

From equation (12), the above equation is written as,

$\begin{array}{c}{g}_3\left(t\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t\\ +0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t\\ +\left(\frac{2}{6\left(3\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(3\right)\left(\frac{\pi }{3}\right)-\sin 2\left(3\right)\left(\frac{\pi }{3}\right)\right]\cos \left(3\right)\left(\frac{\pi }{3}\right)t\right)\\ +\left(\frac{2}{6\left(3\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(3\right)\left(\frac{\pi }{3}\right)-\cos 3\left(3\right)\left(\frac{\pi }{3}\right)\right]\sin \left(3\right)\left(\frac{\pi }{3}\right)t\right)\\ =1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t\\ +0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t\\ +\left(\frac{1}{3\pi }\left[\sin \left(3\pi \right)-\sin \left(2\pi \right)\right]\cos \left(\pi \right)t\right)\\ +\left(\frac{1}{3\pi }\left[\cos \left(2\pi \right)-\cos \left(3\pi \right)\right]\sin \left(\pi \right)t\right)\\ =1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t\\ +0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t\\ +\left(\frac{1}{3\pi }\left[0-0\right]\cos \left(\pi \right)t\right)+\left(\frac{1}{3\pi }\left[1+1\right]\sin \left(\pi \right)t\right)\end{array}$

$\begin{array}{c}{g}_3\left(t\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t\\ +0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t+0.212\sin \left(\pi \right)t\end{array}$        (13)

Similarly, for $n=5$,

$g_5\left(t\right)=f\left(t\right)$ have the Fourier coefficients $a_o$, $a_1$, $a_2$, $a_3$, $a_4$, $a_5$ and $b_1$, $b_2$, $b_3$, $b_4$, $b_5$.

Therefore, equation (11) will be as follows,

$\begin{array}{l}{g}_5\left(t\right)=1.667+\left(\frac{2}{6\left(1\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(1\right)\left(\frac{\pi }{3}\right)-\sin 2\left(1\right)\left(\frac{\pi }{3}\right)\right]\cos \left(1\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(1\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(1\right)\left(\frac{\pi }{3}\right)-\cos 3\left(1\right)\left(\frac{\pi }{3}\right)\right]\sin \left(1\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(2\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(2\right)\left(\frac{\pi }{3}\right)-\sin 2\left(2\right)\left(\frac{\pi }{3}\right)\right]\cos \left(2\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(2\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(2\right)\left(\frac{\pi }{3}\right)-\cos 3\left(2\right)\left(\frac{\pi }{3}\right)\right]\sin \left(2\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(3\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(3\right)\left(\frac{\pi }{3}\right)-\sin 2\left(3\right)\left(\frac{\pi }{3}\right)\right]\cos \left(3\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(3\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(3\right)\left(\frac{\pi }{3}\right)-\cos 3\left(3\right)\left(\frac{\pi }{3}\right)\right]\sin \left(3\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(4\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(4\right)\left(\frac{\pi }{3}\right)-\sin 2\left(4\right)\left(\frac{\pi }{3}\right)\right]\cos \left(4\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(4\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(4\right)\left(\frac{\pi }{3}\right)-\cos 3\left(4\right)\left(\frac{\pi }{3}\right)\right]\sin \left(4\right)\left(\frac{\pi }{3}\right)t\right)\\ \left(\frac{2}{6\left(5\right)\left(\frac{\pi }{3}\right)}\left[\sin 3\left(5\right)\left(\frac{\pi }{3}\right)-\sin 2\left(5\right)\left(\frac{\pi }{3}\right)\right]\cos \left(5\right)\left(\frac{\pi }{3}\right)t\right)+\\ \left(\frac{2}{6\left(5\right)\left(\frac{\pi }{3}\right)}\left[\cos 2\left(5\right)\left(\frac{\pi }{3}\right)-\cos 3\left(5\right)\left(\frac{\pi }{3}\right)\right]\sin \left(5\right)\left(\frac{\pi }{3}\right)t\right)\end{array}$

From equation (13), the above equation is written as,

$\begin{array}{c}{g}_5\left(t\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t\\ +0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t+0.212\sin \left(\pi \right)t\\ +\left(\frac{1}{4\pi }\left[\sin \left(4\pi \right)-\sin \left(\frac{8\pi }{3}\right)\right]\cos \left(\frac{4\pi }{3}\right)t\right)\\ +\left(\frac{1}{4\pi }\left[\cos \left(\frac{8\pi }{3}\right)-\cos \left(4\pi \right)\right]\sin \left(\frac{4\pi }{3}\right)t\right)\\ +\left(\frac{1}{5\pi }\left[\sin \left(5\pi \right)-\sin \left(\frac{10\pi }{3}\right)\right]\cos \left(\frac{5\pi }{3}\right)t\right)\\ +\left(\frac{1}{5\pi }\left[\cos \left(\frac{10\pi }{3}\right)-\cos \left(5\pi \right)\right]\sin \left(\frac{5\pi }{3}\right)t\right)\\ =1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t\\ +0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t+0.212\sin \left(\pi \right)t\\ +\left(\frac{1}{4\pi }\left[0-0.866\right]\cos \left(\frac{4\pi }{3}\right)t\right)+\left(\frac{1}{4\pi }\left[-0.5-1\right]\sin \left(\frac{4\pi }{3}\right)t\right)\\ +\left(\frac{1}{5\pi }\left[0+0.866\right]\cos \left(\frac{5\pi }{3}\right)t\right)+\left(\frac{1}{5\pi }\left[-0.5+1\right]\sin \left(\frac{5\pi }{3}\right)t\right)\end{array}$

$\begin{array}{c}{g}_5\left(t\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t+0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t\\ +0.212\sin \left(\pi \right)t-0.069\cos \left(\frac{4\pi }{3}\right)t-0.119\sin \left(\frac{4\pi }{3}\right)t+0.055\cos \left(\frac{5\pi }{3}\right)t\\ +0.0318\sin \left(\frac{5\pi }{3}\right)t\end{array}$

MATLAB code to sketch for $g_2\left(t\right)$, $g_3\left(t\right)$, and $g_5\left(t\right)$:

t=-5:0.01:5;

g2=1.667-0.275*cos(3.141*t/3)+0.159*sin(3.141*t/3)+0.137*cos(2*3.141*t/3)-0.238*sin(2*3.141*t/3);

g3=1.667-0.275*cos(3.141*t/3)+0.159*sin(3.141*t/3)+0.137*cos(2*3.141*t/3)-0.238*sin(2*3.141*t/3)+0.212*sin(pi*t);

g5=1.667- 0.275*cos(3.141*t/3)+0.159*sin(3.141*t/3)+0.137*cos(2*3.141*t/3)-0.238*sin(2*3.141*t/3)+0.212*sin(pi*t)-0.069*cos(4*pi*t/3)-0.119*sin(4*pi*t/3)+0.055*cos(5*pi*t/3)+0.0318*sin(5*pi*t/3);

plot(t,g2,t,g3,t,g5)

legend({'g2','g3','g5'},'Location','best')

xlabel('Time t in sec')

ylabel('The values g2, g3, and g5')

title('Plots for g2,g3, and g5')

MATLAB output: The MATLAB output shown in Figure 1.

MATLAB code to sketch for $f\left(t\right)$:

t=linspace(-5,5,1000); % vector for time over 1000 points.

T=6; % Period

w0=2*pi/T; % natural frequency, is w0=2*pi.

f0=1.667; % constant.

N=40;

for i=1:1000;

sum=0;   

    for n=1:N;     

      sum=sum+(1/n*pi)*(sin(n*pi) -sin(2*n*pi/3))*cos(n*pi*t(i)/3) + (1/n*pi)*(cos(2*n*pi/3) -cos(n*pi))*sin(n*pi*t(i)/3);

    end

f40(i)=f0+sum;

end

plot(t,f40)

xlabel('Time t in seconds')

ylabel('Value of function f(t)')

plot_ttle = ['Fourier Series representation of function f(t) for N = ',num2str(N)];

title(plot_ttle);

MATLAB output: The MATLAB output shown in Figure 2.

MATLAB code to sketch for $g_2\left(t\right)$, $g_3\left(t\right)$, and $g_5\left(t\right)$ along with $f\left(t\right)$:

t=linspace(-5,5,1000); % vector for time over 1000 points.

T=6; % Period

w0=2*pi/T; % natural frequency, is w0=2*pi.

f0=1.667; % constant.

N=40; % consider N=40 for instant.

for i=1:1000;

   g2=1.667-0.275*cos(3.141*t(i)/3)+0.159*sin(3.141*t(i)/3)+0.137*cos(2*3.141*t(i)/3)-0.238*sin(2*3.141*t(i)/3);

   g3=1.667-0.275*cos(3.141*t(i)/3)+0.159*sin(3.141*t(i)/3)+0.137*cos(2*3.141*t(i)/3)-0.238*sin(2*3.141*t(i)/3)+0.212*sin(pi*t(i));

   g5=1.667-0.275*cos(3.141*t(i)/3)+0.159*sin(3.141*t(i)/3)+0.137*cos(2*3.141*t(i)/3)-0.238*sin(2*3.141*t/3)+0.212*sin(pi*t)-0.069*cos(4*pi*t(i)/3)-0.119*sin(4*pi*t(i)/3)+0.055*cos(5*pi*t(i)/3)+0.0318*sin(5*pi*t(i)/3);

end

for i=1:1000;

sum=0;  

    for n=1:N;     

      sum=sum+(1/n*pi)*(sin(n*pi) -sin(2*n*pi/3))*cos(n*pi*t(i)/3) + (1/n*pi)*(cos(2*n*pi/3) -cos(n*pi))*sin(n*pi*t(i)/3);  

    end

f40(i)=f0+sum;

end

plot(t,g2,t,g3,t,g5,t,f40)

legend({'g2','g3','g5','f40'},'Location','best')

xlabel('Time t in sec')

ylabel('The values g2, g3, g5 and f40')

title('Plots for g2, g3, g5 and f40')

MATLAB output:

Conclusion:

Thus, the sketch for $g_2\left(t\right)$, $g_3\left(t\right)$, $g_5\left(t\right)$ is drawn as shown in Figure 1, and the sketch for $f\left(t\right)$ is drawn as shown in Figure 2, and the sketch for $g_2\left(t\right)$, $g_3\left(t\right)$, and $g_5\left(t\right)$ along with $f\left(t\right)$ is drawn as shown in Figure 3.

To determine

Find the function $f\left(2.5\right)$, $g_2\left(2.5\right)$, $g_3\left(2.5\right)$, and $g_5\left(2.5\right)$.

Answer to Problem 8E

The value of $f\left(2.5\right)$, $g_2\left(2.5\right)$, $g_3\left(2.5\right)$, and $g_5\left(2.5\right)$ is determined.

Explanation of Solution

Given data:

Refer to Figure 17.30 in the textbook.

Calculation:

From Part (a), the function $f\left(t\right)$ is,

$\begin{array}{l}f\left(t\right)=1.667+{\displaystyle \sum _{n=1}^{\infty }\left(\frac{2}{6n\left(\frac{\pi }{3}\right)}\left[\sin 3n\left(\frac{\pi }{3}\right)-\sin 2n\left(\frac{\pi }{3}\right)\right]\cos n\left(\frac{\pi }{3}\right)t\right)}+\\ {\displaystyle \sum _{n=1}^{\infty }\left(\frac{2}{6n\left(\frac{\pi }{3}\right)}\left[\cos 2n\left(\frac{\pi }{3}\right)-\cos 3n\left(\frac{\pi }{3}\right)\right]\sin n\left(\frac{\pi }{3}\right)t\right)}\left\{\because {\omega }_0=\frac{\pi }{3}\right\}\end{array}$

Finding $f\left(2.5\right)$,

$\begin{array}{l}f\left(2.5\right)=1.667+{\displaystyle \sum _{n=1}^{\infty }\left(\frac{2}{6n\left(\frac{\pi }{3}\right)}\left[\sin 3n\left(\frac{\pi }{3}\right)-\sin 2n\left(\frac{\pi }{3}\right)\right]\cos 2.5n\left(\frac{\pi }{3}\right)\right)}+\\ {\displaystyle \sum _{n=1}^{\infty }\left(\frac{2}{6n\left(\frac{\pi }{3}\right)}\left[\cos 2n\left(\frac{\pi }{3}\right)-\cos 3n\left(\frac{\pi }{3}\right)\right]\sin 2.5n\left(\frac{\pi }{3}\right)\right)}\end{array}$

From Part (a),

$g_2\left(t\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t+0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t$

Finding $g_2\left(2.5\right)$,

$\begin{array}{c}{g}_2\left(2.5\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)\left(2.5\right)+0.159\sin \left(\frac{\pi }{3}\right)\left(2.5\right)+0.137\cos \left(\frac{2\pi }{3}\right)\left(2.5\right)\\ -0.238\sin \left(\frac{2\pi }{3}\right)\left(2.5\right)\\ =1.667-0.3437+0.3442-0.17125-0.5152\\ =0.9811\end{array}$

From Part (a),

$\begin{array}{c}{g}_3\left(t\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t\\ +0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t+0.212\sin \left(\pi \right)t\end{array}$

Finding $g_3\left(2.5\right)$,

$\begin{array}{c}{g}_3\left(2.5\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)\left(2.5\right)+0.159\sin \left(\frac{\pi }{3}\right)\left(2.5\right)\\ +0.137\cos \left(\frac{2\pi }{3}\right)\left(2.5\right)-0.238\sin \left(\frac{2\pi }{3}\right)\left(2.5\right)+0.212\sin \left(2.5\pi \right)\\ =1.667-0.3437+0.3442-0.17125-0.5152+0.212\\ =1.1931\end{array}$

From Part (a),

$\begin{array}{c}{g}_5\left(t\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)t+0.159\sin \left(\frac{\pi }{3}\right)t+0.137\cos \left(\frac{2\pi }{3}\right)t-0.238\sin \left(\frac{2\pi }{3}\right)t\\ +0.212\sin \left(\pi \right)t-0.069\cos \left(\frac{4\pi }{3}\right)t-0.119\sin \left(\frac{4\pi }{3}\right)t+0.055\cos \left(\frac{5\pi }{3}\right)t\\ +0.0318\sin \left(\frac{5\pi }{3}\right)t\end{array}$

Finding $g_5\left(2.5\right)$,

$\begin{array}{c}{g}_5\left(t\right)=1.667-0.275\cos \left(\frac{\pi }{3}\right)\left(2.5\right)+0.159\sin \left(\frac{\pi }{3}\right)\left(2.5\right)+0.137\cos \left(\frac{2\pi }{3}\right)\left(2.5\right)\\ -0.238\sin \left(\frac{2\pi }{3}\right)\left(2.5\right)+0.212\sin \left(\pi \right)\left(2.5\right)-0.069\cos \left(\frac{4\pi }{3}\right)\left(2.5\right)\\ -0.119\sin \left(\frac{4\pi }{3}\right)\left(2.5\right)+0.055\cos \left(\frac{5\pi }{3}\right)\left(2.5\right)+0.0318\sin \left(\frac{5\pi }{3}\right)\left(2.5\right)\\ =1.667-0.3437+0.3442-0.17125-0.5152+0.212\\ +0.0862+0.2576+0.06875-0.0688\\ =1.5368\end{array}$

Conclusion:

Thus, the value of $f\left(2.5\right)$, $g_2\left(2.5\right)$, $g_3\left(2.5\right)$, and $g_5\left(2.5\right)$ is determined.