EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 8220100254147
Author: Chapra
Publisher: MCG
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Chapter 19, Problem 6P

Construct amplitude and phase line spectra for Prob. 19.4.

Expert Solution & Answer
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To determine

To graph: The amplitude and phase line spectra for the sawtooth wave as shown in the following figure,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 19, Problem 6P , additional homework tip  1

Explanation of Solution

Given Information: The sawtooth wave given in the following figure,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 19, Problem 6P , additional homework tip  2

Formula used:

Consider f(t) is a periodic function with period T defined in the interval 0xT

then the Fourier series expansion of the function,

f(t)=a0+k=1[akcos(kω0t)+bksin(kω0t)] with ω0=2πT.

And the coefficients are defined by,

a0=1T0Tf(t)dt;ak=2T0Tf(t)cos(kω0t)dt;bk=2T0Tf(t)sin(kω0t)dt

Alternatively, the Fourier series can also be written as, f(t)=a0+k=1[ckcos(kω0tϕk)]

Here, the amplitude ck and phase ϕk for each term is defined as,

ck2=ak2+bk2,ϕk=tan1(bkak)

Plot ck and ϕk in the frequency domain to plot the amplitude and phase line spectra.

Graph:

Consider the sawtooth wave given in the following figure,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 19, Problem 6P , additional homework tip  3

Therefore, the sawtooth wave is a periodic function f(t) with period T in the interval 0tT and from the graph,

f(t) is a straight line joining the points (0,0) and (T2,1) in 0tT2.

f(t) is a straight line joining the points (T2,1) and (T,0) in T2tT.

Therefore, the sawtooth wave,

f(t)={2Tt0tT222TtT2tT

Therefore, the Fourier series expansion of this function is,

f(t)=a0+k=1[akcos(kω0t)+bksin(kω0t)];ω0=2πT

In the above expression, the coefficients are defined by,

Now, find ak.

ak=2T0Tf(t)cos(2kπTt)dt=2T0T2(2Tt)cos(2kπTt)dt+2TT2T(22Tt)cos(2kπTt)dt=4T20T2tcos(2kπTt)dt+4TT2Tcos(2kπTt)dt4T2T2Ttcos(2kπTt)dt

Consider,

I=tcos(2kπTt)dt=tcos(2kπTt)dt{ddt(t)cos(2kπTt)dt}dt=tT2kπsin(2kπTt)T2kπsin(2kπTt)dt=Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)

Hence,

0T2tcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]0T2=T2kπT2sin(2kπTT2)+T24k2π2cos(2kπTT2)T24k2π2cos(0)=T24kπsin(kπ)+T24k2π2cos(kπ)T24k2π2=T24k2π2(1)kT24k2π2

Further,

T2Tcos(2kπTt)dt=T2kπsin(2kπTt)|T2T=T2kπ{sin(2kπ)sin(kπ)}=0

T2Ttcos(2kπTt)dt=[Tt2kπsin(2kπTt)+T24k2π2cos(2kπTt)]T2T=T22kπsin(2kπ)+T24k2π2cos(2kπ)T24kπsin(kπ)T24k2π2cos(kπ)=T24k2π2T24k2π2(1)k

Therefore,

ak=4T2[T24k2π2(1)kT24k2π2]+4T×04T2[T24k2π2T24k2π2(1)k]=1k2π2(1)k+1k2π21k2π2+1k2π2(1)k=0

Now, find bk.

bk=2T0Tf(t)sin(2kπTt)dt=2T0T2(2Tt)sin(2kπTt)dt+2TT2T(22Tt)sin(2kπTt)dt=4T20T2tsin(2kπTt)dt+4TT2Tsin(2kπTt)dt4T2T2Ttsin(2kπTt)dt

Consider,

I=tsin(2kπTt)dt=tsin(2kπTt)dt{ddt(t)sin(2kπTt)dt}dt=tT2kπcos(2kπTt)+T2kπcos(2kπTt)dt=Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)

Hence,

0T2tsin(2kπTt)dt=[Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]0T2=T24kπcos(kπ)+T24k2π2sin(kπ)=T24kπ(1)k

T2Tsin(2kπTt)dt=T2kπcos(2kπTt)|T2T=T2kπcos(2kπ)+T2kπcos(kπ)=T2kπ+T2kπ(1)k

Further,

T2Ttsin(2kπTt)dt=[Tt2kπcos(2kπTt)+T24k2π2sin(2kπTt)]T2T=T22kπcos(2kπ)+T24k2π2sin(2kπ)+T24kπcos(kπ)T24k2π2sin(kπ)=T22kπ+T24kπ(1)k

Thus,

bk=4T2[T24kπ(1)k]+4T[T2kπ+T2kπ(1)k]4T2[T22kπ+T24kπ(1)k]=1kπ(1)k2kπ+2kπ(1)k+2kπ1kπ(1)k=2kπ(1)k

Hence, the coefficients of the Fourier series expansions are,

a0=0,ak=0,bk=2kπ(1)k

That is,

bk={2kπ>0k even2kπ<0k odd

Consider,

ck=ak2+bk2=0+4k2π2(1)2k=4k2π2=2kπ

Thus, the amplitude of the kth term is 2kπ, that is, the amplitudes for k=1,2,3,4,5,... are,

2π,1π,23π,12π,25π,...

Furthermore, consider,

ϕk=tan1(bkak)

As ak=0 and bk>0 for even k thus,

ϕk=π2

As ak=0 and bk<0 for odd k thus,

ϕk=π2

Therefore,

ϕk={π2k evenπ2k odd

Thus, the phases corresponding to k=1,2,3,4,5,... are π2,π2,π2,π2,π2,...

Use the following MATLAB code to construct the amplitude plot.

function Code_97924_19_6P_a()

for k = 1: 8

f(k) = k;

% define the amplitude

A(k) = 2/(k*pi);

end

% plot the values thus obtained

stem(f,A,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[A(8) A(4) A(2) A(1)]);

set(gca,'YTickLabel',{'1/4\pi' '1/2\pi' '1/\pi' '2/\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

end

Execute the above code to obtain the amplitude plot as,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 19, Problem 6P , additional homework tip  4

Interpretation: The above plot shows the amplitude plot for the sawtooth wave as shown in the figure provided.

Use the following MATLAB file can be used to construct the phase plot.

function Code_97924_19_6P_b()

for k = 1: 8

f(k) = k;

% define the phase

P(k) = (-1)^k*(pi/2);

end

% plot the values thus obtained

stem(f,P,'filled','LineWidth',4,'Color','k','Marker', 'none');

% define geometric properties

set(gca,'XTickLabel',{' ' 'f_0' 'f_1' 'f_2' 'f_3' 'f_4' 'f_5' 'f_6' 'f_7'});

set(gca,'YTick',[-pi -pi/2 0 pi/2 pi]);

set(gca,'YTickLabel',{'-\pi' '-\pi/2' '0' '\pi/2' '\pi'});

set(gca,'Fontname','Times New Roman','FontSize',12);

xlim([0, 9]); grid on

ylim([-pi-0.1, pi+0.1]); grid on

end

Execute the above code to obtain the plot as,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 19, Problem 6P , additional homework tip  5

Interpretation: The above plot shows the phase line spectra for the sawtooth wave as shown in the figure provided.

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