Question

4. Consider the periodic function 0, -1<t<- f(t) cos(#(t + 1)), } <t< 0 cos(at), 0<t< 0, }<t<i with f(t) = f(t+2). (a) Determine a general expression for the Fourier series of f. (b) Use MATLAB to plot both f and the sum of the first 5 non-zero terms of the Fourier series for f on the same set of axes for -1<t<3.

Answer 1

Given that Considee the periodic function akta. Cascailt+1)] 12 ctca ostat 2 <tel Solution f(+) Cos (Tt) o atktk - 1 f(t) caseti (H+). Sto Cositit), 이사를 글사시 And f(x+2) = f(t) 0 fouriel series f(t) = aots an (2n TL t) nal Cos (2014+)+ba sin of the function Since period is ī=2 Hence flt): 20+S , Bri Cos (nt t) + bn Sin intit) where aos J'Afc+)dt = 3 [5% od + odt + scos (tolf+i) dt 172 of cos(Tit) drejo dt ₂ ܟܘ S - +++ to co

© an: I'flt) cos in 17) et odt + 2 2 5 +3 cos (T(+41)eus (n tr t) dt + s'adas ( 9 L) cos (not) dit 11 cos[ 97] -² odt tot cas [ ] דמ + to so (1702) T-n2 T an=0, nzl bn = 27 J' f(t) Sin (nt) at . 를 odt -₂ ces(+6C++i) sin(nt)dF+J"cosLT1 H) sin(n + 4) It tlodt ½ sot n-Sin [07] (-1+ n2) 7 + n-sin(??] (-1 +92) 71 +0=2 h-Sin ( (-1 tn 2) Ku's

Hence fowied series ft) =2soo no --Sin [22] (-1+n2) 7 nal Sin int.) TI Since sin (1997.), of his add and co 암 nis Even So when n=1, bizo hence n=2 to 6 will be non Zeo bn through ܬܩܝܗ

1 2 3 f=0; t=linspace(-1,3); for n=2:1:6 4 5 - f=f+2*((n-sin(n*pi/2))/((-1+n^2) *pi))*sin(n*pi*t); end 6 7 00 plot(t,f);
output

matlab code

Figure 1 x + 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0 1 3 N

Plot in matlab

f=0;
t=linspace(-1,3);
for n=2:1:6
  
f=f+2*((n-sin(n*pi/2))/((-1+n^2)*pi))*sin(n*pi*t);
end

plot(t,f);

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