Question

Consider the following signal that is periodic with period T=2

$f(t)=(1-t^2)cos(3\pi t), -1\leqslant t \leqslant 1$

a) Draw a picture of this signal for $-3\leqslant t \leqslant 3$ .

b) For which values of t will the Fourier series provide a good approximation for this signal? Justify based on part a

c) Find the complex Fourier coefficients Cn from the definition. (integrate through using a table of integrals)

d) Find the real Fourier coefficients an, bn. You may use your knowledge of Cn to do this, without integrating. Alternatively, you may choose instead to find an and bn by integration, and use this knowledge to find Cn.

T/2 nwt

Answer 1

f (t) (1- t2) cos(3nt),-1 sts1 graph of f in the interval [-3, 3] period T-2 f(t 0 2.75 0.309359 2.5 2.25 0.66291 1.4E-16 f(t) 1.2 1.75 0.66291 1.4E-16 1.25 0.309359 0.6 0 0.75 0.309359 1.4E-16 0.25 -0.66291 0.2 0.5 0 0.25 -0.66291 0.5 0.75 0.309359 1.4E-16 0 1.25 0.309359 1.5 1.75 0.66291 1.4E-16 2.25 0.66291 2.5 2.75 0.309359 1.4E-16 0

Answer: (b) Consider the following signal that is periodic with period T-2 f (t) (1-t2) cos(3Tt), - 1 s ts 1 For the values of t = 0, the Fourier series provide a good approximation for this signal. c) Complex Fourier coefficient f (t) (1-t2) cos(3Tt), - 1 s ts 1 which is an even function. Then the complex Fourier series is given by 2π1nt Cne(rint) Where f(t) exp(-πǐnt) dt But f (t) is even function so from the integral only cosine part of exponential function will survive. Hence f (t) cos(Trnt) dt 2 x 1 =| ((1-t2) cos(Ist) ) cos(mt) dt

Observe that cos(3mt) cos(nt)CO(3+n)t) +cos(T(3 -n)t)] Thus ((1- t2) cos(3Tt)) cos(nt) dt (1-t') sin( π(3 + n)t)-(-2)-cos(T(3 + n)t) + (-2) sin(T(3 + n)t) (T(3+n) sin(T(3- -(-21) 3-n)t) 3 -n)t) sin (T (3 - n)t) The sine terms are zero and hence -cos(n(3 + n) COS(T(3 -n) t(3-n (r(3+n) cos T(3 n)) cos(T(3 -n) (r(3+n) (r(3-n)