Question

Signals and systems

Problem 2 (20 points) Let -S2t+1, Osts1 x(t) = -t +4, 1sts 3 be a periodic signal with fundamental period T=3 and Fourier coefficients ar a. Determine the value of ao. b. Determine ak, k = 0, by: 1. first finding the Fourier coefficients of ii.then using the appropriate property of the continuous-time Fourier series. c. Use the result of part (b) to express the Fourier transform of x().

Answer 1

a) The value of a₀ would be

$a_0 = \frac23\int_0^3 x(t) dt \\ = \frac23\int_0^1 (2t+1)dt + \frac23\int_1^3 (4-t)dt \\ = \frac23[t^2+t]_0^1 + \frac23[4t-t^2/2]_1^3 = \frac43 + \frac23(8-4) = 4$

b) Note that

$x'(t) = \begin{cases} 2 & \text{ if } 0 \le t < 1\\ -1 & \text{ if } 1 \le t \le 3 \end{cases}$

b) i)Now, the Fourier coefficients for this function would be

$a_n' = \frac23\int_0^3 x'(t)\sin\left ( 2\pi n t/3 \right ) dt \\ = \frac23\left [\int_0^1 2\sin(2\pi n t/3)dt - \int_1^3 \sin(2\pi n t/3)dt \right ] \\ = \frac23\left [ -3\cos(2\pi nt/3)/\pi n \right ]_0^1-\frac23[-3\cos(2\pi nt/3)/2\pi n]_1^3 \\ =\frac{2(1-\cos(2\pi n/3))-(-\cos(2\pi n)&plus;\cos(2\pi n/3))}{\pi n} \\ = \frac{3(1-\cos(2\pi n/3))}{\pi n}$

$b_n' = \frac23\int_0^3 x'(t)\cos\left ( 2\pi n t/3 \right ) dt \\ = \frac23\left [\int_0^1 2\cos(2\pi n t/3)dt - \int_1^3 \cos(2\pi n t/3)dt \right ] \\ = \frac23\left [ 3\sin(2\pi nt/3)/\pi n \right ]_0^1-\frac23[3\sin(2\pi nt/3)/2\pi n]_1^3 \\ =\frac{2\sin(2\pi n/3)&plus;\sin(2\pi n/3)}{\pi n} \\ = \frac{3\sin(2\pi n/3)}{\pi n}$

Therefore, we get

$x'(t) = a_n'\sin(2\pi nt/3) &plus; b_n'\cos(2\pi nt/3)$

where the coefficients are as found above.

ii) From this fourier series expansion of the derivative, we get

$x(t) = \frac{a_0}{2} &plus; \frac{a_n'}{2\pi n/3}\cos(2\pi nt/3) - \frac{b_n'}{2\pi n/3}\sin(2\pi nt/3)$

Hence, the Fourier coefficients of x(t) are

$\\ b_n = \frac{a_n'}{2\pi n/3} = \frac{9(1-\cos(2n\pi/3))}{2(n\pi)^2}\\ a_n = -\frac{b_n'}{2\pi n/3} = -\frac{9\sin(2n\pi/3)}{2(n\pi)^2}$

c) We know that

$\\ \mathcal F(\cos(\omega_0 t)) = \pi[\delta(\omega -\omega_0)&plus;\delta(\omega&plus;\omega_0)] \\ \mathcal F(\sin(\omega_0 t)) = -i\pi [\delta(\omega -\omega_0)-\delta(\omega&plus;\omega_0)] \\ \mathcal F(1) = 2\pi\delta(\omega)$

Hence,

$\mathcal F(x(t)) = a_0\pi\delta(\omega) &plus; \sum_n (a_n\mathcal F(\sin(2n\pi t/3))&plus;b_n\mathcal F(\cos(2n\pi t/3))) \\ =a_0\pi\delta(\omega) &plus; \sum_n [-ia_n\pi [\delta(\omega -2n\pi/3)-\delta(\omega&plus;2n\pi/3)] \\ &plus; b_n\pi[\delta(\omega -2n\pi/3)&plus;\delta(\omega&plus;2n\pi/3)]] \\ =a_0\pi\delta(\omega) &plus; \pi\sum_n[ (b_n-ia_n)\delta(\omega-2n\pi/3)&plus;(b_n&plus;ia_n)\delta(\omega&plus;2n\pi/3)] \\ = 4\pi\delta(\omega) &plus; \frac{9}{2\pi}\sum_n[ (1-e^{-2in\pi/3})\delta(\omega-2n\pi/3)&plus;(1-e^{2in\pi/3})\delta(\omega&plus;2n\pi/3)]$