Question

Problem 1. x(t) = 2 cos(210.8t) + 3cos(270.2t) 1) Sketch x(t) for 0<t<2 2) Find the Fourier Series coefficients for x(t)

Answer 1

(1)

The given signal is:

x(t) = 2 cos(210.8t) + 3 cos(210.2t)

Putting t values in the x(t) and find the x(t).

t	X(t)
0	5
0.5	0.809
0.7	0
1	1.55
1.5	-0.309
2	-4

5 datal 4 3 2 1 X(t) 00 -1 N -3 4 -50 0 0.5 1 1.5 2 t

(2)

x(t) = 2 cos(210.8t) + 3 cos(210.2t)

x(t) = 2 cos (210 x 8 2 t) + 3 cos(211 XT 10 10 1. t)

21 X(t) = 2 cos ($T ) Ft) +3 cos (1)

When a signal is a combination of two or more individual periodic signals with fundamental frequencies w_o1, w_o2, w_o3,……………then the fundamental frequency of the resultant signal is given by:

w = Greatest Common Divisor (W01,W02, Wo3 -

In the above signal we find the fundamental frequencies w:

8п W01 5

Wo2 2л 5

w = GCD (W01,W02)

w = GCD 81 211 5 5

211 W 5

So that the fundamental frequency w:

211 W 5

The Fourier series coefficients a_k is given by:

x(t). Σ αχ eikwof ... ... (1) k=- 0

Consider the given signal:

21 X(t) = 2 cos ($T ) Ft) +3 cos (1)

cos(wot) = -(ewot+ (ejwot + e-jwot)

877 -t л -t Levent x(t) = 2 te is 54 te vitit + 3 2 2

3 2T 811 3 jest t 2T -t x(t) 8T -t 5 = e + e test te je .(2) 2 2

Expend the equation (1)

x(t) = a ejwot+a-e-jwot + 24e4wot + a_4e-j4wot ....(3)

Comparing the equation (2) and (3)

The non zero Fourier series coefficients are:

3 ar 2

3 a-1 2

04 = 1

a 1 -4