Question 8.1: Find the FS for the signal x(t) = 2 − 2 sin(2 2π/5t+π/6) +4 ...
Find the FS for the signal
x(t) = 2 − 2 sin\left(2\frac{2π}{5}t + \frac{π}{6} \right) + 4 cos\left(4\frac{2π}{5}t -\frac{π}{3} \right).
One period of the signal is shown in Fig. 8.1a. As the signal is given in terms of cosine and sine functions, we just have to express these functions in terms of complex exponentials using Euler’s formula. The fundamental frequency of the waveform is w_{0} = 2\frac{2π}{5}, as the DC component is periodic with any period. The period is 2.5. All the components of x(t) are rewritten using cosine waveform with positive amplitude for each frequency.
x(t) = 2+ 2 cos\left(2\frac{2π}{5}t + \frac{2π}{3} \right) + 4 cos\left(4\frac{2π}{5}t -\frac{π}{3} \right).
Using Euler’s formula, we get
x(t) = 2+ 1e^{j\left(2\frac{2π}{5}t +\frac{2π}{3} \right) } + 1e^{-j\left(2\frac{2π}{5}t +\frac{2π}{3} \right) } + 2e^{j\left(4\frac{2π}{5}t -\frac{π}{3} \right)} +2e^{-j\left(4\frac{2π}{5}t -\frac{π}{3} \right)}.
Comparing this expression with the definition, Eq. (8.1) x(t) =\sum\limits_{k=−∞}^{∞}{X(k)e^{jkw_{0} t} },we get the exponential form of the FS
coefficients as \left\{X(0) = 2, X(1) = 1\angle \frac{2π}{3}, X(−1) = 1\angle -\frac{2π}{3}, X(2) = 2\angle -\frac{2π}{3}, X(−2) = 2\angle \frac{π}{3} \right\} .
Figure 8.1b shows the three frequency components of the signal. The FS magnitude spectrum and the phase spectrum of the signal are shown, respectively, in Fig. 8.1c, d.
