Question 17.10: Find the exponential Fourier series expansion of the periodi......
Find the exponential Fourier series expansion of the periodic function
f(t) = e^{t} , 0 < t < 2π with f(t + 2 π) = f(t) .Since T = 2 π , ω_{o} = 2 π /T = 1 . Hence
c_{n} = \frac{1}{T} \int^{T}_{0}{f(t) e^{-jnω_{0}t} dt} = \frac{1}{2π} \int^{2π}_{0}{e^{t} e^{-jnt}dt}
= \frac{1}{2π} \frac{1}{1 – jn} e^{(1 – jn)t} \mid^{2π}_{0} = \frac{1}{2π(1 – jn) } [e^{2π} e^{-j2πn} – 1]
But by Euler’s identity,
e^{- j2πn} = \cos 2πn – j \sin 2π n = 1 – j 0 = 1
Thus,
c_{n} = \frac{1}{2π (1 – jn)} [e^{2π} – 1] = \frac{85}{1 – nj}
The complex Fourier series is
f(t) = \sum\limits^{∞}_{n = -∞}{\frac{85}{1 – jn } e^{jnt}}
We may want to plot the complex frequency spectrum of f(t) . If we let c_{n} = \left|c_{n}\right| \underline{/ θ_{n} } , then
\left|c_{n}\right| = \frac{85}{\sqrt{1 + n²}} , θ_{n} = \tan^{-1} n
By inserting in negative and positive values of n, we obtain the amplitude and the phase plots of c_{n} versus nω_{0} = n , as in Fig. 17.30.
