Question 17.11: Find the complex Fourier series of the sawtooth wave in Fig.......

From Fig. 17.9,f(t)  =  t,  0 <  t  <  1 ,  T =  1   so that  ω_{0} = 2π/T = 2 π . Hence,

c_{n} = \frac{1}{T} \int^{T}_{0}{f(t) e^{-jnω_{0}t} dt } = \frac{1}{1} \int^{1}_{0}{te^{-j2nπt} dt}                              (17.11.1)

But

\int{te^{at} dt} = \frac{e^{at}}{a^{2}} (ax  –  1)  +  C

Applying this to Eq. (17.11.1) gives

\begin{matrix} c_{n} = \frac{e^{-j2nπt}}{(-j2nπ)^{2}}(-j2nπt  –  1) \mid^{1}_{0} \\ = \frac{e^{-j2nπ}(-j2nπ  –  1)  +  1}{-4n^{2}π^{2}}\end{matrix}\quad \quad (17.11.2)

Again,

e^{-j2πn}  =  \cos  2πn  –  j  \sin  2πn  =  1  –  j 0 =  1

so that Eq. (17.11.2) becomes

c_{n} = \frac{-j2nπ}{-4n^{2}π^{2}} = \frac{j}{2nπ}                      (17.11.3)

This does not include the case when n = 0 . When n = 0 ,

c_{0} = \frac{1}{T} \int^{T}_{0}{f(t) dt} = \frac{1}{1} \int^{1}_{0}{t  dt } = \frac{t^{2}}{2} \mid^{0}_{1} = 0.5                      (17.11.4)

 Hence,

f(t) = 0.5  +  \sum\limits^{∞}_{\begin{matrix}n = – ∞\\ n≠0\end{matrix}}{\frac{j}{2nπ} e^{j2nπt}}                    (17.11.5)

and

\left|c_{n}\right| = \begin{cases} \frac{1}{2nπ} ,  & n ≠ 0 \\ 0.5 ,  &  n = 0 \end{cases} , θ_{n} = 90° ,                 n  ≠  0              (17.11.6)

By plotting  \left|c_{n}\right|  and  θ_{n}    for different n, we obtain the amplitude spectrum and the phase spectrum shown in Fig. 17.31.