Question 8.3: Find the FT of the rectangular pulse v(t) = u(t) − u(t − a),...

X(jω) =\int\limits_{0}^{a}e^{-jwt}dt=\frac{2 sin(0.5ωa)}{w}e^{−j0.5ωa}

u(t) − u(t − a) ↔\frac{2 sin(0.5ωa)}{w} e^{−j0.5ωa}.
The pulse and its FT are shown, respectively, in Fig. 8.5a, b with a = π.Fourier transform is the most generalized version of the Fourier analysis. The FS of signal, which is a repetition of an aperiodic signal, can be deduced from the FT. The relationship is given by
X(k) = \frac{1}{T} X(jkω0).
The discrete samples of the FT spectrum at intervals of the fundamental frequency, w_{0} , divided by the period are the FS spectrum. The reason for the FT spectrum to be continuous is that it has to reconstruct the pulse, in addition to the infinite extent zeros on either side of the pulse. Let us determine the FS for the example from the corresponding FT. Since X(k) = \frac{1}{T} X(jkω0).with T = 2π and ω = kω0 =k\frac{2\pi }{2\pi } = k, we get
X(k) =\frac{2}{2\pi }-\frac{sin(0.5πk)}{k}e^{−j0.5\pi k}=\frac{1}{\pi }\frac{sin(\frac{\pi }{2} k)}{k}e^{−j\frac{\pi }{2} k} , k ≠ 0 and X(0)=\frac{1}{2}
The FS is
x(t) =\frac{1}{2}+\sum\limits_{k=-\infty }^{\infty }{\frac{1}{\pi }\frac{sin(\frac{\pi }{2} k)}{k}e^{−j\frac{\pi }{2} k}e^{jkt} , k ≠ 0 }
as found earlier. The even-indexed coefficients are zero and the coefficients of the sine component of the complex exponentials are 2/(kπ ). For example, with k = 1 and k = −1 and leaving the factor π in the denominator, we get
(1)(−j1)e^{jt} = −je^{jt} = \frac{e^{jt} }{j} , (1)(j1)e^{-jt} = je^{−jt} = −\frac{e^{−jt} }{j}.

\frac{e^{jt}-e^{-jt}}{j}= 2 sin(t).
Since t or ω is zero, the values X(j0) and v(0) can be easily determined and can be used to verify the closed-form expressions for X(jω) or v(t).