Question

Example 8.3 Calculate the Fourier transform of the rectangular pulse The graph of f() is shown in Figure 8.6, and since the area under it is finite, a Fourier transform exists. From the definition (8.15), we have Solution f0) 24 -T O = 2AT sinc (oT Figure 8.6 The rectangular pulse where sincx is defined, as in Example 7.22, by sinx (x 0) 0 > T) (x = 0)

please explain the part where the question mark is located. how did we get sinc. this is the Fourier transform properties.

Answer 1

Consider $\dpi{100} \left [ -\frac{A}{j\omega }e^{-j\omega t} \right ]_{-T}^{T}$

This equals

$\dpi{100} \left [ \frac{A}{j\omega }e^{j\omega T} -\frac{A}{j\omega }e^{-j\omega T} \right ]$

This equals

$\dpi{100} \frac{A}{j\omega }\left [ e^{j\omega T}-e^{-j\omega T} \right ]$

As $\dpi{100} e^{j\omega T}=\cos(\omega T)+j\sin(\omega T),\,e^{-j\omega T} =\cos(\omega T)-j\sin(\omega T)$ so we have

$\dpi{100} e^{j\omega T}-e^{-j\omega T}=2j\sin(\omega T)$

And so $\dpi{100} \frac{A}{j\omega }\left [ e^{j\omega T}-e^{-j\omega T} \right ]=\frac{A}{j\omega }\times 2j\sin(\omega T)$

That is, $\dpi{100} \frac{2A\sin(\omega T)}{\omega}$

Which we can write as $\dpi{100} \frac{2A\sin(\omega T)}{\omega T}\times T$

As $\dpi{100} \frac{\sin(\omega T)}{\omega T}=\text{sinc}(\omega T)$ is the sinc function defined as $\dpi{100} \frac{\sin(x)}{x}=\text{sinc}(x)$

And so $\dpi{100} \frac{2A\sin(\omega T)}{\omega T}\times T=2AT\text{sinc}(\omega T)$

Thus the expression is equivalent to $\dpi{100} 2AT\text{sinc}(\omega T)$ for $\dpi{100} \omega\neq 0$ and for

$\dpi{100} \omega =0$ it is $\dpi{100} 2A$

Using the convention $\dpi{100} \text{sinc}(0)=1$ we can write the above succintly as

$\dpi{100} 2AT\text{sinc}(\omega T)$ for all $\dpi{100} \omega$

$\dpi{100} \blacksquare$

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