Question 7.7: Inverse DTFT of a periodically repeated rectangle Find the i......

Since we can choose to integrate over any interval in F of width one, let’s choose the simplest one

\mathbf{x}[n]=\int\limits_{-1/2}^{1/2}\operatorname{rect}(w F)\ast\delta_{1}(F)e^{j2\pi F n}\,d F

In this integration interval, there is exactly one rectangle function of width 1/w (because w > 1) and

X[n]=\int\limits_{-1/2w}^{1/2w}e^{j2\pi F n}\,d F=2\int\limits_{0}^{1/2w}\cos(2\pi F n)\,d F={\frac{\sin(\pi n/w)}{\pi n}}={\frac{1}{w}}sinc\left\lgroup\frac{n}{w}\right\rgroup .      (7.21)

From this result we can also establish the handy DTFT pair (which appears in the table of DTFT pairs),

\mathrm{sinc}(n/w)\overset{\mathcal F}{\longleftrightarrow } w\,\mathrm{rect}(w F)^{\ast}\,\delta_{1}(F),\quad w\gt 1

or

\mathrm{sinc}(n/w)\overset{\mathcal F}{\longleftrightarrow } w\,\sum\limits_{k=-\infty }^{\infty }{} \mathrm{rect}(w(F-k)),\quad w\gt 1

or, in radian frequency form, using the convolution property,

\mathbf{y}(t)=\mathbf{x}(t)\ast\mathbf{h}(t)\Rightarrow\mathbf{y}(a t)=|a|\mathbf{x}(a t)*\mathbf{h}(a t)

we get

\mathrm{sinc}(n/w)\overset{\mathcal F}{\longleftrightarrow } \mathrm{w\,rect}(w\Omega/2\pi)\ast\delta_{2\pi}(\Omega),\quad w\gt 1

or

\mathrm{sinc}(n/w)\stackrel{\mathcal F}{\longleftrightarrow }w\sum_{k=-\infty}^{\infty}\mathrm{rect}(w(\Omega-2\pi k)/2\pi),\quad w\gt 1.

(Although these Fourier pairs we derived under the condition w > 1 to make the inversion integral (7.21) simpler, they are actually also correct for w ≤ 1.)