I'm looking at a Signals and Systems problem, and the question asks to show the inverse fourier transform of \$f(w)=\operatorname{rect}(\frac{w-10}{2\pi})\$ is:

$$\mathcal{F}_{t}^{-1}[f(w)]=\operatorname{sinc}(\pi t)e^{10jt}$$

I can see that \$\operatorname{rect}(x)\$ becomes \$1\$ over the interval \$-1/2\$ to \$1/2\$, but how is it affected by a more complex function? In the solutions we were given the bounds become \$10-\pi\$ to \$10+\pi\$. How did they get to these bounds?


2 Answers 2


\$\operatorname{rect}(\cdot)\$ is a function that has value \$1\$ if whatever appears inside those parentheses (it's called the argument of the rect function and I used \$\cdot\$ instead of some algebraic variable as a place holder) has value between \$-\frac 12\$ and \$+\frac 12\$. Otherwise, when the argument is strictly smaller than \$-\frac 12\$ (or strictly larger than \$+\frac 12\$), \$\operatorname{rect}(\cdot)\$ has value \$0\$. In your instance, you need to ask

For what values of \$\omega\$ does \$\displaystyle \frac{\omega - 10}{2\pi}\$ equal a number between \$-\frac 12\$ and \$\frac 12\$?

and a little thought will show, I hope, that \$\omega\$ must be in the interval from \$10-\pi\$ to \$10 + \pi\$. If you have trouble deriving this, try and find the value of \$\omega\$ that makes \$\displaystyle \frac{\omega - 10}{2\pi}\$ equal exactly \$-\frac 12\$ and then, lather, rinse and repeat for exactly \$\frac 12\$.

  • \$\begingroup\$ That makes sense. In other words, set the function inside of rect equal to -1/2 then 1/2. \$\endgroup\$
    – jfa
    Apr 21, 2014 at 0:49

In general, we have $$ F^{-1}[rect(\frac{\omega}{A})]=\frac{A}{2\pi}sinc(\frac{At}{2}) $$

Applying the frequency shift property of fourier transforms $$ f(t)e^{j\omega_0t}\Longleftrightarrow F(\omega-\omega_0) $$

we get

$$ F^{-1}[rect(\frac{\omega-\omega_0}{A})]=e^{j\omega_0t}\frac{A}{2\pi}sinc(\frac{At}{2}) $$

In your case $$A=2\pi,\omega_0=10$$


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