# How does the function inside the rectangular function work?

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?

$\operatorname{rect}(\cdot)$ 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 $\frac 12$.

• That makes sense. In other words, set the function inside of rect equal to -1/2 then 1/2. – JFA Apr 21 '14 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 $$A=2\pi,\omega_0=10$$