# Fourier Transform of triangle function $x(t)=\Delta\left(\frac{t-1}{2}\right)$

Can you please tell me if my working is right for the Fourier Transform of this function: $$x(t)=\Delta\left(\frac{t-1}{2}\right)$$

My workings are:

I have used the fourier transform standard results. Please could you tell me if I which I have to apply:

1. First the $F(t-1)$, then $F(t/2)$
2. First the $F(t/2)$, then $F(t-1)$
3. First the $F(t/2)$, then $F(-1/2)$

Please tell me where I have gone wrong.

$$x(t)=\mathrm{sinc}^2(\omega/2)e^{-j\pi f}$$

I will not go into the details of the specific function that you need to transform. This is for two reasons: first, with a bit of effort you will be able to do it yourself (after having read this answer), and second, because you didn't define your functions. Note that it is not immediately clear how $\Delta(t)$ is defined; also note that there are two frequently used definitions of $\text{sinc}(x)$ (either with or without an implicit factor of $\pi$).

I will discuss the way to derive the Fourier transform of a shifted and scaled version of a function $f(t)$ with Fourier transform $F(\omega)$. For this you need to know the two following Fourier transform properties (derive them yourself and you'll never forget them!):

$$f(at)\Longleftrightarrow \frac{1}{|a|}F\left(\frac{\omega}{a}\right)\\ f(t-a)\Longleftrightarrow F(\omega)e^{-ja\omega}\tag{1}$$

The Fourier transform of $f((t-a)/b)$ can be derived from $(1)$ in two different ways:

1. Find the transform of $f(t-\frac{a}{b})$, then substitute $t\rightarrow t/b$
2. Find the transform of $f(\frac{t}{b})$, then substitute $t\rightarrow t-a$

The first way is then $$f\left(t-\frac{a}{b}\right)\Longleftrightarrow F(\omega)e^{-ja\omega/b}\\ f\left(\frac{t-a}{b}\right)\Longleftrightarrow |b|F(b\omega)e^{-ja\omega}$$

The second way of course gives the same result: $$f\left(\frac{t}{b}\right)\Longleftrightarrow |b|F(b\omega)\\ f\left(\frac{t-a}{b}\right)\Longleftrightarrow |b|F(b\omega)e^{-ja\omega}$$