# 10/09 - Fourier Transform

 $\left \langle e^{j2\pi n t/T} \mid e^{j2\pi m t/T} \right \rangle$ $=\int_{-\infty}^{\infty}e^{j2\pi n t/T} e^{-j2\pi m t/T}\,dt$ $=\int_{-\infty}^{\infty}e^{j2\pi (n-m) t/T}\,dt$ $=\int_{-T/2}^{T/2}e^{j2\pi (n-m) t/T}\,dt$ Assuming the function is perodic with the period T $=T\delta_{m,n}\,\!$

## Fourier Transform

Remember from 10/02 - Fourier Series

• $\alpha_n = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j\,2\,\pi \,n\,t/T}\, dt$
• $x(t) = x(t+T) = \sum_{n=-\infty}^\infty \alpha_n e^{j\,2\pi \,n/T}$

If we let $T \rightarrow \infty$

 $\frac{1}{T}$ $\rightarrow df$ $\frac{n}{T}$ $\rightarrow f$ Remember $f=\frac{2\pi n}{T} \,\!$ $T\,\!$ $\rightarrow \infty$ $\sum_{n=-\infty}^{\infty} \frac{1}{T}$ $\rightarrow \int_{-\infty}^{\infty}() df$

## Definitions

 $F[x(t)]\,\!$ $=X(f)\,\!$ $=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}dt$ $=\left \langle x(t) \mid e^{j2\pi ft}\right \rangle_t$ $F^{-1}[X(f)]\,\!$ $=x(t)\,\!$ $=\int_{-\infty}^{\infty} X(f) e^{j2\pi ft}df$ $=\left \langle X(f) \mid e^{-j2\pi ft}\right \rangle_f$

## Examples

 $\int_{-\infty}^{\infty}e^{j2\pi ft}e^{-j2\pi f\lambda}df$ $=\left\langle e^{j2\pi ft}\mid e^{j2\pi ft}\right\rangle_f$ $=\delta(t-\lambda)\,\!$ $\int_{-\infty}^{\infty}e^{j2\pi tf}e^{-j2\pi tf_0}dt$ $=\left\langle e^{j2\pi tf}\mid e^{j2\pi tf_0}\right\rangle_t$ $=\delta(f-f_0)\,\!$
 $F^{-1}[F[x(t)]]\,\!$ $=\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty} x(\lambda) e^{-j2\pi f\lambda}d\lambda\right ]e^{j2\pi ft} df$ $=\int_{-\infty}^{\infty}X(f) e^{j2\pi ft} df$ $=x(t)\,\!$ $=\int_{-\infty}^{\infty}x(\lambda) \int_{-\infty}^{\infty}e^{j2\pi f(t-\lambda)} df d\lambda$ $=\int_{-\infty}^{\infty}x(\lambda) \delta(t-\lambda) d\lambda$ $=x(t)\,\!$ $=\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty} x(\lambda) e^{-j\omega\lambda}d\lambda\right ]e^{j\omega t} \frac{1}{2\pi}d\omega$ $=\int_{-\infty}^{\infty}x(\lambda) \left [ \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{j(t-\omega) \lambda}d\omega\right ] d\lambda$ $=\int_{-\infty}^{\infty}x(\lambda) \delta(t-\omega) d\lambda$ $=x(t)\,\!$

### Sifting property of the delta function

The dirac delta function is defined as any function, denoted as $\delta(t-u)\,\!$, that works for all variables that makes the following equation true: $x(t)=\int_{-\infty}^{\infty} x(u)\delta(t-u) du$

• When dealing with $\omega\,\!$, it behaves slightly different than dealing with $f\,\!$. When dealing with $=\int_{-\infty}^{\infty}x(\lambda) \left [ \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{j(t-\omega) \lambda}d\omega\right ] d\lambda$, note that the delta function is $\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{j(t-\omega) \lambda}d\omega$. The $\frac{1}{2\pi}$ is tacked onto the front. Thus, when dealing with $\omega\,\!$, you will often need to multiply it by $2\pi\,\!$ to cancel out the $\frac{1}{2\pi}$.

### More properties of the delta function

$\delta(a\,t) = \frac{1}{\left | a \right |}$

$\delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f)$

 $\int_{-\infty}^{\infty}\delta(a\,t)\,dt$ $=\int_{-\infty}^{\infty}\delta(u\,t)\,\frac{du}{\left|a\right|}$ Let $a\,t=u$ and $du=a\,dt$ $=\frac{1}{\left|a\right|}$