Fourier Transform

Fourier Series

We can write any periodic function over a period $T$ as $f(t)=a_0+{\sum }_{k=1}^{\infty }{a}_k\cos \left(\frac{2\pi kt}{T}\right)+{\sum }_{k=1}^{\infty }{b}_k\sin \left(\frac{2\pi kt}{T}\right)$

• If $T=2\pi$, it simplifies the expression

with coefficients are given by solving the following integrals: $a_0=\frac{{\int }_0^{2\pi }f(t)(dt)}{2\pi }$ $a_k=\frac{{\int }_0^{2\pi }f(t)\cos (kt)dt}{{\int }_0^{2\pi }{\cos }^2(kt)dt}$ $b_k=\frac{{\int }_0^{2\pi }f(t)\sin (kt)dt}{{\int }_0^{2\pi }{\sin }^2(kt)dt}$

Now, given the above sinusoidal expression of a function $f(t)$ (with period $2\pi$) $f(t)=a_0+{\sum }_{k=1}^{\infty }{a}_k\cos (kt)+{\sum }_{k=1}^{\infty }{b}_k\sin (kt)$

we can express it more concisely using Euler’s Formula as $f(t)={\sum }_{k=-\infty }^{\infty }{c}_k{e}^{ikt}$ where the $c_k$ coefficients are complex numbers.

But can you express c_k as a sum of a_k and b_k?

Yup.

Approximation

An approximation of a function could be achieved by truncating the series to a finite number of sinusoids: $f(t)\approx {\sum }_{k=-M}^M{c}_k{e}^{ikt}$

For $N$ points (assume $N$ is even), we will use $N$ degrees of freedom (i.e., $N$ coefficients) to exactly interpolate the data.

$f(t)\approx {\sum }_{k=-\frac{N}{2}+1}^{N/2}{c}_k{e}^{\frac{(2\pi i)kt}{T}}$

Plugging in each of our $N$ data points $(t_n,f_n)$ into the expression will give us $N$ equations, involving unknowns coefficients, $c_k$. This leads us to the Discrete Fourier Transform!

From MATH213

See L2 Function.

We want to write $f(t)={\sum }_{n=-\infty }^{\infty }{c}_n{e}^{\frac{2\pi n}{\tau }jt}$ for $t\in [-\tau /2,\tau /2]$

Our goal is to find $c_n$.

Definition 3: Fourier Series - Complex Form

If $f\in L^2([-\tau /2,\tau /2])$ then the Fourier series in complex form of $f(t)$ is ${\sum }_{n=-\infty }^{\infty }{c}_n{e}^{\frac{2\pi n}{\tau }jt}$ where the $c_n$ are found by projecting $f$ into the basis of complex exponentials.

Theorem 3: Fourier Coefficients for Series in Complex Form

If $f\in L^2([-\tau /2,\tau /2])$ then the Fourier coefficients $c_n$ of $f(t)$ are

$c_n=<f(t),e^{\frac{2\pi n}{\tau }jt}>$ $c_n=\frac{1}{\tau }{\int }_{\frac{\tau }{2}}^{\frac{\tau }{2}}f(t)e^{-\frac{2\pi n}{\tau }jt}dt$

If $f$ is real valued than $c_n=\overline{c_{-n}}$.

• I was so confused for a second, but $<a,b>$ is the Inner Product

Theorem

For a real valued, $\tau$ periodic function $f\in L^2([-\tau /2,\tau /2])$:

• If $f$ is an even then the Fourier series can be simplified to a sum of $\cos$ waves.
• If $f$ is an odd then the Fourier series can be simplified to a sum of $\sin$ waves.

Theorem 2: Fourier Sine and Cosine Coefficients

If $f$ is a real valued function that is in $L^2([-\tau /2,\tau /2])$ then

If $f$ is even then the Fourier cosine series for $f$ is ${\sum }_{n=0}^{\infty }{c}_n\cos \left(\frac{2\pi n}{\tau }t\right)$ where $c_n=f(x)=\{\begin{array}{ll}⟨f(t),1⟩& n=0\\ 2⟨f(t),\cos \left(\frac{2\pi n}{\tau }t\right)⟩& n>0\end{array}$

If $f$ is odd then the Fourier sine series for $f$ is ${\sum }_{n=1}^{\infty }{s}_n\sin \left(\frac{2\pi n}{\tau }t\right)$

where $s_n=2⟨f(t),\sin \left(\frac{2\pi n}{\tau }t\right)⟩$

• It’s a little hard to memorize this. Instead, memorize theorem 3 and derive it

Convergence of Fourier Series

See PWC1.

Theorem 5: Convergence of Fourier series

Let $f_p$ be the Periodic Extension of a function $f\in L^2([-\tau /2,\tau /2])$.

• The Fourier series of $f$ converges in the $L^2$ norm (also in the mean and almost everywhere) to $f$ (and also $f_p$) on any finite subinterval of $[-\tau /2,\tau /2]$.
• If$f_p$ is piecewise $C^1$ then the Fourier series of $f$ converges pointwise to $f_p$ for all $x\in R$
• If $f_p$ is piecewise $C^1$ and continuous then the Fourier series of $f$ converges uniformly $f_p$ on any finite interval of $R$.