## Thursday, 19 January 2017

### Fourier series and Fourier transform

The concept of Fourier series is based on the orthogonality of sine and cosine functions. So, it is a perfect thing to start from these properties.
A sine function with time period $2\pi$ radians is given by $sin(t)$ and the integral of the function over its time period is zero. In general, for an arbitrary sine function with Time period $T_0$, the function is given by $sin (\frac{2n\pi{t}}{T_0})$, where the time integral over this new time period is zero. Everything applies to "cosine" function as much as the same.

In functional form, $$sin\left(\frac{2n\pi}{T_0}(t+T_0)\right) = sin(\frac{2n\pi{t}}{T_0})$$ and $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) = \int_{t_0}^{t_0+T_0} \,dt cos(\frac{2m\pi{t}}{T_0}) = 0 \\ for \,\,\,\,n,m > 0$$
From our trigonometric identities, we can write $$sinA cosB = \frac{sin(A+B)+ sin(A-B)}{2}$$ Using this in our case, we will get (n+m) and (n-m) where n and m are integers. This gives, $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0$$
Similarly, with the other identity $$sinA sinB = cos (A-B) - cos(A+B) \\ cosA cosB = cos(A+B) + cos(A-B)$$ we can prove $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0$$ except now, there arise a problem for the case n = m as we will get an extra non-zero term $$\int_{t_0}^{t_0+T_0} \,dt \frac{cos(\frac{2(n-m)\pi{t}}{T_0})}{2} = \int_{t_0}^{t_0+T_0} \frac{1}{2}\,dt = \frac{T_0}{2}$$
Finally we get the following identities,
$$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0$$ and
$$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) sin(\frac{2m\pi{t}}{T_0}) = \int_{t_0}^{t_0+T_0} \,dt cos(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = \delta_{nm}\frac{T_0}{2}$$
This is the fundamental orthogonality relation we require for the sine and cosine function to serve as the basis vectors.
Now, the Fourier series is defined as, $$f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n cos(\frac{2n\pi{t}}{T_0}) + \sum_{n=1}^\infty b_n sin(\frac{2n\pi{t}}{T_0})$$
The only unknown terms in this series are the Fourier coefficients $a_0, a_n, b_n$ which can be determined using our orthogonality relations as, $$a_n = \frac{2}{T_0}\int_{t_0}^{t_0+T_0} \,dt cos(\frac{2n\pi{t}}{T_0}) f(t) \,\,\,\, n = 0,1,2..\\ b_n = \frac{2}{T_0}\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) f(t)$$
Once we are able to write a function in terms of sine and cosine, we proceed to write it in terms of exponential functions using Euler's formula. And we can contract the series in a simple form as, $$f(t) = \sum_{n=-\infty}^\infty c_n e^{i\frac{2n\pi{t}}{T_0}}$$
Now, the only unknown in this series is $c_n$ and it can also be determined using the orthogonality relation of exponentials $$\int_{t_0}^{t_0+T_0}\,dt\,e^{i\frac{2n\pi{t}}{T_0}}e^{-i\frac{2m\pi{t}}{T_0}} = \delta_{nm}T_0$$ which gives finally, $$c_m = \frac{1}{T_0} \int_{t_0}^{t_0+T_0} f(t) e^{\frac{i2n\pi{t}}{T_0}} \,dt$$

Fourier Transform:

Substituting for $\omega_n = \frac{2n\pi}{T_0}$ we get $$f(t) = \sum_{n= -\infty}^\infty c_n e^{i\omega_nt}$$ and $$c_m = \frac{1}{T_0} \int_{-\frac{T_0}{2}}^{\frac{T_0}{2}} f(t) e^{-i\omega_nt}\,dt$$ (from the substitution $t_0 = -\frac{T_0}{2}$)

It is easily seen from the above equation that, it fails for $T_0 = \infty$ For a function with infinite period, the coefficient cannot be calculated from above equation.

Since, many of the real function used in Physics has this structure (Time period infinity), it is necessary to look our for an alternative to deal with these functions. This is where the integral comes about to play the significant role and known as fourier transform.

For very large time period, the change in $\omega$ - the step size becomes negligible i.e. $$\omega_{n+1} -\omega_n = \frac{2\pi}{T_0} \\ \rightarrow \,\,\,\, \delta\omega = d\omega\\ \frac{1}{T_0} = \frac{d\omega}{2\pi}$$ and there is no need for the index 'n'.

Combining everything together, $T_0\rightarrow\infty$ and substituting for $c_n$ with some dummy variable, $$f(t) = \sum_{n= -\infty}^\infty\frac{d\omega}{2\pi} \left[\int_{-\infty}^{\infty}f(\tau) e^{-i\omega_n\tau}\,d\tau\right]e^{i\omega_n{t}}$$ As "n" changes $d\omega$ is very small and the sum can be replaced by Integral,

$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega_nt}\,d\omega \left[\int_{-\infty}^\infty f{\tau} e^{-i\omega_n\tau}\,d\tau\right]$$ the bracketed term depends only on $\omega$. Thus, we have finally arrived at our fourier and inverse fourier transform. It doesn't matter which one you call the fourier transform and which one is the inverse one. By splitting $\frac{1}{2\pi}$ factor in two different ways, the final equations are, $$f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\,d\omega e^{i\omega{t}}F(\omega) \\ F(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\,dt \,e^{-i\omega{t}}\,f(t) \,\,\,\,\\or\,\,\,\,\\ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty}\,d\omega e^{i\omega{t}}F(\omega) \\ F(\omega) = \int_{-\infty}^{\infty}\,dt e^{-i\omega{t}}\,f(t)$$ or you can put the $2\pi$ factor in the other way.

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