Fourier Transform

• Fourier Series

A PERIODIC function ${f(x)}$ can be represented as a sum of sine and cosine functions with some coefficients. Such a representation is known as Fourier series. The condition of periodicity is necessary for the Fourier series. There are few more conditions, known as Dirichlet’s conditions.

Consider such a function ${f(x)}$ of period ${2L}$ and defined over ${(A, A+2L)}$. It can be written as

${f(x) = \frac {a_0}{2}+ \sum \limits_{n=1}^{\infty} a_n cos \Big( \frac {n \pi x}{L} \Big) + b_n sin \Big( \frac {n \pi x}{L} \Big)}$

The constants ${a_0, a_n}$ and ${b_n}$ are known as Fourier coefficients. They are calculated as:

${a_0 = \frac {1}{L} \int \limits_{A}^{A+2L} f(x)dx}$

${a_n = \frac {1}{L} \int \limits_{A}^{A+2L} f(x) cos \Big( \frac {n \pi x}{L} \Big) dx}$

${b_n = \frac {1}{L} \int \limits_{A}^{A+2L} f(x) sin \Big( \frac {n \pi x}{L} \Big) dx}$

This is the Fourier series.

• Fourier Series : Complex Representation

Using ${e^{i \theta} = cos (\theta) + i \ sin(\theta)}$ and ${e^{-i \theta} = cos (\theta) - i \ sin(\theta)}$, we can write

${cos (\theta) = \frac {e^{i \theta} + e^{-i \theta}}{2} \ and \ sin (\theta) = \frac {e^{i \theta} - e^{-i \theta}}{2i }}$

Substituting this in the Fourier series for ${f(x)}$,

${f(x) = \frac {a_0}{2} + \sum \limits_{n=1}^{\infty} \Big[ \Big (\frac {a_n - i b_n}{2} \Big ) e^{i n \pi x /L} + \Big (\frac {a_n + i b_n}{2} \Big ) e^{- i n \pi x /L} \Big]}$

Let ${c_0 = \frac {a_0}{2}}$, ${c_n = \frac {a_0 - i b_n}{2}}$ and ${c_{-n} = \frac {a_0 + i b_n}{2}}$.

Using this, we can write

${c_n = \frac {1}{2L} \int \limits_{-L}^{L} f(u) e^{- i n \pi x / L} dx}$

and

${f(x) = \sum \limits_{n = - \infty}^{\infty} c_n e^{i n \pi x / L}}$

• Fourier Integral

We know that the definite integral is actually a sum. Further, in reality, the function ${f(x)}$ may not be always periodic (e.g. an unrepeated pulse in a circuit). In such cases, the sum is replaced by an integral. Thus, we now allow the function to be aperiodic or ${L \to \infty}$. We introduce a new term ${\lambda = \frac {n \pi}{L}}$, so ${\Delta \lambda = \frac {\pi}{L}}$.

Clearly, as ${L \to \infty}$, ${\Delta \lambda \to zero}$. Thus, ${f(x)}$ can be written as

${f(x) = \sum \limits_{n = - \infty}^{\infty} \Big ( \frac {1}{2 \pi} \int \limits_{-L}^{L} f(u) e^{- i \lambda (u-x)}du \Big) \Delta \lambda}$

Taking limit, as ${L \to \infty}$, we can apply the fundamental theorem of integral calculus,

${f(x) = \frac {1}{2 \pi} \int \limits_{\lambda = - \infty}^{\infty} \int \limits_{u= - \infty}^{\infty} f(u) e^{- i \lambda (u-x)}dud \lambda}$

This is the Fourier integral representation of ${f(x)}$ for ${- \infty < x < \infty}$.

• Simplified Fourier Integral

I) Separating ${u}$ and ${\lambda}$,

${f(x) = \frac {1}{2 \pi} \int \limits_{- \infty}^{\infty} e^{i \lambda x} d \lambda \int \limits_{- \infty}^{\infty} f(u) e^{- i \lambda u} du}$

II) Using Euler’s identity,

${f(x) = \frac {1}{2 \pi} \int \limits_{\lambda = - \infty}^{\infty} \int \limits_{u= - \infty}^{\infty} f(u) cos [\lambda (u-x)] du d \lambda - \frac {i}{2 \pi} \int \limits_{\lambda = - \infty}^{\infty} \int \limits_{u= - \infty}^{\infty} f(u) sin [\lambda (u-x)] du d \lambda}$

III) We can now split the integrals of ${\lambda}$ and ${u}$. Sine being an odd function, ${\int_{- \infty}^{\infty} sin [\lambda (u-x)] d \lambda}$ is always ${0}$, hence the 2nd integral is always ${0}$. Hence,

${f(x) = \frac {1}{2 \pi} \int \limits_{\lambda = - \infty}^{\infty} \int \limits_{u= - \infty}^{\infty} f(u) cos [\lambda (u-x)] du d \lambda}$

• Fourier Transform

${f(x) = \frac {1}{2 \pi} \int \limits_{\lambda = - \infty}^{\infty} \int \limits_{u= - \infty}^{\infty} f(u) e^{- i \lambda (u-x)}dud \lambda}$

${f(x) = \frac {1}{2 \pi} \int \limits_{\lambda = - \infty}^{\infty} \Bigg ( \int \limits_{u= - \infty}^{\infty} f(u) e^{- i \lambda u} du \Bigg ) e^{i \lambda x} d \lambda}$

The inner integral is known as the Fourier Transform of ${f(x)}$ and is denoted by ${F(\lambda)}$. Thus,

${F(\lambda) = \int \limits_{u= - \infty}^{\infty} f(u) e^{- i \lambda u} du}$

The inverse Fourier transform is given by

${f(x) = \frac {1}{2 \pi} \int \limits_{\lambda = - \infty}^{\infty} F(\lambda) e^{i \lambda x} d \lambda}$

• Fourier Cosine Transform (For even functions)

${F_c (\lambda) = \int \limits_0^{\infty} f(u) cos (\lambda u)du}$

and

${f(x) = \frac {2}{\pi} \int \limits_0^{\infty} F_c (\lambda) cos (\lambda x) d \lambda}$

• Fourier Sine Transform (For odd functions)

${F_s (\lambda) = \int \limits_0^{\infty} f(u) sin (\lambda u)du}$

and

${f(x) = \frac {2}{\pi} \int \limits_0^{\infty} F_s (\lambda) sin (\lambda x) d \lambda}$