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}

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