Mathematical Tools for Physics - Department of Physics - University

15—Fourier Analysis 375

This example has four main peaks in the frequency spectrum. The real part ofgis an even

function and the imaginary part is odd.

f

Real

g Imag

This is another example with four main peaks.
In either case, if you simply look at the function of time on the left it isn’t obvious what sort of
frequencies are present. That’s why there are standard, well-developed computer programs to do the
Fourier analysis.

15.4 Derivatives
There are a few simple, but important relations involving differentiation. What is the Fourier transform
of the derivative of a function? Do some partial integration.

F(f ̇) =

∫

dteiωt

df

dt

=eiωtf(t)

∣∣

∣

∞

−∞

−iω

∫

dteiωtf(t) =−iωF(f) (15.12)

Here I’ve introduced the occasionally useful notation thatF(f)is the Fourier transform off. The

boundary terms in the partial integration will go to zero if you assume that the functionfapproaches

zero at infinity.

Thenthtime derivative simply give you more factors:(−iω)non the transformed function.

15.5 Green’s Functions
This technique showed up in the chapter on ordinary differential equations, section4.6, as a method to
solve the forced harmonic oscillator. In that instance I said that you can look at a force as a succession
of impulses, as if you’re looking at the atomic level and visualizing a force as many tiny collisions by
atoms. Here I’ll get to the same sort of result as an application of transform methods. The basic
technique is to Fourier transform everything in sight.
The damped, forced harmonic oscillator differential equation is

m

d^2 x

dt^2

+b

dx

dt

+kx=F 0 (t) (15.13)

Multiply byeiωtand integrate over all time. You do the transforms of the derivatives by partial

integration as in Eq. (15.12).
∫∞

−∞

dteiωt[Eq. (15.13)]=−mω^2 x ̃−ibω ̃x+kx ̃=F ̃ 0 , where x ̃(ω) =

∫∞

−∞

dteiωtx(t)

This is an algebraic equation that is easy to solve for the function ̃x(ω).

x ̃(ω) =

F ̃ 0 (ω)

−mω^2 −ibω+k

Now use the inverse transform to recover the functionx(t).

x(t) =

∫∞

−∞

dω

2 π

e−iωtx ̃(ω) =

∫

dω

2 π

e−iωt

F ̃ 0 (ω)

−mω^2 −ibω+k

=

∫

dω

2 π

e−iωt

−mω^2 −ibω+k

∫

dt′F 0 (t′)eiωt

′

=

∫

dt′F 0 (t′)

∫

dω

2 π

e−iωt

−mω^2 −ibω+k

eiωt

′

(15.14)