6Fourier transforms

IB Methods

6.6 Linear systems and response functions

Suppose we have an amplifier that modifies an input signal

I

(

t

) to produce an

output

O

(

t

). Typically, amplifiers work by modifying the amplitudes and phases

of specific frequencies in the output. By Fourier’s inversion theorem, we know

I(t) =

1

2π

Z

e

iωt

˜

I(ω) dω.

This

˜

I(ω) is the resolution of I(t).

We specify what the amplifier does by the transfer function

˜

R

(

ω

) such that

the output is given by

O(t) =

1

2π

Z

∞

−∞

e

iωt

˜

R(ω)

˜

I(ω) dω.

Since this

˜

R

(

ω

)

˜

I

(

ω

) is a product, on computing

O

(

t

) =

F

−1

[

˜

R

(

ω

)

˜

I

(

ω

)], we

obtain a convolution

O(t) =

Z

∞

−∞

R(t − u)I(u) du,

where

R(t) =

1

2π

Z

∞

−∞

e

iωt

˜

R(ω) dω

is the response function. By plugging it directly into the equation above, we see

that R(t) is the response to an input signal I(t) = δ(t).

Now note that causality implies that the amplifier cannot “respond” before

any input has been given. So we must have R(t) = 0 for all t < 0.

Assume that we only start providing input at t = 0. Then

O(t) =

Z

∞

−∞

R(t − u)I(u) du =

Z

t

0

R(t − u)I(u) du.

This is exactly the same form of solution as we found for initial value problems

with the response function R(t) playing the role of the Green’s function.