6Fourier transforms

IB Methods

6.7 General form of transfer function

To model the situation, suppose the amplifier’s operation is described by the

ordinary differential equation

I(t) = L

m

O(t),

where

L

m

=

m

X

i=0

a

i

d

i

dt

i

with

a

i

∈ C

. In other words, we have an

m

th order ordinary differential equation

with constant coefficients, and the input is the forcing function. Notice that

we are not saying that

O

(

t

) can be expressed as a function of derivatives of

I

(

t

). Instead,

I

(

t

) influences

O

(

t

) by being the forcing term the differential

equation has to satisfy. This makes sense because when we send an input into

the amplifier, this input wave “pushes” the amplifier, and the amplifier is forced

to react in some way to match the input.

Taking the Fourier transform and using F

dO

dt

= iω

˜

O(ω), we have

˜

I(ω) =

m

X

j=0

a

j

(iω)

j

˜

O(ω).

So we get

˜

O(ω) =

˜

I(ω)

a

0

+ iωa

1

+ ··· + (iω)

m

a

m

.

Hence, the transfer function is just

˜

R(ω) =

1

a

0

+ iωa

1

+ ··· + (iω)

m

a

m

.

The denominator is an

n

th order polynomial. By the fundamental theorem of

algebra, it has

m

roots, say

c

j

∈ C

for

j

= 1

, ··· , J

, where

c

j

has multiplicity

k

j

. Then we can write our transfer function as

˜

R(ω) =

1

a

m

J

Y

j=1

1

(iω −c

j

)

k

j

.

By repeated use of partial fractions, we can write this as

˜

R(ω) =

J

X

j=1

k

j

X

r=1

Γ

rj

(iω −c

j

)

r

for some constants Γ

rj

∈ C.

By linearity of the (inverse) Fourier transform, we can find

O

(

t

) if we know

the inverse Fourier transform of all functions of the form

1

(iω −α)

p

.

To compute this, there are many possible approaches. We can try to evaluate

the integral directly, learn some fancy tricks from complex analysis, or cheat,

and let the lecturer tell you the answer. Consider the function

h

0

(t) =

(

e

αt

t > 0

0 otherwise

The Fourier transform is then

˜

h

o

(ω) =

Z

∞

−∞

e

−iωt

h

0

(t) dt =

Z

∞

0

e

(α−iω)t

dt =

1

iω −α

provided Re(α) < 0 (so that e

(α−iω)t

→ 0 as t → ∞). Similarly, we can let

h

1

(t) =

(

te

αt

t > 0

0 otherwise

Then we get

˜

h

1

(ω) = F[th

0

(t)] = i

d

dω

F[h

0

(t)] =

1

(iω −α)

2

.

Proceeding inductively, we can define

h

p

(t) =

(

t

p

p!

e

αt

t > 0

0 otherwise

,

and this has Fourier transform

˜

h

p

(ω) =

1

(iω −α)

p+1

,

again provided

Re

(

α

)

<

0. So the response function is a linear combination

of these functions

h

p

(

t

) (if any of the roots

c

j

have non-negative real part,

then it turns out the system is unstable, and is better analysed by the Laplace

transform). We see that the response function does indeed vanish for all

t <

0.

In fact, each term (except

h

0

) increases from zero at

t

= 0 to rise to some

maximum before eventually decaying exponentially.

Fourier transforms can also be used to solve ordinary differential equations

on the whole real line

R

, provided the functions are sufficiently well-behaved.

For example, suppose y : R → R solves

y

00

− A

2

y = −f(x)

with y and y

0

→ 0 as |x| → ∞. Taking the Fourier transform, we get

˜y(k) =

˜

f(k)

k

2

+ A

2

.

Since this is a product, the inverse Fourier transform will be a convolution of

f

(

x

)

and an object whose Fourier transform is

1

k

2

+A

2

. Again, we’ll cheat. Consider

h(x) =

1

2µ

e

−µ|x|

.

with µ > 0. Then

˜

f(k) =

1

2µ

Z

∞

−∞

e

−ikx

e

−µ|x|

dx

=

1

µ

Re

Z

∞

0

e

−(µ+ik)x

dx

=

1

µ

Re

1

ik + µ

=

1

µ

2

+ k

2

.

Therefore we get

y(x) =

1

2A

Z

∞

−∞

e

−A|x−u|

f(u) du.

So far, we have done Fourier analysis over some abelian groups. For example,

we’ve done it over

S

1

, which is an abelian group under multiplication, and

R

,

which is an abelian group under addition. We will next look at Fourier analysis

over another abelian group, Z

m

, known as the discrete Fourier transform.