1

Fourier Transforms

Background

While the Fourier series/transform is very important for representing a signal in the frequency domain, it is also important for calculating a system’s response (convolution).

• A system’s transfer function is the Fourier transform of its impulse response

• Fourier transform of a signal’s derivative is multiplication in the frequency domain: jX(j)

• Convolution in the time domain is given by multiplication in the frequency domain (similar idea to log transformations)

3

Fourier Series in exponential formConsider the Fourier series of the 2T periodic function:

Due to the Euler formula

It can be rewritten as

With the decomposition coefficients calculated as:

(1)€

˜ f (x) =a0

2+ (an cos

nxπ

T+ bn sin

nxπ

T)

n =1

∞

∑

€

e iθ = cosθ + isinθ

€

˜ f (x) = cneinx

n =−∞

∞

∑

€

cn =1

2Te

−inπ

Ttf (t)

−T

T

∫ (2)

4

Fourier transformThe frequencies are and

Therefore (1) and (2) are represented as

Since, on one hand the function with period T has also the periods kT for any integer k, and

on the other hand any non-periodic function can be considered as a function with infinite

period, we can run the T to infinity, and obtain the Riemann sum with ∆w→∞, converging to

the integral:

(3)€

wn =nπ

T

€

˜ f (x) =1

2πe iwx e−iwt f (t)dt

−T

T

∫ ⎛

⎝ ⎜

⎞

⎠ ⎟

n =−∞

∞

∑ Δw€

Δw =π

T

€

˜ f (x) =1

2πe iwx e−iwt f (t)dt

−T

T

∫ ⎛

⎝ ⎜

⎞

⎠ ⎟dw

−∞

∞

∫ (4)

5

Fourier transform definitionThe integral (4) suggests the formal definition:

The funciotn F(w) is called a Fourier Transform of function f(x) if:

The function

Is called an inverse Fourier transform of F(w). €

F(w) := F{ f (t)} := e−iwt f (t)dt−∞

∞

∫

€

F −1{F(w)} :=1

2πe iwxF(w)dw

−∞

∞

∫ (6)

(5)

6

Example 1The Fourier transform of

is

The inverse Fourier transform is€

f (t) :=1

0

⎧ ⎨ ⎩

| t |≤1

| t |>1

€

F{ f (t)} = e−iwtdt−1

1

∫ = 2sinw

ww ≠ 0

2 w = 0

⎧ ⎨ ⎪

⎩ ⎪

€

2

2πe iwx sinw

wdw

−∞

∞

∫ =2

πcoswx

sinw

wdw

0

∞

∫ =

=1

π

sin(w(x +1))

wdw −

0

∞

∫ 1

π

sin(w(x −1))

wdw =

1 | x |<1

1/2 x =1

0 x >1

⎧

⎨ ⎪

⎩ ⎪0

∞

∫

7

Fourier IntegralIf f(x) and f’(x) are piecewise continuous in every finite interval, and f(x) is absolutely

integrable on R, i.e.

converges, then

Remark: the above conditions are sufficient, but not necessary.

€

1

2[ f (x−) + f (x+)] =

1

2πe iwx e−iwt f (t)dt

−∞

∞

∫ ⎛

⎝ ⎜

⎞

⎠ ⎟dw

−∞

∞

∫

8

Properties of Fourier transform1 Linearity:

For any constants a, b the following equality holds:

Proof is by substitution into (5).

2 Scaling:

For any constant c, the following equality holds:

€

F{af (t) + bg(t)} = aF{ f (t)} + bF{g(t)} = aF(w) + bG(w)

€

F{ f (ct)} =1

| c |F(

w

c)

9

Properties of Fourier transform 23 Time shifting:

Proof:

4. Frequency shifting:

Proof:

€

F{ f (t − t0)} = e−iwt0 F(w)

€

F{ f (t − t0)} = f (t − t0)−∞

∞

∫ e−iwtdt = e−iwt0 f (u)−∞

∞

∫ e−iwudu

€

F{e iwt0 f (t)} = F(w − w0)

€

F{e iw0t f (t)} = e−iwt0 f (t)e−iwtdt−∞

∞

∫ = F(w − w0)

10

Properties of Fourier transform 35. Symmetry:

Proof:

The inverse Fourier transform is

therefore

€

f (t) = F −1{ f (w)} =1

2πF(w)e iwtdw

−∞

∞

∫€

F{F(t)} = 2πf (−w)

€

2πf (−w) =1

2πF(t)e−itwdt

−∞

∞

∫ = F{F(t)}

11

Properties of Fourier transform 4

6. Modulation:

Proof:

Using Euler formula, properties 1 (linearity) and 4 (frequency shifting):

€

F{ f (t)cos(w0t)} =1

2[F(w + w0) + F(w − w0)]

F{ f (t)sin(w0t)} =1

2[F(w + w0) − F(w − w0)]

€

F{ f (t)cos(w0t)} =1

2[F{e iw0t f (t)} + F{e−iw0t f (t)}]

=1

2[F(w + w0) + F(w − w0)]

12

Differentiation in time7. Transform of derivatives

Suppose that f(n) is piecewise continuous, and absolutely integrable on R. Then

In particular

and

Proof:

From the definition of F{f(n)(t)} via integrating by parts.

€

F{ f (n )(t)} = (iw)n F(w)

€

F{ f '(t)} = iwF(w)

€

F{ f ''(t)} = −w2F(w)

13

Example 2

The property of Fourier transform of derivatives can be used for solution of differential

equations:

Setting F{y(t)}=Y(w), we have

€

′ y − 4y = H(t)e−4 t

€

H(t) =0 t < 0

1 t ≥ 0

⎧ ⎨ ⎩

€

F{ ′ y } − 4F{y} = F{H(t)e−4 t} =1

4 + iw

€

iwY (w) − 4Y (w) =1

4 + iw

14

Example 2

Then

Therefore

€

Y (w) =1

(4 + iw)(−4 + iw)= −

1

16 + w2

€

y(w) = F −1{Y (w)} = −1

8e−4| t |

15

Frequency Differentiation

In particular and

Which can be proved from the definition of F{f(t)}.€

F{t n f (t)} = inF (n )(w)

€

F{tf (t)} = i ′ F (w)

€

F{t 2 f (t)} = − ′ ′ F (w)

Fourier Transform

A CT signal x(t) and its frequency domain, Fourier transform signal, X(j), are related by

This is denoted by:

For example:

Often you have tables for common Fourier transformsThe Fourier transform, X(j), represents the frequency content of x(t).

It exists either when x(t)->0 as |t|->∞ or when x(t) is periodic (it generalizes the Fourier series)

dejXtx

dtetxjX

tj

tj

)()(

)()(

21

)()( jXtxF

jatue

Fat

1

)(

analysis

synthesis

Linearity of the Fourier Transform

The Fourier transform is a linear function of x(t)

This follows directly from the definition of the Fourier transform (as the integral operator is linear) & it easily extends to an arbitrary number of signals

Like impulses/convolution, if we know the Fourier transform of simple signals, we can calculate the Fourier transform of more complex signals which are a linear combination of the simple signals

1 1

2 2

1 2 1 2

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

F

F

F

x t X j

x t X j

ax t bx t aX j bX j

Fourier Transform of a Time Shifted Signal

We’ll show that a Fourier transform of a signal which has a simple time shift is:

i.e. the original Fourier transform but shifted in phase by –t0

Proof

Consider the Fourier transform synthesis equation:

but this is the synthesis equation for the Fourier transform

e-j0tX(j)

0

0

12

( )10 2

12

( ) ( )

( ) ( )

( )

j t

j t t

j t j t

x t X j e d

x t t X j e d

e X j e d

)()}({ 00 jXettxF tj

Example: Linearity & Time Shift

Consider the signal (linear sum of two time shifted rectangular pulses)

where x1(t) is of width 1, x2(t) is of width 3, centred on zero (see figures)

Using the FT of a rectangular pulse L10S7

Then using the linearity and time shift Fourier transform properties

€

x(t) = 0.5x1(t − 2.5) + x2(t − 2.5)

€

X1( jω) = 2sin(ω /2)ω

€

X( jω) = e− j 5ω / 2 sin(ω /2) + 2sin(3ω /2)( )ω

⎛

⎝ ⎜

⎞

⎠ ⎟

€

X2( jω) = 2sin(3ω /2)ω

t

t

t

x1(t)

x2(t)

x (t)

Fourier Transform of a Derivative

By differentiating both sides of the Fourier transform synthesis equation with respect to t:

Therefore noting that this is the synthesis equation for the Fourier transform jX(j)

This is very important, because it replaces differentiation in the time domain with multiplication (by j) in the frequency domain.

We can solve ODEs in the frequency domain using algebraic operations (see next slides)

)()( jXj

dt

tdx F

12

( )( ) j tdx t

j X j e ddt

Convolution in the Frequency DomainWe can easily solve ODEs in the frequency domain:

Therefore, to apply convolution in the frequency domain, we just have to multiply the two Fourier Transforms.

To solve for the differential/convolution equation using Fourier transforms:

• Calculate Fourier transforms of x(t) and h(t): X(j) by H(j)

• Multiply H(j) by X(j) to obtain Y(j)

• Calculate the inverse Fourier transform of Y(j)

H(j) is the LTI system’s transfer function which is the Fourier transform of the impulse response, h(t). Very important in the remainder of the course (using Laplace transforms)

This result is proven in the appendix

)()()()(*)()( jXjHjYtxthtyF

Example 1: Solving a First Order ODECalculate the response of a CT LTI system with impulse response:

to the input signal:

Taking Fourier transforms of both signals:

gives the overall frequency response:

to convert this to the time domain, express as partial fractions:

Therefore, the CT system response is:

0)()( btueth bt

0)()( atuetx at

jajX

jbjH

1)(,

1)(

))((

1)(

jajbjY

)(

1

)(

11)(

jbjaabjY

assumeba

)()()( 1 tuetuety btatab

h(t)

t0

Consider an ideal low pass filter in frequency domain:

The filter’s impulse response is the inverse Fourier transform

which is an ideal low pass CT filter. However it is non-causal, so this cannot be manufactured exactly & the time-domain oscillations may be undesirable

We need to approximate this filter with a causal system such as 1st order LTI system impulse response {h(t), H(j)}:

Example 2: Design a Low Pass Filter

1 | |( )

0 | |

( ) | |( )

0 | |

c

c

c

c

H j

X jY j

H(j)

cc

t

tdeth ctjc

c

)sin()( 2

1

1 ( ) 1( ) ( ), ( )

Faty t

a y t x t e u tt a j

24

ConvolutionThe convolution of two functions f(t) and g(t) is defined as:

Theorem:

Proof:

€

f * g ≡ f (u)g(t − u)du−∞

∞

∫

€

F{ f * g} ≡ F{ f }F{g}

F{ f (t)g(t)} ≡1

2π[F *G](w)

€

F{ f * g} = e−iwt f (u)g(t − u)du−∞

∞

∫ ⎛

⎝ ⎜

⎞

⎠ ⎟

−∞

∞

∫ dt =

e−iwu f (u) e−iw(t −u)g(t − u)d(t − u)−∞

∞

∫ ⎛

⎝ ⎜

⎞

⎠ ⎟

−∞

∞

∫ du = F{ f }F{g}

Appendix: Proof of Convolution Property

Taking Fourier transforms gives:

Interchanging the order of integration, we have

By the time shift property, the bracketed term is e-jH(j), so

€

y(t) = x(τ )h(t −τ )dτ−∞

∞

∫

€

Y ( jω) = x(τ )h(t −τ )dτ−∞

∞

∫( )e− jωtdt

−∞

∞

∫

€

Y ( jω) = x(τ ) h(t −τ )−∞

∞

∫ e− jωtdt( )dτ−∞

∞

∫

€

Y ( jω) = x(τ )e− jωτ H( jω)dτ−∞

∞

∫

= H( jω) x(τ )e− jωτ dτ−∞

∞

∫

= H( jω)X( jω)

Summary

The Fourier transform is widely used for designing filters. You can design systems with reject high frequency noise and just retain the low frequency components. This is natural to describe in the frequency domain.

Important properties of the Fourier transform are:

1. Linearity and time shifts

2. Differentiation

3. Convolution

Some operations are simplified in the frequency domain, but there are a number of signals for which the Fourier transform does not exist – this leads naturally onto Laplace transforms. Similar properties hold for Laplace transforms & the Laplace transform is widely used in engineering analysis.

)()( jXj

dt

tdx F

)()()()(*)()( jXjHjYtxthtyF

)()()()( jbYjaXtbytaxF

Lecture 11: ExercisesTheory1. Using linearity & time shift calculate the Fourier transform of

2. Use the FT derivative relationship (S7) and the Fourier series/transform expression for sin(0t) (L10-S3) to evaluate the FT of cos(0t).

3. Calculate the FTs of the systems’ impulse responses

a) b)

4. Calculate the system responses in Q3 when the following input signal is applied

Matlab/Simulink1. Verify the answer to Q1 using the Fourier transform toolbox in Matlab2. Verify Q3 and Q4 in Simulink3. Simulate a first order system in Simulink and input a series of

sinusoidal signals with different frequencies. How does the response depend on the input frequency (S12)?

)()(3)(

txtyt

ty

)()( 5 tuetx t

)2(7)1(5)( )2(3)1(3 tuetuetx tt

( )3 ( ) ( )

y ty t x t

t