Fourier Transforms
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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