Transcript Document

Lecture 9: Fourier Transform
Properties and Examples
3. Basis functions (3 lectures): Concept of basis
function. Fourier series representation of time
functions. Fourier transform and its properties.
Examples, transform of simple time functions.
Specific objectives for today:
• Properties of a Fourier transform
–
–
–
–
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Linearity
Time shifts
Differentiation and integration
Convolution in the frequency domain
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Lecture 9: Resources
Core material
SaS, O&W, C4.3, C4.4
Background material
MIT Lectures 8 and 9.
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Reminder: Fourier Transform
A signal x(t) and its Fourier transform X(jw) are related by

1
x(t )  2  X ( jw )e jwt dw


X ( jw )   x(t )e  jwt dt

This is denoted by:
F
x(t )  X ( jw )
For example (1):
1
e u (t ) 
a  jw
 at
F
Remember that the Fourier transform is a density
function, you must integrate it, rather than summing up
the discrete Fourier series components
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Linearity of the Fourier Transform
If
F
x(t )  X ( jw )
F
and
Then
y (t )  Y ( jw )
F
ax(t )  by(t )  aX ( jw )  bY ( jw )
This follows directly from the definition of the Fourier
transform (as the integral operator is linear). It is easily
extended to a linear combination of an arbitrary number
of signals
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Time Shifting
F
If
x(t )  X ( jw )
F
Then
x(t  t0 )  e  jwt0 X ( jw )

x(t )   X ( jw )e jwt dw
Proof

Now replacing t by t-t0
1
2


  e
x(t  t0 ) 
1
2

1
2



X ( jw )e jw (t t0 ) dw
 jwt0

X ( jw ) e jwt dw
Recognising this as
F{x(t  t0 )}  e jwt0 X ( jw)
A signal which is shifted in time does not have its Fourier
transform magnitude altered, only a shift in phase.
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Example: Linearity & Time Shift
Consider the signal (linear sum of two time
shifted steps)
x(t )  0.5x1 (t  2.5)  x2 (t  2.5)
where x1(t) is of width 1, x2(t) is of width 3,
centred on zero.
Using the rectangular pulse example
X 1 ( jw )  2 sin(w / 2)
w
X 2 ( jw )  2 sin(3w / 2)
w
Then using the linearity and time shift
Fourier transform properties
X ( jw )  e  j 5w / 2  sin(w / 2)  2 sin( 3w / 2)  
w

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Differentiation & Integration
By differentiating both sides of the Fourier transform
synthesis equation:
dx(t ) 1
 2
dt


jwX ( jw)e jwt dw

Therefore:
dx (t ) F
 jwX ( jw )
dt
This is important, because it replaces differentiation in the
time domain with multiplication in the frequency
domain.
Integration is similar:

t

x( )d 
1
X ( jw )  X (0) (w )
jw
The impulse term represents the dc or average value that can
result from integration
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Example: Fourier Transform of a Step Signal
Lets calculate the Fourier transform X(jw)of x(t) = u(t), making use
of the knowledge that:
F
g (t )   (t )  G( jw )  1
and noting that:
t
x(t )   g ( )d

Taking Fourier transform of both sides
X ( jw ) 
G ( jw )
 G (0) (w )
jw
using the integration property. Since G(jw) = 1:
X ( jw ) 
1
  (w )
jw
We can also apply the differentiation property in reverse
 1

du(t ) F
 (t ) 
 jw
  w   1
dt
 jw

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Convolution in the Frequency Domain
With a bit of work (next slide) it can show that:
F
y (t )  h(t ) * x(t )  Y ( jw )  H ( jw ) X ( jw )
Therefore, to apply convolution in the frequency
domain, we just have to multiply the two functions.
To solve for the differential/convolution equation using
Fourier transforms:
1. Calculate Fourier transforms of x(t) and h(t)
2. Multiply H(jw) by X(jw) to obtain Y(jw)
3. Calculate the inverse Fourier transform of Y(jw)
Multiplication in the frequency domain corresponds to
convolution in the time domain and vice versa.
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Proof of Convolution Property

y(t )   x( )h(t   )d

Taking Fourier transforms gives:



Y ( jw )     x( )h(t   )d e  jwt dt
 


Interchanging the order of integration, we have



Y ( jw )   x( )  h(t   )e jwt dt d
 


By the time shift property, the bracketed term is e-jwH(jw), so
Y ( jw ) 

 jw
x
(

)
e
H ( jw )d



 H ( jw )  x( )e  jw d

 H ( jw ) X ( j w )
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Example 1: Solving an ODE
Consider the LTI system time impulse response
h(t )  ebt u(t )
b0
to the input signal
x(t )  eat u(t )
a0
Transforming these signals into the frequency domain
H ( jw ) 
1
,
b  jw
X ( jw ) 
1
a  jw
and the frequency response is
1
Y ( jw ) 
(b  jw )(a  jw )
to convert this to the time domain, express as partial fractions:

1 
1
1


Y ( jw ) 


b  a  (a  jw ) (b  jw ) 
ba
Therefore, the time domain response is:
y(t )  b1a eat u(t )  ebt u(t )
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Example 2: Designing a Low Pass Filter
Lets design a low pass filter:
1 | w | wc
H ( jw )  
0 | w | wc
H(jw)
wc
wc
w
The impulse response of this filter is the inverse Fourier transform
w c j wt
sin(wc t )
1
h(t )  2  e dw 
wc
t
which is an ideal low pass filter
– Non-causal (how to build)
– The time-domain oscillations may be undesirable
How to approximate the frequency selection characteristics?
Consider the system with impulse response:
F
1
 at
e u (t ) 
a  jw
Causal and non-oscillatory time domain response and performs a degree
of low pass filtering
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Lecture 9: 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:
F
1. Linearity and time shifts ax(t )  by(t )  aX ( jw )  bY ( jw )
2. Differentiation
dx (t ) F
 jwX ( jw )
dt
F
3. Convolution
y (t )  h(t ) * x(t )  Y ( jw )  H ( jw ) X ( jw )
Some operations are simplified in the frequency domain, but there
are a number of signals for which the Fourier transform do not
exist – this leads naturally onto Laplace transforms
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Lecture 9: Exercises
Theory
SaS, O&W,
Q4.6
Q4.13
Q3.20, 4.20
Q4.26
Q4.31
Q4.32
Q4.33
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