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 .
Objectives :
1. Properties of a Fourier transform
– Linearity & time shifts
–
Differentiation
– Convolution in the frequency domain
2. Understand why an ideal low pass filter cannot be manufactured
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Core material
• SaS, O&W, Chapter 4.3, C4.4
• SaS, HvV, Chapter 3.6
• SaLSA, C, Chapter 5.4, 6.1
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 : j w
X ( j w
)
• Convolution in the time domain is given by multiplication in the frequency domain (similar idea to log transformations)
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A CT signal x ( t ) and its frequency domain, Fourier transform signal,
X ( j w
), are related by
X ( j w
)
 


 x ( t ) e
 j w t dt analysis


 x ( t )

This is denoted by: x ( t )
F

For example:
X e
 at u ( t )
F

(
1
2
 j a w

)
1

X ( j w j w
) e j w t d w synthesis
Often you have tables for common Fourier transforms
The Fourier transform, X ( j w
), 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)
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The Fourier transform is a linear function of x ( t ) x t
1
F

X
1
( j w
) x t
2
F

X
2
( j w
) ax t
1

2
( )
F
 aX
1
( j w
)
 bX
2
( j w
)
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
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We’ll show that a Fourier transform of a signal which has a simple time shift is:
F { x ( t
 t
0
)}
 e
 j w t
0 X ( j w
) i.e. the original Fourier transform but shifted in phase by – w t
0
Proof
Consider the Fourier transform synthesis equation:

(
 t
0
)


2
1

2
1

2
1

 

( w
) e d w
 

( w
) e j w
( t t
0
) d w
 

 e
 w
0 ( w
)
 e d w but this is the synthesis equation for the Fourier transform e j w
0 t X ( j w
)
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Consider the signal (linear sum of two time shifted rectangular pulses) x ( t )

0 .
5 x
1
( t

2 .
5 )
 x
2
( t

2 .
5 ) x
1
( t ) where x
1
( t ) is of width 1, x
2
( t ) is of width 3, centred on zero (see figures)
Using the FT of a rectangular pulse L10S7
X
1
( j w
)
 2 sin( w
/ 2 ) w
X
2
( j w
)
 2 sin( 3 w
/ 2 ) w
Then using the linearity and time shift
Fourier transform properties
X ( j w
)
 e
 j 5 w
/ 2
 sin( w
/ 2 )

2 sin( 3 w
/ 2 )
 w x
2
( t ) x ( t ) t t t
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By differentiating both sides of the Fourier transform synthesis equation with respect to t : dt

2
1



 w
( w
) e d w
Therefore noting that this is the synthesis equation for the
Fourier transform j w
X ( j w
) dx ) F
 dt
( t j w
X ( j w
)
This is very important, because it replaces differentiation in the time domain with multiplication (by j w
) in the frequency domain .
We can solve ODEs in the frequency domain using algebraic operations (see next slides)
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We can easily solve ODEs in the frequency domain: y ( t )
 h ( t ) * x ( t )
F

Y ( j w
)

H ( j w
) X ( j w
)
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:
1. Calculate Fourier transforms of x ( t ) and h ( t ): X ( j w
) by H ( j w
)
2. Multiply H ( j w
) by X ( j w
) to obtain Y ( j w
)
3. Calculate the inverse Fourier transform of Y ( j w
)
H ( j w ) 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
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Calculate the response of a CT LTI system with impulse response: h ( t )
 e
 bt u ( t ) b

0 to the input signal: x ( t )
 e
 at u ( t ) a

0
Taking Fourier transforms of both signals:
H ( j w
)
 b

1 j w
, X ( j w
)
 a

1 j w gives the overall frequency response:
Y ( j w
)

( b
 j w
1
)( a
 j w
) to convert this to the time domain, express as partial fractions :
Y ( j w
)

1 b
 a


( a

1 j w
)

( b

1 j w
)

 assume b
 a
Therefore, the CT system response is: y ( t )
 b
1
 a
 e
 at u ( t )
 e
 bt u ( t )

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Consider an ideal low pass filter in frequency domain:
( w
)
1 | w w c
 0 | w w c
H ( j w
)
( w
)

( w
) | w w c
0 | w w c
w c w c w
The filter’s impulse response is the inverse Fourier transform h ( t )

1
2

 w c
 w c e j w t d w  sin( w c
 t t ) h ( t )
0 t 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 1 st order LTI system impulse response { h ( t ), H ( j w
)}:
EE-2027 SaS 06-07, L11 a

1

 t

( )

( ),
 at e ( )
F

1 a
 j w
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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 ( j w
)
 bY ( j w
)
2. Differentiation
3. Convolution dx ( dt t ) F
 y ( t )
 j w
X ( j w
) h ( t ) * x ( t )
F

Y ( j w
)

H ( j w
) X ( j w
)
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.
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Theory
1.
Using linearity & time shift calculate the Fourier transform of x ( t )

5 e

3 ( t

1 ) u ( t

1 )

7 e

3 ( t

2 ) u ( t

2 )
2.
Use the FT derivative relationship (S7) and the Fourier series/transform expression for sin( w
0 t ) (L10-S3) to evaluate the FT of cos( w
0 t ).
3.
Calculate the FTs of the systems’ impulse responses a)
 y ( t
 t
)

3 y ( t )
 x ( t ) b)
3

 t

( )

( )
4.
Calculate the system responses in Q3 when the following input signal is applied x ( t )
 e

5 t u ( t )
Matlab/Simulink
5.
Verify the answer to Q1 using the Fourier transform toolbox in Matlab
6.
Verify Q3 and Q4 in Simulink
7.
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)?
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This will be combined with the Laplace Tutorial L16
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y ( t )
 


 x ( t
) h ( t
 t
) d t
Taking Fourier transforms gives:
Y ( j w
)



 



 x ( t
) h ( t
 t
) d t e
 j w t dt
Interchanging the order of integration, we have

Y ( j w
)






 x ( t
)



 h ( t
 t
) e
 j w t dt d t
By the time shift property, the bracketed term is e

Y ( j w
)
 x ( t
) e
 j wt
H ( j w
) d t
j wt
H ( j w
), so


H (
H ( j w
)



 x ( t j w
) X (
) e
 j w
) j wt d t
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