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1. EE-2027 SaS, L9 1/14
Lecture 8: 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
– Linearity
– Time shifts
– Differentiation and integration
– Convolution in the frequency domain

2. EE-2027 SaS, L9 2/14
Lecture 9: Resources
Core material
SaS, O&W, C4.3, C4.4
Background material
MIT Lectures 8 and 9.

3. EE-2027 SaS, L9 3/14
Reminder: Fourier Transform
A signal x(t) and its Fourier transform X(jω) are related by
This is denoted by:
For example (1):
Remember that the Fourier transform is a density
function, you must integrate it, rather than summing up
the discrete Fourier series components
∫
∫
∞
∞−
−
∞
∞−
=
=
dtetxjX
dejXtx
tj
tj
ω
ω
π
ω
ωω
)()(
)()( 2
1
)()( ωjXtx
F
↔
ωja
tue
F
at
+
↔− 1
)(

4. EE-2027 SaS, L9 4/14
Linearity of the Fourier Transform
If
and
Then
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
)()( ωjXtx
F
↔
)()( ωjYty
F
↔
)()()()( ωω jbYjaXtbytax
F
+↔+

5. EE-2027 SaS, L9 5/14
Time Shifting
If
Then
Proof
Now replacing t by t-t0
Recognising this as
A signal which is shifted in time does not have its Fourier
transform magnitude altered, only a shift in phase.
)()( ωjXtx
F
↔
)()( 0
0 ωω
jXettx tj
F
−
↔−
∫
∞
∞−
= ωω ω
π dejXtx tj
)()( 2
1
( )∫
∫
∞
∞−
−
∞
∞−
−
=
=−
ωω
ωω
ωω
π
ω
π
dejXe
dejXttx
tjtj
ttj
)(
)()(
0
0
2
1
)(
2
1
0
)()}({ 0
0 ωω
jXettxF tj−
=−

6. EE-2027 SaS, L9 6/14
Example: Linearity & Time Shift
Consider the signal (linear sum of two
time shifted steps)
where x1(t) is of width 1, x2(t) is of width 3,
centred on zero.
Using the rectangular pulse example
Then using the linearity and time shift
Fourier transform properties
)5.2()5.2(5.0)( 21 −+−= txtxtx
ω
ωω )2/sin(2)(1 =jX
( )2.5 /2 sin( / 2) 2sin(3 / 2)
( ) j
X j e ω ω ω
ω ω
− + 
=  ÷
 
ω
ωω )2/3sin(2)(2 =jX

7. EE-2027 SaS, L9 7/14
Differentiation & Integration
By differentiating both sides of the Fourier transform
synthesis equation:
Therefore:
This is important, because it replaces differentiation in the
time domain with multiplication in the frequency
domain.
Integration is similar:
The impulse term represents the dc or average value that can
result from integration
)(
)(
ωω jXj
dt
tdx F
↔
∫
∞
∞−
= ωωω ω
π dejXj
dt
tdx tj
)(
)(
2
1
)()0()(
1
)( ωδπω
ω
ττ XjX
j
dx
t
+=∫ ∞−

8. EE-2027 SaS, L9 8/14
Example: Fourier Transform of a Step Signal
Lets calculate the Fourier transform X(jω)of x(t) = u(t), making use
of the knowledge that:
and noting that:
Taking Fourier transform of both sides
using the integration property. Since G(jω) = 1:
We can also apply the differentiation property in reverse
1)()()( =↔= ωδ jGttg
F
∫ ∞−
=
t
dgtx ττ)()(
)()0(
)(
)( ωδπ
ω
ω
ω G
j
jG
jX +=
)(
1
)( ωπδ
ω
ω +=
j
jX
( ) 1
1)(
)( =





+↔= ωπδ
ω
ωδ
j
j
dt
tdu
t
F

9. EE-2027 SaS, L9 9/14
Convolution in the Frequency Domain
With a bit of work (next slide) it can show that:
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(jω) by X(jω) to obtain Y(jω)
3. Calculate the inverse Fourier transform of Y(jω)
Multiplication in the frequency domain corresponds to
convolution in the time domain and vice versa.
)()()()(*)()( ωωω jXjHjYtxthty
F
=↔=

10. EE-2027 SaS, L9 10/14
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-jωτ
H(jω), so
∫
∞
∞−
−= τττ dthxty )()()(
∫ ∫
∞
∞−
−
∞
∞−




 −= dtedthxjY tjω
τττω )()()(
∫ ∫
∞
∞−
−
∞
∞−




 −= τττω ω
ddtethxjY tj
)()()(
)()(
)()(
)()()(
ωω
ττω
τωτω
ωτ
ωτ
jXjH
dexjH
djHexjY
j
j
=
=
=
∫
∫
∞
∞−
−
∞
∞−
−

11. EE-2027 SaS, L9 11/14
Example 1: Solving an ODE
Consider the LTI system time impulse response
to the input signal
Transforming these signals into the frequency domain
and the frequency response is
to convert this to the time domain, express as partial fractions:
Therefore, the time domain response is:
0)()( >= −
btueth bt
0)()( >= −
atuetx at
ω
ω
ω
ω
ja
jX
jb
jH
+
=
+
=
1
)(,
1
)(
))((
1
)(
ωω
ω
jajb
jY
++
=






+
−
+−
=
)(
1
)(
11
)(
ωω
ω
jbjaab
jY b≠a
( ))()()( 1
tuetuety btat
ab
−−
− −=

12. EE-2027 SaS, L9 12/14
Example 2: Designing a Low Pass Filter
Lets design a low pass filter:
The impulse response of this filter is the inverse Fourier transform
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:
Causal and non-oscillatory time domain response and performs a degree
of low pass filtering



>
<
=
c
c
jH
ωω
ωω
ω
||0
||1
)(
ω
H(jω)
ωc−ωc
t
t
deth ctjc
c π
ω
ω
ω
ω
ω
π
)sin(
)( 2
1
== ∫−
ωja
tue
F
at
+
↔− 1
)(

13. EE-2027 SaS, L9 13/14
Lecture 8: 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 do not
exist – this leads naturally onto Laplace transforms
)(
)(
ωω jXj
dt
tdx F
↔
)()()()(*)()( ωωω jXjHjYtxthty
F
=↔=
)()()()( ωω jbYjaXtbytax
F
+↔+

14. EE-2027 SaS, L9 14/14
Lecture 8: Exercises
Theory
SaS, O&W,
Q4.6
Q4.13
Q3.20, 4.20
Q4.26
Q4.31
Q4.32
Q4.33