Chapter 5
The Fourier Transform
Basic Idea
• We covered the Fourier Transform which to represent
periodic signals
• We assumed periodic continuous signals
• We used Fourier Series to represent periodic continuous
time signals in terms of their harmonic frequency
components (Ck).
• We want to extend this discussion to find the frequency
spectra of a given signal
Basic Idea
• The Fourier Transform is a method for representing signals
and systems in the frequency domain
• We start by assuming the period of the signal is T= INF
• All physically realizable signals have Fourier Transform
• For aperiodic signals Fourier Transform pairs is described
as
Fourier
Transforms of f(t)
Inverse Fourier
Transforms of F(w)
Remember:
notes
Example – Rectangular Signal
• Compute the Fourier Transform
of an aperiodic rectangular pulse
of T seconds evenly distributed
about t=0.
V
-T/2
T/2
f (t )  Vu(t  T / 2)  Vu(t  T / 2)

• Remember this the same
rectangular signal as we worked
before but with T0 infinity!
F (w )   f (t )e  jwt dt

 TV sin c(wT / 2)
notes
thus

Vrect (t / T ) 
TV sin c(wT / 2)
All physically realizable signals have Fourier Transforms
Fourier Transform of Unit Impulse Function
f (t )  A (t  t0 )

F (w )   f (t )e  jwt dt


  A (t  t0 )e  jwt dt  Ae jwt0

if : f (t )  A (t  0)  F (w )  A
Example:
F (w )   (w  w0 )
1 
 {F (w )}  f (t ) 
F (w )e jwt dw

2 
1 

 (w  w0 )e jwt dw

2 
e jw0t

2
thus
thus

 (t ) 
A
1

e jw0t 
2 (w  w0 )
Plot magnitude and phase of f(t)
Fourier Series Properties
Make sure how to use these properties!
Fourier Series Properties - Linearity
f (t )  B cos(w0t )
 B / 2( e j w 0 t  e  j w 0 t )
B
B
 {e jw0t }   {e  jw0t }
2
2
Find F(w)
Re member : e jw0t 
2 (w  w0 )
 { f (t )}  F (w ) 

B
B
2 (w  w0 )  2 (w  w0 )
2
2
Fourier Series Properties - Linearity
f (t )  B cos(w0t )
 B / 2( e j w 0 t  e  j w 0 t )
B
B
 {e jw0t }   {e  jw0t }
2
2

Re member : e jw0t 
2 (w  w0 )
 { f (t )}  F (w ) 

B
B
2 (w  w0 )  2 (w  w0 )
2
2
Due to linearity
Fourier Series Properties - Time Scaling
g (t )  rect (2t / T1 )
Re member :

f (t )  Vrect (t / T1 ) 
F (w )  T1V sin c(T1w / 2)
Thus
rect(t/T)

g (t ) 
G ( w)
g (t ) 
1

f (at ) 
G ( w)  1 / | a | F (w / a )
V
a 2
Then
G (w ) 
1 1
( )(T1V ) sin c(T1w / 4)
V 2
T
 ( 1 ) sin c(T1w / 4)
2
rect(t/(T/2))
Due to Time
Scaling Property
Remember:
sinc(0)=1;
sinc(2pi)=0=sinc(pi)
Fourier Series Properties - Duality or Symmetry
Example:
Find the time-domain waveform for

if : f (t ) 
F (w )
F (t )  2f (w )

 {F (t )}  2f (w )
where : f (w )  f (t ) t  w
F (w )  Au (w  B )  Au (w  B )
Re member :

Vrect (t / T ) 
TV sin c(wT / 2)
We _ have :
Arect(w/2B)
F (w )  Au (w  B )  Au (w  B )  Arect (w / 2 B )
U sin g _ Duality _ find _ F (t )
2f (w )  2Arect (w / T )
 2Arect (w / 2 B )
Thus
Remember we had:
F (t )  2 BA sin c( Bt )
Refer to FTP
Table

BA sin c( Bt ) 
Arect (w / 2 B )
FTP: Fourier Transfer Pair
Fourier Series Properties - Duality or Symmetry
Example: find the frequency response
Of y(t)
Fourier Series Properties - Duality or Symmetry
Example: find the frequency response
Of y(t)
We know
Using Fourier Transform Pairs
Using duality

if : f (t ) 
F (w )

F (t ) 
2f (w )
 {F (t )}  2f (w )
where : f (w )  f (t ) t  w
Fourier Series Properties - Convolution
Proof
Proof
Fourier Series Properties - Convolution
Example:
Find the Fourier Transform of x(t)=sinc2(t)
Re member :
BA sin c( Bt )  Arect (w / 2 B )


sin c(t ) 
rect (w / 2)
f (t )  u (t  1)  u (t  1)

sin c(t ) 
rect (w / 2)

F (w )   f (t )e  jwt dt

 TV sin c(wT / 2)
thus

Vrect (t / T ) 
TV sin c(wT / 2)
w
X1(w)
w
X2(w)
Refer to Schaum’s
Prob. 2.6
In this case we
have B=1, A=1
Fourier Series Properties - Convolution
Example:
Find the Fourier Transform of x(t)= sinc2(t) sinc(t)
We need to find the convolution of a rect and a triangle function:
w
Refer to Schaum’s
Prob. 2.6
Fourier Series Properties - Frequency Shifting

x(t )e jw0t 
X (w  w0 )
Example:
Find the Fourier Transform of g3(t) if g1(t)=2cos(200t), g2(t)=2cos(1000t);
g3(t)=g1(t).g2(t) ; that is [G3(w)]
g 3 (t )  5 cos( 200t )e j1000t  5 cos( 200t )e  j1000t
 G3 (w )  5 (w  200  1000 )  5 (w  200  1000 )
 5 (w  200  1000 )  5 (w  200  1000 )
 5 (w  800 )  5 (w  1200 )
 5 (w  1200 )  5 (w  800 )
Remember: cosa . cosb=1/2[cos(a+b)+cos(a-b)]
Fourier Series Properties - Time Differentiation

f (t ) 
F (w )
Example:

g (t )  df (t ) / dt 
jwF (w )
Also
t
1
g (t )   f ( )d 
F (w )  F (0) (w )  G (w )
j
w


Note

F ( 0) 
 f (t )e

 jw t

dt   f (t )dt

f (t )  sgn( t )
g (t )  u (t )  dg (t ) / dt   (t )
df (t ) / dt  2 (t )

 (t ) 
1
thus

sgn( t )  df (t ) / dt 
1
More…
• Read your notes for applications of Fourier Transform.
• Read about Power Spectral Density
• Read about Bode Plots
Schaums’ Outlines Problems
• Schaum’s Outlines:
– Do problems 5.16-5.43
– Do problems 5.4, 5.5, 5.6. 5.7, 5.8, 5.9, 5.10, 5.14
• Do problems in the text