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ENGR 3324: Signals and Systems
Ch6
Continuous-Time Signal Analysis
Engineering and Physics
University of Central Oklahoma
Dr. Mohamed Bingabr
Outline
• Introduction
• Fourier Series (FS) representation of
Periodic Signals.
• Trigonometric and Exponential Form of FS.
• Gibbs Phenomenon.
• Parseval’s Theorem.
• Simplifications Through Signal Symmetry.
• LTIC System Response to Periodic Inputs.
Sinusoidal Wave and phase
x(t) = Asin(t) = Asin(250t)
x(t)
A
t
T0 = 20 msec
x(t-0.0025)= Asin(250[t-0.0025])
= Asin(250t-0.25)= Asin(250t-45o)
A
t
td = 2.5 msec
Time delay td = 25 msec correspond to phase shift =45o
Representation of Quantity using Basis
• Any number can be represented as a
linear sum of the basis number {1, 10,
100, 1000}
Ex: 10437 =10(1000) + 4(100) + 3(10) +7(1)
• Any 3-D vector can be represented as a
linear sum of the basis vectors {[1 0 0],
[0 1 0], [0 0 1]}
Ex: [2 4 5]= 2 [1 0 0] + 4[0 1 0]+ 5[0 0 1]
Basis Functions for Time Signal
• Any periodic signal x(t) with fundamental frequency
0 can be represented by a linear sum of the basis
functions {1, cos(0t), cos(20t),…, cos(n0t),
sin(0t), sin(20t),…, sin(n0t)}
Ex:
x(t) =1+ cos(2t)+ 2cos(2 2t)+ 0.5sin(23t)+ 3sin(2t)
x(t) =1+ cos(2t)+ 2cos(2 2t)+ 3sin(2t)+ 0.5sin(23t)
+
+
+
=
Purpose of the Fourier Series (FS)
FS is used to find the frequency components and
their strengths for a given periodic signal x(t).
The Three forms of Fourier Series
• Trigonometric Form
• Compact Trigonometric (Polar) Form.
• Complex Exponential Form.
Trigonometric Form
• It is simply a linear combination of sines and
cosines at multiples of its fundamental
frequency, f0=1/T.


n 1
n 1
xt   a0   an cos2f 0 nt    bn sin 2f 0 nt 
• a0 counts for any dc offset in x(t).
• a0, an, and bn are called the trigonometric
Fourier Series Coefficients.
• The nth harmonic frequency is nf0.
Trigonometric Form
• How to evaluate the Fourier Series Coefficients
(FSC) of x(t)?


n 1
n 1
xt   a0   an cos2nf 0t    bn sin 2nf 0t 
To find a0 integrate both side of the equation over a full period
1
a0   xt dt
T0 T0
Trigonometric Form


n 1
n 1
xt   a0   an cos2nf 0t    bn sin 2nf 0t 
To find an multiply both side by cos(2mf0t) and then integrate
over a full period, m =1,2,…,n,…
2
an   xt  cos2nf 0t dt
T0 T0
To find bn multiply both side by sin(2mf0t) and then integrate
over a full period, m =1,2,…,n,…
2
bn   xt sin 2nf 0t dt
T0 T0
Example

f t   a0   an cos2nt   bn sin 2nt 
f(t)
1
n 1
e-t/2

0

• Fundamental period
T0 = 
• Fundamental
frequency
f0 = 1/T0 = 1/ Hz
0 = 2/T0 = 2 rad/s
a0 
an 
1

2



0


0




2
2
e dt    e  1  0.504


e

t
2

t
2
 2 
cos2nt  dt  0.504 
2 
 1  16n 
 8n 
bn   e sin 2nt  dt  0.504 
2 
0

 1  16n 
an and bn decrease in amplitude as n  .
2


t
2

2


cos2nt   4n sin 2nt 
f t   0.5041  
2
 n 1 1  16n

To what value does the FS converge at the point of discontinuity?
Dirichlet Conditions
•
A periodic signal x(t), has a Fourier series if
it satisfies the following conditions:
1. x(t) is absolutely integrable over any period,
namely

x(t ) dt  
T0
2. x(t) has only a finite number of maxima and
minima over any period
3. x(t) has only a finite number of
discontinuities over any period
Compact Trigonometric Form
• Using single sinusoid,
xt  
C0

dc component

  Cn cos2nf 0t   n 









n 1
nth harmonic
C0  a0
• Cn , and  n are related to the trigonometric coefficients an
and bn as:
Cn  an  bn
2
2
and
 bn 
 n   tan  
 an 
1
The above relationships are obtained from the
trigonometric identity
a cos(x) + b sin(x) = c cos(x + )
Role of Amplitude in Shaping Waveform

xt   C0   Cn cos2nf 0t   n 
n 1
Role of the Phase in Shaping a
Periodic Signal

xt   C0   Cn cos2nf 0t   n 
n 1
Compact Trigonometric

f t   C0   Cn cos2nt   n 
f(t)
n 1
1
a0  0.504
e-t/2

0
 2 
an  0.504 
2 
 1  16n 
 8n 
bn  0.504 
2 
 1  16n 
C0  ao  0.504

• Fundamental period
T0 = 
• Fundamental frequency
f0 = 1/T0 = 1/ Hz
0 = 2/T0 = 2 rad/s

f t   0.504  0.504
n 1

2

Cn  a  b  0.504
2
 1  16n
1   bn 
   tan 1 4n
 n  tan 
 an 
2
n
2
1  16n
2
2
n

cos 2nt  tan 1 4n





Line Spectra of x(t)
• The amplitude spectrum of x(t) is defined
as the plot of the magnitudes |Cn|
versus 
• The phase spectrum of x(t) is defined as
the plot of the angles Cn  phase(Cn )
versus 
• This results in line spectra
• Bandwidth the difference between the
highest and lowest frequencies of the
spectral components of a signal.
Line Spectra
f(t)

2

Cn  0.504
2
1

16
n

C0  0.504
1
e-t/2

 n   tan 1 4n

0





f t   0.504  0.504
n 1
2
1  16n
2

cos 2nt  tan 1 4n

f(t)=0.504 + 0.244 cos(2t-75.96o) + 0.125 cos(4t-82.87o) +
o) + 0.063 cos(8t-86.24o) + …
0.084
cos(6t-85.24
C
n
n
0.504
0.244

0.125
0.084
0
2
4
6
0.063
8
10

-/2
Line Spectra

f t   0.504  0.504
n 1
2
1  16n 2

cos 2nt  tan 1 4n

f(t)=0.504 + 0.244 cos(2t-75.96o) + 0.125 cos(4t-82.87o) +
o) + 0.063 cos(8t-86.24o) + …
0.084
cos(6t-85.24
C
n
n
0.504
0.244

0.125
0.084
0
2
4
6
0.063
8
10

-/2
HW8_Ch6: 6.1-1 (a,d), 6.1-3, 6.1-7(a, b, c)
Exponential Form
• x(t) can be expressed as
xt  

j 2f 0 nt
D
e
 n
n  
 j 2f 0 nt
To find Dn multiply both side by e
over a full period, m =1,2,…,n,…
and then integrate
1
Dn   xt e  j 2f 0 nt dt , n  0,  1,  2,....
To To
Dn is a complex quantity in general Dn=|Dn|ej
D-n = Dn*
|Dn|=|D-n|
Even
Dn = -
D-n
Odd
D0 is called the constant or dc component of x(t)
Line Spectra of x(t) in the Exponential
Form
• The line spectra for the exponential form has
negative frequencies because of the
mathematical nature of the complex exponent.
x(t )  ... | D 2 | e  j 2 e  j 20t  | D1 | e  j1 e  j0t  D0 
| D1 | e j1 e j0t  | D2 | e j 2 e j 20t  ...
x(t )  C0  C1 cos(0t  1 )  C2 cos( 20t   2 )  ...
|Dn| = 0.5 Cn
Dn =
Cn
Example
Find the exponential Fourier Series for the squarepulse periodic signal.
f(t)
 /2
1
1
 jnt
Dn 
e
dt

2  / 2
sin n / 2

 0.5sinc( n / 2)
n
1
D0 
2
n even
0
Dn  
 1 / n n odd
 0
n  
 
for all n  3,7,11,15, 
n  3,7,11,15,
2
 /2
/2

2
• Fundamental period
T0 = 2
• Fundamental frequency
f0 = 1/T0 = 1/2 Hz
0 = 2/T0 = 1 rad/s
Exponential Line Spectra
|Dn|
1
1
Dn
1
1
Example
The compact trigonometric Fourier Series
coefficients for the square-pulse periodic signal.
f(t)
1
C0 
2
 0 n even
Cn   2
n odd
 n
 0 for all n  3,7,11,15, 
n  
n  3,7,11,15,
 
1
2
 /2
/2

2
Relationships between the Coefficients
of the Different Forms
Dn  0.5an  jbn 
D n  D n  0.5an  jbn 

Dn  0.5Cn  n  0.5Cn e
D0  a0  C0
j n
Relationships between the Coefficients
of the Different Forms
an  Dn  D n  2 ReDn 
bk  j Dn  D n   2 ImDn 
an  Cn cos n 
bn  Cn sin  n 
a0  D0  c0
Relationships between the Coefficients
of the Different Forms
Cn  an  bn
2
2
 bn 
 n   tan  
 an 
Cn  2 Dn
1
 n  Dn
C0  a0  D0
Example
Find the exponential Fourier Series and sketch the
corresponding spectra for the impulse train shown
below. From this result sketch the trigonometric
spectrum and write the trigonometric Fourier Series.
 T (t )
Solution
0
Dn  1 / T0
1
 T0 (t ) 
T0

jn0t
e

n  
Cn  2 | Dn | 2 / T0
C0 | D0 | 1 / T0
1
 T0 (t ) 
T0



1  2 cos( n0t )
n 1


-2T0 -T0
T0
2T0
Rectangular Pulse Train Example
Clearly x(t) satisfies the Dirichlet conditions.
x(t)
1
2
 /2
/2
The compact trigonometric form is


2

1  2


( n 1) / 2
x(t )    cos nt  (1)
1 
2 n 1 n
2

n odd
Does the Fourier series converge to x(t) at every point?
Gibbs Phenomenon
• Given an odd positive integer N, define the
N-th partial sum of the previous series


1 N 2


( n 1) / 2
x N (t )    cos nt  (1)
1 
2 n 1 n
2

n odd
• According to Fourier’s theorem, it should be
lim | xN (t )  x(t ) | 0
N 
Gibbs Phenomenon – Cont’d
x3 (t )
x9 (t )
Gibbs Phenomenon – Cont’d
x21 (t )
x45 (t )
overshoot: about 9 % of the signal magnitude
(present even if N  )
Parseval’s Theorem
• Let x(t) be a periodic signal with period T
• The average power P of the signal is defined as
1
P
T

T /2
T / 2
2
x(t ) dt
• Expressing the signal as

xt   C0   Cn cos( n0t   n )
n 1
it is also

P  C0   0.5Cn
2
n 1
2

P  D  2 Dn
2
0
n 1
2
Simplifications Through Signal Symmetry
• If x (t) is EVEN: It must contain DC and
Cosine Terms. Hence bn = 0, and Dn =
an/2.
• If x(t) is ODD: It must contain ONLY
Sines Terms. Hence a0 = an = 0, and
Dn=-jbn/2.
LTIC System Response to Periodic
Inputs
e
j 0 t
H(s)
H(j)
H ( j0 )e j0t
A periodic signal x(t) with period T0 can be expressed as

x(t ) 
jn0 t
D
e
 n
n  
For a linear system
x(t ) 

jn0 t
D
e
 n
n  
H(s)
H(j)
y (t ) 

jn0t
D
H
(
jn

)
e
 n
0
n  
Fourier Series Analysis of DC Power
Supply
A full-wave rectifier is used to obtain a dc signal from a
sinusoid sin(t). The rectified signal x(t) is applied to the
input of a lowpass RC filter, which suppress the timevarying component and yields a dc component with
some residual ripple. Find the filter output y(t). Find
also the dc output and the rms value of the ripple
voltage.
R=15
sint
Full-wave
rectifier
x(t)
C=1/5 F
y(t)
Fourier Series Analysis of Full-Wave
Rectifier
2
Dn 
 (1  4n 2 )
D0  2 / 

2
j 2 nt
x(t )  
e
2

(
1

4
n
)
n  
1
H ( j ) 
3 j  1
y (t ) 

 D H ( jn )e
n  

n
0
jn0t
PDC  4 /  2

Pripple  2 | Dn |2
n 1
Dn 
2
 (1  4n ) 36n  1
2
2
Pripple  0.0025
ripple rms  Pripple  0.05
2
j 2 nt
y (t )  
e
2
Ripple rms is only 5%

(
1

4
n
)( j 6n  1)
n  
of the input amplitude
HW9_Ch6: 6.3-1(a,d), 6.3-5, 6.3-7, 6.3-11, 6.4-1, 6.4-3
Fourier Series Analysis of Full-Wave
Rectifier- Matlab Code
clear all
t=0:1/1000:3*pi;
for i=1:100
n=i;
yp=(2*exp(j*2*n*t))/(pi*(1-4*n^2)*(j*6*n+1));
This Matlab code will
n=-i;
plot y(t) for -100  n 
yn=(2*exp(j*2*n*t))/(pi*(1-4*n^2)*(j*6*n+1));
100 and find the ripple
y(i,:)=yp+yn;
power according to the
end
equations below

yf = 2/pi + sum(y);
2
j 2 nt
y (t ) 
e
plot(t,yf, t, (2/pi)*ones(1,length(yf)))
2

(
1

4
n
)( j 6n  1)
n


axis([0 3*pi 0 1]);


Pripple  2 | Dn |2  0.0025
Power=0;
for n=1:50
Power(n) = abs(2/(pi*(1-4*n^2)*(j*6*n+1)));
end
TotalPower = 2*sum((Power.^2));
figure; stem( Power(1,1:20));
n 1