Transcript ELEC 360: Signals and Systems - UCO

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.
x t   a 0 

a
n 1
n
cos  2  f 0 nt  

b
n
sin  2  f 0 nt 
n 1
• 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)?
x t   a 0 

a
n
cos  2  nf 0 t  
n 1

b
n
sin  2  nf 0 t 
n 1
To find a0 integrate both side of the equation over a full period
a0 
1
T0
 x t dt
T0
Trigonometric Form
x t   a 0 

a
n
cos  2  nf 0 t  
n 1

b
n
sin  2  nf 0 t 
n 1
To find an multiply both side by cos(2mf0t) and then integrate
over a full period, m =1,2,…,n,…
an 
2
T0
 x t  cos 2 nf t dt
0
T0
To find bn multiply both side by sin(2mf0t) and then integrate
over a full period, m =1,2,…,n,…
bn 
2
T0
 x t  sin 2 nf t dt
0
T0
Example
f(t)
f t   a 0 
1
0
a
n
cos  2 nt   b n sin  2 nt 
n 1
e-t/2



• Fundamental period
T0 = 
• Fundamental
frequency
f0 = 1/T0 = 1/ Hz
0 = 2/T0 = 2 rad/s
a0 
an 
bn 
1

2

2



0

 

2
2
2
e dt    e  1   0 . 504

 



t

t
2
e
0



e
0
t
2
2


cos  2 nt  dt  0 . 504 

2
 1  16 n 
8n


sin  2 nt  dt  0 . 504 

2
 1  16 n 
a n and b n decrease in amplitude

f t   0 . 504 1 


as n   .

 1  16 n 2 cos 2 nt   4 n sin 2 nt 
n 1

2
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


cos  2  nf t   
 C 
   

n
n 1
dc component
0
n
nth harmonic
C 0  a0
• C n , and  n are related to the trigonometric coefficients an
and bn as:
Cn 
2
a n  bn
2
and
 n   tan
1
 bn 


a 
 n 
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   C 0 

C
n 1
n
cos  2  nf 0 t   n 
Role of the Phase in Shaping a
Periodic Signal
x t   C 0 

C
n 1
n
cos  2  nf 0 t   n 
Compact Trigonometric
f t   C 0 
f(t)
C
n
cos  2 nt   n 
n 1
1
e-t/2


0
a 0  0 . 504
2


a n  0 . 504 

2
 1  16 n 

• Fundamental period
T0 = 
• Fundamental frequency
f0 = 1/T0 = 1/ Hz
0 = 2/T0 = 2 rad/s
8n


b n  0 . 504 

2
 1  16 n 
C 0  a o  0 . 504
Cn 
a b
2
n
 n  tan
f t   0 . 504  0 . 504


n 1
2
1  16 n
1
2
n
  bn

 a
 n


2

 0 . 504

2
 1  16 n

   tan


cos 2 nt  tan
2
1
1
4n
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  C n  phase (C n )
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

C n  0 . 504

2
1

16
n

C 0  0 . 504
1
e-t/2

 n   tan

0
f t   0 . 504  0 . 504


n 1
2
1  16 n
1
4n

cos 2 nt  tan
2




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

2

1  16 n
n 1

cos 2 nt  tan
2
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  

D
n
n  
j 2  f 0 nt
e
 j 2  f 0 nt
To find Dn multiply both side by e
over a full period, m =1,2,…,n,…
Dn 
1
To

x t e
 j 2  f 0 nt
dt
and then integrate
, n  0 ,  1,  2 ,....
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
| D1 | e
 j 1
j 1
e
e
 j 0 t
j 0 t
 D0 
 | D2 | e
j 2
x ( t )  C 0  C 1 cos(  0 t   1 )  C 2 cos( 2  0 t   2 )  ...
|Dn| = 0.5 Cn
Dn =
Cn
e
j 2 0t
 ...
Example
Find the exponential Fourier Series for the squarepulse periodic signal.
f(t)
Dn 

D0 
Dn
 /2
1
e
2
1
 jnt
dt
 / 2
sin n  / 2
n
2
2
 0
n  
 
/2

2
 0 . 5 sinc( n  / 2 )
1
0

 1 / n
 /2
n even
n odd
for all n  3 , 7 ,11 ,15 , 
n  3 , 7 ,11 ,15 , 
• 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)
C0 
1
1
2
 0
Cn   2
  n
 0
n  
 
n even
n odd
for all n  3 , 7 ,11 ,15 , 
n  3 , 7 ,11 ,15 , 
2
 /2
/2

2
Relationships between the Coefficients
of the Different Forms
D n  0 . 5  a n  jb n 
Dn  D

n
 0 . 5  a n  jb n 
D n  0 .5C n   n  0 .5C n e
D0  a0  C 0
j n
Relationships between the Coefficients
of the Different Forms
a n  D n  D  n  2 Re D n 
b k  j  D n  D  n    2 Im D n 
a n  C n cos  n 
b n   C n sin  n 
a0  D 0  c0
Relationships between the Coefficients
of the Different Forms
Cn 
2
a n  bn
 n   tan
1
2
 bn 


a 
 n 
Cn  2 Dn
 n  Dn
C 0  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
D n  1 / T0
 T (t ) 
0
1
T0


e
jn  0 t
n  
C n  2 | D n | 2 / T 0
C 0  | D 0 | 1 / T 0

1 

 T0 ( t ) 
1  2  cos( n  0 t ) 
T0 
n 1

-2T0 -T0
T0
2T0
Rectangular Pulse Train Example
Clearly x(t) satisfies the Dirichlet conditions.
x(t)
1
2
 /2
/2

2
The compact trigonometric form is
x (t ) 
1
2



n 1
n odd


 

( n 1 ) / 2
cos  nt  (  1)
1 
n
2

2
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
x N (t ) 
1
2
N


n 1
n odd


 

( n 1 ) / 2
cos  nt  (  1)
1 
n
2

2
• According to Fourier’s theorem, it should be
lim | x N ( t )  x ( t ) |  0
N 
Gibbs Phenomenon – Cont’d
x3 (t )
x9 (t )
Gibbs Phenomenon – Cont’d
x 21 ( t )
x 45 ( 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
P 
1
T

T /2
T / 2
2
x ( t ) dt
• Expressing the signal as
x t   C 0 

C
n
cos( n  0 t   n )
n 1
it is also

2
P  C0 
 0 .5C
n 1
2
n

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
H(s)
H(j)
j 0 t
H ( j 0 ) e
j 0 t
A periodic signal x(t) with period T0 can be expressed as

x (t ) 

Dne
jn  0 t
n  
For a linear system

x (t ) 

n  
Dne
jn  0 t
H(s)
H(j)

y (t ) 
D
n  
n
H ( jn  0 ) e
jn  0 t
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  4 n )
2

x (t ) 
 (1  4 n )
2
n  
e
Pripple  2  | D n |
j 2 nt
Dn 
3 j  1
D
n
H ( jn  0 ) e
y (t ) 

n  
2
 (1  4 n ) 36 n  1
ripple rms 
2
 (1  4 n )( j 6 n  1)
2
2
2
Pripple  0 . 0025
jn  0 t
n  

2
n 1
1

y (t ) 
PDC  4 / 

2

H ( j ) 
D0  2 / 
e
Pripple  0 . 05
j 2 nt
Ripple rms is only 5%
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
2
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));
n=-i;
yn=(2*exp(j*2*n*t))/(pi*(1-4*n^2)*(j*6*n+1));
y(i,:)=yp+yn;
end
yf = 2/pi + sum(y);
y (t ) 
plot(t,yf, t, (2/pi)*ones(1,length(yf)))
axis([0 3*pi 0 1]);
This Matlab code will
plot y(t) for -100  n 
100 and find the ripple
power according to the
equations below

2

 (1  4 n )( j 6 n  1)
2
n  

Pripple  2  | D n |  0 . 0025
2
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
e
j 2 nt