Transcript No Slide Title

FOURIER ANALYSIS TECHNIQUES
LEARNING GOALS
FOURIER SERIES
Fourier series permit the extension of steady state analysis to general periodic
signal.
FOURIER TRANSFORM
Fourier transform allows us to extend the concepts of frequency domain to
arbitrary non-periodic inputs
FOURIER SERIES
The Fourier series permits the representation of an arbitrary periodic signal as
a sum of sinusoids or complex exponentials
Periodic signal
The signal f (t ) is periodic iff there exists T  0 such that
f (t )  f (t  T ), t
The smallest T that satisfies the previous condition is called the (fundamental)
period of the signal
FOURIER SERIES RESULTS
If f (t ) is periodic, with period T0 , then f (t ) can be expressed in one of the following
equivalent forms
Phasor for n-th harmonic
Cosine expansion


n 1
n 1

f (t )  a0   Dn cos(n o t   n )  a0   Re Dn ne jn o t
f (t ) 

 cne jn0t
Complex exponential expansion

e j  cos  j sin 

f (t )  a0   an cos n o t   bn sin n o t
n 1
c0  a0
2
T0
Dn n  2cn  an  jbn
n  

0 
Trigonometric series
n 1
e j  e  j  2 cos
e j  e  j  2 j sin 
Relationship between exponential and trigonometric expansions
For n  0
cne jn9t  cne  jn0t  (cn  cn ) cos n0t  j (cn  cn ) sin n0t
an
bn
 2cn  an  jbn
2cn  an  jbn
If f (t ) is real - valued then cn  cn 
GENERAL STRATEGY:
. Approximate a periodic signal using a Fourier series
. Analyze the network for each harmonic using phasors or complex exponentials
. Use the superposition principle to determine the response to the periodic signal
*
Approximation with 4 terms
Original Periodic Signal
EXAMPLE OF
QUALITY OF
APPROXIMATION
f N (t ) 
N
jn  0 t
c
e
 n
n N
Approximation with 2 terms
2
2
n 1
n 1
f 2 (t )  a0   an cos n o t   bn sin n o t
Approximation with 100 terms
100
100
n 1
n 1
a0   an cos n o t   bn sin n o t
EXPONENTIAL FOURIER SERIES
Any “physically realizable” periodic signal, with period To, can be represented over
the interval t1  t  t1  T0 by the expression
f (t ) 
n 
 cne jn0t ; 0 
n
2
T0
The sum of exponential functions is always a continuous function. Hence, the right
hand side is a continuous function.
Technically, one requires the signal, f(t), to be at least piecewise continuous. In that
case, the equality does not hold at the points where the signal is discontinuous
Computation of the exponential Fourier series coefficients
t1 T0
f (t ) 

 f ( t )e
 jk 0 t
t1
1
ck 
T0
 cne jn t
0
n  
t1
t1 T0
n
dt 
t1 T0

t1
 e  jk0t
t1 T0
 0 for n  k
j ( n  k ) 0 t
e
dt



T0 for n  k
t1
 n jn0t   jk0t
dt
  cn e
e
 n

t1 T0

f (t )e  jk 0t dt
t1
t1 is arbitray and can be chosen to make computations simpler
LEARNING EXAMPLE
Determine the exponential Fourier series
n 
f (t ) 
0

cn   2V
n
sin
 n
2
 cne jn0t ; 0 
n
n even
n
odd
1
cn 
T0
2
T0
t1 T0

f (t )e  jn0t dt
t1
A strategy :
t1  T / 2
cn 
1
T
T
2
 jn  t
 v (t )e 0 dt 
T

2
1. Determine T0 and  0
2
T0  T   0 
T

1
T
T
4


T
2
(V )e  jn 0 t dt 
T
4
2. Select a convenient t1
3. Do the integratio n
T
2
1
1
 jn  0 t
Ve
dt

Ve  jn 0 t dt


T T
TT

4
4
V
 jn 0 t T / 4
 jn 0 t T / 4
 jn 0 t T / 2  This is for n  0!

cn 
e
e
e
T / 2
T / 4
T /4
 c0  0 in this case
T ( jn 0 ) 
e j  e  j
 2 sin 
j
T0  2
 T
 T
T 
 T
T 
T 
V   jn0   4   jn0   2   jn0  4   jn0   4   jn0  2   jn0  4  
e

cn 
e
e
e
e
e
jTn0 


n
n
 V 
j
V  j2

 n 
 jn 
jn 
2
cn 
 2e
e
e  
4
sin

2
sin(
n

)
2e




j 2n 
 2 

 2n 
LEARNING EXTENSION
T0  2
Determine the exponential Fourier series
0  
n 
f (t ) 
 cne jn0t ; 0 
n
1
cn 
T0
t1  0
t1 T0

f (t )e  jn0t dt
t1
A strategy :
1. Determine T0 and  0

12
1  1  jnt 
1
cn   v (t )dt    e
dt  
 e  jnt
20
2 0
 2 jn

1
0
1  e  jn

;
2 jn
For n  0
c0 
1
2
2. Select a convenient t1
For n  0
3. Do the integratio n
2
T0
LEARNING EXTENSION
T0  6
Determine the exponential Fourier series
0 

3
n 
f (t ) 
0
n 
1
cn 
T0
t1  2
c0 
2
1
v (t )dt  2
6 2
n
n
n
1
2
j t
j t 
14
1  1  j 3 t
cn   v (t )dt    4e
dt   2e 3 dt   4e 3 dt 
6 2
6 2

1
1
n
n
n
n
2 n
 j 2 n

j
j
j
j
j
3
3
3
3
3
3
 4e
 2e
 2e
 4e
 4e
4e



2n
n 

cn  n 4 sin
 2 sin 
3
3

n
cn 
2j
 cne jn t ;
t1 T0
 f ( t )e
0 
 jn 0 t
2
T0
dt
t1
A strategy :
1. Determine T0 and  0
2. Select a convenient t1
3. Do the integratio n
e j  e  j
 2 sin 
j
TRIGONOMETRIC FOURIER SERIES


f 2 (t )  a0   an cos n o t   bn sin n o t
n 1
c0  a0
0 
n 1
2
T0
e j  cos  j sin 
e j  e  j  2 cos
e j  e  j  2 j sin 
Relationship between exponential and trigonometric expansions
For n  0
cne jn0t  cne  jn0t  (cn  cn ) cos n0t  j (cn  cn ) sin n0t
an
cn 
1
T0
t1 T0
 f ( t )e
t1
 jn 0 t
dt
bn
1
a0 
T0
t1 T0
 2cn  an  jbn
2cn  an  jbn
If f (t ) is real - valued then cn  cn 
*
 f (t )dt
The trigonometric form permits the
use of symmetry properties of the
t1
function to simplify the computation
t1 T0
2
an 
 f (t ) cos n 0t dt of coefficients
T0
2
bn 
T0
Even function symmetry
Odd function symmetry
t1
t1 T0
 f (t ) sin n0t dt

t1
f ( t )  f ( t )
f (t )   f (t )
x
0
x
x
0
 f (t )dt   f (t )dt
x

0
x
x
0
 f (t )dt    f (t )dt
TRIGONOMETRIC SERIES FOR FUNCTIONS WITH EVEN SYMMETRY
a0 
1
T0
2
an 
T0
2
bn 
T0
t1 T0
 f (t )dt
t1
t1 T0

T0
t


f (t ) cos n 0 t dt 1
2
4
an 
T0
T0
2

0
t1
t1 T0

f (t ) sin n 0 t dt
bn 
t1
f (t ) cos n 0 t dt , n  1,2,
T0
2
2
T0
2
T0
 f (t )dt
0
2cn  an  jbn
2cn  an  jbn  cn  cn  an
 f (t ) sin n 0t dt  0,

a0 
T0
2
n  1,2,
T0
2
a0  2
2n
n 

cn  n 4 sin
 2 sin   an , n  1,2,
3
3

TRIGONOMETRIC SERIES FOR FUNCTIONS WITH ODD SYMMETRY
an 
bn 
2
T0
4
T0
T0
2
 f (t ) cos n 0t dt  0,

T0
2
T0
2
 f (t ) sin n 0t dt ,
0
n  0,1,
c0  0
cn  c* n   jbn , n  1,2,
n  1,2,
FUNCTIONS WITH HALF-WAVE SYMMETRY
 T 
f (t )   f  t  0 , t
2

Examples of signals with half-wave symmetry
Each half cycle is an inverted copy of the adjacent half cycle
a0  0
an  bn  0
an 
bn 
4
T0
4
T0
T0
2
 f (t ) cos n0t dt
for n even
for n odd
0
T0
2
 f (t ) sin n0t dt
0
for n odd
There is further simplification if the function
is also odd or even symmetric
LEARNING EXAMPLE
Find the trigonometric Fourier series coefficients
This is an even function with half-wave symmetry
even  bn  0, n  1,2,
half - wave symmetry  an 
4
T0
T0
2
 f (t ) cos n 0t dt
0
an  0 for n even
for n odd
0 
2
  0T  2
T
T
T
T

2
4 4
 84
a2 k 1   V cos2k  1 0 t dt  V cos2k  1 0 t dt   V cos2k  1 0 t dt
T 0
T0
T




4
8
T
4V
(2k  1)
a2 k 1 
sin( 2k  1) 0 
sin
T (2k  1) 0
4 (2k  1)
2
LEARNING EXAMPLE
Find the trigonometric Fourier series coefficients
This is an odd function with half-wave symmetry
odd  an  0; n  0,1,2,
half - wave symmetry  bn 
4
T0
T0
2
 f (t ) sin n 0t dt
for n odd
0
bn  0 for n odd
T0
T0

4
2
4  4V
4V
T

b2 k 1   
t sin[( 2k  1) 0 t ] dt   
(t  0 ) sin[( 2k  1) 0 t ] dt 
T0 0 T0
2
T0 T0




4
b2 k 1 
8  4V
T02
32V
b2 k 1  2
T0
T0
4
 t sin[( 2k  1) 0t ]dt
 0T0
4

Use change of variable
to show that the two
integrals have the
same value
 t sin  t dt  

2
t cos t

0
T0
t ] 4
 t cos[(2k  1) 0 t sin[( 2k  1) 0


2 
(
2
k

1
)



(
2
k

1
)

0

0
0
b2 k 1 
8V
(2k  1)
sin
2
[(2k  1) ]2

sin  t
2
LEARNING EXAMPLE
Find the trigonometric Fourier series coefficients
f (t ) 
3
is odd with half - wave symmetry
2
odd  an  0; n  0,1,2,
half - wave symmetry  bn 
4
T0
T0
2
 f (t ) sin n 0t dt
0
bn  0 for n odd
T0
2
T
0
4 1
2
 cos[(2k  1) 0t ]02
b2 k 1 
sin[( 2k  1) 0 t ]dt 

T0 0 2
T0 (2k  1) 0
b2 k 1 
1
2
(1  cos[(2k  1) ]) 
(2k  1)
(2k  1)
3 
2
 f (t )   
sin( 2k  1) 0 t
2 k 0 (2k  1)
for n odd
LEARNING EXTENSION
Determine the type of symmetry of the signals
LEARNING EXTENSION
Determine the trigonometric Fourier series expansion
even function  bn  0
T0  6
0 
an 

3
1
a0   (2  4)  2
3
21
22
an   2 cos n0t dt   4 cos n0t dt
30
31
an 
4
T0
a0 
T0
2
2
T0
 f (t ) cos n 0t dt ,
0
4 
2n
n 
2
sin

sin


n 
3
3 
T0
2
 f (t )dt
0
n  1,2,
LEARNING EXTENSION
Determine the trigonometric Fourier series expansion
Half-wave symmetry
a0  0
an  bn  0
T0  4
0 

an 
2
bn 
2
n
n
an   2 cos t dt   cos t dt , for n odd
2
2
0
1
1
2
n
n
bn   2 sin t dt   sin t dt , for n odd
2
2
0
1
1
a2 k 1 
b2k 1 
4
T0
4
T0
T0
2
 f (t ) cos n0t dt
for n even
for n odd
0
T0
2
 f (t ) sin n0t dt
for n odd
0
2
(2k  1)
sin
(2k  1)
2
2
(2  cos(2k  1) )
(2k  1)
Fourier Series Via PSPICE Simulation
1.
2.
3.
4.
1. PSPICE schematic
Create a suitable PSPICE schematic
Create the waveform of interest
Set up simulation parameters
View the results
LEARNING EXAMPLE
Determine the Fourier Series of this
waveform
VPWL_FILE in PSPICE library
piecewise linear periodic voltage source
File specifying waveform
Text file defining corners of piecewise
linear waveform
Use transient analysis
comments
Fundamental
frequency (Hz)
Schematic used for Fourier series example
To view result:
From PROBE menu
View/Output File
and search until you find the
Fourier analysis data
Accuracy of simulation is affected by
setup parameters.
Decreases with number of cycles
increases with number of points
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(V_Vs)
DC COMPONENT = -1.353267E-08
RELEVANT SEGMENT OF
OUTPUT FILE
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE
NO
(HZ)
COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
1.000E+00
2.000E+00
3.000E+00
4.000E+00
5.000E+00
6.000E+00
7.000E+00
8.000E+00
9.000E+00
1.000E+01
1.100E+01
1.200E+01
1.300E+01
1.400E+01
1.500E+01
1.600E+01
1.700E+01
1.800E+01
1.900E+01
2.000E+01
4.222E+00
1.283E+00
4.378E-01
3.838E-01
1.079E-04
1.703E-01
8.012E-02
8.016E-02
5.144E-02
1.397E-04
3.440E-02
3.531E-02
2.343E-02
3.068E-02
3.379E-04
2.355E-02
1.309E-02
1.596E-02
1.085E-02
2.994E-04
NORMALIZED
PHASE (DEG)
1.000E+00 2.969E-07 0.000E+00
3.039E-01 1.800E+02 1.800E+02
1.037E-01 -1.800E+02 -1.800E+02
9.090E-02 4.620E-07 -7.254E-07
2.556E-05 1.712E-03 1.711E-03
4.034E-02 1.800E+02 1.800E+02
1.898E-02 -9.548E-06 -1.163E-05
1.899E-02 5.191E-06 2.816E-06
1.218E-02 -1.800E+02 -1.800E+02
3.310E-05 1.800E+02 1.800E+02
8.149E-03 -1.112E-04 -1.145E-04
8.364E-03 1.800E+02 1.800E+02
5.549E-03 1.800E+02 1.800E+02
7.267E-03 -3.545E-05 -3.960E-05
8.003E-05 -3.208E-03 -3.212E-03
5.579E-03 -1.800E+02 -1.800E+02
3.101E-03 2.905E-04 2.854E-04
3.781E-03 -5.322E-05 -5.856E-05
2.569E-03 -1.800E+02 -1.800E+02
7.092E-05 1.800E+02 1.800E+02
TOTAL HARMONIC DISTORTION = 3.378352E+01 PERCENT
* file pwl1.txt
* example 14.5
* BECA 7
* ORCAD 9.1
* By J.L. Aravena
0,0
0.1,1
0.2,3
0.3,6
0.4,3
0.5,0
0.6,-3
0.7,-6
0.8,-3
0.9,-1
1.0,0
It is easier to study the effect of time-shift with the exponential
series expansion
TIME-SHIFTING
f (t ) 
n
 cn e
jn  0 t
n  
f (t  t0 ) 
n
 cn e
jn  0 ( t  t 0 )

n  
 cne  jn t
n 
0 0
n  
e
jn  0 t

Time shifting the function only changes the phase of the coefficients
t0
LEARNING EXAMPLE
 time shift for f (t )
 n 0 t0  phase shift for cn
0

cn   2V
n
sin
 n
2
f (t )
v (t ) 
T
v (t )  f (t  0 )
4
0T0  2
T  T0
n even
n
n 
0


n
cvn   2V
n  j 2
 n sin 2 e
 cvne jn t
0
n  
e
 jn

2
 cos n

2


 sin n   1
2

2
odd
 n 0 t 0   n 0
 j sin n
T0
n

4
2
n even
n
odd

2
cv , 2 k 1   j
2V
n
Time shifting and half-wave periodic signals
Every half - wave periodic signal, f (t ), with
period, T0 , can be expressed as
 T 
f (t )  f1 (t )  f1  t  0 
2

T0

f
(
t
)
0

t


2
with f1 (t )  
T0
 0
 t  T0

2
Assume f1 (t ) 
n
 cne jn t
0
n  
0T0  2
T
n  
 jn  0 0 
T0
2 e jn  0 t 
f1 (t  )    cne


2
n   

e  jn   1 , n
n
 f (t ) 

 2c2k 1e  j ( 2k 1) t
0
k  
Only the odd coefficients of f1 are used
LEARNING EXTENSION
1
T0  2   0 
v (t  1)
2

T0
It was shown before c0 
2
that for v(t)
1  e  jn
cn 
,n  0
j 2n
compute the coefficients for v(t-1 ) and show that
they differ by a πn angle from those of v(t).
For the delayed function
c0d
12
1
  v (t  1)dt 
20
2
For n  0
cnd


e  jn
12
1 2  jnt  1 e  jn  e  j 2 n
 jnt

1  e  jn
  v (t  1)e
dt   e
dt
j 2n
j 2n
20
21
cnd  e  jn cn


WAVEFORM GENERATION
Assume f (t ) 
n 
 cn e
jn  0 t
f ( t ) 
n  
f (t  t0 ) 
n 
 cn e
jn  0 ( t  t 0 )
n  
f ( t  t0 ) 
 cne  jn t
0 0
n  
 c e
n 
n  

n 
n
 jn  0 t 0
e 
j n 0 t
n 
Time scaling does not change the
 cne j n0t values of the series expansion
n  
e
jn  0 t
Time-shifting modifies the phase of
the coefficients
If the Fourier series for f(t) is known then one can
easily determine the expansion for any time-shifted
and time-scaled version of f(t)
The coefficients of a linear combination of signals are the linear combination of
the coefficients
cn3  cn1  cn2 n
One can tabulate the expansions for some basic waveforms and use them to determ
the expansions or other signals
Signals with
Fourier series
tabulated in
BECA 8
Use the table of Fourier series to determine the expansions
of these functions
LEARNING EXTENSION
c0 
4
3
From the table of Fourier series


A
n jn0  t  2 
f (t )  
sin
e
T0
n  n

c0 
For v2 (t )
 T  
v2 (t )  f  t  0

2 

For v1 (t )
v1 (t )  f (t   / 2)
A  4, T0  3,   1, 0 
A  2, T0  6,   2

2
2
n jn 3 t
c

v1 (t )  
sin
e
0
3
n

3
n  

Strictly speaking the value for n=0 must
be computed separately.
e  jn  (1)n
2
3
4
n jn
v2 (t )  
sin
e
3
n   n

2  31 1 
 
 t
3 
2 2
n0
4
n  jn jn
v2 (t )  
sin
e
e
n

3
n 

n 0
2
t
3
A
T0
FREQUENCY SPECTRUM
The spectrum is a graphical display of the coefficients of the Fourier series.
The one-sided spectrum is based on the representation


n 1
n 1

f (t )  a0   Dn cos(n o t   n )  a0   Re Dn ne jn o t

The amplitude spectrum displays Dn as the function of the frequency.
The phase spectrum displays the angle  n as function of the frequency.
The frequency axis is usually drawn in units of fundamental frequency
The two-sided spectrum is based on the exponential representation
f (t ) 

 cne jn t
0
n  
In the two-sided case, the amplitude spectrum plots |cn| while the phase spectrum
plots cn versus frequency (in units of fundamental frequency)
Dn n  2cn  an  jbn
cn  cn*
Both spectra display equivalent information
LEARNING EXAMPLE
The Fourier series expansion, when A=5,
is given by
v (t ) 


n 1
n odd
20
40
sin n 0 t  2 2 cos n 0 t )
n
n
Determine and plot the first four terms
of the spectrum
Dn n  2cn  an  jbn
D11  
40
2
j
20

 7.5  122 D3  
40
20

j
 2.2  102
2
3
9
D55  1.3  97, D7 7  0.92  95
Phase spectrum
Amplitude spectrum
LEARNING EXTENSION
Determine the trigonometric Fourier series and plot the
first four terms of the amplitude and phase spectra
1  1
v (t )   
sin 2 n t
2 1 n
From the table of
series
A  A
f (t )    sin n 0 t
2 n1n
1
1
D0  ; Dn   j ; n  0
2
n
n
0
1
2
3
4
|Dn|
0.5
0.318309
0.159155
0.106103
0.079577
Arg(Dn)
0
-90
-90
-90
-90
Dn n  2cn  an  jbn
0.6
|Dn|
0.5
0.4
0.3
|Dn|
0.2
0.1
0
1
2
3
4
5
STEADY STATE NETWOK RESPONSE TO PERIODIC INPUTS
1. Replace the periodic signal by its Fourier series
2. Determine the steady state response to each harmonic
3. Add the steady state harmonic responses
H ( j ) | H ( j ) | e j ( ) | H ( j ) |  ( )
H ( j )
|c|e
j (1t  )
| D | cos(1t   )
Re{| D |  e j1t }
| c || H ( j1 ) | e j (1t   (1 ))
| D || H ( j1 ) | cos(1t     (1 ))
Re{| D |  | H ( j1 ) |  (1 )e1t }
LEARNING EXAMPLE
I ( j )
v (t ) : A  5, T0  
(2 ||
V1 ( j )
2
1
) 
1  2 j
j
1
V0  V1
2
1
j
v (t ) 


n 1
n odd
40
 20

sin 2n t  2 2 cos 2n t 

n
 n

| 4  j 4 | 16  16 2
(4  j 4 )  tan1 
H ( j ) 
40
20
Dn  an  jbn   2 2  j
n
n
2
2
2
1  2 j
V1 ( j ) 
V ( j ) 
V ( j )
2
4

4
j

2
1  2 j
n  2n
1
1

  tan 1 
4  j 4
16  16 2
2
2
 40   20 
| Dn |  2 2   

n

n


 

n
n
Dn  180  tan 1
 (180  tan 1
)
2
2
 40   20 
von (t )   2 2   

n

n


 

2
1
16  64n2
cos(2nt  tan 1
n
 tan 1 2n  180)
2
LEARNING EXTENSION Find the steady state expression for the current i (t )
v S (t ) 
D0 
2
j
V ( j )
I ( j )  S
 Y ( j )VS ( j )
Z ( j )
Z  1 2 ||
I ( j 2n)  Y ( j 2n) Dn 
Y (0) 
1  j 2n
Dn
3  j 2n
20

20


 40
cos 2nt
2

(
4
n

1
)
n1

; Dn 
40
180, n  1;  n  2n
2
 (4n  1)
4
4
6  j 2 3  j
j
Z ( j )  1 
1


2
2

j
2

2

j
2

1  j
2
j
1  j
Y ( j ) 
phasor for n - th harmonic
3  j
1
3
1  4n 2
Y ( j 2 n) 
( tan 1 2n   tan 1 (2n / 3))
2
9  4n

i (t )  Y (0) D0   | Yn || Dn | cos(2nt  Yn )
n 1
AVERAGE POWER
In a network with periodic sources (of the same period) the steady state voltage
across any element and the current through are all of the form

v (t )  Vdc  Vn cos(n 0 t   vn ) The average power is the
n 1

i (t )  I dc   I n cos(n 0 t   in )
sum of the average powers
for each harmonic
n 1
1
Average Power P 
T
t 0 T

t o T
 v (t )i (t )dt
to


n 1
n 1
v (t )i (t )  Vdc I dc  Vdc  I n cos(n 0 t   in )  I dc Vn cos(n 0 t   vn )
t0



 Vn I n
n1 1n2 1
1
2
cos(n1 0 t   vn1 ) cos(n 0 t   in2 )
Vn1I n 2 cos((n1  n2 )0t  ( vn1   in2 )) 
 Vn1 I n2 cos(n10t   vn1 ) cos(n0t   in2 )    2  cos((n  n ) t  (   ))
n1 1n2 1
n1 1n2 1

1
2
0
vn1
in2 



2
0 

T

t 0 T
0
 cos(k0t   )dt  T

t0
k 0

 T cos( vn   in ) for n1  n2

k 0
V I
Average Power P  Vdc I dc   n n cos( vn  in )
n 1 2
LEARNING EXTENSION
Determine the average power
v (t )  64  36 cos(377 t  60)  24 cos(754t  102)[V ]
 cos  cos(  180)
i (t )  1.8 cos(377 t  45)  1.2 cos(754t  100)[ A]
P  0.5(36 1.8 cos(15)  24 1.2 cos(102  180  100))
P
62.59  28.78
 16.91[W ]
2
LEARNING EXAMPLE
Determine the current, i (t ), and the average power absorbed
by the network
v (t )  42  16 cos(377t  30)  12 cos(754t  20)[V ]
Z ( j )  R  jL 
1
jC
I ( j ) 
  377
V ( j 377)  1630,
Z ( j 377)  16  j 0.020  377  j
I ( j 377) 
  754
1
10 4377
i (t )  0.64 sin( 377t  79.88)  0.75 cos(754t  26.49)
1630

 0.6479.88
V I
16  j 7.54  j 26.53
Average Power P  Vdc I dc   n n cos( vn  in )
n 1 2
V ( j 754)  12  20,
Z ( j 754)  16  j 0.020  754  j
I ( j 754) 
V ( j )
Z ( j )
 0
capacitor acts as open circuit (Z  )
I ( j 0)  0
P  42(0) 
1
10 4  754
12  20
 0.75  26.49
16  j15.08  j13.26
16  0.64 cos(49.88)  12  0.75 cos(6.49)
2
FOURIER TRANSFORM
F ( ) 


f ( t )e
 jt
f (t ) 
dt



F ( )e j t

F ( )  F[ f (t )]
d
2
f (t )  F-1 [F ( )]
A heuristic view of the Fourier transform
A non-periodic function can be viewed as the limit of a periodic function when the
period approaches infinity
f p (t ) 

 cn e
jn
2
t
T
f p (t ) 
n 
2

 cnTe
2
t
T

1
T
2
 
T
2
 jn t
f (t )e T dt n   n
n  
T /2
 jn t
1 T /2
cn 
f (t )e T dt cnT  

T T / 2
T / 2
jn
T 

F ( )e jt


 f ( t )e

j t
d
2
dt
LEARNING EXAMPLE
Determine the Fourier transform
For comparison we show the spectrum
of a related periodic function

V ( )   v (t )e
 j t
dt  V
 /2
 j t
e
dt

 / 2
 /2

j / 2
 j / 2
 1  j t 
e

e
V ( )  V 
e
V

j

j

  / 2
V ( )  V
sin


V sin 2
cn 
T0 
2
2

2
Spectrum for T0  5
LEARNING EXAMPLES
Determine the Fourier transform of
Determine the Fourier transform of the
unit impulse function
F ( ) 


f (t )e  jt dt
F ( )  F[ f (t )]
Sifting or samplingproperty of
the impulse

  (t  a ) f (t )dt  f (a )
F[ (t )]    (t  a )e
dt  e


F ( )e j t

the corresponding function of time


2 (  0 )e j t

 j a

d
2
Consider F ( )  2 (   0 ) and find
f (t ) 

 j t
t
o
f (t ) 


f (t )  e j
d
2
f (t )  e j0t
F ( )  F[e jω0t ]  2πδ(ω-ω0 )
LEARNING EXTENSION
Determine F ( )  F[sin 0t]
2 (  0 )  2 (  0 )
e j 0t  e  j 0t
F
(

)

f (t )  sin  0 t 

2j
2j
F ( )  j  (  0 )   (  0 )
Proof of the convolution property

f (t )  f1 (t )  f 2 (t ) 
 f1 ( x) f 2 (t  x)dx




F ( )     f1 ( x ) f 2 (u)e  j ue  j x dudx

  

F ( ) 


f (t )e  j t dt




    f1 ( x ) f 2 (t  x )dx  e  j t dt

  
Exchanging orders of integration



F ( )     f1 ( x ) f 2 (t  x )e  j t dt dx

  
Change integration variable
ut  x t u x
And limits of integration remain the same



F ( )     f1 ( x ) f 2 (u)e  j ( u x )dudx

  


F ( )   f1 ( x )e
dx   f 2 (u)e  j udu
 


F ( )  F1( )F2 ( )
 j x
A Systems application of the convolution property

vo (t )   h(t  x )vi ( x )dx
vi (t )
h(t )

LEARNING EXTENSION
The output (response) of a network can
be computed using the Fourier transform
Use Fourier transform to determine vo (t )
From the table of transforms
vi (t )  e t u(t )
h(t )  e 2t u(t )
(And all initial conditions are zero)
A1  ( s  2)Vo ( s) |s 2  1
Vo ( )  H ( )Vi ( )
A2  ( s  1)Vo ( s) |s 1  1
Use partial fraction
1
1

expansion!
j  2 j  1
1
A
A
 1  2
( s  2)( s  1) s  2 s  1
Vo ( ) 

1
1

j  1 j  2

v o ( t )  e  t  e  2 t u( t )
PARSEVAL’S THEOREM

 | f (t ) | dt 
2




| F ( ) |2
Think of f(t) as a voltage applied to a one Ohm resistor
d
2
p(t )  v (t )i (t ) | f (t ) |2
By definition, the left hand side is the energy of the signal
| f (t ) |2  power (or energy density in time)
| F ( ) |2  Energy density in the frequency domain
Parseval’s theorem permits the determination of the energy of a signal in a
given frequency range
Intuitively, if the Fourier transform has a large magnitude over a frequency range
then the signal has significant energy over that range
And if the magnitude of the Fourier transform is zero (or very small) then the
signal has no significant energy in that range
LEARNING BY APPLICATION
Examine the effect of this low-pass filter in the
quality of the input signal
Vo ( )  H ( )Vi ( )
20
Vi ( ) 
j  20
1
1
jC
H ( ) 

1
 R jRC  1
jC
One can use Bode plots to visualize
the effect of the filter
High frequencies in the input signal are attenuated
in the output
The effect is clearly visible in the time domain
The output signal is slower and with less energy than the input signal
EFFECT OF IDEAL FILTERS
Effect of band-pass filter
Effect of low-pass filter
Effect of band-stop filter
Effect of high-pass filter
EXAMPLE
AMPLITUDE MODULATED (AM) BROADCASTING
Audio signals do not propagate well in atmosphere – they get attenuated very quick
Original Solution: Move the audio signals to a different frequency range for broadca
The frequency range 540kHz – 1700kHz is reserved for AM modulated broadcasting
Carrier signals
Broadcasted
signal
Audio signal
AM receivers pick a faint copy of v(t
f  1000kHz ; f  900kHz
a
c
v( t )  (1  s( t ))cos 2 f t
c
Nothing in audio range!
f  1000kHz ; f  900kHz
a
c
v( t )  (1  s( t ))cos 2 f t
c
Audio signal has been AM modulated to the radio frequency range
EXAMPLE
HARMONICS IN POWER SYSTEMS
R  0.2
line
Harmonics account for
301.8W or 9.14% of the
total power
LEARNING BY DESIGN
“Tuning-out” an AM radio station
signal from station 1: v1 (t )  [1  s1 (t )] cos1t
signal from station 2: v2 (t )  [1  s2 (t )] cos 2 t
For simplicity assume s1 (t )  s2 (t )
f1  900kHz, f 2  960kHz
Assumingboth stations reach the receiver
with equal strength;
Fourier transform of signal broadcast
signal at antenna : v r ( t )  K [v1 ( t )  v 2 ( t )]
by two AM stations
Proposed tuning circuit
Ideal filter to tune out one AM station
vo ( s )
R
( s / L)


vr ( s ) R  Ls  1 ( s / L)
Cs
Fourier transform of received signal
R
s
Next we show how to design the tuning circuit G ( s ) 
L
v
R 1
by selecting suitable R,L,C
s2  s 
L LC
Gv ( s ) 
Ideal filter to tune out one AM station
Designing the tuning circuit
R
L
Gv ( s ) 
R 1
s2  s 
L LC
s
Design equations
center frequency : f o 
bandwidth: BW 
1
2 LC
1 R
2 L
Frequency response of circuit
tuned to 960kHz
Design specifications
BW  60kHz
f o  900kHz, or 960kHz
More unknowns than equations. Make
some choices
BW  10kHz
R  10
f0 
BW 
1 R
 L  159.2H
2 L
C  196.4 pF
1

2 LC
C  172.6 pF
Fourier transform of received signal
f o  900kHz
f o  960kHz
LEARNING EXAMPLE
An example of band-pass filter
v S (t )  0.1sin[ 2 103 t ]  0.01sin[ 2 104 t ]  0.1sin[ 2 105 t ]
noise
signal
noise
Design requirement: Make the signal 100 times stronger than the noise
A1 ( j ) 
Vo
VS
V  0
Proposed
two-stage
band-pass
with two
identical
stages
Vout
2
Vo
VA
A
(
j

)

  A1 ( j )
KCL@ V- :

0
VS
1
R2
 1 
jC

j

V
V  VA Vo  VA VA
 CR1 
A1 ( j )  o 
KCL@ VA : S


0
1
1
VS
 2 
1
R1
2  
j



jC
jC
C 2 R1 R2
 CR2 
General form of a band- pass
In a log scale, this filter
is symmetric around
 o  center frequency
 0 
Ao   j
the center frequency.
A

gain
at
center
frequency
o
Q 
Hence, focus on 1kHz
G
 
Q  quality factor of the filter
  2   0  j   o2
Q 
Ao o2
j
 1 
o
10Q

j

Gain
at


:
CR1 
Vo
10 2
1
 o2

A1 ( j ) 

 o (1 
) j
VS


100
10Q
2
1
2  
j



C 2 R1 R2
Design constraint
 CR2 
Equating terms one gets a set of
equations that can be used for design
j
10Q
1
 o2
2
 o (1 
) j
100
10Q
Design equations
o 
Q
1
C R1R2
1 R2
2 R1
o
Q

2
CR2
Ao  
R2
2 R1
 o2
After two stages
the noise gain is
1

1000 1000 times
smaller
Since center frequency is given
this equation constrains the quality factor
Solving for Q
1kHz, 100kHz are noise frequencies
10kHz is the signal frequency
We use the requirements to constraint Q

R2
 40
R1
1  100Q 2  1000  Q  10
Picking R1  1k  R2  40k
o  2 104  C  2.5nF
Resulting Bode plot obtained with PSPICE AC sweep
Filter output
obtained with
PSPICE
Fourier transform of
output signal obtained
with PSPICE
EXAMPLE
MULTIPLE FREQUENCY “TRAP” CIRCUIT
 
O
BASIC NOTCH FILTER
1
V ( )
j C
H ( ) 

V ( ) R  jL  1
j C
jL 
O
in
1   LC

1   LC  j RC
L  L  L  10 H
1
2
3
1
LC
2
2
Tuned for 10kHz
Multiple frequency trap
Eliminates 10kHz, 20kHz and 30kHz
Fourier