TUTORIAL ON HARMONICS
Theory and Computation
Techniques
C.J. Hatziadoniu: hatz@siu.edu
AC Drive Harmonics
Prime
Mover
DC Link
Synchronous
Generator
Inverter
Rectifier
Induction
Motor
Load
N
S
Generator
Controller
PWM
PWM
Motor
Controller
Coordination
Harmonic Sources:
 Power converter switching action
 Motor own generated harmonics (spatial distribution of windings, stator
saturation)
 Transformer/inductor iron core saturation
 Harmonics flowing between generator and motor sides
Potential Problems due to Harmonics

Power losses and heating: reduced efficiency,
equipment de-rating

Over-voltage and voltage spiking, due to
resonance: insulation stressing, limiting the
forward and reverse blocking voltage of power
semiconductor devices, heating, de-rating

EMI: noise, control inaccuracy or instability

Torque pulsation: mechanical fatigue, start-up
limitation
Power Loss and Heating
I1+Ih
B
RI12+RIh2
V1+Vh
Losses into the resistive and magnetic components
•Resistive losses: skin effect
•Magnetic losses: Eddy currents and hysterisis losses increase with
frequency
~V12
+
~Vh2
H
Over-Voltage, Over-Current (due to
resonance)
n 
1
LC
+
n L 
1
nC
nC 
-
1
n L
+
+
-
Series resonance tank
Capacitor loss due to
harmonics (insulation loss)
Parallel resonance tank
PH 

h
C tan( h )   hVh2
Interference with Control, EMI
Rad., EMI with
other processes
Rad
Common
modes
Grid
Control
Measurements
Common or
diff. modes
Common (ground)
modes
What Are Harmonics?
Technical
of an AC Voltage Component
Non-periodic Distortion
Steady
DC State:
CurrentSubharmonic
Ripple
Non-Harmonic
AC InputDisturbance/Distortion
from an Inverter
Description
1
1
100
1
0.5
100
0.5
0.5
0
0
0
80
50
-0.5
-0.5
A high frequency sinusoidal current or voltage produced by certain
non-linear and switching processes in the system during normal
periodic operation (steady state);

The harmonic frequency is an integer multiple of the system operating
Actual
-1
frequency (fundamental).
-1
Desired
60
-1
0.5
1is the harmonic
1.5
The non-sinusoidal part 000in a periodic voltage
or
current
0.5
1
1.5
0.5
1
1.5
0
ripple or harmonic distortion—comprised
of
harmonic
frequencies.
Non-periodic Steady State: Interharmonic Component
Volts
%%
Current

40
1
1
1
20
-50
0.5
Transient
DC Input
from aResponse
Diode Rectifier
Mathematical Definition
0
0
0
Specified
 Sine and cosine functions
of time with frequencies that are integer
0.5
-0.5
Actual
multiples of a fundamental
frequency
-100
Actual
Distortion
-1
-1
Specified
Actual
Ripple
 Harmonic sine and cosine
functions
sum
up
to
a
periodic
(non-20
Desired
0.5
1
1.5
00
sinusoidal) function
0
0.5
1
1.5
0
0.5
1
1.5
Time
Time
 Terms of the Fourier series expansion of a periodic
function;
Time
Harmonic Analysis

What is it?

Principles, properties and methods for expressing periodic
functions as sum of (harmonic) sine and cosine terms:




Fourier Series
Fourier Transform
Discrete Fourier Transform
Where is it used?

Obtain the response of a system to arbitrary periodic inputs;
quantify/assess harmonic effects at each frequency

Framework for describing the quality of the system input and
output signals (spectrum)
Superposition

A LTI system responds linearly to its inputs
 ui1uo1, ui2uo2


aui1+bui2auo1+buo2
For sinusoidal inputs:
ui1  V1 cos(1t )  uo  Vo1 cos(1t  1 )
ui 2  V2 cos(2t )  uo  Vo 2 cos(2t   2 )
ui1  ui 2  uo  Vo1 cos(1t  1 )  Vo 2 cos(2t   2 )
Application preview: DC Drive
io
Find the armature current io(t) below
+
is
vo
u s  2  240  sin(2 60  t ), V
-
E=150 V
Ra=1 W
La=5 mH
DC Voltage Approximation
Input Voltage
io
+
s
v,V
200
is
0
vo
-200
0
0.005
0.01
0.015
0.02
0.025
0.02
0.025
-
Output Voltage
200
o
v,V
300
100
0
0
0.005
0.01
0.015
Time, s
uo  216.1  [144.1  cos(2  120  t )  28.8  cos(2  240  t )  12.3  cos(2  360  t )] V
Source Superposition
uo  216.1  [144.1  cos(2  120  t )  28.8  cos(2  240  t )  12.3  cos(2  360  t )] V
io
I o ,dc 
Vo ,dc  216.1 V
+
u o , 2  144.1  cos( 2  120  t ) V
-
Vo ,dc  E
Ra
La= 0.005 H
Ra=1 W
+
u o, 4  28o.,8dc
 cos( 2  240  t ) V
-
Ra=1 W
+
E=150 V
V
-
 216.1 V
 66.1 A
E=150 V
u o,6  12.3  cos( 2  360  t ) V
Output Response
io  66.1  36.9 cos( 2 120  t  75)  3.78 cos(2  240  t  82.4)  1.08 cos( 2  360  t  85) A
Output Voltage
200
o
v,V
300
100
0
0
0.005
0.01
0.015
0.02
0.025
0.02
0.025
Output Current
o
i ,A
100
50
0
0
0.005
0.01
0.015
Time, s
Procedure to obtain response
Step 1: Obtain the harmonic composition
of the input (Fourier Analysis)
Step 2: Obtain the system output at each
input frequency (equivalent circuit,
T.F. frequency response)
Step 3: Sum the outputs from Step 2.
Fundamental Theory Outline

Harmonic Fundamental Theory—Part a:
Periodic Signals—sinusoidal function
approximation
 Fourier Series—definition, computation
 Forms of the Fourier Series
 Signal Spectrum
 Applications of the FS in LTI
 Wave Form Quality of Periodic Signals

Measures Describing the Magnitude
of a Signal

Amplitude and Peak Value

Average Value or dc Offset

Root Mean Square Value (RMS) or
Power
Amplitude and Peak Value
y
y
A
A
t
-A

t
-A
Peak of a Symmetric Oscillation
Non-Symmetric Signals
y
y
A
A
A+B
t
-B
t

Peak-to-peak variation
Average Value
y
Yav
1
Yav 
T
T
 y (t )  dt
0
T

t
Signal=(constant part) + (oscillating terms)
Examples
y
y
A
A
Yav
Yav=0
Yav  kA
t
kT
y
y
A
A
2

t
T
Yav 
A
Yav
t
-B
A B
2
t
AC Signals
y
y
A
A
t
t
y
t

Zero Average Value
DC and Unidirectional Signals
y
y
Dc offset
Dc offset
t
t
y
y
A
2

A
t
t
Root Mean Square Value (RMS)
y2
y
Yrms
1

T
T
 y dt
2
0
Yrms
T
Yrms
1

T
t
T
T
2
y
 dt
0
For periodic signals, time window
equals one period
t
Remarks on RMS

RMS is a measure of the overall
magnitude of the signal (also Treferred
1 2
to as norm or power ofI the
signal).

i dt , V 
T


0
The rms of current and voltage is
directly related to power.
V2
PR  R  I 2 

R
Electric equipment rating and size is
given in voltage and current rms
S V  I
values.
T
1 2
u dt

T 0
Examples of Signal RMS
y
y
A
A
A
2
t
t
y
y
A
Yrms
Yav
A
A k
A
Yav  Ak
120º
2
3
t
kT
T
t
Effect of DC Offset
y
y
2
 A 
2
 d

 2
A2  d 2
d
d
t
New RMS=
SQRT [ (RMS of Unshifted)2+(DC offset)2]
t
Examples of Signals with equal
RMS
y
y
A
A
A
2
A
2
t
t
y
A
2
y
A
A
t
2
3
t
RMS and Amplitude
y
A
2
y
A
t
y
t
t

Amplitude: Local effects in time; Device insulation, voltage
withstand break down, hot spots

RMS: Sustained effects in time; Heat dissipation, power output
Harmonic Analysis: Problem Statement

Approximate the square pulse function by a
sinusoidal function in the interval [–T/2 , T/2]
f(t)
A cos(
-T
2
t)
T
1
B sin(
-T/2
2
t)
T
T/2
-1
T
3T/2
General Problem

Find a cosine function of period T that best
fits a given function f(t) in the interval [0,T]

Assumptions:
f(t) is periodic of period T
F (t )  A cos(t )
2

T
f (t )  F (t )
Approximation Error
A
f(t)
1

Error:
e(t )  f (t )  F (t )  f (t )  A cos(t )
-T/2
T/2
-1
Objective:
Minimize the error e(t)
e(t)
1
Method:
Find value of A that gives the
Least Mean Square Error
-T/2
T/2
-1
Procedure
T
1
E
T
Define the average square error as

e(t ) 2 dt
0
T
E
1
T

T

f (t )
T
 f (t )  A cos(t )2 dt  1
0
:

2
A
2
E
A
2
T
T

0
2
0
T
1
f (t ) cos(t )dt 
T


 A2 cos(t ) 2  2 Af (t ) cos(t ) dt
f (t ) 2 dt
0
E is a quadratic function of A. The optimum choice of A is
the one minimizing E.
Optimum Value of A
E
2
A
2
A
2
T
T

T
f (t ) cos(t )dt 
1
T
0

f (t ) 2 dt
0
T
Find dE/dA:
dE
2
 A
dA
T
 f (t) cos(t)dt.
0
T
Set dE/dA equal to zero
dE
2
 0  Aopt 
dA
T

0
f (t ) cos(t )dt
EXAMPLE: SQUARE PULSE
1
1
0
-1
-1.5
1
0
0
-1
-1
-1.5
-1.5
0
-1
-0.5
0
Error
0.5
1
1.5
-1
-1.5
-1
-1
-1
-0.5
-0.5
-0.5
0
0.5
0
0
1
0.5
0.5
1.5
1
1
1.5
1.5
Y-Axis
A geometrical interpretation
z
E  zx 
z2  x2
x
X-Axis
Norm of a function, error, etc is defined as:
f
2
1

T

T
2
f (t ) dt
Shifted Pulse
f(t-T/4)
f(t-b)
B
B
1
1
A
T/2
-1
T
f (t  )  B sin( t )
4
T
2
T
B   f (t  ) sin( t )dt
T 0
4
T
T/2
T
-1
f (t  b)  A cos(t )  B sin( t )
2
A
T
T
2
B
T
T
 f (t  b) cos(t )dt
0
 f (t  b) sin( t )dt
0
A3cos(3t)
Approximation with
many harmonic
terms
f(t)
T
E
1
T

0
A
Average Square Error :
2c
os
(2
t
)
A1cos(t)
2
 f (t )   A1 cos t  B1 sin t      AN cos( Nt )  BN sin( Nt ) dt
Harmonic Basis
The terms
An cos( nt ), Bn sin( nt ), n  0,1,
From an orthogonal basis
Orthogonality property:
T
1
T

T
cos( mt ) cos( kt )dt 
1
T
0
0
T
1
T

0

cos( mt ) sin( kt )dt  0
0, k  m

sin( mt ) sin( kt )dt  
1 , k  m
 2
Optimum coefficients
2
T
E
1
T
  f (t)  A cos t  B sin t     A cos( Nt)  B sin( Nt) dt 
1
1
N
N
0

1
T
f
2

 2 f  A1 cos t  2 f  B1 sin t  A12 cos 2 t  B12 sin 2 t   dt 
T

1 2 1 2
2
A1  B1  A1
2
2
T

f  cos t  dt  B1
T


2
T

f  sin t  dt 
T
1
T

f 2 dt  
T
The property of orthogonality eliminates the
cross harmonic product terms from the Sq. error
For each n, set
E
0
Ai
Optimum coefficients

Obtain the optimum expansion
coefficients:
T
2
An 
T

f (t ) cos( nt )dt
0
T
2
Bn 
T

0
f (t ) sin( nt )dt
Example—Square Wave Pulse
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5

T /2
A1 
2
T

f (t ) cos( 2t )dt 
T / 2


2


1


f (t ) cos( 2t )d (2t )

 /2
(.) 
0
4


cos( 2t )d (2t ) 
4

0
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5

B1 

f (t ) sin( 2t )d (2t ) 
f (t ) sin( 2t )d (2t )  0




1
2

0
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
A2  B2  0
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
 /2
A3 
4


cos 3(2t )d (2t ) 
4
sin
3
3
2

4
3
0
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
 /2
An 
4


4
n
4
cos n(2t )d (2t ) 
sin(
)
n
2
n
0
1
n
A
B
1
4/
0
2
0
0
3
-4/3
0
4
0
0
5
4/5
0
6
0
0
n
±4/n
0
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
Waveform Recovery
n=1-3
n=1-9
n=1-7
n=1-5
n=1
1
0.5
0
-0.5
-1
-0.5
1
0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.2
0.3
0.4
0.5
E  0 as N  
0
-0.5
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
Example: Sawtooth
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
1
0.5
0
-0.5
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
1
0.5
0
-0.5
-1
-0.5
Odd Symmetry
T /2
Bn 
2
T

2t  sin( 2nt )dt 
T / 2

T /2
4
T

0
()dt 
2


0

2
 sin( n )  d ( )  (1) n 1

n
1
0.5
0
-0.5
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
1
0.5
0
-0.5
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
n
An
Bn
0
0
0
1
0
2/
2
0
-1/
3
0
2/3
4
0
-1/2
5
0
2/5
6
0
-1/3
7
0
2/7
n
0
(1) n1 2 n 
Periodic Approximation
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Approximation of the Rectified sine
v(t)
Vm
v (t )  Vm sin(t )

2
T
A periodic signal= (constant part)+
(oscillating part)
t
vo(t)
Vm
d=T/2
T/2
T
t
Average Value
vo(t)
Vm
d=T/2
T/2
d
1
2
A0   Vm sin( t )  dt 
d 0
T
Vm


t
T
T /2
V
m
sin( t )  dt 
0
 sin( t )  dt    cos(t ) 
0
Vm
t 
t 0

2

Vm
Harmonic Terms
vo(t)
Vm
d=T/2
T/2
T
t
2
2
Ak   Vm sin( t )  cos(k
t )  dt 
d 0
d
d
4Vm
T
Vm


T /2
 sin( t )  cos(2kt )  dt 
0
 sin( 2k  1)t  sin( 2k  1)t  dt  
0
4Vm
 (2k  1)( 2k  1)
Summary
v(t)
Vm
vo (t ) 
v (t )  Vm sin(t )
2

T
t
Vm
T/2
T
t

Vm 
4Vm
1
cos( 2kt )

 k 1, 2, (2k  1)( 2k  1)
( n  2k ) 
vo (t ) 
vo(t)
2
2

Vm 
4Vm
1
cos( nt )
 n 2, 4, (n  1)( n  1)

Numerical Problem: DC Drive
io
E=150 V
Ra=1 W
La=5 mH
+
Input Voltage
is
s
v,V
200
0
vo
-200
0
0.005
0.01
0.015
0.02
0.025
0.02
0.025
Output Voltage
300
200
-
o
v,V
u s  2  240  sin(2 60  t ), V
100
0
0
0.005
0.01
0.015
Time, s
Input Harmonic Approximation
Input Voltage
us  2  240  sin( 2 60  t )  339.4  sin( 377  t )
s
v,V
200
Average or dc component
0
-200
Vo ,dc 
0
0.005
2

0.01
Vo ,dc 
Vm  216.1 V
0.015
0.02
0.025
2

Vm  216.1 V
Output Voltage
Harmonic Expansion
200
o
v,V
300
100
0
uo  Vo,dc 
0
0.005
0.01
0.015
0.02
4Vm
0.025

1
 cos( 2 60n  t )
 n2,4, (n  1)  (n  1)

Time, s
Truncated Approximation (n=2, 4, and 6)
uo  216.1  (144.1 cos(2 120  t )  28.8  cos(2  240  t )  12.3  cos(2  360  t ))
Equivalent Circuit
io
Vo ,dc  216.1 V
+
io
+
i
s
u o , 2  144.1  cos(
2  120  t ) V
+
u o, 4  28.8  cos( 2  240  t ) V
+
voRa=1 W
E=150 V
u o,6  12.3  cos( 2  360  t ) V
u s  - 2  240  sin(2 60  t ), V
E=150 V
La= 0.005 H R =1 W
a
La=5 mH
-
uo  216.1  (144.1 cos(2 120  t )  28.8  cos(2  240  t )  12.3  cos(2  360  t ))
Superimpose Sources: DC Source
io
Vo ,dc  216.1 V
+
u o , 2  144.1  cos( 2  120  t ) V
I o ,dc 
Vo ,dc  E
Ra
 66.1 A
La= 0.005 H
+
u o, 4  28.8  cos( 2  240  t ) V
-
Ra=1 W
+
E=150 V
Vo ,dc  216.1 V
-
u o,6  12.3  cos( 2  360  t ) V
Ra=1 W
E=150 V
Superposition: n=2, f=120 Hz
io 2  36.9 cos( 2 120  t    75 )  36.9 cos( 2 120  t  75 )
io
Vo ,dc  216.1 V
+
I o, 2 
Vo , 2
Za
u o , 2  144.1  cos( 2  120  t ) V
+
+ u o, 4  28.8  cos( 2  240  t ) V
-
Vo , 2  144.1 V
-+
-
u o,6  12.3  cos( 2  360  t ) V

144.1
  75  36.9(  75) A
La= 0.005 H
3.9
iX a  i (2 120) La
 i3.77 W
Ra=1 W
Ra=1 W
E=150 V
Z a (2)  Ra  iX a
 1  i3.77  3.9 75 W
Superposition: n=4, f=240 Hz
io
Vo ,dc  216.1 V
+
I o, 4 
u o , 2  144.1  cos( 2  120  t ) V
+
u o, 4  28.8  cos( 2  240  t ) V
+
- V  28.8 V
Vo, 4
Za

28.8
  3.78
  82L.a4=0.005
H  (  82.4) A
7.61
iX a  i (2 240) La
 i 7.54 W
Ra=1 W
Ra=1 W
E=150 V
 1  i 7.54  7.6182.4 W
o,4
+-
u o,6  12.3  cos( 2  360  t ) V
-
Z a (4)  Ra  iX a
Superposition: n=6, f=360 Hz
io
Vo ,dc  216.1 V
I o,6 
+
u o , 2  144.1  cos( 2  120  t ) V
++
u o, 4  28.8  cos( 2  240  t ) V
- Vo , 6  12.3 V
-
+
-
Vo , 6
u o,6  12.3  cos( 2  360  t ) V
Za

12.3
  85  1.08(  85) A
11.35
La= 0.005 H
iX a  i (2 360) La
 i11.31 W
Ra=1 W
Ra=1 W
E=150 V
Z a (4)  Ra  iX a
 1  i11.31  11.3585 W
Summary
Freq.,
Hz
Vo ampl, V
216.1
66.1
120
144.1
36.9
240
28.8
3.78 P
1
3.9
h,2
RMS
12.3
240
1.08
71.1
Power
loss, W
2
Po,dc  Ra I o2,dc  1W 66.1A
0 (dc)
360
Za magn, W
Io ampl, A
4,369.2
680.8
2
2

36.9 A
 I o,2  7.61I o, 2
 Ra 
 1W 
  Ra
2
2
2


2
11.35
Total
Power
Loss
Output
Power
(66.1A)(150V)
7.14
.583
5,057.7
9,915
Output Time and Frequency
Response
250
300
 66.1300
 36.9 cos( 2
200
o
-20050
oh
0
oh
100
-100
V , V and I , A
s
0
s
120  t  75)  3.78 cos(2
Voltage
 240  t  82.4)  1.08Current
cos( 2
 360  t  85) A
vo
200
100
100 0
v , V and i , A
+
is
200
o
v,V

i ,A
io
io
Output Voltage
Output Spectra
Input Voltage and Current
150
0.005
0.01
0.015
0.02
0.025
0.01
0.015
0.02
0.01
0.02
2 0.015
4
Time,
s
Time,
sHarmonic
Number
0.025
0.025
6
Output Current
100
50
-300
0
00
0
0.005
0.005
0
-
Generalization: Fourier Series
The Fourier theorem states that a bounded periodic function f(t) with
limited finite number of discontinuities can be described by an infinite
series of sine and cosine terms of frequency that is the integer multiple
of the fundamental frequency of f(t):

f (t )  A0 
 A cos(nt)  B sin( nt)
n
n
1
T
Where
A0 
1
T

0
f (t )dt
is the zero frequency or average value of f(t).
Waveform Symmetry
Half Wave Symmetry
 Quarter Wave Symmetry

Odd
 Even

Half Wave Symmetry
 T
f (t )   f  t  
2

•Half-wave symmetry is independent of
the function shift w.r.t the time axis
•Even harmonics have zero coefficient
A0  A2  B2  A4  B4   B2   0
Square Wave
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Triangular
0.5
0.4
0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
-0.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Saw Tooth—Counter Example
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Quarter Wave Symmetry

Half wave and odd symmetry
 T
f (t )   f  t   and f (t )   f (t )
2


Half wave and even symmetry
 T
f (t )   f  t   and f (t )  f (t )
2

Half-wave: odd and even
f (t ) 
1
 A cos(nt)
n
n 1, 3,
0.5
0
-0.5
-1
-1.5
-1
-0.5
0
0.5
1
1.5
1
f (t ) 
0.5
0
n
n 1, 3,
-0.5
-1
-1.5
 B sin( nt)
-1
-0.5
0
0.5
1
1.5
Quarter Wave Symmetry Simplification
 /2
An 
4


f (t ) cos( nt )dt
0
 /2
Bn 
4


0
n  1,3, 
f (t ) sin( nt )dt
Forms of the Fourier Transform

Trigonometric

Combined Trigonometric

Exponential
•Trigonometric form

f (t )  A0 
 A cos(nt)  B sin( nt)
n
n
n 1
•Combined Trigonometric
An cos( nt )  Bn sin( nt ) 
f (t )  A0 

2
n
An  Bn cos( nt   n )
2
n 1, 2 ,
  Bn
 n  arctan 
 An

  Bn

A  B cos nt  arctan 
 An

2
n



2

 


•Exponential
An2  Bn2 j ( nt  n )
e j  e  j
2
2
cos  
 An  Bn cos( nt   n ) 
e
 e  j ( nt  n ) 
2
2
 A2  B 2

 A2  B 2

n
n
n
n
j n
 j n
jnt
  e  jnt  C n  e jnt  C  n  e  jnt

e e

e
2
2







1

f (t )e
T
C n e jnt
f (t ) 
n  
Cn
T
C n  C *n
 jnt
1
dt 
T

T
f (t )cos( nt )  j sin( nt ) dt

Im
Relations between the different
forms of the FS
C0  A0 ,
-jBn
j n
2Cn  An  jBn  2 Cn e2|C | ,
n
n
An
2 Cn 
Re
An  Bn ,
2
2
arg( Cn )   arg( Cn )   n , n  1,2,
Summary of FS Formulas
Series Type
f(t)=
A0 
A cos(nt )  B sin( nt )
1
A 
f (t ) dt
T 
2
A 
f (t ) cos( nt ) dt
T 
2
B 
f (t ) sin( nt ) dt , n  1,2 ,
T 
n
n
n 1, 2 ,
0
T
Trigonometric
n
T
n
Combined
Trigonometric (Phasor
form, Harmonic
amplitude and phase)
A0 

n 1, 2 ,
  Bn
An2  Bn2 cos( nt   n ),  n  arctan 
 An


C n e jnt , C n 
n  
Exponential
C 0  A0
Cn 
1
2
1
T
 f (t )e
,
T
An  j
C  n  C n*
 jnt



1
2
Bn 
1
2
An2  Bn2 e j n , n  1,2, 
 n changes
 n  na
Im
Time Shift
f (t ) shifts

 f (t  a)
C n changes
 C n e jna
C n , An2  Bn2 remain unchanged
n
a
-jBn
n|
2|C
n
An
Re
Example: SQP -90° Shift
1
1
0.8 0.8
original
shifted
0.6 0.6
n
2|Cn|
n
n/2
0.4 0.4
1
4/
0
/2
0.2 0.2
2
0
0
-
0
3
4/3

/2
-0.2 -0.2
4
0
0
-
-0.4 -0.4
5
4/5
0
/2
6
0
0
-
7
5/7

/2
n
4/n
(n-1)/2
-/2
0
-0.6
-0.8
-0.6
-0.8
-1
-1
-1.5
-1
-1.5
-0.5
-1
0
-0.5
0.5
0
1
0.5
1.5
1
1.5
Example: SQP -60° Shift
1
1
0.8
0.6
0.8
0.6
original
shifted
0.4 0.4
n
2|Cn|
n
n/3
0.2 0.2
1
4/
0
/3
0
2
0
0
-
-0.2 -0.2
3
4/3

0
-0.4 -0.4
4
0
0
-
-0.6
5
4/5
0
/3
-0.8
6
0
0
-
-1
7
5/7

2/3
n
4/n
1.5
(n-1)/2
(n-3)/6
0
-0.6
-0.8
-1
-1.5
-1.5
-1
-1
-0.5
-0.5
0
0
0.5
0.5
1
1
1.5
SPECTRUM: SQ. Pulse
(amplitude=1)
n
n
Magnitude,
|C | 2|C |
1.5
0.8
0.6
1
0.4
0.5
0.2
)
Harmonicarg(C
phase
n angle, rad
0
0
-20
1
2
-15
3
2
-15
3
4
-10
5
4
-10
5
6
-5
7
80
9
105 11 12
10 13 14
15 15
20
7
80 9 105 11 12
10 13 14
15 15
Harmonic
Freq.Order
20
0
4
-1
2
-2
0
-3
-2
-4
-4
-20
1
6
-5
42
n
SPECTRUM: Sawtooth
(amplitude=1)
n
Magnitude, 2|C |
0.8
2
n
0.6
0.4
0.2
Harmonic phase angle, rad
0
1
2
3
4
5
6
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15
5
4
3
2
1
0
7
8
9 10 11 12 13 14 15
Harmonic Order
SPECTRUM: Triangular wave
(amplitude=1)
n
Magnitude, 2|C |
1
0.8
0.4
0.2
0
Harmonic phase angle, rad
8
n 2
0.6
1
2
3
4
5
6
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15
1
0.5
0
-0.5
-1
7
8
9 10 11 12 13 14 15
Harmonic Order
SPECTRUM: Rectified SINE (peak=1)
4
 (n 2  1)
n
0.5
0.4
0
Magnitude: C and 2|C |, n=1,2..
0.6
0.3
0.2
0.1
0
0
2
4
6
8
Harmonic Order
10
12
14
Using FS to Find the Steady State Response of an LTI System
Input periodic, fundamental freq.=f1=60
Hz
Voltage
Division
+
+
+
+
5kW
5kW
1mF
uin
uout
Uout[n]=
Uin[n]
106/jn(2f1)
-
-
-
1
 U in [n]
1  j 0.005(2f1 )  n
-
=0.005 s
11
UH
out[ n]  U 1
U
[
n
]

[
n
]

 U in [n] 
Hout
( ) 


in
U in 1  jRC 1 1 jnj
1
At harmonic frequencies:
U out [n]  H [n]  U in [n] 
H [ n] 
U in [n]
1  (n1 )
2
U out [n]
1
1


U in [n] 1  jn1 RC 1  jn1
,


arg U out [n]  arg U in [n]  arg H [n]  arg U [n]  arctan( n  )
in
1
Square Pulse Excitation
Harm.
Order
u (t ) 
Circ. TF
|H(n)|,
out <H(n)
15
| U
1, 3,
| cos<Uin(n)
2n  f1  t  arg(|Uout(n)|,
Uout [n<Uout(n)
]) 
|Uin(n)|,
out [n ]
Inp.
Out.
0
0
0
0
0
0.4686
-1.083
1.2732
0
0.5967
-1.083
2
0.2564
-1.3115
0
0
0
0
3
0.1741
-1.3958
0.4244
-3.1416
0.0739
1.7458
4
0.1315
-1.4389
0
0
0
0
5
0.1055
-1.4651
0.2546
0
0.0269
-1.4651
6
0.0881
-1.4826
0
0
0
0
7
0.0756
-1.4952
0.1819
-3.1416
8
0.0662
-1.5046
0
0
0
0
9
0.0588
-1.5119
0.1415
0
0.0083
-1.5119
10
0.053
-1.5178
0
0
0
0
11
0.0482
-1.5226
0.1157
-3.1416
0.0056
1.619
12
0.0442
-1.5266
0
0
0
0
13
0.0408
-1.53
0.0979
0
0.004
-1.53
14
0.0379
-1.5329
0
0
0
0
15
0.0353
-1.5354
0.0849
-3.1416
0.003
1.6061
0
1
1
0.59670.0137
 cos(2 1 f1 1.6464
t  1.083)
SQUARE PULSE Excitation
input
output
1
0.5
0
0
5
10
15
1
0.5
0
-0.5
-1
-0.025 -0.02 -0.015 -0.01 -0.005
0
0.005
time, s
0.01
0.015
0.02
0.025
Rectified SINE Wave
Harm. Order
Inp.
|Uin(n)|, <Uin(n)
Circ. TF
|H(n)|, <H(n)
Out.
|Uout(n)|, <Uout(n)
0
1
0
0.6366
0
0.6366
0
1
0.4686
-1.083
0
0
0
0
2
0.2564
-1.3115
0.0849
3.1416
0.0218
1.8301
3
0.1741
-1.3958
0
0
0
0
4
0.1315
-1.4389
0.0202
3.1416
0.0027
1.7027
5
0.1055
-1.4651
0
0
0
0
6
0.0881
-1.4826
0.0089
3.1416
0.0008
1.659
7
0.0756
-1.4952
0
0
0
0
8
0.0662
-1.5046
0.005
3.1416
0.0003
1.637
9
0.0588
-1.5119
0
0
0
0
10
0.053
-1.5178
0.0032
3.1416
0.0002
1.6238
11
0.0482
-1.5226
0
0
0
0
12
0.0442
-1.5266
0.0022
3.1416
0.0001
1.615
13
0.0408
-1.53
0
0
0
0
14
0.0379
-1.5329
0.0016
3.1416
0.0001
1.6087
15
0.0353
-1.5354
0
0
0
0
Rect. SINE wave
0.6
input
output
0.4
0.2
0
0
5
10
15
1
0.8
0.6
0.4
0.2
0
-0.025 -0.02 -0.015 -0.01 -0.005
0
0.005
time, s
0.01
0.015
0.02
0.025
Total RMS of A Signal
Rewrite the FS as:
f (t )  A0 

2 Fn cos( nt   n )
n 1, 2
Nth harmonic rms
(except for n=0)
Total rms of the wave form:
Fn 
Frms 
An  Bn
2
2
1
2
f
(
t
)
dt

TT
2  2 Cn
Total RMS and the FS Terms
Frms 
2
1
T

T
f (t ) 2 dt 
1
T

T
2


 A0 
2 Fn cos( nt   n ) dt


n 1, 2



Using the orthogonality
between the terms:
Frms  A0 
2
2

Fn
n 1, 2
For ac wave forms (A0=0) it is convenient to write:
2
Frms
 F12 

Fn2  F12  FH2
n  2 , 3,
2
Waveform Quality-AC Signals

Fn2  F12  FH2
FH-Axis
2
Frms
 F12 
n  2 , 3,
F rm s
FH=√(F22+F32+…)
B
THDà increases
A
F1
F1-Axis
Total Harmonic Distortion Index
THD 
F22  F32  
F1
FH


F1
2
Frms
 F12
F1
Waveform Quality-DC SIgnals
(A0≠0)
2
Frms
 A02  F22  F42    Fdc2  Fac2
Fdc  A0
Fac 

Fn2
n  2 , 4 ,
F rm s
Fdc
Ripple Factor
F
RF  ac 
Fdc
2
Frms
 Fdc2
Fdc
Fac=√(F22+F42+…)
Example: W.F.Q. of the circuit driven by a Sq.P.
Inp. Rms:
|Uin(n)|/√2
Harm. Order
Out. Rms:
|Uout(n)|/√2
0
0
0
1
0.900288
0.421931
2
0
0
3
0.300096
0.052255
4
0
0
5
0.180029
0.019021
6
7

12  0.9 2  0.4359

2 – 0.42192)=0.0575{
0
√(0.4258
0
0.128623
0.009687
8
0
0
9
0.100056
0.005869
10
0
0
11
0.081812
0.00396
12
0
0
13
0.069226
0.002828
14
0
0
15
0.060033
0.002121
RMS
1.00
(excact)
0.4258
%THD
0.4359
100%  48.43% 48.43
(exact)
0.9
0.0575
100%  13.6%
0.4219
13.6
Example: WFQ of the circuit driven by a rect. sine
Inp. Rms:
|Uin(n)|/√2
Harm. Order
Out. Rms:
|Uout(n)|/√2
0
0.6366=Uin(0)
0.6366=Uout(0)
1
0
0
2
0.060033
0.015415
3
0
0
4
0.014284
0.001909
5
0
0
6
0.006293
0.000566
0.7707 2  0.6366 2  0.3078
0
0.63682  0.6366 2  0.01596
0
8
0.003536
0.000212
9
0
0
10
0.002263
0.000141
11
0
0
12
0.001556
7.07E-05
13
0
0
14
0.001131
7.07E-05
15
0
0
RMS
0.707
(exact)
0.6368
%RF
0.3078
100%  48.35%
0.6366
48.35
(exact)
0.01596
100%  2.5%
0.6366
2.5