Download Waveforms and definitions class notes

EE462L, Spring 2014
Waveforms and Definitions
1
Instantaneous power p(t) flowing into the box
i (t )
+
v (t )
p (t )  v(t )  i (t )
Circuit in a box,
two wires
−
i (t )
ia (t )
+
p(t )  va (t )  ia (t )  vb (t )  ib (t )
va (t )
−
Any wire can be the
voltage reference
ib (t )
+
vb (t )
Circuit in a box,
three wires
−
ia (t )  ib (t )
Works for any circuit, as long as all N wires are accounted for. There must
2
be (N – 1) voltage measurements, and (N – 1) current measurements.
Average value of
periodic instantaneous power p(t)
1 t o T
Pavg  
p(t )dt
T to
3
Two-wire sinusoidal case
v(t )  V sin( ot   ),
i(t )  I sin( ot   )
p(t )  v(t )  i(t )  V sin( ot   )  I sin( ot   )
zero average
 cos(   )  cos( 2ot     ) 
p(t )  VI 

2


1 t o T
VI
V I
Pavg  
p(t )dt  cos(   ) 
cos(   )
T to
2
2 2
Pavg  Vrms I rms cos(   )
Average power
Displacement power factor
4
Root-mean squared value of a
periodic waveform with period T
2
Vrms
1 t o T 2
 
v (t )dt
T to
Compare to the average power
expression
1 t o T
Pavg  
p(t )dt
T to
The average value of the squared voltage
compare
Apply v(t) to a resistor
1 t o T
1 to T  v 2 (t ) 
1 t o T 2
Pavg  
p(t )dt  
v (t )dt

 dt 

t
t
t
T o
T o  R 
RT o
2
Vrms
Pavg 
R
rms is based on a power concept, describing the
equivalent voltage that will produce a given
average power to a resistor
5
Root-mean squared value of a periodic
waveform with period T
1 t o T 2
2
Vrms  
v (t )dt
T to
For the sinusoidal case v(t )  V sin( ot   ),
2
Vrms
1 t o T 2 2
 
V sin (ot   )dt
T to
2 t T
2
V
V
o
2
Vrms

1  cos 2(ot   ) dt 
2T to
2T



t T
 sin 2(ot   )  o
t 

2

o

 to
2
V
V
2
Vrms 
, Vrms 
2
2
6
RMS of some common periodic waveforms
Duty cycle controller
V
0<D<1
By inspection, this is
the average value of
the squared
waveform
0
DT
T
T
DT
0
0
1 2
1
2
Vrms   v (t )dt 
T
T
2
V
2
2
V
dt


DT

DV

T
Vrms  V D
7
RMS of common periodic waveforms, cont.
Sawtooth
V
0
T
T
2
2T
2
1
V
V
V


2
2
3T
Vrms    t  dt 
t dt 
t

3
3
0
T T 
T
3
T
0
0
V
Vrms 
3
8
RMS of common periodic waveforms, cont.
Using the power concept, it is easy to reason that the following waveforms
would all produce the same average power to a resistor, and thus their rms
values are identical and equal to the previous example
V
V
0
0
0
-V
V
V
V
0
0
0
V
0
V
Vrms 
3
9
RMS of common periodic waveforms, cont.
Now, consider a useful example, based upon a waveform that is often seen in
DC-DC converter currents. Decompose the waveform into its ripple, plus its
minimum value.
i (t )
 Imax  Imin 
the ripple
i (t )
Imax
0
I avg
=
Imin
+
the minimum value
I avg 
Imax  Imin 
2
Imin
0
10
RMS of common periodic waveforms, cont.

2
I rms
 Avg i (t )  I min 2


2
2
I rms
 Avg i2 (t )  2i (t )  I min  I min

 
2
2
I rms
 Avg i2 (t )  2 I min  Avg i (t ) I min
2
I rms

I max  I min 2

 2I
3
I max  I min   I 2

min
min
2
Define I PP  I max  I min
2
I PP
2
2
I rms 
 I min I PP  I min
3
11
RMS of common periodic waveforms, cont.
I
Recognize that I min  I avg  PP
2
2
I rms
2
I PP
I 
I 



  I avg  PP  I PP   I avg  PP 
3 
2 
2 

2
2
2
2
I PP
I PP
I PP
2
2
I rms 
 I avg I PP 
 I avg  I avg I PP 
3
2
2
2
I PP
I PP
2
2
I rms 

 I avg
3
4
i (t )
4
2
I PP
2
2
I rms  I avg 
I avg
I avg 
I max  I min 
2
I PP  I max  I min
12
12
RMS of segmented waveforms
Consider a modification of the previous example. A constant value exists
during D of the cycle, and a sawtooth exists during (1-D) of the cycle.
i (t )
I avg
In this example, I avg is defined as
the average value of the sawtooth
portion
I PP
Io
DT (1-D)T
t T
to T
 t  DT

1 o
1o
2
2
2
2
I rms 
i ( t )dt 
i ( t )dt   i ( t )dt 



T
T
to
to  DT
 to


1
1
2
I rms
  DT 
T

DT
to  DT

to
to T

1
2
i ( t )dt  (1  D )T 
i ( t )dt 


(1  D )T
to  DT

2
13
RMS of segmented waveforms, cont.

1
1
2
I rms
  DT 
T

2
I rms

to  DT
DT


to
to T

1
2
i ( t )dt  (1  D )T 
i ( t )dt 


(1  D )T
to  DT

2
 
 
1
DT  Avg i 2 ( t ) over DT  (1  D)T  Avg i 2 ( t ) over (1-D)T
T
 

 
2
I rms
 D  Avg i 2 ( t ) over DT  (1  D)  Avg i 2 ( t ) over (1-D)T
2 
 2
I PP
2
2
I rms  D  I o  (1  D )   I avg 


12 
a weighted average
So, the squared rms value of a segmented waveform can be
computed by finding the squared rms values of each segment,
weighting each by its fraction of T, and adding
14
Practice Problem
The periodic waveform shown is applied to a 100Ω resistor.
What value of α yields 50W average power to the resistor?
150V
0V
0

T
2
T
T
15
Fourier series for any physically realizable
periodic waveform with period T
i ( t )  I avg 
T


k 1
k 1
 I k sin( kot   k )  I avg   I k cos(kot   k  90o )
2
o

2
1

2f o f o
1 t T
Iavg   o i ( t )dt
T to
I k  ak2  bk2
sin( k ) 
ak 
2 T
i ( t ) coskot dt
T 0
cos( k ) 
bk 
2 T
i ( t ) sinkot dt
T 0
tan( k ) 
ak
ak2  bk2
bk
ak2  bk2
sin( k ) ak

cos( k ) bk
When using arctan, be careful
to get the correct quadrant
16
Two interesting properties
Half-wave symmetry,
T
i(t  )   i(t )
2
then no even harmonics
(remove the average
value from i(t) before
making the above test)
Time shift,
i ( t  T ) 

 I k sin( ko t  T    k )
k 1


 I k sinko t   k  k o 
k 1
where the fundamental angle
shift is  o  oT .
Thus, harmonic k is shifted by k
times the fundamental angle shift
17
Square wave
T/2
V
–V
T
v(t ) 

1
4V
sinko t  
 k 1, k odd k

4V

1
1








sin
1

t

sin
3

t

sin
5

t


o
o
o


3
5
18
Triangle wave
T/2
V
–V
T
v(t ) 

8V


1
 k 1, k odd k
2
2
cosko t 
8V 
1
1







c
os
1

t

cos
3

t

cos
5

t


o
o
o


9
25
2 
19
Half-wave rectified cosine wave
T
I
T/2
T/2

I
2I
k / 21 1


i ( t )   coso t  

1
cosko t 

2
 2
 k  2,4,6,
k 1
I

I
2I  1
1
1







 coso t  
cos
2

t

cos
4

t

cos
6

t


o
o
o

 2
  3
15
35
I
20
Triac light dimmer waveshapes
(bulb voltage and current waveforms are identical)
Current
α = 30º
0
30 60
90 120 150 180 210 240 270 300 330 360
Angle
Current
α = 90º
0
30 60 90 120 150 180 210 240 270 300 330 360
Angle
Current
α = 150º
0
30
60
90
120
150
180
210
240
270
300
330
360
Angle
21
Fourier coefficients for light dimmer waveform
a1 
Vp
 1

sin 2  , b1  V p 1  
sin
2

  2


Vp  1
1





ak 
cos(
1

k
)


cos(
1

k
)


cos(
1

k
)


cos(
1

k
)

 , k  3,5,7,...
 1  k
1 k
bk 
Vp  1
1





sin(
1

k
)


sin(
1

k
)


sin(
1

k
)


sin(
1

k
)

, k  3,5,7,...


 1  k
1 k

Vp is the peak value of the underlying AC waveform
22
RMS in terms of Fourier Coefficients

2
V
2
Vrms 2  Vavg
 k
2
k 1
which means that Vrms
 Vavg
and that
Vrms 
Vk
2
for any k
23
Bounds on RMS
From the power concept, it is obvious that the rms voltage or
current can never be greater than the maximum absolute value
of the corresponding v(t) or i(t)
From the Fourier concept, it is obvious that the rms voltage or
current can never be less than the absolute value of the
average of the corresponding v(t) or i(t)
24
Total harmonic distortion − THD
(for voltage or current)

2
V
 k
THDV 2  k  22
V1
25
4
40
2
20
Amperes
Amperes
Some measured current waveforms
0
-2
-20
-4
-40
240V residential air conditioner
THDi = 10.5%
Refrigerator
THDi = 6.3%
0.2
0.15
0.1
0.05
0
-0.05
-0.1
-0.15
-0.2
277V fluorescent light (electronic ballast)
THDi = 11.6%
0.3
0.2
Amperes
Amperes
0
0.1
0
-0.1
-0.2
-0.3
277V fluorescent light (magnetic ballast)
THDi = 18.5%
26
Some measured current waveforms, cont.
12
8
Amperes
Amperes
4
0
-4
-8
-12
Vacuum cleaner
THDi = 25.9%
25
20
15
10
5
0
-5
-10
-15
-20
-25
Microwave oven
THDi = 31.9%
Amperes
5
0
-5
PC
THDi = 134%
27
Resulting voltage waveform at the service panel
for a room filled with PCs
THDV = 5.1%
(2.2% of 3rd, 3.9% of 5th, 1.4% of 7th)
200
150
100
Volts
50
0
-50
-100
-150
-200
THDV = 5% considered to be the upper limit before problems are
noticed
THDV = 10% considered to be terrible
28
Some measured current waveforms, cont.
Bad enough to cause
many power electronic
loads to malfunction
5000HP, three-phase, motor drive
(locomotive-size)
29
Now, back to instantaneous power p(t)
i (t )
+
v (t )
p (t )  v(t )  i (t )
Circuit in a box,
two wires
−
i (t )
ia (t )
+
p(t )  va (t )  ia (t )  vb (t )  ib (t )
Any wire can be the
voltage reference
ib (t )
va (t )
−
+
vb (t )
Circuit in a box,
three wires
−
ia (t )  ib (t )
30
Average power in terms of Fourier coefficients
v ( t )  Vavg 
i ( t )  I avg 

Vk sin( kot   k )
k 1

 I k sin( kot   k )
k 1



 

p( t )  Vavg  Vk sin( ko t   k )   I avg   I k sin( ko t   k )
k 1
k 1

 

Pavg 
Messy!
1 t o T
p( t )dt

t
T o
31
Average power in terms of Fourier coefficients,
cont.
Pavg 
1 t o T
p( t )dt
T to
 V
I
Pavg  Vavg  I avg   k  k cos( k   k )
2
k 1 2
Not wanted in an AC
system
Pavg  Pdc
Due to
the DC

Cross products disappear because
the product of unlike harmonics are
themselves harmonics whose
averages are zero over T!
Vk ,rms I k ,rms
P1
Due to
the 1st
harmonic

P2
Due to
the 2nd
harmonic

P3
Due to
the 3rd
harmonic


Harmonic power – usually
small wrt. P1
32
Consider a special case where one single harmonic is
superimposed on a fundamental frequency sine wave
120
120
100
100
80
60
Fund. freq
+
Harmonic
80
Combined
60
40
40
20
20
0
0
-20
-20
-40
-40
-60
-60
-80
-80
-100
-100
-120
-120
Using the combined waveform,
• Determine the order of the harmonic
• Estimate the magnitude of the harmonic
• From the above, estimate the RMS value of the waveform,
• and the THD of the waveform
33
Single harmonic case, cont.
Determine the order of the harmonic
• Count the number of cycles of the harmonic,
or the number of peaks of the harmonic
120
100
80
60
40
20
17
0
-20
-40
-60
-80
-100
-120
T
34
Single harmonic case, cont.
Estimate the magnitude of the harmonic
• Estimate the peak-to-peak value of the harmonic where
the fundamental is approximately constant
Viewed near the peak of the underlying fundamental (where the fundamental is reasonably
constant), the peak-to-peak value of the harmonic appears to be about 30
120
100
Imagining the underlying fundamental, the peak
value of the fundamental appears to be about 100
80
60
40
20
0
-20
-40
-60
-80
-100
-120
Thus, the peak value of the
harmonic is about 15
35
Single harmonic case, cont.
Estimate the RMS value of the waveform
2
2
 V2
V
V
2
Vrms 2  Vavg
  k  02  1  17
2
2
k 1 2
1002 152 10225



 5113V 2
2
2
2
Vrms  71.5V
Note – without the harmonic, the rms value would
have been 70.7V (almost as large!)
36
Single harmonic case, cont.
Estimate the THD of the waveform
 V2
k
2
V
17
 2
2
V
THD2  k  2
 2  17
V12
V12 V12
2
2
V17 15
THD 

 0.15
V1 100
37
Given single-phase v(t) and i(t) waveforms for a load
• Determine their magnitudes and phase angles
• Determine the average power
• Determine the impedance of the load
• Using a series RL or RC equivalent, determine the R and L
or C
100
80
60
40
20
Voltage
0
0
30
60
90
120
150
180
210
240
270
300
330
360
Current
-20
-40
-60
-80
-100
38
Determine voltage and current magnitudes and phase angles
Voltage sinewave has peak = 100V, phase angle = 0º
Current sinewave has peak = 50A, phase angle = -45º
Using a sine reference,
~
~
V  1000 V , I  50  45 A
100
80
60
40
20
Voltage
0
0
30
60
90
120
150
180
210
240
270
300
330
360
Current
-20
-40
-60
-80
-100
39
The average power is
 V
Ik
k
Pavg  Vavg  I avg  

cos( k   k )
2
k 1 2
V1 I1
Pavg  0  0 

cos( 1  1 )
2
2
Pavg 
100 50

cos0  ( 45)
2
2
Pavg  1767W
40
The equivalent series impedance is inductive
because the current lags the voltage
~
V
1000
Z eq  ~ 
 245  Req  jLeq
I 50  45
Req  2 cos(45 )  1.414
Leq  2 sin( 45)  1.414
where ω is the radian frequency (2πf)
If the current leads the voltage, then the impedance angle is
negative, and there is an equivalent capacitance
41
C’s and L’s operating in periodic steady-state
Examine the current passing through a capacitor that is operating
in periodic steady state. The governing equation is
i(t )  C
dv ( t )
dt
which leads to
t
1 o t
v ( t )  v ( to ) 
i ( t )dt

C
to
Since the capacitor is in periodic steady state, then the voltage at
time to is the same as the voltage one period T later, so
v ( to  T )  v ( to ), or
t
1 o T
v ( to  T )  v ( to )  0 
i ( t )dt
C 
to T
The conclusion is that
 i (t )dt  0
to
which means that
to
the average current through a capacitor operating in periodic
steady state is zero
42
Now, an inductor
Examine the voltage across an inductor that is operating in
periodic steady state. The governing equation is
v(t )  L
di ( t )
dt
which leads to
t
1 o t
i ( t )  i ( to ) 
v ( t )dt

L
to
Since the inductor is in periodic steady state, then the voltage at
time to is the same as the voltage one period T later, so
i ( to  T )  i ( to ), or
t
1 o T
i ( to  T )  i ( to )  0 
v ( t )dt
L 
to T
The conclusion is that
 v( t )dt  0
to
which means that
to
the average voltage across an inductor operating in periodic
steady state is zero
43
KVL and KCL in periodic steady-state
Since KVL and KCL apply at any instance, then they must also be valid
in averages. Consider KVL,
 v(t )
 0, v1 ( t )  v2 ( t )  v3 ( t )    v N ( t )  0
Around loop
t
t
t
t
t
1 o T
1 o T
1 o T
1 o T
1 o T
v1 ( t )dt 
v2 ( t )dt 
v3 ( t )dt   
v N ( t )dt 
(0)dt  0
T 
T 
T 
T 
T 
to
to
to
V1avg  V2avg  V3avg    VNavg  0
to
to
KVL applies in the average sense
The same reasoning applies to KCL
 i (t )
 0,
i1 ( t )  i2 ( t )  i3 ( t )    i N ( t )  0
Out of node
I1avg  I 2avg  I 3avg    I Navg  0
KCL applies in the average sense
44
KVL and KCL in the average sense
Consider the circuit shown that has a constant duty cycle switch
+ VSavg −
+ VRavg −
0A
R1
Iavg
+
L
V
Iavg
VLavg = 0
R2
−
A DC multimeter (i.e., averaging) would show
and would show V = VSavg + VRavg
45
KVL and KCL in the average sense, cont.
Consider the circuit shown that has a constant duty cycle switch
+ VSavg −
+ VRavg −
R1
Iavg
+
C
V
0
Iavg
VCavg
R2
−
A DC multimeter (i.e., averaging) would show
and would show V = VSavg + VRavg + VCavg
46
Practice Problem
iin
+
Vin
–
id
iL
L
C
iC
–
Vout
+
Iout
4a.
Assuming continuous conduction in L, and ripple free Vout and Iout , draw the “switch
V
closed” and switch open” circuits and use them to develop the out equation.
Vin
4b.
Consider the case where the converter is operating at 50kHz, Vin = 40V, Vout = 120V, P =
240W. Components L = 100µH, C = 1500µF. Carefully sketch the inductor and capacitor
currents on the graph provided.
4c.
Use the graphs to determine the inductor’s rms current, and the capacitor’s peak-to-peak
current.
4d.
Use the graphs to determine the capacitor’s peak-to-peak ripple voltage.
47
12
10
8
6
4
2
0
−2
−4
0
T/2
T
48