2 3 EE462L Waveforms Definitions PPT
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1
EE462L, Spring 2014Waveforms and Definitions
2
Instantaneous power p(t) flowing into the box
)()()( titvtp Circuit in a box, two wires
)(ti
)(tv+
−
)(ti
)()()()()( titvtitvtp bbaa )(tvaCircuit in a box,
three wires
)(tia+
−
)(tib
+
−)(tvb
)()( titi ba Any wire can be the voltage reference
Works for any circuit, as long as all N wires are accounted for. There must be (N – 1) voltage measurements, and (N – 1) current measurements.
3
Average value ofperiodic instantaneous power p(t)
Tot
otavg dttp
TP )(1
4
Two-wire sinusoidal case
)sin()sin()()()( tItVtitvtp oo
)cos(22
)cos(2
)(1 IVVIdttp
TP
Tot
otavg
),sin()( tVtv o )sin()( tIti o
2
)2cos()cos()( tVItp o
)cos( rmsrmsavg IVP Displacement power factor
Average power
zero average
5
Root-mean squared value of a periodic waveform with period T
Tot
otavg dttp
TP )(1
RVP rms
avg
2
Tot
otrms dttv
TV )(1 22
Apply v(t) to a resistor
Tot
otTot
otTot
otavg dttv
RTdt
Rtv
Tdttp
TP )(1)(1)(1 2
2
Compare to the average power expression
rms is based on a power concept, describing the equivalent voltage that will produce a given average power to a resistor
The average value of the squared voltage
compare
6
Root-mean squared value of a periodic waveform with period T
Tot
otorms dttV
TV )(sin1 222
Tot
oto
oTot
otorms
ttTVdtt
TVV
2
)(2sin2
)(2cos12
222
,2
22 VVrms
Tot
otrms dttv
TV )(1 22
For the sinusoidal case
2VVrms
),sin()( tVtv o
7
RMS of some common periodic waveforms
22
0
2
0
22 1)(1 DVDTTVdtV
Tdttv
TV
DTT
rms
DVVrms
Duty cycle controller
DT
T
V
0
0 < D < 1By inspection, this is the average value of
the squared waveform
8
RMS of common periodic waveforms, cont.
TTT
rms tTVdtt
TVdtt
TV
TV
03
3
2
0
23
2
0
22
31
T
V
0
3VVrms
Sawtooth
9
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
0
V
0
V
0
0
-V
V
0
3VVrms
V
0
V
0
10
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.
minmax II
0
)(tithe ripple
+
0
minI
the minimum value
)(timaxI
minI=
2
minmax IIIavg
avgI
11
RMS of common periodic waveforms, cont. 2
min2 )( ItiAvgIrms
2minmin
22 )(2)( IItitiAvgIrms
2minmin
22 )( 2)( ItiAvgItiAvgIrms
2min
minmaxmin
2minmax2
22
3I
III
IIIrms
2minmin
22
3IIIII PP
PPrms
minmax IIIPP Define
12
RMS of common periodic waveforms, cont.
2minPP
avgIII
222
223
PP
avgPPPP
avgPP
rmsIIIIIII
423
22
222 PP
PPavgavgPP
PPavgPP
rmsIIIIIIIII
222
243 avgPPPP
rms IIII
Recognize that
12
222 PPavgrms
III
avgI
)(ti
minmax IIIPP
2
minmax IIIavg
13
RMS of segmented waveformsConsider 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.
DTot
ot
Tot
DTot
Tot
otrms dttidtti
Tdtti
TI )()(1)(1 2222
avgI PPI
)(ti
DT (1-D)T
oI
DTot
ot
Tot
DTotrms dtti
TDTDdtti
DTDT
TI )(
)1(1)1()(11 222
avgIIn this example, is defined as the average value of the sawtooth portion
14
RMS of segmented waveforms, cont.
ToverDToverrms tiAvgTDtiAvgDTT
I D)-1( 2
22 )( )1()( 1
ToverDToverrms tiAvgDtiAvgDI D)-1( 2
22 )( )1()(
12)1(
2222 PPavgorms
IIDIDI a weighted average
DTot
ot
Tot
DTotrms dtti
TDTDdtti
DTDT
TI )(
)1(1)1()(11 222
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
15
150V
T T2 T 0
0V
Practice ProblemThe periodic waveform shown is applied to a 100Ω resistor. What value of α yields 50W average power to the resistor?
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Fourier series for any physically realizable periodic waveform with period T
11)90cos()sin()(
k
okokavg
kkokavg tkIItkIIti
ooo ffT 1
222
Tt
tavgoo
dttiT
I )(1
Tok dttkti
Ta 0 cos)(2
Tok dttkti
Tb 0 sin)(2
22)sin(
kk
kk
ba
a
22)cos(
kk
kk
ba
b
kk
kk
k ba
)cos()sin()tan(
22kkk baI
When using arctan, be careful to get the correct quadrant
17
Two interesting properties
Half-wave symmetry,
)()2
( tiTti
then no even harmonics
(remove the average value from i(t) before making the above test)
1)sin()(
kkok TtkITti
Time shift,
1sin
kokok ktkI
Thus, harmonic k is shifted by k times the fundamental angle shift
where the fundamental angle shift is .Too
18
Square wave
V
–V
T
T/2
tttVt
kVtv
oddkkooo
,1o 5sin
513sin
311sin4ksin14)(
19
Triangle wave
V
–V
T
T/2
oddkktos
k
Vtv ,1
o22 kc18)(
tttosV
ooo2 5cos2513cos
911c8
20
Half-wave rectified cosine wave
I
T/2
T
T/2
tkk
ItIIti ok
ko
cos
1112cos
2)(
,6,4,22
12/
tttItII
oooo
6cos3514cos
1512cos
312cos
2
21
Triac light dimmer waveshapes(bulb voltage and current waveforms are identical)
0 30 60 90 120 150 180 210 240 270 300 330 360
Angle
Cur
rent
α = 30º
0 30 60 90 120 150 180 210 240 270 300 330 360
Angle
Cur
rent
α = 90º
0 30 60 90 120 150 180 210 240 270 300 330 360
Angle
Cur
rent
α = 150º
22
Fourier coefficients for light dimmer waveform
,sin21
pVa
2sin
2111 pVb
,...7,5,3,)1cos()1cos(1
1)1cos()1cos(1
1
kkk
kkk
kV
a pk
,...7,5,3,)1sin()1sin(1
1)1sin()1sin(1
1
kkk
kkk
kV
b pk
Vp is the peak value of the underlying AC waveform
23
RMS in terms of Fourier Coefficients
1
222
2k
kavgrms
VVV
avgrms VV which means that
and that
2k
rmsV
V for any k
24
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)
25
21
2
2
2V
VTHD k
k
V
Total harmonic distortion − THD(for voltage or current)
26
Some measured current waveforms
-4
-2
0
2
4A
mpe
res
RefrigeratorTHDi = 6.3%
240V residential air conditionerTHDi = 10.5%
277V fluorescent light (magnetic ballast)
THDi = 18.5%
277V fluorescent light (electronic ballast)THDi = 11.6%
27
Some measured current waveforms, cont.
Microwave ovenTHDi = 31.9%
PCTHDi = 134%
Vacuum cleanerTHDi = 25.9%
28
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
-50
0
50
100
150
200
Volts
THDV = 5% considered to be the upper limit before problems are noticed
THDV = 10% considered to be terrible
29
Some measured current waveforms, cont.
5000HP, three-phase, motor drive
(locomotive-size)
Bad enough to cause many power electronic loads to malfunction
30
Now, back to instantaneous power p(t)
)()()( titvtp Circuit in a box, two wires
)(ti
)(tv+
−
)(ti
)()()()()( titvtitvtp bbaa )(tvaCircuit in a box,
three wires
)(tia+
−
)(tib
+
−)(tvb
)()( titi ba Any wire can be the voltage reference
31
Average power in terms of Fourier coefficients
1)sin()(
kkokavg tkVVtv
1)sin()(
kkokavg tkIIti
11)sin()sin()(
kkokavg
kkokavg tkIItkVVtp Messy!
Tt
tavgoo
dttpT
P )(1
32
Average power in terms of Fourier coefficients, cont.
Tt
tavgoo
dttpT
P )(1
1)cos(
22kkk
kkavgavgavg
IVIVP
rmskV ,
Cross products disappear because the product of unlike harmonics are themselves harmonics whose averages are zero over T!
rmskI ,
321 PPPPP dcavg
Due to the DC
Due to the 1st
harmonic
Due to the 2nd
harmonic
Due to the 3rd
harmonic
Harmonic power – usually small wrt. P1
Not wanted in an AC system
33
-120
-100
-80
-60
-40
-20
0
20
40
60
80
100
120
-120
-100
-80
-60
-40
-20
0
20
40
60
80
100
120
• 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
Fund. freq
Harmonic+
Consider a special case where one single harmonic is superimposed on a fundamental frequency sine wave
Using the combined waveform,
Combined
34
-120
-100
-80
-60
-40
-20
0
20
40
60
80
100
120
• Count the number of cycles of the harmonic, or the number of peaks of the harmonic
T
17
Single harmonic case, cont.Determine the order of the harmonic
35
• Estimate the peak-to-peak value of the harmonic where the fundamental is approximately constant
-120
-100
-80
-60
-40
-20
0
20
40
60
80
100
120
Single harmonic case, cont.Estimate the magnitude of the harmonic
Imagining the underlying fundamental, the peak value of the fundamental appears to be about 100
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
Thus, the peak value of the harmonic is about 15
36
Single harmonic case, cont.Estimate the RMS value of the waveform
22
02
217
21
1
22
22 VVVVVk
kavgrms
VVrms 5.71
222
51132
102252
152
100 V
Note – without the harmonic, the rms value would have been 70.7V (almost as large!)
37
Single harmonic case, cont.Estimate the THD of the waveform
21
217
21
217
21
2
2
2
2
2
2
2
V
V
V
V
V
V
THD k
k
15.010015
117 VVTHD
38-100
-80
-60
-40
-20
0
20
40
60
80
100
0 30 60 90 120 150 180 210 240 270 300 330 360
VoltageCurrent
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
39-100
-80
-60
-40
-20
0
20
40
60
80
100
0 30 60 90 120 150 180 210 240 270 300 330 360
VoltageCurrent
Determine voltage and current magnitudes and phase angles
Voltage sinewave has peak = 100V, phase angle = 0º
Current sinewave has peak = 50A, phase angle = -45º
, 0100~ VV AI 4550~
Using a sine reference,
40
The average power is
1)cos(
22kkk
kkavgavgavg
IVIVP
)cos(22
00 1111 IVPavg
)45(0cos2
502
100avgP
WPavg 1767
41
The equivalent series impedance is inductive because the current lags the voltage
eqeqeq LjRIVZ
45245500100
~~
414.1)45cos(2eqR
414.1)45sin(2eqL
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
42
C’s and L’s operating in periodic steady-stateExamine the current passing through a capacitor that is operating in periodic steady state. The governing equation is
dttdvCti )()( which leads to
tot
oto dtti
Ctvtv )(1)()(
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
),()( oo tvTtv
The conclusion is that
Tot
otoo dtti
CtvTtv )(10)()(or
0)( Tot
otdtti
the average current through a capacitor operating in periodic steady state is zero
which means that
43
Now, an inductorExamine the voltage across an inductor that is operating in periodic steady state. The governing equation is
dttdiLtv )()( which leads to
tot
oto dttv
Ltiti )(1)()(
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
),()( oo tiTti
The conclusion is that
Tot
otoo dttv
LtiTti )(10)()(or
0)( Tot
otdttv
the average voltage across an inductor operating in periodic steady state is zero
which means that
44
KVL and KCL in periodic steady-state
,0)(
loopAroundtv
,0)(
nodeofOutti
0)()()()( 321 tvtvtvtv N
Since KVL and KCL apply at any instance, then they must also be valid in averages. Consider KVL,
0)()()()( 321 titititi N
0)0(1)(1)(1)(1)(1321
dt
Tdttv
Tdttv
Tdttv
Tdttv
T
Tot
ot
Tot
otN
Tot
ot
Tot
ot
Tot
ot
0321 Navgavgavgavg VVVV
The same reasoning applies to KCL
0321 Navgavgavgavg IIII
KVL applies in the average sense
KCL applies in the average sense
45
KVL and KCL in the average sense
Consider the circuit shown that has a constant duty cycle switch
R1
LV
+
VLavg = 0
−
+ VRavg −+ VSavg −
Iavg
A DC multimeter (i.e., averaging) would show
R2
0 A
Iavg
and would show V = VSavg + VRavg
46
KVL and KCL in the average sense, cont.
Consider the circuit shown that has a constant duty cycle switch
R1
CV
+
VCavg
−
+ VRavg −+ VSavg −
Iavg
A DC multimeter (i.e., averaging) would show
R2
0
and would show V = VSavg + VRavg + VCavg
Iavg
47
+
Vin –
– Vout
+ iL L C iC
Iout
id iin
4a. Assuming continuous conduction in L, and ripple free Vout and Iout , draw the “switch
closed” and switch open” circuits and use them to develop the in
outVV
equation.
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.
Practice Problem
48
12
10
8
6
4
2
0
−2
−4 0 T/2 T
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