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?

16

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