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

R

VP rmsavg

2

Tot

otrms dttv

TV )(

1 22

Apply v(t) to a resistor

Tot

ot

Tot

ot

Tot

otavg dttv

RTdt

R

tv

Tdttp

TP )(

1)(1)(

1 22

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 )(sin

1 222

Tot

oto

oTot

otorms

tt

T

Vdtt

T

VV

2

)(2sin

2)(2cos1

2

222

,2

22 V

Vrms

Tot

otrms dttv

TV )(

1 22

For the sinusoidal case

2

VVrms

),sin()( tVtv o

7

RMS of some common periodic waveforms

22

0

2

0

22 1)(

1DVDT

T

VdtV

Tdttv

TV

DTT

rms

DVVrms

Duty cycle controller

DT

T

V

0

0 < D < 1

By inspection, this is the average value of

the squared waveform

8

RMS of common periodic waveforms, cont.

TTT

rms tT

Vdtt

T

Vdtt

T

V

TV

0

33

2

0

23

2

0

22

3

1

T

V

0

3

VVrms

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

3

VVrms

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

)(ti

maxI

minI=

2

minmax IIIavg

avgI

11

RMS of common periodic waveforms, cont.

2min

2 )( ItiAvgIrms

2minmin

22 )(2)( IItitiAvgIrms

2minmin

22 )( 2)( ItiAvgItiAvgIrms

2min

minmaxmin

2minmax2

22

3I

III

IIIrms

2minmin

22

3III

II PP

PPrms

minmax IIIPP Define

12

RMS of common periodic waveforms, cont.

2minPP

avgI

II

222

223

PP

avgPPPP

avgPP

rmsI

III

II

I

423

22

222 PP

PPavgavgPP

PPavgPP

rmsI

IIII

III

I

222

2

43 avgPPPP

rms III

I

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 Problem

The 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

2

22

Tt

tavgo

odtti

TI )(

1

Tok dttkti

Ta

0cos)(

2

Tok dttkti

Tb

0sin)(

2

22)sin(

kk

kk

ba

a

22)cos(

kk

kk

ba

b

k

k

k

kk b

a

)cos(

)sin()tan(

22kkk baI

When using arctan, be careful to get the correct quadrant

17

Two interesting properties

Half-wave symmetry,

)()2

( tiT

ti

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

ttt

Vt

k

Vtv

oddkkooo

,1o 5sin

5

13sin

3

11sin

4ksin

14)(

19

Triangle wave

V

–V

T

T/2

oddkktos

k

Vtv

,1o22

kc18

)(

tttos

Vooo2

5cos25

13cos

9

11c

8

20

Half-wave rectified cosine wave

I

T/2

T

T/2

tkk

It

IIti o

k

ko

cos

1

11

2cos

2)(

,6,4,22

12/

ttt

It

IIoooo

6cos

35

14cos

15

12cos

3

12cos

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

Cu

rre

nt

α = 30º

0 30 60 90 120 150 180 210 240 270 300 330 360

Angle

Cu

rre

nt

α = 90º

0 30 60 90 120 150 180 210 240 270 300 330 360

Angle

Cu

rre

nt

α = 150º

22

Fourier coefficients for light dimmer waveform

,sin21

pV

a

2sin2

111 pVb

,...7,5,3,)1cos()1cos(1

1)1cos()1cos(

1

1

kkk

kkk

k

Va

pk

,...7,5,3,)1sin()1sin(1

1)1sin()1sin(

1

1

kkk

kkk

k

Vb

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

2

1

2

2

2

V

V

THD kk

V

Total harmonic distortion − THD(for voltage or current)

26

Some measured current waveforms

-4

-2

0

2

4A

mp

eres

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

Vo

lts

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

tavgo

odttp

TP )(

1

32

Average power in terms of Fourier coefficients, cont.

Tt

tavgo

odttp

TP )(

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 VVVVV

k

kavgrms

VVrms 5.71

222

51132

10225

2

15

2

100V

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

15.0100

15

1

17 V

VTHD

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

Voltage

Current

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

Voltage

Current

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 IV

Pavg

)45(0cos2

50

2

100avgP

WPavg 1767

41

The equivalent series impedance is inductive because the current lags the voltage

eqeqeq LjRI

VZ

4524550

0100~

~

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-state

Examine the current passing through a capacitor that is operating in periodic steady state. The governing equation is

dt

tdvCti

)()( which leads to

tot

oto dtti

Ctvtv )(

1)()(

Since the capacitor is in periodic steady state, then the voltage at time t o 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

ot

dtti

the average current through a capacitor operating in periodic steady state is zero

which means that

43

Now, an inductor

Examine the voltage across an inductor that is operating in periodic steady state. The governing equation is

dt

tdiLtv

)()( which leads to

tot

oto dttv

Ltiti )(

1)()(

Since the inductor is in periodic steady state, then the voltage at time t o 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

ot

dttv

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

)(1

321

dtT

dttvT

dttvT

dttvT

dttvT

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

out

V

V 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