1) Be able to perform a phasor transform and its inverse ... · Sinusoid Steady-State Analysis The...

Sinusoid Steady-State AnalysisObjectives:

1) Be able to perform a phasor transform

and its inverse;

2) Be able to phasor-transform a circuit;

3) Solve arbitrarily complex circuits with

sinusoidal sources using phasor

method.

Sinusoid Steady-State AnalysisReview the terminology of sinusoids

v(t) = Vmcos(t + ) V

•Vm : Amplitude or magnitude

• : angular frequency [rad/s]

•t : time [s]

• : phase angle [deg]

•f = /2 : frequency [Hz]

•T = 1/f : period [s]

What is the frequency of the sinusoid

given below?

A. 4t

B. 4 Hz

C. 4 rad/s

i(t) = 36 cos(4t + 45) mA

What is the phase angle of the

sinusoid given below?

A. 60 deg

B. 60 rad

C. 60 deg

v(t) = 50 cos(3000t 60) V

Suppose we want to find the value of the

sinusoidal current described below at

t = 5 ms. When we substitute this value

for t into the sinusoid, the argument for

the cosine function has what units?

A. degrees

B. radians

C. A mix of degrees

and radians

i(t) = 0.2 cos(50t + 45) A

Sinusoid Steady-State AnalysisEvaluating a sinusoid at a specified time

i(0.005) = 0.2 cos(50t + 45) A50 has the units rad/s

50t has the units radians

45 has the units degrees

Need to convert 50t to degrees, or 45 to

radians, and set your calculator

appropriately!

2/360 = 50(0.005)/x x = 45

Thus, i(0.005) = 0.2 cos(45 + 45) A

= 0.2 cos(90) = 0 A

Sinusoid Steady-State AnalysisSuppose v(t) = Vmcos(t + ) V in the circuit

below:

The equation for i(t) is )cos()()(

tVtRidt

tdiL m

The solution of this equation for i(t) is

)(tan

)cos()(

)cos()(

)(

1

22

)(

22

RL

tLR

Ve

LR

Vti mtLRm

where

Consider the equation for the current

in the RL circuit with the sinusoidal

source:

Which term(s) go to 0 as t ∞?

A. The first term

B. The second term

C. Both terms

D. Neither term

)cos()(

)cos()(

)(22

)(

22

t

LR

Ve

LR

Vti mtLRm

Sinusoid Steady-State Analysis

•The first term is the transient term – it

decays to 0 as t goes to infinity

•The second term is the steady-state term –

it persists for all time greater than 0. This

term

• Is sinusoidal

• Has the same frequency as the input voltage

• Has a different magnitude and phase angle

compared to the input voltage

)cos()(

)cos()(

)(22

)(

22

t

LR

Ve

LR

Vti mtLRm

Sinusoid Steady-State AnalysisThe circuit analysis techniques we are

studying will allow us to calculate the

sinusoidal steady-state response of the

circuit – that is, the circuit’s response

to a sinusoidal input once the transient

response has effectively decayed to 0.

This is also called the AC Steady-State

(ACSS) response.

Sinusoid Steady-State AnalysisPhasor

• A complex number in polar form, with a

magnitude and phase angle

• Derived from a sinusoid using the phasor

transform

jtj

m

jtj

m

tj

m

m

j

j

eeV

eeV

eV

tVtv

e

je

Re

Re

Re

Re

)(

)cos()(

cos

sincos

Now

:identitys Euler'

Sinusoid Steady-State AnalysisPhasor transform

• Extracts a sinusoid’s magnitude and phase

angle

• Transforms a function of time into a function of

frequency

m

j

mm VeVtV )cos(PV

Inverse phasor transform

• Turns a phasor back into a sinusoid, if

someone tells you the frequency

)cos()( tVVtv mm

-1-1 PP V

Suppose

i(t) = 36 cos(4t + 45) mA

Then the phasor transform of i(t) is

A. 36 mA

B. 3645 mA

C. 364t mA

Sinusoid Steady-State AnalysisAdding or subtracting sinusoids in the time

domain is hard (you need trig identities!) But

if you phasor-transform the sinusoids, it is

easy to combine them. For example,

A

A

A

)57.26300cos(18.11)(

57.2618.1113.531087.365

)13.53300cos(10)87.36300cos(5)(

1

tti

ttti

I

I

P


Consider the equations that relate

voltage and current in a resistor,

inductor, and capacitor in the time

domain, and in the phasor domain.


Time domainPhasor (frequency)

domain

P

)cos()()(

)cos()(

tRItRitv

tIti

m

m

m

m

RIR

I

IV

I

Note that the voltage and current phasors for a resistor

have the same phase angle – therefore we say that the

voltage and current in a resistor are “in phase”.

Resistors:


Time domain

Phasor (frequency) domain

P

)90cos(

)sin(

)()(

)cos()(

tLI

tLI

dttdiLtv

tIti

m

m

m

I

V

I

LjeLIj

jeLI

eeLI

eLI

I

j

m

j

m

jj

m

j

m

m

)(

90

)90(

In an inductor, the voltage leads the current by 90

Inductors:


Time domain

Phasor (frequency) domain

P

)90cos(

)sin(

)()(

)cos()(

tCV

tCV

dttdvCti

tVtv

m

m

m

V

I

V

CjeCVj

jeCV

eeCV

eCV

V

j

m

j

m

jj

m

j

m

m

)(

90

)90(

In a capacitor, the voltage lags the current by 90

Capacitors:

The impedance of a resistor is

A. Always a real number

B. Always an imaginary number

C. A complex number with real

and imaginary parts

The impedance of an inductor is

A. A real number

B. A positive imaginary

number

C. A negative imaginary

number

The impedance of a capacitor has the

units

A. Farads [F]

B. Henries [H]

C. Ohms []

Impedance is a phasor.

A. True

B. False

Sinusoid Steady-State AnalysisSummary:

• In the time domain

• Resistor: v(t) = Ri(t)

• inductor: v(t) = Ldi(t)/dt

• Capacitor: i(t) = Cdv(t)/dt

• In the phasor domain V = ZI

• Z is impedance, defined as the ratio of V to I

• Z has the units Ohms []

• Resistor: ZR = R

• Inductor: ZL = jL

• Capacitor: ZC = 1/jC = -j/C

Sinusoid Steady-State AnalysisSummary:

• In the time domain

• Ohm’s law (only for resistors): v = Ri

• KVL (around a loop): v1 + v2 + … + vn = 0

• KCL (at a node): i1 + i2 + … + in = 0

• These three laws lead to all other time-domain

circuit analysis techniques

• In the phasor domain

• Ohm’s law (for R, L, C): V = ZI

• KVL (around a loop): V1 + V2 + … + Vn = 0

• KCL (at a node): I1 + I2 + … + In = 0

• These three laws mean we can use all time-domain

circuit analysis techniques in the phasor domain!


Circuit in the

time domain

Circuit in the

phasor domain

Solution in the

time domain

Solution in the

phasor domain

P

P -1

Easy –

algebra!

Hard –

calculus!

Sinusoid Steady-State AnalysisSteps in ACSS Analysis:

1. Redraw the circuit (the phasor transform does not

change the components or their connections).

2. Phasor transform all known v(t) and i(t).

3. Represent unknown voltages and currents with V

and I.

4. Replace component values with impedance (Z)

values.

5. Use any circuit analysis method(s) to write

equations and solve them with a calculator.

6. Inverse-transform the result, which is a phasor,

back to the time domain.

Sinusoid Steady-State AnalysisExample 9.7

is(t) = 8cos(200,000t) A.

Find v(t), i1(t), and v2(t)

in the steady-state.

1. Redraw the circuit.

2. Phasor transform all

known voltages and

currents.

3. Represent unknown

voltages and

currents with phasor

symbols

The next step is to calculate the

impedance of all resistors, inductors,

and capacitors. Actually, the

impedance of the resistors doesn’t

require calculation.

A. True

B. False

The impedance of inductors and

capacitors is calculated using

A. The source frequency

B. The component value

C. Both A and B


is(t) = 8cos(200,000t) A.



4. Calculate the

impedances of all

resistors, inductors,

and capacitors.

5)1)(000,200(

8)40)(000,200(

jj

C

jZ

jjLjZ

C

L


is(t) = 8cos(200,000t) A.



5. Use DC circuit analysis techniques to find the

phasor V.

• In this problem, we’ll simplify the circuit by replacing

all impedances by one equivalent impedance, and

use Ohm’s law for phasors.

V)2532()08)(34(

)34(5||)86(||10

jjZ

jjjZeq

IV


is(t) = 8cos(200,000t) A.



6. Inverse phasor-transform the result to get back

to the time domain.

V

V V

)87.36000,200cos(4087.3640)(

87.3640)2432(

1

ttv

j

P

V

The most direct way to calculate the

phasor I1 is

A. Voltage division

B. Current division

C. Source transform


is(t) = 8cos(200,000t) A.




phasor I1.

• Use current division.

A )4.22.3(

)08(10

)34(

1

1

j

j

Z

Zs

eq

II


is(t) = 8cos(200,000t) A.




to the time domain.

A

A A

)87.36000,200cos(487.364)(

87.364)4.22.3(

1

1

1

tti

j

P

I

The most direct way to calculate the

phasor V2 is

A. Voltage division

B. Current division

C. Node voltage method


is(t) = 8cos(200,000t) A.




phasor V2.

• Use voltage division.

V )032(

)2432(86

82

j

jj

j

ZZ

Z

LR

L

VV


is(t) = 8cos(200,000t) A.




to the time domain.

V

V V

ttv

j

000,200cos32032)(

032)032(

1

2

2

P

V


Suppose we can vary the frequency of the current

source. What frequency will cause is(t) and v(t) to

be in-phase in the steady-state?

If is(t) and v(t) are in-phase in the

steady-state, this means

A. Their phase angles are the same.

B. The equivalent impedance seen by the

current source has a phase angle of 0.

C. The equivalent impedance seen by the

current source is purely resistive.

D. All of the above.


rad/s 000,50)40(

3614036

0)36(10100)36(10

)36(10101

)36(10

)36(10106036

)6)(6(10

)6)(6)((10

)6)(6(10

)6(10

)6)(6(10

)6)(6(

6

1

10

11

22

2

22

22

22

22

2222

L

CL

LCLL

LCL

Z

L

LCjLjL

LjLj

LjLjCj

LjLj

Lj

LjLj

LjLj

CjLjZ

eq

eq

Im


Suppose we can vary the frequency of the current

source. What frequency will cause is(t) and v(t) to

be in-phase in the steady-state?

420||)26(||10

20)1)(000,50(

2)40)(000,50(

000,50

jjZ

jj

C

jZ

jjLjZ

eq

C

L

rad/s For

P

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1) Be able to perform a phasor transform and its inverse ... · Sinusoid Steady-State Analysis The...

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