Phasors

• A phasor is a complex number that represents the magnitude and phase of a sinusoid:

tX M cos

MXX

Complex Exponentials• A complex exponential is the mathematical tool

needed to obtain phasor of a sinusoidal function.• A complex exponential is

ejt = cos t + j sin t• A complex number A = z can be represented

A = z = z ej = z cos + j z sin • We represent a real-valued sinusoid as the real

part of a complex exponential.• Complex exponentials provide the link between

time functions and phasors.• Complex exponentials make solving for AC

steady state an algebraic problem

Complex Exponentials (cont’d)

Aejt = z ej ejt = z ej(t+

z ej(t+= z cos (t++ j z sin (t+

Re[Aejt] = z cos (t+

What do you get when you multiple A with ejwt

for the real part?

Sinusoids, Complex Exponentials, and Phasors

• Sinusoid:

z cos (t+Re[Aejt] • Complex exponential:

Aejt = z ej(t+z ej

• Phasor for the above sinusoid:

V = z

Phasor Relationships for Circuit Elements

• Phasors allow us to express current-voltage relationships for inductors and capacitors much like we express the current-voltage relationship for a resistor.

I-V Relationship for a Capacitor

Suppose that v(t) is a sinusoid:

v(t) = VM ej(t+

Find i(t).

C v(t)

+

-

i(t)

dt

tdvCti

)()(

Computing the Current

dt

edVC

dt

tdvCti

jtjM

)(

)(

)()( tCvjeCVjti jtjM

Phasor Relationship

• Represent v(t) and i(t) as phasors:

V = VM

I = jC V

• The derivative in the relationship between v(t) and i(t) becomes a multiplication by j in the relationship between V and I.

I-V Relationship for an Inductor

V = jL I

L v(t)

+

-

i(t)

dt

tdiLtv

)()(

Kirchhoff’s Laws

• KCL and KVL hold as well in phasor domain.

Impedance

• AC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm’s law:

V = I Z

• Z is called impedance.

Impedance (cont’d)

• Impedance depends on the frequency .• Impedance is (often) a complex number.• Impedance is not a phasor (why?).• Impedance allows us to use the same solution

techniques for AC steady state as we use for DC steady state.

• Impedance in series/parallel can be combined as resistors

Impedance Example:Single Loop Circuit

20k+

-1F10V 0 VC

+

-

= 377

Find VC

Impedance Example (cont’d)

• How do we find VC?

• First compute impedances for resistor and capacitor:

ZR = 20k= 20k 0

ZC = 1/j (377 1F) = 2.65k -90

Impedance Example (cont’d)

20k 0

+

-2.65k -9010V 0 VC

+

-

Impedance Example (cont’d)

Now use the voltage divider to find VC:

0k2090-k65.290-k65.2

0 10VCV

46.82 V31.154.717.20

9065.20 10VCV

Analysis Techniques

• All the analysis techniques we have learned for the linear circuits are applicable to compute phasors– KCL&KVL– node analysis/loop analysis– superposition– Thevenin equivalents/Notron equivalents– source exchange

• The only difference is that now complex numbers are used.

• Phasors can then converted to corresponding sinusoidal functions to get the time-varying function.

Impedance

• V = I Z, Z is impedance, measured in ohms ()

• Resistor:

– The impedance is R

• Inductor:

– The impedance is jL

• Capacitor:

– The impedance is 1/jC

Analysis Techniques

• All the analysis techniques we have learned for the linear circuits are applicable to compute phasors– KCL&KVL– node analysis/loop analysis– superposition– Thevenin equivalents/Notron equivalents– source exchange

• The only difference is that now complex numbers are used.

• Phasors can then converted to corresponding sinusoidal functions to get the time-varying function.

Admittance

• I = YV, Y is called admittance, the reciprocal of impedance, measured in siemens (S)

• Resistor:

– The admittance is 1/R

• Inductor:

– The admittance is 1/jL

• Capacitor:

– The admittance is jC

Phasor Diagrams

• A phasor diagram is just a graph of several phasors on the complex plane (using real and imaginary axes).

• A phasor diagram helps to visualize the relationships between currents and voltages.

An Example

2mA 40

–

1F VC

+

–

1k VR

+

+

–

V

• I = 2mA 40• VR = 2V 40

• VC = 5.31V -50• V = 5.67V -29.37

Phasor Diagram

Real Axis

Imaginary Axis

VR

VC

V

Homework #9

• How to determine initial conditions for a transient circuit. When a sudden change occurs, only two types quantities will definitely remain the same before and after the change. – IL(t), inductor current– Vc(t), capacitor voltage

• Find these two types of the values before the change and use them as initial conditions of the circuit after change