INC 111 Basic Circuit Analysis

Week 11Force Response of

a Sinusoidal Input andPhasor Concept

Two Types of Analysis

• non-periodic electric source

• periodic electric source

(Transient response analysis of a step input)

(Steady state response analysis of a sinusoidal input)

Forced Response of Sinusoidal Input

In this part of the course, we will focus on finding the force responseof a sinusoidal input.

• Start oscillate from stop

input

displacement Period that have transient

• Have oscillated for a long time

input

displacement

We will only be interested in this case for force response (not count the transient)

TheoryForce response of a sinusoidal input is also a sinusoidal signalwith the same frequency but with different amplitude and phase shift.

AC

+

-

5sin(3t+π/3)

i(t)

3H

2Ω

v2(t) Sine wave

Sine wave

v1(t) Sine wave

vL(t) Sine wave

Input

Output

Phase shift

Amplitude

AC

+

-

sin(t)

i(t)

RC

What is the relationship between sin(t) and i(t) ?

sin(t)

i(t)

Phase shift

AC

+

-10sin(2πt + π/4) 2Ω

i(t)

Find i(t)

)4/2sin(5)(2

)4/2sin(10)(

2)()4/2sin(10)()(

tti

tti

titRtitv

Note: Only amplitude changes, frequency and phase still remain the same.

R circuit

AC

+

-

Asin(ωt)

i(t)

L Find i(t)

)2

(sin)cos(

cossin

sin1)(1)(

)()(

tLAt

LA

tLAtdt

LA

tdtAL

dttvL

ti

dttdiLtvfrom

L circuit

AC

+

-

Asin(ωt) R

i(t)

AC

+

-

Asin(ωt)

i(t)

L

)2

(sin)(

tLAti )(sin)( t

RAti

ωL เรยก ความตานทานเสมอน (impedance) Phase shift -90

Phasor Diagram of an inductor

v

i

Phasor Diagram of a resistor

vi

Note: No power consumed in inductorsi lags v 90o

AC

+

-

Asin(ωt)

i(t)

C Find i(t)

)2

sin(1

)(cos

)sin()()(

t

C

AtCA

dttAdC

dttdvCti

ความตานทานเสมอน (impedance) Phase shift +90

C circuit

Phasor Diagram of a capacitor

v

i

Phasor Diagram of a resistor

vi

Note: No power consumed in capacitorsi leads v 90o

Kirchhoff's Law with AC Circuit

AC

+

-

v(t)

i(t)

RC

vRi

vC

i

)13.53sin(5

)90sin(4)sin(3)(o

o

t

tttv

v(t)

)sin(3)( ttvR

)90sin(4)( oC ttv

KCL,KVL still hold.

)90sin(4)sin(3

)cos(4)sin(3)13.53sin()cos(5)13.53cos()sin(5

)13.53sin(5)(

o

oo

o

tt

tttt

ttv

This is similar to adding vectors.Therefore, we will represent sine voltage with a vector.

4

3

5

Vector QuantityComplex numbers can be viewed as vectors whereX-axis represents real partsY-axis represents imaginary parts

There are two ways to represent complex numbers.

• Cartesian form 3+j4

• Polar form 5∟53o

Operation add, subtract, multiply, division?

Interchange Rectangular, Polar form

jba

a

b

r

θ

r

sincos

arctan

22

rbra

abbar

Complex Number Forms(Rectangular, Polar Form)

s = 4 + j3

jω

σ

3

4

Rectangular form: 4 + j3

Polar form magnitude=5, angle = 37

บวก ลบ คณ หาร vector ??

Rectangular form

Add, Subtraction

Polar form

45)1()34( jjj

Multiplication

2510)1237(25

122375

Division

4925

122375

ImpedanceCompare to ohm’s law, impedance is a ratio of V/I in when V and I is in the vector format.

AC

+

-

Asin(ωt)

i(t)

L

Inductor

LjL

LAA

IVZ

LAt

LAti

AtAtv

o

o

o

o

o

9090

0

90)2

(sin)(

0)(sin)(

AC

+

-

Asin(ωt)

i(t)

C

CjCCAA

IVZ

CAtCAti

AtAtv

oo

o

o

o

190

190

0

90)2

(sin)(

0)(sin)(

Capacitor

1Ω

3H

Example: Find impedance in form of polar value for ω = 1/3 rad/sec

Note: Impedance depends on frequency and R,L,C values

45213131 jjLjR

Rules that can be used inPhasor Analysis

• Ohm’s law

• KVL/KCL

• Nodal, Mesh Analysis

• Superposition

• Thevenin / Norton

Summary of Procedures• Change voltage/current sources in to phasor form

• Change R, L, C value into phasor form

R L C R jωL 1/jωC

• Use DC circuit analysis techniques normally, but the value of voltage, current, and resistance can be complex numbers

• Change back to the time-domain form if the problem asks.

Example

AC

+

-

2sin(t/3)

i(t)

3H

1Ω

+ vR(t) -

+vL(t)

-

Find i(t), vR(t), vL(t)

45245202

totalZVI

4521452 IRVR

452901452 LjIIZV LL

Phasor form

)4

3/sin(2)(

)4

3/sin(2)(

)4

3/sin(2)(

ttv

ttv

tti

L

R

452 I

452 RV

452LV

V

I

Example

AC

+

-

5sin(3t+π/3)

i(t)

3H

2Ω

+ vR(t) -

+vL(t)

-

Find i(t), vL(t)

AC

+

-

5∟60

i(t)

j9

2

47.1754.047.7722.9

605

totalZVI

53.7288.490947.1754.09 jIIZV LL

AC

+

-

5∟60

i(t)

j9

247.7722.992 jZ total

)3.03sin(54.0)( tti

)27.13sin(88.4)( ttvL 47.1708.10247.1754.02 IIRVR

V

IVR

VL47.1754.0 I

53.7288.4 LV

47.1708.1 RV

Phasor Diagram