INC 112 Basic Circuit Analysis Week 9 Force Response of a Sinusoidal Input and Phasor Concept.

INC 112 Basic Circuit Analysis

Week 9

Force Response of

a Sinusoidal Input and

Phasor Concept

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)

Input

Output

Phase shift

Amplitude

Theory

Force 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

AC

+

-

sin(t)

i(t)

R

C

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

tit

Rtitv

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

AC

+

-

Asin(ωt)

i(t)

L Find i(t)

)2

(sin)cos(

cossin

sin1

)(1

)(

)()(

tL

At

L

A

t

L

Atdt

L

A

tdtAL

dttvL

ti

dt

tdiLtvfrom

AC

+

-

Asin(ωt) R

i(t)

AC

+

-

Asin(ωt)

i(t)

L

)2

(sin)(

t

L

Ati )(sin)( t

R

Ati

ωL เรยก ความตานทานเสมอน (impedance) Phase shift -90

Phasor Diagram of an inductor

v

i

Power = (vi cosθ)/2 = 0

Phasor Diagram of a resistor

v

i

Power = (vi cosθ)/2 = vi/2

Note: No power consumed in inductorsi lags v

AC

+

-

Asin(ωt)

i(t)

C Find i(t)

)2

sin(1

)(cos

)sin()()(

t

C

A

tCAdt

tAdC

dt

tdvCti

ความตานทานเสมอน (impedance) Phase shift +90

Phasor Diagram of a capacitor

v

i

Power = (vi cosθ)/2 = 0

Phasor Diagram of a resistor

v

i

Power = (vi cosθ)/2 = vi/2

Note: No power consumed in capacitorsi leads v

Kirchhoff's Law with AC Circuit

AC

+

-

v(t)

i(t)

R

C

vR

i

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

tt

tt

ttv

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

4

3

5

Vector Quantity

Complex 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

sin

cos

arctan

22

rb

ra

a

b

bar

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

492

5

122375

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

45213

131 jjLjR

Cartesian form

Polar form

jba

dc

a

bphased

bamagnitudec

1

22

tan

Rules that can be used inPhasor Analysis

• Ohm’s law

• KVL/KCL

• Nodal, Mesh Analysis

• Superposition

• Thevenin / Norton

Example

AC

+

-

2sin(t/3)

i(t)

3H

1Ω

+ vR(t) -

+vL(t)

-

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

452452

02

totalZ

VI

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

In an RLC circuit with sinusoidal voltage/current source,voltages and currents at all points are in sinusoidal wave form toobut with different amplitudes and phase shifts.

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

+

-

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

totalZ

VI

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

Resistor consumes power2

254.0 22

RIP rms

Inductor consumes no power P = 0