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Module M2-1Electrical Engineering

T U T O R I A L 8 C I R C U I T A N A L Y S I S I V

N O V E M B E R 2 0 1 3

Topics

Phasor

Impedance

2

Impedance

Nodal Analysis

Mesh Analysis

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Sinusoids3

A sinusoid is a signal that has the form of the sine or cosine function.

)sin()( tVtv m

where Vm = the amplitude of the sinusoidω = the angular frequency in radians/sФ = the phase

Sinusoids

A periodic function is one that satisfies v(t) = v(t + nT), for all t and for all integers n.

4

g

HzT

f1

f 22

T

2sin ( ) sin sin 2 sinm m m mV t T V t V t V t

sin( ) sin cos cos sinA B A B A B

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Sinusoids

Graphical addition

5

cos sin cos( )A t B t C t 3cos 4sin 5cos( 53 1)t t t cos sin cos( )A t B t C t 3cos 4sin 5cos( 53.1)t t t

2 2 1, tanB

C A BA

Sinusoids: Example6

Given a sinusoid, , calculate its )604sin(5 ot amplitude, phase, angular frequency, period, and frequency.

Solution:

Amplitude = 5, phase = –60o, angular frequency = 4π rad/s, Period = 0.5 s, frequency = 2 Hz.

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Phasors7

A phasor is a complex number that represents the amplitude p pand phase of a sinusoid.

It can be represented in one of the following three forms:

)i( jj a Rectangular

rzjrez

)sin(cos jrjyxz a. Rectangular

b. Polar

c. Exponential

22 yxr

x

y1tan where 

Phasor Properties8

• Addition & subtraction

M ltiplication

)()( 212121 yyjxxzz

• Multiplication

• Division

• Reciprocal

212121 rrzz

212

1

2

1 r

r

z

z

rz

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• Square root

• Complex conjugate

• Euler’s identity

2 rz

jrerjyxz

sincos je j

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Phasors9

phasor domain time domain

( )j tm mV V e V ( ) cosmv t V t

Phasor will be defined from the cosine function

Phasors10

( )j tm mV V e V m m

( )j tm mI V e I

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Phasors: Example11

Transform the following sinusoids to phasors:

i = 6cos(50t – 40o) A

v = –4sin(30t + 50o) V

406

1404

A

= 4cos (30t+50o+90o)  V 

Transform the following phasors to sinusoids:

V 3010 V

A j12) j(5 I

= 10cos(ωt + 210o)  V

22.62 13 )12

5( tan 512 122

= 13cos(ωt + 22.62o)  A

Phasors

Let

12

( ) cosmv t V t

Derivative 90

( )sin cos 90

Re Re

m m

j t j tm m

dv tV t V t

dt

V e j V e

1 1 Integral

90

1 1( ) sin cos 90

1 1 Re Re

m m

j t j tm m

v t dt V t V t

V e V ej

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Phasors13

The differences between v(t) and V:

v(t) is instantaneous or time-domain representationV is the frequency or phasor-domain representation

v(t) is time dependent, V is not

v(t) is always real with no complex term, V is generally complexg y p

Note: Phasor analysis applies only when frequency is constant; when it is applied to two or more sinusoid signals only if they have the same frequency.

Phasors14

Relationship between differential, integral operation   in phasor listed as follow:in phasor listed as follow:

)(tv VV

dt

dvVj

vdtj

V

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Circuit Elements15

Resistor: Inductor: Capacitor:

Circuit Elements16

Summary of voltage current relationshipSummary of voltage‐current relationship

Element Time domain Frequency domain

R

L

Riv RIV

diL LIjV

C

dtLv LIjV

dt

dvCi

Cj

IV

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Impedance and Admittance

Frequency dependence

17

LjZ

Z

Z

;

0;0

CjZ

1

0;

;0

Z

Z

Impedance: Example18

Find Z for

>> ((200 + 8*10*i)*1/(1*10*10^ 3*i))/((200 +

Find Zin for

ω = 10 rad/s

>> ((200 + 8*10*i)*1/(1*10*10^-3*i))/((200 + 8*10*i) + 1/(1*10*10^-3*i)) + 80 + 1/(0.5*10^-3*10*i)

ans =1.2950e+002 -2.9505e+002i

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Nodal Analysis19

Currents entering a node equal to currents l i th t dleaving that node

3 Steps to solve:

1. Select a node as the reference node. Assign voltages with respect to the reference node to the remaining nodes.

2. Apply KCL to each of the non-reference nodes. Use Oh ’ l h b h i f Ohm’s law to express the branch currents in terms of node voltages.

3. Solve the resulting simultaneous equations to obtain the unknown node voltages.

Nodal Analysis: Example20

Find iX in the circuit using nodal analysis.1010

tV4cos20 xI2xI

Solution sradt /40204cos20 We first convert the circuit to the frequency domain

sradt /4,0204cos20 41 jLjH

25.0 jLjH

5.21

1.0 jCj

F

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Nodal Analysis: Example21

Node 1 Node 2

10

V020xI2xI

1V 2V

5.2j

4j

2j

Apply KCL at node 145.210

20 2111

j

VV

j

VV

or 205.2)5.11( 21 VjVj (1)

22

At node 2,24

2 221

j

V

j

VVIx

Nodal Analysis: Example

24 jj

But Ix= V1/-j2.5. Substituting this gives

245.2

2 2211

j

V

j

VV

j

V

By simplifying, we get 01511 21 VV (2)

Equation (1) and (2) can be put in matrix form as

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Nodal Analysis: Example

205.25.11 1Vjj VV 43.1897.181

The current Ix is given by

Aj

VIx

4.10859.79052

43.1897.18

521

01511 2V 3.19891.132V

j 905.25.2

Transforming this to the time domain,

Atix )4.1084cos(59.7

24

The sum of voltages in a mesh equals to zero.

Mesh Analysis

3 Steps to solve:

1. Assign mesh currents in to the n meshes.

2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents.

3. Solve the resulting n simultaneous equations to get the mesh currents.

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Mesh Analysis: Example25

Determine current Io in the circuit using mesh analysis.

4

A05 2j

10jV9020

8 2j

1I

2I

3I 0I

Mesh Analysis: Example26

SolutionApplying KVL to mesh 1,we obtainpp y g ,

010)2()2108( 321 IjIjIjj (1)For mesh 2,

09020)2()2()224( 312 IjIjIjj (2)

For mesh 3 I = 5 Substituting this in Eqs (1) and (2)For mesh 3, I3= 5 .Substituting this in Eqs. (1) and (2)

502)88( 21 jIjIj (3)

1020)44(2 21 jjIjIj (4)

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Mesh Analysis: Example27

Put eqs. (3) and (4) in matrix form as

30

50

442

288

2

1

j

j

I

I

jj

jj

AI 22.3512.62

The desired current isThe desired current is

AII 78.14412.620