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Series & Parallel Circuits • RESISTORS IN SERIES: Two or more circuit elements are said to be connected in series if they carry the same current and not merely equal currents.. - An open

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WORLD'S #1 ACADEMIC OUTLINE

INTRODUCTION TO AC AND DC CIRCUIT ANALYSIS

I Its magnitude is measured in coulombs (C) The

charge carried by an electron is q.=-1.60xI o-19C

A prolon has the same amount of charge, but with

positive polarity

2 Charges are, therefore, of two types: positive and

negative

3 Charges exist in discrete, integral multiples of ±q•

4 Charges are conserved: They can be neither created

nor destroyed

CURRENT In an electric circuit, the charges move

along specified, closed paths The motion of charges

constitutes an electric conduction current Current is

measured in amperes (coulomb/sec) That is, one

ampere is the current that flows when one coulomb of

charge flows per second Or, current i(t) in amperes is:

coulombs, and I is the time in seconds

-Conventions: A negative quantity of charge

traveling from B to A along a conductor is

equivalent to a positive quantity of charge going

from A to B Hence, current is a vector

- Unidirectional transfer of electric charge through a

conductor constitutes the direct electric current, or

dc This is in contrast to ac (alternating current)

which implles a time-reversal of direction of current

such as a sinusoidal wave

·VOLTAGE Voltage difference (also known as

potential difference) between two points is

defined as the work [wet) in joules] required to

move a unit charge from one point to the other

The unit of potential difference is the volt (V)

Voltage (I') can be positive or negative

dw (t)

Note that: v (t) = ~volt

• POWER Power pet) is the rate of doing work or the

rate of change of energy The electrical unit of power

is the watt (W)

dw (t)

p (t) = ~joule/sec or watts

dw (t) dq dw (t)

Note: V(t)l(t) = ~ dt = ~=p(t)

Thus, pet) = vet) i(t) is subject to the following sign

convention: Whellever th e reference direction fi)r the

c urrent ill all elelllent is ill the dire c tion ofthe voltage

drop across the element use the forllluia p=vi That

is, the current must enter the positive terminal to use

the above formula Otherwise use p= - vi If p(t»O ,

then the element absorbs power If p(t)<O, then the

element delivers power See the following examples:

-In Fig ] (a) the element is

Fig.1 consuming/

power,

3A

- FACT: The total power delivered • BRANCH A branch is an -6Vr

(a) Load element (for example, resis

sources,

• NODE A node is a junction

-6V

~ _ f -3A

• GROUND A ground is the voltage

reference point which has zero

(b) Source potential

Ideal Independent Sources

• IDEAL INDEPENDENT VOLTAGE SOURCE

This element maintains a specified voltage between its terminals regardless of the rest of the circuit where it is inserted The voltage is completely independent of the current through the element

The symbol for the element is shown in Fig 2

- Fig 3 shows other variations of Fig 2:

- Fig 3(a) shows an ideal battery which has a constant voltage with respect to time and (b) shows a sinusoidal

Vmcoswt S V(t)

voltage source which

between the peak values

+v and -v ' producing an Fig 4 alternating current

current through it independent of is ampere flow the voltage across it See Fig 4 from b t o

Ideal Dependent Sources

• IDEAL DEPENDENT VOLTAGE SOURCE In a

dependent (or controlled) voltage source, the voltage across the source depends upon the voltage or current across some other element in the network There are two types of controlled voltage sources:

- voltage controlled voltage source

v=av x where Vx is a voltage somewhere else in the circuit and a is a constant

See Fig 5

- current controlled voltage source

v=fJl" where ix is a current and fJ

• IDEAL DEPENDENT

CURRENT SOURCE In a

dependent (or controlled) current source, the current through the source depends upon the voltage or current through another element in the network There are two types

of controlled current sources:

-voltage controlled current

i=rvx where Vx

- current controlled current source

i=,ix where ix is a current and , is a cons

See Fig 8

Fig 2

Point a is Vs volt s

hi g r than point b

Fig 3 O (a) :L »

V

5b

• OHM'S LAW (at a given temperature) is given by Fig to (b) c[ Bv=iR volt, and is based on

the following convention: l l lJ ll

Fig.S - - - - O B

+

V

b

Fig 6

B

+

V -b

Fig 7

B

I

+

V"

-b

Fig 8

I

• RESISTANCE A resistor

is a circuit element used to

model the resistance to the

flow of electrical charge The unit of resistance is the ohm denoted by (D)

• CONDUCTANCE

Conductance (G) is defined

as the reciprocal of resistance; that is, G= I/R Conductance

is measured in siemens (S)

Ohm's Law

reference direction for the current in the resistor is in -the direction of -the voltage v = I R V = - I

Ohm's Law Otherwise use v=-iR

-~

• OPEN CIRCUIT The graphical representation of an "open circuit"

(R-+oo) is shown in Fig II Note that the current can't flow due to infinite resistance The terminals maintain an open circuit voltage

• SHORT CIRCUIT The graphical representation of a "short circuit"

(R=O) is shown in Fig 12 Note that a short circuit can be considered

as a piece of wire assumed as a perfect conductor (voltage = 0 resistance = 0) Short circuit current flows in the loop

Kirchhoff's Laws

(KVL): The algebraic SUlII of' all the voltages around any closed path in a cilt'uit equal, zero (at every instant o/'timej

- Common Convention:

To avoid sign errors regarding KVL, use the voltage polarity sign of the element which you first encounter in passing by that element Example: If

in Fig 13 we choose to take the loop clockwise, we hit the negative sign of the voltage source and the positive sign

of VI ' Thus -v+vl=O

• KIRCHHOFF'S CURI!ENT LAW

(KCL): The algebrai c sum o( all the currents at any node in a circuit equals zero (at every instant o/'time)

- Common Convention:

Currents entering will be negative Currents leaving will be positive (Opposite method is also permitted if you are ~

consistent.) Example: In Fig 14, KCL for the node is -12A+IOA+i2=0

Conductance

Fig

r l rl TI

Resistor Symbols

Fig II

l)t~ , + -< o m

v

Z

- - - -;=0 R oo 0

Fig 12

V=O

Fig 13

Kirchhoff's Voltage Law Fig 14

Kirchhoff's , Current Law ~

C

Series & Parallel Circuits

• RESISTORS IN SERIES: Two or more circuit elements

are said to be connected in series if they carry the same

current (and not merely equal currents) For any nwnber

of resistances in series add the values Mathematically,

R,q= L Rn for n resistors connected in series

n

- In Fig 15: v=iR I+iR2+iR J= iReo

R q =R,+R2+ R 3

Resistors in Series

• RESISTORS IN PARALLEL: Elements are connected

in parallel if they have

across them In other + O-~ r -'

words,

if both of

terminals of

element are connected Resistors in Parallel

to those of the other;

that is, if they share the same pair of nodes

Mathematically, for n resistors connected in parallel

- FACT: Conductances in parallel add

Fig 16 is a special case: For two resistors in parallel,

RIR z

R,q= RI+ R z and G,q= GI+ G z

• FACTS:

- For resistors in parallel, the equivalent resistance R e o

smaller than the smallest resistance of the parallel

combination

- For resistors in series, R,q is larger than the largest of

the resistors

- A short circuit in parallel with n resistors is equivalent

to a short circuit

- An open circuit in series with n resistors is equivalent

Current Divider

• CURRENT DIVIDER: Current division is used for

finding the currents through resistors connected in parallel

Fig 17

Current Divider

'q

-In Ig , IRM = G mperes, were IRM IS e current

in one of the resistors R m, and Geq=GI+G2+···+Gn

• VOLTAGE DIVIDER: Voltage division is used for

finding the voltages across resistors connected in series

Fig IS

Voltage Divider

If resistors are connected in series and a voltage source

v is applied across the combination, the voltage vm

across one of the resistors Rm IS Vm= LRn volts

• NODAL ANALYSIS A general procedure used to solve any circuit of any size The goal is to generate enough equations to solve for a set of node voltages

• CONCEPT: Given a system with N nodes, pick one as the reference node That is, J.Nu=O volt Solve for the other N - I node voltages

• GENERAL PROCEDURE:

- N nodes Pick one as reference This leaves N-I unknown node voltages Thus we need N-I equations

- Consider voltage sources If the circuit has L voltage sources, then represent each voltage source in terms of its node voltages This will yield L equations

- (We now need N-I -L equations) Replace each voltage source by a short circuit This will result in a circuit with (N-I-L) nodes (not including the reference node) Write Kirchhoff's Current Law in terms of node voltages at these (N-I-L) nodes (referring back to the original circuit)

- Solve for the node voltages

Note: For dependent sources, represent controlling variable in terms of node voltages

Example:

Fig 19

I Five nodes One node is the reference node Thus we have four unknown node voltages: VI' V2 • vJ and v4'

2 The circuit has two voltage sources Thus we obtain two equations (equation numbers are in parenthesis):

(1) Iv=vrvi

(2) 2V=V] - 1'2

3 After replacing each voltage source by a short circuit,

we have two nodes (not including the reference node):

Fig 20

We then write KCL at nodes one and two:

(3) i l +i2 =0

(4) -i2 - I + i4 - 2 = 0 ~i4 - i2 = 3

In order to represent these currents in terms of node voltages, we need to look again at Fig 19 from which

we can see that:

(5) i , = (vl)(I)

(6) i2 = (v4 - V 2 )'J,

(7) i4 = (I'J){'/4)

4 Now we solve for the node voltages by plugging equations 5, 6, 7 into equations 3, 4:

(8) VI + .5v 4 - 5v2 =0

(9) 25vJ - 5v4 + .5v2 = 3

VI = 1.286V, v2 = 4.857 V, v] = 6.857 V, V 4 =

• MESH ANALYSIS: This is an analysis technique defined as follows:

- Loop A closed path in a circuit

- Mesh A closed path that does not contain any other

- Loop current

• OBJECTIVE: Write a set ofequarions in terms of these loop currents, solve for the loop currents, and from these find all the branch currents

• GENERAL PROCEDURE:

1 Number of meshes equals the number of unknown loop currents For example, if we have L meshes, then

we have L loop currents

2 Open circuit current sources If there are m current sources, the new circuit has (L-III) loops

3 How to pick loop currents:

a Define loop currents for each loop obtained in step

2 i.e, 1 1 1 2 / -

b Put each current source back in the circuit and define a loop current that flows through the current source This yields III more loop currents

4 Each current source wi 11 yield onc equation, and obtain the other L-nr equations by writing KVL around loops obtained in step 2

NOTE: By thi.~ method each clIrrent sOllrce shollid have only one ClIrrellt throllKh it

EXERCISE

I We have three unknown loop currents; thus, we need three equations Fig 2 I

Fig 21

20

40

2 After open-circuiting the current sources the new circuit will have only one loop Fig 22

Fig 22

°

- 6V

3 We now put current sources

and define loop currents 12 and ll' Fig

Fig 23

20

6A 0

4n

~

/,

Mesh I: -6 + [II +12] 1.0+ 211 + [II +ll]4.o=0

Thus II =-"h

Source Transformations

• PRACTICAL VOLTAGE SOURCE: A practical (real) voltage source is more accurately modeled as an ideal voltage source in series with a finite internal source resistance

• PRACTICAL CURRENT SOURCE: A practical (real) current source is more accurately modeled as an ideal ,urrent

source in parallel with a finite internal source resistance

• FACT: A practical voltage source ean be transfurmed into

an equivalent practical current source or vice versa

3n

4.12 = -SA 1.1 =6 1

Source T ransformation

Thevenin's Theorem

o FACT: If a circuit is linear, then its v versus i curve must

be a straight line Any linear circuit (Fig 25a) can be

represented by its Thevenin equivalent circuit in the

following form (Fig 25b) such that the two circuits are

equivalent, i.e., same v and i relationship

, : 0 +

Circuit

Fig 26

find VTH and RTH

using the methods below . _ _ ' = :0 °

t

To find v TH : The Thevenin Line

equivalent voltage, v TH '

the open circuit voltage voc'

since vTH= voc> find VrH

across which you want Thevenin's

equivalent and calculate voltage using mesh, nodal, etc

Two ways to find R TH :

I Use R TH = ~'c That is, with this method, you must find

l.w·

voc before you can find R TH Also, you must find the

short circuit current, i sc That is, place a short circuit

across the terminals and calculate the resulting short

circuit current (in the direction of the open circuit voltage

drop across the noqes you want Thevenin's equivalent)

Use any method to find i sc , specifically source

transformations

2The second way to

follows: "Kill" all Linear Circuit +

the independent wi th i nde pe en t V=RTH 1A

s ou r c s rcmo \c d _

(not dependent)

sources in the ( a

are shorted and

independent current (b)

sources are opened Method 2 to find RTH

Place a one ampere

source (Fig 27a) or a one volt source (Fig 27b)

across the nodes you want Thevenin's equivalent and

find the corresponding v or i ( RTH = vor R' m = t)

Note: If all the sources are independent, then RTH is just

a series-parallel combination of the resistive network

left over when all the sources are turned off

o THEOREM OF MAXIMUM POWER TRANSFER

Maximum power is transferred

Fig 28

Thevenill

R TH In Fig 28, RL =RTH

condition for vTH

maximum power to R L

o NORTON EQUIVALENT

Po we r Tr a nsf e r

VTH

Th e v en

FACT: For a circuit with two or more independent sources we can find the response by considering each independent source separately (others set to zero) and adding the separate responses

Inductors • Capacitors

o INDUCTOR: An inductor is a two terminal device that consists of a coiled

Fig 30

around a core An +

i(t)

alternating current flowing t

through the device produces

a magnetic flux ~ that V(t) L

changes with time This flux then induces a voltage

V(t) = Ld/(t)

across the coil The cit mathematical model of an

Inductor

di (t)

direction for the current through the inductor is in the direction of the voltage drop across the inductor

di (t)

Otherwise, use v (t) = - L d t

FACTS:

- The unit of inductance is the Henry denoted by (H)

- The inductor acts like a short circuit to a constant, or

dc, current

- Inductors follow the same rule as resistors That is

L e q = L L., for n inductors connected in series

L,q= _1_1_ for n inductors connected in parallel

~L

- The energy stored in an inductor is given by

_ I Li2

o CAPACITOR:

A two terminal device Fig 31

that consists of two +- - - - - - conducting bodies that ~ i(t) 1

are separated by insulating material known V(t)

as a dielectric

Capacitor

other within the device

They must therefore be transported between the conducting bodies via external circuitry connected to the terminals of the capacitor The conducting bodies are flat, rectangular conductors that are separated by the dielectric material The mathematical model of a capacitor is shown in Fig 31

The model above is based on the following convention:

( ) dv (t)

I t = C ~ applies whenever the reference direction for the current through the capacitor is in the direction of the voltage drop across the capacitor

dv (t)

Otherwise, use i (t) = - C ~

- The unit of capacitance is the farad denoted by (F) voltage

- Capacitors follow the same rule as conductances That

is, C eq = _1-1- for n capacitors connected in series

~C

Ceq = LC for n capacitors connected in parallel

- The energy stored in a capacitor is given by:

w=!-Cv 2

o COMPLEX NUMBERS Real numbers such as -\ 3.5 and - j3 are oriented along the horizontal axis; thus they have an angle of either 180" or 0", and normally the angle is omitted when the number is mentioned The set

of imaginary numbers is located along the vertical axis,

in the complex plane Thus, imaginary numbers have an angle of either 90' or -90" This angle is indicated using

the operator j = IL90' or FT

- Note that p = IL180" = -I -j = IL-90' (and that

11= - j) The set of complex numbers is the set of all points in the complex plane Thus, complex numbers have both a real and an imaginary component -3+j4 4+jO.5 0.5-j2 0-j20 and 20+jO are all examples of complex numbers expressed in their rectangular form ; the same numbers are /25 L tan -1( y-3 ) ' /16.25 L tan -1( n !.i)

/4.25 L tan -1(-%5)' 20L-90 and 20LO expressed in their polar form In the rectangular form , a complex number A is written in terms of its real and imaginary components; hence A = x +jy where x is the real

component, y is the imaginary component, and j is by definition A In polar form, a complex number is written in terms of its magnitude and angle Therefore,

A = Ae jfJ = ALe where A is the magnitude 6 is the angle, e is the base of the natural logarithm, and as before, j = A

- The transition from rectangular to polar form makes

use of the geometry of the right triangle, namely

x +jy = (jx2+ y2)ejfJ= Ae jfJ where tane= Ix'

That is, A = (jx2+ y2), e = tan-I~

- Given a complex number A = ALa = Acosa + jAsina A*=AL-a = Acosa-jAsinais the complex

conjugate of A To add or subtract complex numbers,

use the rectangular form; to multiply, divide, and raise to

a power, use the polar form

Fig 32

- It is not obvious in which

resolved by a graphical jy

representation of

- Fig 32 shows a complex

Re

• PHASORS A phasor is a transformation that applies to sinusoids The definition ofthi, transfonnation is as follows:

Acos((I)t + 6) H Ae j9

e.g., 2cos«(I)t + 10°) H 2e jlO

4sin(3t) = 4cos(3t - 90") H

- This transformation is very useful because it replaces derivatives and integrals with algebra Its limitations are that it is only useful in systems where the signals are sinusoids, and the system must be in a steady-state The steady-state is a condition in which circuit values remain essentially constant, occurring after all in itial transients or fluctuating conditions have settled down This condition is valid if there is sinusoidal excitation with only one frequency present

o FACT: Differentiation in the time domain corresponds

to multiplication by j(l) in the phasor domain e

di

- Considering v (t) = L d t ' in phasor domain this equation becomes V = Ljm/

- Considering i (t) = C d~~t), in phasor domain thi s

equation becomes 7 = Cjmv

- Considering v (t) = Ri (t), in phasor domain this

equation becomes V = RI

frequency

h term

Sinusoidal Power

• EFFECTIVE (RMS) VALUE The effective value

of some time-varying voltage or current is that dc

voltage or current value which would dissipate the

same power in a resistor, R, as the time-varying

voltage or current would dissipate on the time

average in R If itt) = Icos (ou), then the effective

value of the current would be leff = /2

• REAL POWER The real power, P (watt), is the

average value of the product of instantaneous voltage

and instantaneous current That is, if the instantaneous

power, p(l) is given by p(/) = v(t);(/) and T is the

period of the voltage, or current, waveform, then the

+

average power, P, is given by: P =

o

It can be shown that P reduces to P = V.ffleffcos8

watt, where V.ff is the effective value of the voltage,

leffis the effective value of the current, and 8 is the

angle between phasor ~ff and phasor ~ff

(i.e.,a = av , O' - a 1 , 0')' Given any physical system,

P (watt) is the only form of power that can be

converted from electric to heat, mechanical,

chemical or any other form (at the expense of some

loss) P is always positive

• REACTIVE POWER Given any circuit that

contains inductance and (or) capacitance, the

reactive power, Q (volt - ampere reactive -VAR), is

given by Q = V.ffleffsin8VAR The reactive power

cannot be converted to any other form, but is

necessary for the conversion of real power Q can be

either positive or negative

• COMPLEX POWER The complex power, (volt ­

ampere -~) is a complex number whose real part is P

(watt) and whose imaginary (reactive) part is Q(VAR)

Therefore, S(VA) = peW) +jQ(VAR) = V.ff~ff *

• APPARENT POWER The apparent power, S(VA) ,

is the magnitude of the complex power: S = V.ffleff'

• POWER FACTOR The number cos 8=p/VI

represents the amount of watt that can be converted

in a system that requires so many volt-ampere; cos 8

is called the power factor; it is normally less than one

(less than 100%) and it can be lagging (I lags V) or

leading (I leads V); since cos8= cost-B) the word

leading or lagging must be

specified after the value of the

F i g 33 power-factor

• THEOREM OF MAXIMUM

POWER TRANSFER

Looking at Fig 33, for maximum ZL

average power transfer to the load

impedance, Z Lmust equal the

conjugate of the_ Thevenin

impedance That is, ZL = Z71l·'

However, a type of restriction occurs when the

magnitude of ~_ can be varied but its phase angle

cannot Under this restriction, the greatest amount of

power is transferred to the load when the magnitude

of ~_ is set equal to the magnitude of Z71I ; that is,

when IZII=IZTIII

• POWER TRIANGLE

( a "",",r T riang l e for la gging ( b "",", rTri ang l e f or lea ding

""'"" F a ct o (ind uctiv e L oad ) !'<Mer fa ct or Ut pa< llI ve Load)

• IMPEDANCE The ratio of phasor voltage across

an element divided by the phasor current through the element is defined as impedance, Z Z is not a phasor, but rather just a complex number e.g., the impedance of an inductor is given by ZL =jOJL; and, the impedance of a capacitor is given by

- I

Zc = jmC' Impedances follow the same laws as resistors, e.g., impedances in series add

• ADMITTANCE The admittance is the inverse

(reciprocal) 0/ impedance Admittances follow the

same laws as conductances

FACT: If the input to a linear system is sinusoidal, then the output in steady state will be sinusoidal with the same frequency, but different amplitude and phase

FACT: With the generalization of Ohm's law to time varying situations, we can now apply voltage and current division, loop and nodal analysis, and Thevenin and Norton equivalent circuits to the design and analysis of linear circuits in the sinusoidal steady state

EXERCISE For the circuit in Fig

35, find the output voltage ,:,",(t) when the input voltage vinet) = 20cos (4001) V

Go to phasor domain

vinet) = 20cos(4001) H

By inspection, OJ =

400 sec Thus, Zz

= jOJL = j(400)(0.5)

as in Fig 36

2 Do the calculations

in phasor domain since calculus becomes algebra

v

3 Go back to time domain

Vout(t) = lO /2 cos (400t - ~) V

Note: Since the input is sinusoidal, output will be sinusoidal with the same frequency, but different amplitude and phase

Frequency Response

In the sinusoidal steady state, the output (response) for a

sinusoidal input is also

"Sinusoidal with the same

frequency but with a different amplitUde and phase

Example (Fig 37):

Excitation is vet)

Response is i(t)

Let vet) = 2cosOJt Thus: j(t)=A(ro)cos(OJt + 9(OJ»

• Determining the frequency response of a circuit is investigating the effect of varying the frequency of the sinusoidal source There are two primary reasons for interest in the frequency response

- First: Circuits that will transmit signals at some frequencies noiiceably better than at other frequencies can be used to filter out, or eliminate,

signals in an unwanted frequency range The ability

to design circuits that are frequency-sel e ctive is what makes radio, telephone and television communication possible

Fig.3S

O.SH

20e jcr = 20LO' = ~n Fig 36

• FIRST-ORDER CIRCUITS: A first order circuit occurs whenever a circuit can be reduced to a Thevenin or Norton equivalent connected to the terminals of an equivalent inductor or capacitor

Note that capacitors and inductors are not present in

a first order circuit Furthermore, if more than one inductor or capacitor exists in the circuit, they must

be interconnected so that they can be replaced by a single equivalent element

- Concept: In electric circuits, switches are often turned from on to off or vice-versa It is therefore necessary to analyze the pcrformance of the circuit before and after the switch is open or closed Usc 0­

to denote the time just prior to switching and 0+ to denote the time immediately following switching

The circuit behavior before and after the switch is operated, has a common link If the switch is at

must equal the voltage across the capacitor at (=(0+ Similarly, the inductor current cannot change instantaneously

Consider circuits that only have DC

I A capacitor is an open circuit to DC only in steady state

2 An inductor is a short circuit to DC only in steady state

3 Steady state is defined as the condition when all the transients have died out That is all the current and voltages have reached a constant value

- Definition: Transient-time between two steady states

The scene: A switch is thrown at 1= 10, Solve the circuit

assuming the circuit contains one capacitor or one coil

Step I: Analyze the circuit for t<to using mesh,

nodal, inspection, etc Remember, capacitors are open circuits and coils are short circuits for DC steady states Whether it's asked for or not, find

vc( 0-) or IL(0

-Step 2:Draw circuits for t>O and represent as

Fig 39

Fig 39

vc(t)

- Second: If

response is known the

response of the c

other input can be predicte

Convert Fig 37 to the phasor domain See Fig 38

V

1 = - - =

21.0

)(1) 2+(m)2 Lian - I Ill=~

It is possible to find i(t) for any ro

Defi th R e .~p{)ll~ e th fi f' - H ( )

me e ratio ExciUlti{)1I as e trans er unCtion OJ

FACTS:

I H(OJ) is independent of the value of the input because

if you change the input (excitation) value the output changes correspondingly and the ratio stays the same

0 )2

W B

ult on

OJ

ca de

w

2 Given H(t»), which is a function of the system,

we can find the output for any specific input

Generalize: In a linear system the phasor voltages

and currents are linearly related, i.e., if X(t») is

the input to a system and Y(t») is the output, then

Y(t») is linearly related to X(t») which leads to

our d<!finition of transfer function:

H((i)) = ~((i)) = ~utput

X( (i) ) Input

Find Thevenin equivalent across the capacitor or

the coil as in Fig 40

3 Write down the answer

[Note that the final value (steady state solution) is

v TH for vc(t) and ~: for IL(t)] Thus: ve(t) =

-(t-to))

v TH ( ve(tO-)-v TH) exp [ RTHC ' t > 0

le(t)

VTH [ (-) VTH] [-(t-tolRTH]

IL(t) = RTH + I L to - RTH up L ,t>O

vL(t) = L

Example:

The switch in the

circuit shown in

Fig 41 has been

closed for' a long

SmH

time At t=O, the

switch is opened

Fig 42

C

For t>O, Fig 43

Fig 44

Voe' Fig 45

, - _ _ _

Voc =

Therefore:

1,(0+) =

IL(oo) =

l' = ~

RTH

.4xI0-3sec

Thus, ~ =

1<0

Fig 42

E E J '" = 8 10(0

Need to find the Thevenin equivalent across the coil

Fig 44

R TH : Kill independent sources Fig 46

Fig 45 Fig 46

1=0

, - 0 +

12'1

SU

loc

Veil = 160V

20A (initial value),

Fig 47 1: = 8A

= 8 X 10-3

20

2500/see

IL(t) = IL(oo) + [h(O+) - h(oo)] exp ~t, t > 0

IL(t) = 8 + [20 - 8]exp-zsoo,

= 8 + 12exp( -2500t)A VL(t) = -d-t­

= (8 x 10-3)(12)(-2500)exp-2500t

= -240exp( -2500t) V

Fig 48 Fig 49

Sketch

ofVdt) ofldt)

-240

t(sec)

Finding the natural response of a parallel RLC circuit consists of finding

Fig 50

the voltage created across the parallel

capacitor, or both t~O

IL(O-) =IL(O+) and vcCO-) =

Goal: Need v(t) for t>O

v(t) dv(t) IRCt) = R' Ic( t) = C~and

,

IL(t) = ILco+) + t f v (a) da

o

At t=O+, veCO+) = vc(O-) [Recall that capacitive

v(oo)= O

Question: How does ve(t) go from ve(O+) to Solution: IR(t) + IL(t) + le(t) = 0, or

~ +! f'v (a) da + C dv(t) = 0

Differentiate both sides with respect to t and divide

through by capacitance C

d 2 v ldv 1

d(2 + RCdt + LC v = 0

Assume a solution of the form: v = Ae", where A and

A e st( s 2 + RC S + LC 1) 0 =

Cannot use A =

equation

=l± RC (:J2 - L~

2

( L) - ! and

2RC LC

2

S2= 2;~ - (~C) - :C

If these two solutions are denoted as VI and Vz

respectively, then their sum is also a solution

v = VI + V z = Ale'" + Aze s" is a solution

Find AI and Azfrom the initial conditions

There are 3 basic forms for v(t)

2

(Jc) > L~ two real negative distinct roots

Solution is given as above: vet) =Ale'.' +Aze'"

2 Critically damped

(2jC) 2 = 2c sl = Sz [s" Sz real and equal]

Solution changes to vet) =A.e'" + Azte'.'

3 Underdamped

(Jc)2 < L~ s =sz'

The solution is vet) = A.e'.' +Aze"'

Resonance

• RESONANT FREQUENCY: The frequency at which the input impedance is purely real

- Resonant frequency is not defined in terms of the transfer function For the parallel RLC circuit, the transfer function equals the input impedance

- At resonant frequency, Z((i)) is real Thus the phase of Z( (i)) is O

- At resonant frequency, the magnitude of Z((i)),

IZ((i))1 does not have to be a maximum although

at many times it is a maximum

- Not all circuits have a finite nonzero resonant frequency

FREQUENCY:

I Set the imaginary part of Z((i)) = 0 and solve for w

OR

2 Set the phase = 0 and solve for

Z((i)) =

G +j

(i)2 LC - I = 0 OR (i) 2 LC = I ~(i) = _1_

Observation: If Z((i)) is real then Y ( (i))

frequencies t»z - ())I where the magnitude of the

value ())2 is called the upper cutoff frequency and

())I is called the lower cutoff frequency For parallel

RLC, BW = ic'

• Q - QUALITY FACTOR

Q ())) =21t' maximum energ~ stored in the system

( energy lost In the system over

one cycle of the itlput signal

Denote Q( t») at t» = t»o as Qo' For the parallel RLC

circuit, Qo= ())oRe Note that BW = (i)Q 0

o Assume v(t) in Fig 51 is sinusoidal Then the circuit

goes to the phasor domain as in Fig, 52

+ ­

L

R es onance - co nt fiv lII page 5

• QUESTION: Find (a) the bandwidth BW, (b) resonant

frequency 000 , and (c)Q

~ (a)For BW, need ll(w)=~ For this example,

I

~

Q

IH(w)1 Fig 53

From the sketch in

= ~ Fig 53, 00 1=0,

and 002= YR .Thus, W('Bd)

BW=~-~=r' T

(b) There is no finite non-zero resonant frequency since

(c Q ( w) = 2n:

)

w(t)= I/zLi2(t)=t/zLl.,2COS2oot Thus maximum value of

00 occurs when cos2oot=1 Therefore, w.,ax(t)=t/2Llm2

Energy lost/cycle computes as:

tIIIII Thus, Q = 2n: 11 2 = RT

"I11III 12 Rl.,T

ZSubstitute T = : => Q(w) = ~L

III Generalization: First l(OO) = R +jX(oo)

Fig 54

• Consider the following two circuits:

l

D Z(s) = Rs +jXs Y(p) = Gp + =-x

) p

=FACT: Q,= I!,I FACT: Q p = l~pl

Qo(system)

Note: Manufacturers of coils

intended for frequency-selective

circuits specify the Qof the coil at specific frequencies The

coil Q and the system Q are not the same To find a coil that

works at/o=106 Hz use a specification sheet Assume we

find Q(coil)=230 and the corresponding L =200JJll

BW = so = 1.2 x 10 rad/sec

It follows from ClIo= k that C;:I.4x 10-loF

From BW = ic it Fig 56

follows that R=60kD

Z

~ Thus the values in

Fig 55 change to that

shown in Fig 56

D U.S.$5.95 CAN.$8.95

Fig 57, shows a more practical model of the coil because it accounts for the losses in the inductor, represented by the resistor R

Recall Q(coil) = Qp = 230

Smce QP=-I-1= wL ' Xp

p

it follows that

Now repeat for the capacitor

357kD Now obtain better model, as in Fig 58,

reduces to the values shown in Fig 59

Bode Plot

FACT: The transfer function relates the steady-state response to the excitation source That is, for sinusoidal input x(t) = Acos(oot + fJ) the output

Definition: The amplitude of ll(jw) in decibels is

EXERCISE Given: i(t)= 2cos2t = input, ll(jw) = jw~2 Objective: Find output and amplitude of H(jw) in

decibels _ 2 I

Thus, IH(2j)1 = Jz and 8(2j) =_45'

Steady state response is given by:

and 8Uoo) = -tan-I~

db = 20log 1H(jw) 1

=2010g~

4 + w

= 20 log ( 2 ( 4 + ill 2)-112 )

= 20 log 2 + 201og(4 +

= 20 log 2 -IOlog(4 +

-Take H(jw) = w

1+)wB

It follows IH(jW)1 = l Ao 2]

I+(~B) 1/2

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In db, db = 2010g [[ Ao 211/2 1= Fig 60

( )

1+ !!L

roB 20logAo -10 log (1 +(:B) 2) ~k

We sketch ll(jw) in dB versus 00

Let Ao = I Then it follows: dB = -IOIOg( I +( ~Br) The actual sketch resembles that shown in Fig 60

Note: For bode plots an approximation is made:

ForOO<OOB,1 + (~Br-I, ForQ»OOB' 1+ (~Br - (~Br This approximation can be used to sketch the result on semilog paper Back to the example:

ForOO<Wn,dB IOlogl =0 For 00> Wn, dB IOIOg(~Br

= -20 log fQ Fig 61

w

For 00 = OOB' -20 log w B = 0

W

For 00 = I0Wn, -20 log w B = -20 tr For 00= 100Wn, -20 log w w

The sketch is as in Fig 61:

Generalization: Bode plot of Fig 62

dB

( 1 rlooks like Fig 62 1+ fQ

)WB

It follows that the Bode plot of

I)m looks like Fig 63

(1+ fQ slope ~ -20 • m dB/decade

)WB

Also, Bode plot of constant Ao is Fig 63

20logAo as shown in Fig 64 slope =\20 • m dB/decade

dB

Note: Application of this Bode approximation - for a more involved H(jw), we can obtain a ~

quick approximation plot of

~a.mple: (jw + 10P

First we need to put terms in the form

J /WB' [ 'w/jJ

1000 1+) /10 ( I+'W/ )

= 20loglO + 20 log II +jo/tol·1 + 20 log II + jwl- 2

+ 2010g ll+ j o/tool-1

= 20logl0 + 60log II + jW;{ol- 40log II + jwl

- 2010gl' + j%)()1 Now each function is in standard form Sketch each term and add as shown in Fig 65 Begin with 20dB straighl

line -40dBldec slope takes us to -20dB at 00=10 The

20dBIdec slope This 20dBIdec slope takes us from -20dB

at 00=10 to OdB at 00= 100 At 00= 100 the -20dB/dec

slope adds to the 20dBIdec slope keeping US at OdB

Fig 65

dB

20 log IH(jc·,)1 "" dB 2010g10

60 d-ecs1ope

40

1000 oo(Iog) -20

-40 :: :::::: ::.~ ~I

Customer Hotline # 1 8002309522 We welcome your feedback so we can 20~

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~ 91r~ jlJll ~~II~ I! rJI~ JIJ 211f1llll! Il il~ I mamtam and exceed your expectations 61,.1,.11" 6