AP2212-Lecture 9-SWT(1).pdf

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AP2212 – Lecture 9

AC analysis 2

Course organization

TSANG, Stephen

E-mail: [email protected]

Tel: 3442-4618Office: P6706

mailto:[email protected]

mailto:[email protected]



2

Lecture 08 - Review Managed several examples of circuits with Resistive

(R ), Capacitive (C ) and inductive elements (L).

Reactance (denoted X, units of ohms) is the ratio (magnitude only)

voltage to current: measures how a component opposes the flow of

electricity

X R = R X L = w L X C = 1/w C

Impedance, Z (magnitude and phase) of the current

Z R = R Z L = jw L Z C = - j X C 1 = 1/jw C

t LV d

d I

)(cos t LI v P L w w

Ri v

)(sin t R I v P R w

dt

dV C i

V

QC

)cos( t C

I v

p

C w

w

-



3

Lecture 08 - Review

Graphical

representation ofcomplex impedance

Impedances combine in the

same way as resistors complex impedances can be

Impedance are added , subtracted , multiplied and divided

in the same way as other complex quantities

They can also be expressed in a range of forms such asthe rectangular , polar and exponential forms



4

Power, voltage, and current gains

Power gain is defined as:

Voltage gain is defined as: (why?)

Current gain is defined as: (why?)

Gain =10log P

out

P in

è

ö

ø

÷dB

Gain = 20log V

out

V in

è

ö

ø

÷dB

Gain = 20log I out

I in

è

ö

ø

÷dB



5

Power, voltage, and current gains

X

RinRout

Pin, Vin Pout, Vout

Proof that power gain = voltage gain if Rin equal to Rout



6

Lecture 09 - Outline

High-pass RC filter

Filter representation

Bode diagram, phasor diagram

Transfer function

Low-pass RL filter

Resonance RLC circuit

Charging and discharging in RCL circuit

Application and engineering manipulation of RCL circuit



7

Application of RCL circuit

Still remember how does a radio works?

Source: National Radio and Astronomy Observatory

1. Convert sound to electrical signal 2. Modulate with carrier frequency

AM

FM

3. Transmit in air4. Received by an antenna

How to extract the sound?

8



8

Application of RCL circuit

FM receiver

FM carrier frequency: 88-108MHz

In human, the audible range of frequency: 20Hzto 20kHz

There are multiple channels in air: e.g. 88.1MHz, 99.7MHz

We have to find a way to select the frequency

A resonance circuit Frequency response

We concern:

The frequency response

The resonance(center) frequency

The width of the spectrum (Bandwidth

The amplification (Gain)

9



9

A High-Pass RC Filter

Consider the following circuit

We want to know what is its function?

?sin:knowweIf output

find

pinput v t V v w

This circuit is commonly

drawn in the following way:

10



10

RC Filter 1/2

vi =V p sin t ® vi =V pe j t

1) Use complex numbers:

)( 1tan:

)/1(/~

1

22

-

-

-

w w

w w

RC where

eC R C j R Z j

2) Find impendence:

)( )sin()/1(

~/~~

222

)(

p p

t j

p

i

I t I i C R

eV Z v i

w w

w

w

3) Calculate current:

11



11

RC Filter 2/2

)(1 22

22

i

o

i

o

v

v

v

v

w w

w This is a high pass

filter!

)sin(1

)sin()/1( 222222

w

w

w w

w

t C R

RC V t

C R

R V iR v

p p

o

Findoutput:

)sin(122

w

w

w

t V

v p

oDefine: = RC , then:

~sin

t j

pi pi eV v t V v w

w

12



12

High-pass Filter (as function of w )

For High-pass RC filter , output is taken from resistor

22

22

1 :Since

w

w

i

o

v

v

passcansignalsf highThen , :If 11

i

o

v

v w

passtcan'signalsf lowThen , :If 01

i

o

v

v w

70702

1

1 .:If

i

o

v

v w

The frequency times the time constant is equal to 1

13



13

High-pass Filter (as function of f )

the gain response has two

asymptotes that meet at the cut-

off frequency

figures of this form are calledBode diagrams(plots)

= RC has units of (1/angular

frequency)

(Remember definition of w : w = 2p f )

then we can write: =1/( 2p f C ), and

w = (2p f )/(2p f C ) = f /f C )

14



14

High-pass Filter (as phasor diagrams)

The behaviour in these three regions can beillustrated using phasor diagrams

15



15

Differentiating Circuit

2)(1

)sin(:beforesawwe As

-

RC

t V iR v p

R

w

w

This is a

differentiatingcircuit!

sin t V v pi w

/2CR1/tan1/RC itTherefore p w w

t RCV RC

t V v p

p

R w w

w

w cos

)(1

)sin(

2

-

dt dv RC v v i

R o

16



16

Transfer function

Consider the potential divider shown here From considerations on this circuit:

21

2

ZZ

Z

i o v v

21

2

ZZ

Z

i

o

v

v

rearranging, the gain of the

circuit is

this is also called the transferfunction of the circuit

17



A Low-Pass RL Network

Low-pass networks can also

be produced using RL circuits

they behave similarly to the

corresponding CR circuit

R

LLR

R

v

v

i

o

w w

j1

1

j

LR

R

ZZ

Z

the voltage gain is

rad/s1

L

R c

w Hz22 L

R f c

c p p

w

the cut-off frequency is

18



A High-Pass RL Network

High-pass networks can also

be produced using RL circuits

these behave similarly to the

corresponding CR circuit

the voltage gain is

the cut-off frequency is

L

R

L

R LR

L

v

v

i

o

w w

w

w

j1

1

j1

1

j

j

-

LR

L

ZZ

Z

rad/s1

L

R c

w Hz22 L

R f

c

c p p

w

19



A Comparison of RC and RL Networks

Circuits using RC and

RL techniques have

similar characteristics

20



Make filters to condition a signal

Low-pass filter

High-pass filter

21



Noise spectrum (From L5)

22



Series RLC Circuits and Resonance

the impedance is given by

)1

( j j

1 j

C LR

C LR

w

w

w

w -Z

C L

w

w 1

LC 12

w

LC

1w

if the magnitude of the reactance

of the inductor and capacitor are

equal, the imaginary part is zero,

and the impedance is simply R

this occurs when

23



Resonance in RLC circuits

The condition is known as resonance

The resonant frequency is

LC f o

p 2

1

LC

1w

in the series resonant circuit,

the impedance is at a minimum

at resonance

the current is at a maximumat resonance

LC

1w We define as the resonant angular frequency

24

l f



Quality factor

The resonant effect can be quantified by the quality

factor, Q

Q is the ratio of the maximum energy stored to the

energy dissipated in each cycle

R

X

R

X Q C L

factor Quality it can be shown that:

What happenswhen R 0?

C L

R Q 1 and:

Q=

max. energy stored

energy dissipated in one cycle

WHY?

25



RLC Resonance

Current resonance occurs

for RLC series circuit.w o is called resonancefrequency.

C

L

R

Q 1

LC

o

1w

LC

f op 2

1

The resonance peak atw w o is prominent and sharp

for lower resistance values.Why?

26

K i



Key points A combination of resistor (R), capacitor (C), and inductor (L) can

be used to construct filters and resonance for electrical signals.

In such frequency dependent system, we concern about the cut-off frequency, phase changes, quality(Q) factor.

Bode plot and phasor diagram are effectively ways to representthe frequency response of the circuit.

In a resonance circuit, the Q fact is maximum when thereactance of C and L are equal.

27

T i t



Transient response

We have looked at the behavior of systems in response to:

Fixed DC signals (L06) “Constant” AC signals (L07-L09)

What happens before these circuits reach “steady-state”?

this is referred to as the transient response

What happens to the circuit on the right

when at t=0 the switch is closed?

1. Was the capacitor charged or discharged

at t=0-?2. What is the value (phase) of V at t=0?

28

Ch i C it (1/2)



Charging Capacitors (1/2)

Kirchhoff’s voltage law:

In a capacitor we have:

(First-order differential equation with constant coefficients) Assuming V C = 0 at t = 0,

this can be solved to give:

Also since i = C( dv /dt )(still assuming V C = 0 at t = 0), then

V v iR

t

v

C i d

d

V v t

v CR

d

d :Therefore

)e()e( --

t -

CR

t -

V V v 11

t -

CR

t -

I I i ee(I=V/R)

29

Ch gi g C it (2/2)



Charging Capacitors(2/2)

Thus both the voltage and current have an

exponential form

)e()e( --

t

-CR

t

-V V v 11

t -

CR

t -

I I i ee

30

Energizing Inductors



Energizing Inductors

A similar analysis of an RL circuit gives

t -L

Rt -

V V v ee )e1()e1( --

t -L

Rt -

I I i

where I = V/R

31

Discharging Capacitors



Discharging Capacitors Consider this circuit for discharging

a capacitor (At t = 0, V C = V )

Kirchhoff’s voltage law:

Then:

Solving this equation as before gives:

(I = V/R )

0v iR

0d

dv

t

v CR

t -

CR

t -

V V v ee

--

t -

CR

t -

I I i ee

32

De energizing Inductors



De-energizing Inductors

A similar analysis of this

circuit gives

(I = V/R)

--

t -

L

Rt -

V V v ee

t -

L

Rt -

I I i ee

33

A comparison of the four circuits



A comparison of the four circuits

34

Response of First Order Systems



Response of First-Order Systems

Initial and final value formulae

Increasing or decreasing exponential waveforms (foreither voltage or current) are given by:

V i and I i are the initial values of the voltage and current

V f and I f are the final values of the voltage and current

1. The first term in each case is the steady-state response

2. The second term represents the transient response

3. The combination gives the total response of the arrangement

-- /e)( t

f i f V V V v -

- /e)( t

f i f I I I i

35

Tutorial



Tutorial

The input voltage of this CR circuit

changes from 5 V to 10 V at t = 0.

What is the output voltage?

36

About exponential curves



About exponential curves

37

Output of first-order systems to a square



Output of first order systems to a square

waves [for different time response (T)]

see

38

Output of first-order systems to a square



Output of first order systems to a square

waves [for different frequencies ( f )]

See

39

Second-Order Systems



Second Order Systems

Circuits with capacitance and inductance result in

second-order differential equations. for example, the circuit:

is described by the equation:

Second order systems also have transients.

They will be more complex than first order systems

Transient solutions depend on the equation’scoefficients useful to find an equation’s “general form”

V v t

v RC

t

v LC

C

C C

d

d

d

d2

2

40

Second order differential equation



Second order differential equation

When a step input is applied to a second-order system,

the form of the resultant transient depends on therelative magnitudes of the coefficients of its differential

equation. The general form of the response is

n is the undamped natural frequency (rad/s)

(Greek Zeta) is the damping factor

x y t

y

t

y

nn

d

d2

d

d1

2

2

2 w

w

41

Response of second-order systems



Response of second order systems

=0 undamped

<1 under damped

=1 critically damped

>1 over damped

Will all responses reach “steady-state”?

Which one reaches “steady-state” faster?

Which one oscillates?

42

Key Points



Key Points

The charging or discharging of a capacitor are each associated

with exponential voltage and current waveforms

(Same for the energizing and de-energizing of an inductor)

Circuits that contain resistance, and either capacitance or

inductance, are termed first-order systems

The increasing or decreasing exponential waveforms of first-

order systems can be described by the initial and final value

formulae

Circuits that contain both capacitance and inductance are

usually second-order systems. These are characterized by their

undamped natural frequency and their damping factor

43

Tutorial



Tutorial

As we increase R, the frequency range over which the dissipative characteristicsdominate the behavior of the circuit increases. In order to quantify this behavior we

define a parameter called theQual i ty Factor Q

which is related to the sharpnessof the peak and it is given by

Q = 2

max. energy stored

total energy lost per cycle at resonance= 2

E S

E D

which represents the ratio of the energy stored to the energy

dissipated in a circuit. The energy stored in the circuit is

E S =

1

2 LI

2+1

2CV

2

For Vc = Asin(ωt ) the current flowing in the circuit is I = C dVc/dt =

ωCAcos(ωt ) . The total energy stored in the reactive elements is

44



E S =

1

2 L

2C

2 A

2cos

2 t ( )+

1

2CA

2sin

2 t ( )

At the resonance frequency where ω = ω0 theenergy stored in the circuit becomes 0

=

1

LC

E S =

1

2CA

2

The energy dissipated per period is equal to the averageresistive power dissipated times the oscillation period.

E D = R I

2= R

0

2C

2 A

2

2

è

ö

ø

÷

2

0

= 2

1

2

RC

0 L

A2

è

ö

ø

And so the ratio Q becomes

Q =

0 L R

=1

0 RC

=1

R LC

45

Tutorial



Find the total impedance, resonance frequency, and

the Q factor of the circuit below:

Z = R

1+ jR

C 1

L

èç

ö

ø÷

0 =

1

LC Q =

0CR

Documents

AP2212-Lecture 9-SWT(1).pdf