department of information engineering104 capacitor a capacitor is a device that stores charge...

1Department of Information Engineering

Capacitor

• A capacitor is a device that stores charge

– Capacitor is like reservoir that stores water

– Capacitors differs in their capacity in storing charges

conductor

Insulator (vacuum, ceramic,plastic, metal oxide)


Charge up a capacitor

• Initially VC = 0, (0)

(0) CV V VI

R R

I

electrons

VC=0V

R


• Because of the insulator in the capacitor, electrons moving into the –ve side are trapped

– the –ve side becomes –ve charged, +ve voltage side becomes +ve charged,

• VC across capacitor charge stored


Capacitor

• Water analogy


Definition of Capacitor

• Tub stores water, capacitor stores charge

– C = Tub size

– Q = Volume of water

– V = Height of water level

– I = Rate of flow of water

• If a capacitor contains Q amount of water (charges), and the height of water (V), then the size is given by

CQ

tub of sizewater of volume

V


Capacitor

• C = size of the capacitor

Q Q

V or CC V

Large cap Small cap

Same charge (Q) but different V

V V


Capacitors in parallel

• What is the combined capacitance C?

– Two tubs in parallel, can store more water, C should be larger

VC1

C2=

C?V


Capacitor in parallel (C = C1+C2 )

• Parallel circuit, same voltage across C1 and C2

– Q1 = C1V, Q2 = C2V

– Total charge Q=Q1+Q2=(C1+C2)V

– C = = C1+ C2

C1

V

C2

V

Q

Same V for C1 and C2

VC1

C2

Q

V


Capacitors in series

• What is the combined capacitance C?

VC1

C2

= C?V


Capacitor in series

• Same current

– Therefore same amount of Q in both capacitors

C1

C2

V1

V2

V

I

Same QV depends on smaller cap

I


Capacitor in series

• 1 2

1 2

1 21 2

1 2

1 2

1 2

,

V 1 1 1

Q

Q QV V

C C

Q QV V V

C C

C C C

C Chence C

C C


• Resistor in series : R = R1 + R2

• Resistor in parallel :

• Capacitor in series :

• Capacitor in parallel : C = C1 + C2

1 2

1 1 1 R R R

1 2

1 1 1 C C C


Capacitor

• Dump Q amount of water to a tub of size C, so that the water level is changed by V

• But

• Hence

– i.e. the rate of change of the height of water is proportional to the current (inflow of water)

CQ

V

Q I t

V dV I

t dt C


Constant I

• Rate of increase of V = I/C,

• If I is constant, voltage increases constantly


DC and AC circuits

• DC (direct current) circuits

– voltage is constant over time

– e.g. battery or lab power supply

• AC (alternate current) circuits

– voltage is varying over time

– Extremely important! Signal is represented as voltage variation

– e.g. your voice, modem signal, data signal


DC circuit

• Initially VC = VC(0), find VC (t)

•

R

V0

VC

I

0 ( )( ) C CV V t dV

I t CR dt


RC circuit

0

( )

(0) 00

( )0 (0)

0

0

/0 0

/0 0

( )( )

1 1

ln( )

( ( ))ln

( (0))

( ) ( (0)) , (where )

( ) ( (0))

C

C

C

C

C C

V t t

CV

C

V tC V

C

C

t CR kC C

t CRC C

dV V V tI t C

dt R

dV dtV V CR

tV V k

CRV V t t

kV V CR

V t V k V V e k e

V t V V V e

( 0, ( ) (0) 1)C Ct V t V k


The general formula for charging/discharging a capacitor

/0 0( ) ( (0)) t CR

C CV t V V V e

Destination

Distance=Destination-Initial


Example - charging a capacitor

• At t=0, VC=0. What is VC(t)?

• Destination=V0

• Distance=V0-0=V0

• VC(t) = V0-V0e-t/RC = V0(1-e-t/RC)

R

V0

VC

I


RC circuit

• Charging curve for R = 1k, C = 1F, RC = 1ms

I=CdVcap/dt ~ 0


How long does it take to charge up the capacitor?

• Depends on RC (the time constant)

• t=RC: VCAP(t) = 0.63 V0

• t=5RC: VCAP(t) = 0.99 V0

• t=5RC, the capacitor is almost fully charged up

– (to fully charge up a capacitor 100% takes infinitely long time, not practical)

• R = 1k, C = 1F, takes around 5ms to charge up


Example

• Initially VC = 5V, V0=15V, find VC(t).

• Destination=15

• Distance=15-5=10

• Answer: VC(t) = 15 - 10e-t/RC

R

V0=15V

VC(0)=5V

I


Example: discharging a capacitor

• Initially VC = 15V, V0= 0V, what is VC(t)?

• Destination=0V

• Distance=0-15=-15V

• Solution: VC(t) = 0 - (-15)e-t/RC = 15e-t/RC

R

V0=0V

VC(0)=15V

I


How long will it take to discharge the capacitor?

• VC(t) = V0 e-t/RC

• t=RC: VC(t) = 0.37 V0

• t=5RC: VC(t) = 0.01 V0

• Again, take 5RC to discharge the capacitor


Example

• Initially VC = 15V, V0= -5V, what is VC(t)?

• Solution: VC(t) = -5 - (-20)e-t/RC = -5 + 20e-t/RC

R

V0=-5V

VC(0)=15V

I

destinationDistance


AC circuit analysis

• If V1 is an AC voltage, what is V2?

R

C V2V1


The importance of sinusoids

• In real world, interesting signals are time-varying

– if we pass the time-varying signal to a linear system, what is the output response?

Linear System(resistor,capacitor,and inductor)

?


Sunglass example

• Sunglass is used to filter sun light

• Sun light is a varying signal, how to analyze it?

Sun light


Principle of superposition!!

• Sun light = sum of light with different wavelengths

• Each individual light has a frequency and a wavelength

Sun light +


Characteristic of a filter

• The filter (sunglass) can be characterized by its frequency response, that depends on only two parameters

– The attenuation factor (V2/V1)

– The phase (angle) change

V1 V2


Input-output (transfer) characteristic of a light filter

• Magnitude plot

• Phase plot

2

1

V

V

frequency

frequency

1.0

0


• Input : divide sun light into sinusoidal light wave

• Output : sum all individual sinusoidal responses

• Apply the same approach to electrical signal

Sun light + + FilteredSun light


Fourier theorem: signal = sum of sinusoids

= sum of sinusoids

Linear Systemsinusoids

decompose

Sinusoidalresponses

sum

Output response

Input

Output = sum of sinusoids’ response

Input


How to represent a sine wave (sinusoid)?

• A sine wave

– (unit in radian)

• Make , so that is a function of time,

– Unit of : Radian/second

• We have

– For sinusoidal signal, one cycle = 2 radian

– If the rate of rotation = f cycle/s, then =2f (rad/s)

( ) sinf t A

t

( ) sinf t A t


Example : square wave = sum of sine waves !!

• Periodic square wave f(o) at frequency o

–

• 1kHz square wave = 1kHz + 3kHz +5kHz + . . . (sine waves)

0 0 0 0

4 1 1( ) (sin sin3 sin5 )

3 5w w w w f t t t

Fundamental freq Harmonics


Your ear

• Why our inner ear has this shape?

Shape of a cochlea A stretch out cochlea


Sinusoidal response of capacitor

• Let V = A cos wt (w = 2f)

• I = C dV/dt = - wC Asin wt

• Impedance = , complex relation because V and I are not in phase

~V

I

I leads V by 900

V

I

V

I


j - complex number notation

• V and I are not in phase

– Apart from the ratio of magnitude between V and I

– Also need to know the phase angle between V and I

• To keep track with the magnitude and phase information

– sin and cosine are tedious to use

– Simpler to use a the j-notation


A mathematician game

• The board

– 2 dimensional complex plane

– x-axis represents the real axis

• The player

– a vector (e.g. 1) on the plane

• The rule

– j represents an operation that rotates a vector anticlockwise by 90o on the complex plane

1 x


j

• j*1 = j (meaning rotate vector 1 by 90o)

• j*j=j2 = -1 (rotates the vector j by 90o)

– Therefore

1

j 1

j2 1

j3 1

1 j


• -j = rotate 900 clockwise

– -j = j3 = rotate 2700 or 900 clockwise

•

–

1

j3 1 = -j

jjj

j1

j1

jj1


j - complex number notation

• a+jb– Sum of vector a and jb– Represent a vector with a magnitude and phase

– Magnitude =

– Phase =

• – a vector with magnitude = 1 making angle with real axis

a

jb22 ba

ab

tan 1

cos sinj 1


Euler formula

•

Euler formula

• Proof by series expansion of ex2 3

2 3

2 4 3 5

12! 3!

( ) ( )1

2! 3!

12! 4! 3! 5!

cos sin

x

j

x xe x

j je j

j

cos sin jj e

1

ej


Complex sinusoid ejwt

•

– Represents a rotating vector that rotates at a rate of f cycles/sec ( )

j te

2 f

1

ejt

t

Direction of rotation


Real part of complex sinusoid

• R{ . } denotes the real part of something

• Example

– V = A cos wt = R { A ejwt } because

• V = R { A ejwt } = R { A(cos wt + j sin wt) }

= A cos wt

• If V = A cos (wt+), then V = R { A ej(wt+ }

– Without loss of generality, we shall assume =0 so as to simplify our calculation


Impedance of a capacitor

•

• To recover V and I, take the real part

– V = R { A ejwt } = A cos wt

– I = R { jwC Aejwt } = R { jwC A (cos wt + j sin wt) }

= - wC A sin wt

1

j t

j t

C

V A e

dVI C j C Ae j C V

dtV

ZI j C

ω

ωω ω

ω(Ohm’s law for capacitor)


Impedance of a capacitor (XC)

•

• magnitude of impedance =

• Small =>low frequency => ZC is large => open circuit

• Large =>high frequency => ZC is small => short circuit

1 C

VZ

I jwC

1CZCw


Circuit analysis

• If V2 is a varying signal, what is V2?

R

C V2V1


Impedance of capacitor

• Definition

– Impedance of capacitor =

• By KVL

• So that

• Hence

2C

VZ

I

1 2 CV IR V IR IZ

1

1

C

I VR Z

2 1 1 1

1/ 1

1/ 1C

CC

Z jwCV IZ V V V

R Z R jwC jwCR


Transfer function T(w)

• Transfer function T(w)= Output / Input

=

• Note that the output response depends on the frequency of the input sinusoid

2

1

( ) 1

( ) 1

V w

V w jwCR


Low pass filter

• Transfer function T(w)=

• High frequency ( ),

– High frequency signals are cut

• Low frequency ( ),

– Low frequency signals are passed

• The circuit is a low pass filter

2

1

( ) 1

( ) 1

V w

V w jwCR

w 2 ( ) 0V w

0w 2 ( ) 1V w


A sketch of the frequency response of the circuit

VINVOUT

V1

C1

C2

R1

R2


• Solve the problem just like any resistors’ network

• KCL

11 11 2

1 1 2 1 2

1

2 2

2 22

1 1 2 2 1 2 1 2 1 2

1 10 (where , )

0

1 ( )

OUTINC C

C C

OUT OUT

C

OUT IN

V VV V VZ Z

R Z Z jwC jwC

V V V

Z R

jwR CV V

jw R C R C R C w R R C C


Useful tricks

• Fast way to compute complex ratio

– a+jb = r1 ej, c+jd = r2 ej,

–

• Hence

–

jdcjba

2j

2

1 er1

)jdc(jdc

1

)21(j

2

1 err

jdcjba

Polar form


Useful trick

• Graphically as

• Magnitude =

• Phase =

)(j

2

1 21err

jdcjba

a

br1

cd

r2

22

22

2

1

dc

barr

1 11 2 tan tan b d

a c


Magnitude and phase response of a low pass filter (LPF)

•

– Magnitude response =

– Phase response =

jwCR11

VV

IN

OUT

2IN

OUT

)wCR(1

1V

V

)wCR(tanV

V 1

IN

OUT


Important parameters of a filter

• What is the cut-off point of the filter?

• How steep is the cut-off slope?

– the steeper the better


Cut-off point

• Defined as the frequency that the output power is halved– Also known as the half-power point

• Power is proportional to the square of voltage,

this is the point where

– i.e.

• Cut-off frequency is the frequency where the output voltage is dropped to about 70% of the input

21

V

V2

IN

OUT

7.02

1V

V

IN

OUT


Decibel (dB)

• dB is the logarithm of a ratio between powers

– 3dB => POUT is 2 times PIN



– -3dB => POUT is half of PIN , and so on

IN

OUT

PP

log10dB


3dB point

• Since

• Therefore,

• Why logarithm is used?– To get the overall power loss of multiple stages, we

simply add up all the dBs of the stages– addition is easier than multiplication

IN

OUT

PP

log10dB

dB321

log1021

PP

ifIN

OUT


• At half power point,

=> CCR = 1

=> C = 1/RC

• since C = 2f , therefore fC = 1/2RC

2

1 1 1

2 21 ( )OUT

IN

V

V wCR

3dB point of a LPF

Don’t forget the2 in exam!


• Phase shift at this frequency = - 45o

3dB point of a LPF

)j1()1(

)jwCR1()1(

VV

IN

OUT

j

0o

1


Slope of the attenuation

• At 3dB point

– |jCR| = 1

• At frequency beyond the 3dB point

– |jCR| >> 1 , hence

• double the frequency , half the output voltage

1 1~

1

1

OUT

IN

OUT

IN

V

V j CR j CR

V

V CR


• The slope falls at a rate of 6dB/octave

– in music, one octave = doubling of frequency

– Doubling the frequency , half the |VOUT/VIN|

=> POUT/PIN = 1/ 4 = -6dB

• Power is falling at a rate of 6dB per doubling of frequency


6dB/octave

• Plot dB vs frequency on semi-log paper

– x-axis is in log scale so that 1,2,4,8,… has uniform spacing

– Y-axis is in dB (linear scale)

– expect to see a straight line of slope = 6dB/octave

6dB/octave


High pass filter (HPF)

• At low freq

– cap is open circuit => VOUT ~ 0

• At high frequency

– cap is short circuit => VOUT ~ VIN

• Pass only high frequency

VinVout

VC


High pass filter (HPF)

• 3dB point of HPF = 1/2RC

• Slope = 6dB/octave

90o

45o

0o

6dB/octave


Transfer function T() of a system

•

• The magnitude and phase plots tell us everything we need to know about the linear circuit

( )( ) ( ) ( )

( )

ww w w

w OUT

IN

VT T

V

|T(w)|

w

1

Low-pass filter High-pass filter Band-pass filter

|T(w)|

w

1

|T(w)|

w

1


Input and output impedance of LPF

• Input impedance

– What is the worst case value?

(hint: ideally ZIN should be as large as possible, so worst case means the minimum ZIN)


Input and output impedance of LPF

• Output impedance

– What is the worst case value?

(hint: ideally ZOUT should be as small as possible, so worst case means maximum ZOUT)


Bandpass filter

• A filter that passes frequency only within the two 3dB points


Bandpass filter

• Use a HPF to get rid of low frequency

f3db = 1/2RC~1600Hz


Bandpass filter

• Use a LPF to get rid of high frequency

f3db = 1/2RC~8300Hz


Bandpass filter

• Can we connect the LPF and HPF together without changing the 3dB point?

• Yes if loading effect is insignificant (10X rule)


• Output impedance of circuit A

• Input impedance of circuit B

• ZIN is 10 times larger than ZOUT, pass 10X rule

ZOUT = 10K (worst case)

ZIN = 100K (worst case)


C blocking capacitor

• This useful circuit (high pass filter) only passes ac signal and blocks DC

– (B) is the equivalent circuit of (A) for ac input

VIN VOUT

4.7F

15V

10k

4.7kvIN

3k

4.7F

vOUT

(A) (B)


For ac signal, all DC voltage sources have zero impedance

• The 15V DC is like a zero impedance path to ground

– So the 4.7k and 10k are in parallel !!

• Why?

– Let the DC voltage at A be V, and the DC currents be I1 and I2

– If the ac signal at A is v, what is the ac current?

A

I1

I2

V V v


•

– I1 => I1-I1 where I1 = v/10k

– I2 => I2+I2 where I2 = v/4.7k

A

V+vI1-I1

I2 +I2

V V v


Superposition

• Divide the signal into DC and AC parts

V+vI1-I1

I2 +I2

VI1

I2

vI1

I2

+


• Impedance sees by the ac signal =

•

• i.e. two resistors in parallel

v

I

1 2

10 * 4.7

10 4.7

v v k k

I I I k k

15V

II1

I2

To ac signal v,10k and 4.7k act as if in parallel


DC blocking capacitor

• 3dB point?

–

– Pass signal with frequency > 11 Hz

6 3

1 1 1~ 11( )

2 2 *4.7*10 *3*10 f HzCR CR

vIN3k

4.7F

vOUT


The design of a good probe

• Good probe has a very large input impedance

• but be careful with the cable’s capacitance !!

cable Large impedancee.g. 1M

probe


Design of a coaxial cable (also known as BNC cable)

signal

ground

Ground clip

probe

shield

0VVin

Tiny capacitance(between signal and ground)


• coaxial cable (coax)– high quality cable, able to carry very high

frequency, used in cable TV (video frequency is very high), and in laboratories

• The shield acts as a Faraday cage

• e.g. like your microwave oven

– external interference (energy) cannot penetrate into the wire (less noise), and signal energy inside cannot leak out


Effect of the tiny stray capacitance

• Cable capacitance is in parallel with input impedance

• Typical value of cable capacitance?– 30pF/foot – 4 feet cable ~ 120pF (pF = 10-12F)

• Impedance of ZC

– At 1kHz, |ZC|=1.3M– At 1MHz, |ZC|~1.3k

• Problem– Scope’s input impedance is reduced at high frequency !

1M


The solution

• Add a 9M resistor and a Cprobe to the probe

– Cprobe is 9 times smaller than the cable capacitance (so that its impedance is 9 times larger)

probe1M


•

=

9X=

X=jwCR1R

jwCR1R9

19 10

OUT

IN

V XV X X


• Overall impedance =

– Impedance is increased by 10 times

• The chain attenuates all frequencies equally without any phase shift !

– i.e. behaves like a pure resistive divider !!

1010

1

RX

j CR


10X input

• Always use 10X mode provides by the probe

– Why? Because input impedance is increased

• But remember the displayed signal is ten times smaller (because of 10:1 attenuation), so you need to multiply the signal by 10

• Probe compensation

– Our method only works if Cprobe is exactly nine times smaller than Cstray

– How to make this adjustment ?


How to adjust the probe?

• square wave = sum of infinitely number of sinusoids

• feeding a square wave signal to the scope is like feeding a full range of sinusoids simultaneously

• square wave is easy to generate

– the scope has a square wave output (1kHz)


How to adjust the probe?

• Use the x10 probe to see the square wave

– if the value of Cprobe is correct, then all the sinusoids are reduced exactly by 10 times smaller

– see a perfect square wave that is 10 times smaller

• If the value of Cprobe is not right, then some frequencies will be attenuated more than the other; the square wave would not be perfect

• Adjust Cprobe so that a perfect square wave can be seen


Using square wave to tune the probe

(to get this perfect square wave,the probe must pass all frequencies uniformly)


Inductor

• Ampere‘s Law

– current generates magnetic field

– direction of magnetic flux : right-hand screw law

currentMagnetic flux

Current flowsinto paper


Inductor

• Straight wire generates a weak magnetic field

• Core generates a stronger field

– The magnetic field is aligned in the same direction

– Weak forces add up to a stronger force

current


Inductor

• Sudden increase of current

– Change of current disturbs the magnetic field

– Inductor reacts by producing a voltage in a direction that try to reduce the increase of current

• Lenz’s Law

– Voltage is induced in a direction to oppose the change

– The induced voltage is linearly proportional to the rate of change of current


Inductor

• Induced voltage

• Induced voltage is given by

– the larger the change in current, the larger the induced voltage

Increasing current

Induced emf to oppose the increase in flux caused by the increasing current

dIV L

dt

+ -


Inductor

• Example

– Switch closed at t=0, what is I?

– At t<0, I=0 (because of open circuit)

V

L

R

I


Inductor

• After the switch is closed, I goes up, so voltage VL is induced in a direction to oppose the flow of I

V

L

R

I

VL


• KVL gives

• Initial condition, I(0)=0, therefore k=1

• Hence

0 0

( )

0 00

0 0

(where )

1

( )

RL

I t t

t

dIV L IR

dtL dI V

I I IR dt R

RdI dt

I I L

I t I kI e

0 0( )

RL

tI t I I e


Inductor

• An application – choke

– lightning can cause a sudden surge of current, which may damage your equipments

– an inductor can be used to smooth the surge

• Time constant = R/L

• After 5 , the change of I is negligible, , the inductor is like a short circuit

I

t

I0

0L

dIV L

dt


Impedance of inductor

•

• impedance ZL =

• Note

– voltage leads current by 90o

– at low frequency (small ), inductor ~ short-circuit

– at high frequency (large ), inductor ~ open-circuit

w

ww

j t

j t

I Ae

dIV L j LAe

dt

wVj L

I


CIVIL

• Capacitor: I leads V by 90o (I=jC*V)

• Inductor: V leads I by 90o (V=jL*I)

CIVIL


LC circuit

• Quick analysis

– at low freq, C is open circuit

• VOUT = VIN

– at high freq, L is open circuit

• VOUT = VIN

– a notch filter !

VOUT/VIN

freq

1


LC circuit

• By calculation

•

2

2

11

OUT C L

IN C L

V X X w LCV R X X jwCR w LC

2

0, 1

1, 0

, 1

OUT

IN

OUT

IN

OUT

IN

Vw

V

Vw

LC V

Vw

V


Another LC resonant circuit (lab work)

• Frequency response?

• at low frequency

– L is short circuit, C is open circuit, VOUT = 0

• at high frequency

– L is open circuit, C is short circuit, VOUT = 0


LC resonant circuit

• The resonant circuit is highly selective

– only pass a narrow range of frequencies

– Usage: tuner circuit in radio receiver

• Q-factor

– a measure of the narrowness of the peak

– the narrower, the better

– Q = f0/f3dB

• (f3dB is the width of the two 3dB points)

• (f0 is the peak frequency)


To verify the Fourier components of a square wave

• In the lab, centered the tune circuit at 16kHz

• Drive the circuit with a square wave ranging from 0Hz to 16kHz, at what frequencies would the tune circuit detect large outputs?

–

– 16kHz (fundamental freq of 16kHz)

– 16/3 kHz (2nd harmonic of 16/3kHz = 16kHz)

– 16/5 kHz (3rd harmonic of 16/5 kHz = 16kHz)

0 0 0 0

4 1 1( ) (sin sin3 sin5 )

3 5w w w w f t t t


Transmission line

• Medium that transmits signal/energy from one place to another is called an transmission line

– e.g. coaxial cable, telephone wires, optical fibre, wave guides, power lines


A cable of length L

• Why the impedance of coaxial cable is 50/75 ?

• What is the propagation speed of the voltage signal?

• How long does it take for the signal to reach RLoad?

• What is the voltage and current along the cable?

• Why some time you hear echo in phone call?

• How to eliminate the echo?

VS

RS

RLoadV(z,t)

I(z,t)

z

L


Uniform lossless transmission line

• The simplest model, the cable has

– Zero resistance, but some inductance and capacitance

– Let L be the inductance of the line per meter (H/m)

– Let C be the capacitance of the line per meter (F/m)

• A distributed circuit

is electricallymodeled by


• Consider an small section of an infinitely long line


Consider a small section of line of length z

• V(z)=V, V(z+z)=V+V, what is V+V?

V V+V

V

z z+z

I



• What is V?

–

V V+V

V

z z+z

I

IV L z

t



• ( )V zV V V z

z

z z+z

V(z)

( )V z

z

V V

V

z


(1)

V V V V

I VL z V z

t z

V IL

z t

Apply KVL to the section

V V+V

V

z z+z


Apply KCL to the section

•

(2)

I I I I

V IC z I z

t z

I VC

z t

I I+II


• Fundamental equations for the line

• To solve, differentiate (1) with respect to z

• differentiate (2) with respect to t

)2(tV

CzI

,)1(tI

LzV

tzI

LzV 2

2

2

2

22

tV

Ctz

I


• Hence

• recast the voltage wave equation into the form

• where is the characteristic velocity of the line

)equationwavecurrent(tI

LCzI

)equationwavevoltage(tV

LCzV

2

2

2

2

2

2

2

2

2

2

22

2

tV

u1

zV

LC1

u


• What does the equation mean?

• If our derivation is correct, then the voltage signal V(z,t) must satisfy the voltage wave equation

• So what are the solutions to this voltage wave equation?

– Any functions that have the form V=f(z-ut) are solutions, where f(.) is an arbitray function

– e.g. cos(z-ut), (z-ut)3.5, e-(z-ut), …

2

2

22

2

tV

u1

zV


• To verify that V=f(z-ut) is a solution of the wave equation

2 22 '' ''

2 2

2 2

2 2 2

( ), ( )

1

V Vu f z ut f z ut

t z

V Vz u t

and


f(z-ut)

• f(z) is a stationary function

– e.g. cos(z)

• f(z-ut) shift the function by ut in the +ve z direction

– e.g. cos(z-ut)

• f(z-ut) describes a traveling wave moving in the +z direction

• Similarly, g(z+ut) describes a traveling wave in the –z dir

ut


• General solution for voltage wave equation

• From (1)

( , ) ( ) ( )V z t f z ut g z ut

wave travellingin +z direction

wave travellingin -z direction

' '

( ) ( )

I VL

t z

I z zL f t g t

t u u


• Integrating partially with respect to time

– h(z) is a constant depending on initial conditions

– not interested, only concerned with the dynamic motion of the wave

• So

1( , ) [ ( ) ( )] ( )I z t f z ut g z ut h z

Lu

1( , ) [ ( ) ( )] z zI z t f z ut g z ut I I

Lu

Voltage wave in +z direction

Voltage wavein -z directionImpedance!


• Overall voltage = f(z-ut) + g(z+ut)

V=f(z-ut)

+z direction

V=g(z-ut)

( , ) ( ) ( )V z t f z ut g z ut

Resultant voltage= f(z-ut)+g(z+ut)


0

1( , ) [ ( ) ( )] z zI z t f z ut g z ut I I

Z •

• Overall current I = I+z – I-z

0

( )z

f z utI

Z

V=f(z-ut)

+z direction

V=g(z-ut)

0

( )z

g z utI

Z

z zI I I


• characteristic impedance of the line

• so this is the impedance looking into an infinitely long line !!

• signal propagation velocity

CL

LuZ0

LC1

u


Example - Category 5 cable

• Category 5 cable is high-grade telephone cable

– commonly used for connecting Ethernet

• Data sheet

– impedance: 100 ohms +/- 15%, 1MHz to 200 MHz

– capacitance: 5.6nF /100 m max

– DC resistance: 9.38 ohms/100 m max

– Propagation delay: 538 ns/100 m @ 100 MHz


Example - Cat 5 cable

• What is L?

• Does it agree with the propagation velocity in the spec?

• Speed quoted in the spec =

m100/H5610CL

)caseourin(100CL

Z

4

0

s/m108.1

1056

1

1056106.5

1LC1

u

8

202118

s/m1085.1s/m10538

100 89


An old problem (?) revisited

• What is V?

• Initially, the source see the source impedance in series with Z0 !!

– voltage (Vi) transmitted into the cable is given by

0

iiS

0S

0i Z

VI ,V

ZR

ZVV


• So V will be the voltage we see -- if we have an infinitely long cable

• But the length of our cable is finite, the traveling voltage wave will reach the end of cable typically at a speed of 2/3 of speed of light (2 x 108 m/sec)

• Suppose the cable is terminated by a load RL, and if

then we have a discontinuity at the boundary,RIV

Li

i


• Reflection and transmission at the boundary

• incident voltage and current relation:

• boundary condition:

0i

i ZIV

LRIV

V


• To satisfy the boundary condition, a portion of voltage and current waves must be reflected

– Vr and Ir are the reflected waves

• And the incident and reflected waves both satisfy

–

such that i r i rL

i r i r

V V V V VVR

I I I I I I

0i r

i r

V VZ

I I


Reflection coefficient

• Let be the reflecting coefficient

• Since

• rearranging the above equation, we have

0 0

0 0

, ,

11

i r i rL i r

i r

i rL

i r

V V V VVR but I I

I I I Z Z

V VR Z Z

V V

i

r

VV

0

0

Lr

i L

R ZV

V R Z


Matched termination RL = Z0

• Important special case

– If RL = Z0, then =0

– No reflection (good!)

– This situation is called matched termination

• In communication, reflection is not good (this is why you can hear echo)

• Always match the line by making RL = Z0 if possible


Transmission coefficient

• How much voltage is transmitted to the load?

– Hence

• If the line is matched (RL=Z0), then maximum energy is transmitted to the load (=0, =1)

1V

VVVV

i

ri

i

0L

L

ZRR2


A matched cable of length (L)

• Suppose a step waveform (1V) is sent at t=0

– what is V at t=0?

• V= ( Z0 / (Z0+Zin) )*1V = 0.5 V

• the incident wave is completely absorbed by RL because the line is matched (RL = Z0)

• no reflected wave

V

L


A shorted cable of length (L)

• What if RL =0 (short circuit)?

– You know the answer, V=0, but NOT at the beginning!

• At t=0

– V= ( Z0 / (Z0+Zin) )*1V = 0.5 V


Cable of finite length (L)

• At t = L / u, the voltage wave reaches the load

–

– so that a voltage wave (Vr) is reflected, where

• t=2L/u– the reflected wave returns to the source– V = Vi + Vr = 0.5V - 0.5 V = 0 !!

• Your expected result happened only after 2L/u sec– no more reflection at the source because the RS=Z0

)0Rce(sin1ZR

ZRL

0L

0L

V5.0VVV iir


• Pulse duration

– let’s say the cable is 30m long and u=2/3 c (velocity of light)

– t= 2L/u = 0.3 s

– A technique to produce very short pulses !


General case

• The case where the source, the line and the load impedances are all different

– reflection coefficient at the load = L

– reflection coefficient at the source = S

0L

0L

ZR

ZR

0S

0S

ZR

ZR


• Voltage at the source side

– at t=0: SS0

0i V

RZ

ZVV

VS

RS

RLoad

L

ViV



– at t=2L/u: iLSiLi VVVV

VS

RS

RLoad

L

ViV pLVi

pSpLVi



– at t=0:

– at t=2L/u:

– at t=4L/u:

SS0

0i V

RZ

ZVV

iLSiLi VVVV

iLSLSiLSLiLSiLi VVVVVV


• Add everything up

–

• since

• therefore

))1()(1(V

)1(V

)1(VV

LSL2

LSLSi

LSLLLSLSLSi

LSLSLSLLSLi

1|x|forxx1x1

1 2

11 1

1 1 1

(the line impedance disappeared !!)

Li L i

S L S L S L

LS

S L

V V V

RV

R R


Numerical example

• RS=100, Z0=50, RL=100 so that S=L=1/3

• Expected result (without considering the cable effect)

100

100 100

1

2

LS

S L

S

RV V

R R

V


A sketch of the waveform of V versus time

• e.g. cable length=1m, u=2*108 m/s

– t=0, V=(50/150)VS = 0.33 VS

– t=2L/u=20ns, V = 1/3 VS ( 1 + L + L S )

= 1/3 VS (1 + 1/3 + 1/9 ) = 0.48 VS

– t=4L/u=40ns, V = 1/3VS (1 + 1/3 + 1/9 + 1/27 + 1/81 )

= 0.498 VS

• So after 20ns, the voltage is very close to the steady state solution (0.5VS)


V versus time at z=0

• Expected

• What actually happened

tt=0

0.5V

t0 20ns

0.33V

0.48V0.5V

40ns


Settling time

• Let t be the settling time (20ns in our case)

• if t << T (period of the pulse)

– mismatch at the terminal is not much of a problem

• The problem becomes critical if

– the frequency is high (small T)

– the cable is long (large t)

• IEEE802.3 (100BaseT)

– Cable length < 100m for 100Mbps network


• What is the use of these two resistors?


Transmission Line

• 100 ohm

– used to kill any oscillation by reducing the phase shift caused by the cable capacitance

• 51 ohm

– use to terminate the 50 ohm coaxial cable (use 51 ohm because there is no 50 ohm resistor )


Applications

• The bus in computer’s backplane must be terminated

– the bus line has approx 100 ohms characteristic impedance

Capacitor Open circuit for DCShort circuit for AC

Eliminate DC power loss

100ohm to eliminatesignal reflection


Applications

• To connect a number of video terminals in series

Un-terminated(open circuit) terminated

by 75 ohmsresistor

department of information engineering104 capacitor a capacitor is a device that stores charge...

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