6.002 Fall 03 1

6.002 CIRCUITS ANDELECTRONICS

Damped Second-Order Systems

6.002 Fall 03 2

Damped Second-Order Systems

CA B

5V

+–

5V

CGS

largeloop

2KΩ50Ω

2KΩ

Remember this Demo

Our old friend, the inverter, driving another.The parasitic inductance of the wire and the gate-to-source capacitance of the MOSFET are shown

[Review complex algebra appendix in Agarwal & Lang for next class]

S

6.002 Fall 03 3

Damped Second-Order Systems

CA B

5V

+–

5V

CGS

largeloop

2KΩ50Ω

2KΩ

+–5V CGS

2KΩB

LRelevant circuit:

S

6.002 Fall 03 4

Now, let’s try to speed up our inverter by closing the switch S to lower the effective resistance

tvA

5

0

vB

0 t

vC

0 t

Observed Output

2kΩ

2kΩ

6.002 Fall 03 5

tvA

5

0

vB

0 t

vC

0 t

Observed Output ~50Ω

50Ω

Huh!

6.002 Fall 03 6

+– CL +

–)(tv

)(tvI

)(ti

In the last lecture, we started by analyzing the simpler LC circuit to build intuition

6.002 Fall 03 7

And for initial conditionsv(0) = 0 i(0) = 0 [ZSR]

IvIV

0 t

We solved

For input

IvLC

vLCdt

vd 112

2

=+

In the last lecture…

6.002 Fall 03 8

toVVtv II ωcos)( −= where

LC1

o =ω

Total solution

In the last lecture…

LC

)(tv

t

IV

IV2

0Iv

+– CL +

–)(tv

)(tvI

)(ti

6.002 Fall 03 9

See A&L Section 13.6

+– C

R L

+

–)(tv)(tvI

)(ti

Damped sinusoids with R – remember demo!

Today, we will close the loop on our observations in the demo by analyzing the RLC circuit

LC

)(tv

t

IV

IV2

0Iv add R

6.002 Fall 03 10

v, i state variables

+– CL +

–)(tv

)(tvI

)(ti

Node method:Recall

element rules

Let’s analyze the RLC network

R

dtdiLvL =

idtvL

t

L =∫∞−

1

L:

dtdvCi C

C =C:

Av

IvLC

vLCdt

dvLR

dtvd 112

2

=++

Rvvdt)v(v

L1 A

t

AI−

=∫ −∞−

dtdvC

RvvA =

−

:Av

:v

6.002 Fall 03 11

+– CL +

–)(tv)(tvI

)(ti

Node method:

Let’s analyze the RLC network

R

Av

2

2

)(1dt

vdvdtdvRCv

LC I =−−

IvLC

vLCdt

dvLR

dtvd 112

2

=++

Rvvdt)v(v

L1 A

t

AI−

=∫ −∞−

dtdvC

RvvA =

−

:Av

:v

vdtdvRCvA +=

2

2

)(1dt

vdCvvL AI =−

2

2

)(1dt

vdvvLC AI =−

6.002 Fall 03 12

Recall, the method of homogeneous andparticular solutions:

( ) ( )tvtvv HP +=

1

2

3

Find the particular solution.

Find the homogeneous solution.

4 steps

The total solution is the sum of theparticular and homogeneous.Use initial conditions to solve for theremaining constants.

Solving

6.002 Fall 03 13

And for initial conditionsv(0) = 0 i(0) = 0 [ZSR]

Iv

0 t

IV

Let’s solve

For input

IvLC

vLCdt

dvLR

dtvd 112

2

=++

6.002 Fall 03 14

1 Particular solution

IP Vv = is a solution.

IPPP V

LCv

LCdtdv

LR

dtvd 112

2

=++

6.002 Fall 03 15

0vLC1

dtdv

LR

dtvd

HHH =++2

2

Solution to

Homogeneous solution2

Recall, vH : solution to homogeneousequation (drive set to zero)

Four-step method:

Dtsts

H eAeAv 2121 +=

General solution

Assume solution of the formA?s,A,Aev st

H ==

B Form the characteristic equationf(s)

Find the roots of the characteristic equationC

21 ss ,

6.002 Fall 03 16

LR

2=α

0vLC1

dtdv

LR

dtvd

HHH =++2

2

Solution to

Homogeneous solution2

Dtt

Hoo

eAeAv⎟⎠⎞⎜

⎝⎛ −−−⎟

⎠⎞⎜

⎝⎛ −+−

+=2222

21

ωααωαα

General solution

LC1

o =ω

Assume solution of the formA?s,A,Aev st

H ==

characteristicequationB 02 =++

LC1s

LRs

0=++ o22 s2s ωα

RootsC o1s22 ωαα −+−=

os 222 ωαα −−−=

so, 012 =++ ststst AeLC

AseLReAs

6.002 Fall 03 17

Total solution3

Find unknowns from initial conditions.

)()()( tvtvtv HP +=

:0)0( =v 210 AAVI ++=

:0)0( =i

dtdvCti =)(

tt

Ioo

eAeAVtv⎟⎠⎞⎜

⎝⎛ −−−⎟

⎠⎞⎜

⎝⎛ −+−

++=2222

21)(ωααωαα

( ) +−+−=⎟⎠⎞⎜

⎝⎛ −+− t

oo

eCA22

221

ωααωαα

( ) to

o

eCA⎟⎠⎞⎜

⎝⎛ −−−

−−−22

222

ωααωαα

so, ( ) ( )oo AA 222

2210 ωααωαα −−−+−+−=

Mathematically: solve for unknowns, done.

6.002 Fall 03 18

Let’s stare at this a while longer…

ttttI

oo

eeAeeAVtv⎟⎠⎞⎜

⎝⎛ −−−⎟

⎠⎞⎜

⎝⎛ −− ++=

2222

21)(ωααωαα

3 cases:

ttI eAeAVtv 21

21)( αα −− ++=

tjttjtI

oo

eeAeeAVtv⎟⎠⎞⎜

⎝⎛ −−−⎟

⎠⎞⎜

⎝⎛ −− ++=

2222

21)(αωααωα

tjttjtI

dd eeAeeAV ωαωα −−− ++= 2122 αωω −= od

t

IvIV

vOverdampedoωα >

Underdampedoωα <

teKteKV dt

dt

I ωω αα sincos 21−− ++= tjte dd

tj d ωωω sincos +=

Critically dampedoωα =Later…

6.002 Fall 03 19

Let’s stare at underdamped a while longer…

Underdamped contd…oωα <

teKteKVtv dt

dt

I ωω αα sincos)( 21−− ++=

:0)0( =v

:0)0( =i dtdvCti =)(

IVK −=1

dKK ωα 210 +−=

d

VKωα1

2 −=

teVteVVtv dt

dId

tII ω

ωαω αα sincos)( −− −−=

tVVtv oII ωcos)( −=

Same as LC as expected

Note: For 00 == ⇒αR

teCKteCK

teCKteCK

dt

ddt

dt

ddt

ωωωα

ωωωααα

αα

cossin

sincos

21

21−−

−−

+−

−−=

6.002 Fall 03 20

Underdamped contd…oωα <

⎟⎟⎠

⎞⎜⎜⎝

⎛−−= −−

dd

t

d

oII teVVtv

ωαω

ωω α 1tancos)(

Remember, scaled sum of sines (of the same frequency) are also sines! -- Appendix B.7

teVteVVtv dt

dId

tII ω

ωαω αα sincos)( −− −−=

add R

Let’s stare at underdamped a while longer…

LC

)(tv

t

IV

IV2

0Iv

6.002 Fall 03 21

Underdamped contd…oωα <

⎟⎟⎠

⎞⎜⎜⎝

⎛−−= −−

dd

t

d

oII teVVtv

ωαω

ωω α 1tancos)(

Remember, scaled sum of sines (of the same frequency) are also sines! -- Appendix B.7

teVteVVtv dt

dId

tII ω

ωαω αα sincos)( −− −−=

LC

)(tv

t

IV

IV2

0Iv add R

tCritically dampedoωα =

Section 13.2.3

underdamped

overdamped

criticallydampedv

6.002 Fall 03 22

tvA

5

0

vB

0 t

vC

0 t

Remember this? Closed the loop…

50Ω

See example 12.9 on page 664 of the A&L textbookfor inverter-pair analysis

6.002 Fall 03 23

Underdamped

:2αωoQ = Quality factor (approximately

the number of cycles of ringing)

:IV Final value :)0(v Initial value

)(tv

t

IV

0

:dω Oscillation frequency

dωπ2

te α−

:α Governs rate of decay

“ringing”

02 =++LC1s

LRs

0=++ o22 s2s ωα

Characteristicequation

22 αωω −= od

⎟⎟⎠

⎞⎜⎜⎝

⎛−−= −−

dd

t

d

oII teVVtv

ωαω

ωω α 1tancos)(

Intuitive AnalysisSee Sec. 12.7 of A&L textbook

6.002 Fall 03 24

Intuitive Analysis

)(tv

t

IV

0

IV

dωπ2

period

QRinging stopsafter cycles

02 =++LC1s

LRs

0=++ o22 s2s ωα

Characteristicequation

αω2

oQ =

22 αωω −= od

)0(v ?

+– CL +

–)(tvIv

)(ti

R

)0(iis –veso v(t) must drop

)0(i)0(v

given-ve+ve

See Sec. 12.7of A&L textbook