35 EE394J Spring11 Notes on Second Order Systems

8/13/2019 35 EE394J Spring11 Notes on Second Order Systems

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The simplest second-order system is one for which the Ls can be combined, the s can be

combined, and the !s can be combined into one L, one , and one !. Let "s consider sit"ations

where there is no so"rce, or a # so"rce.

For an RLC Series Circuit

$f the circ"it is a series circ"it, then there is one c"rrent.

%ote that in the abo&e circ"it, the ' and ! co"ld be con&erted to a %orton e("i&alent, permittin)the incorporation of a c"rrent so"rce into the problem instead of a &olta)e so"rce.

*sin) +'L,

0,

=+++=+++ dtiCdtdi

LiRVvvvV CLR .

Tain) the deri&ati&e,

02

2

=++

C

i

dt

idL

dt

diR .

%ote that the constant so"rce term ' is )one, lea&in) "s with the characteristic equationi.e.,

so"rce free/, which yields the natural responseof the circ"it. !ewritin),

02

2

=++LC

i

dt

di

L

R

dt

id. /

"ess the nat"ral response sol"tion stAei = and try it,

0

,, 22

=

++=++ LCLR

ssAeAeLCAseL

R

eAs stststst

,

1nd we reason that the only possibility for a )eneral sol"tion is when

0,2 =++

LCL

Rss .

a)e of 20

'

&

i

L

!

&L &!


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This leads "s to

LCL

R

L

RLCL

R

L

R

s ,

222

42

2

=

=.

#efinin)

L

R

2= as the dampin) coefficient neperssec/ for the series case 2/

LCo

,= as the nat"ral resonant fre("ency radianssec/ 3/

then we ha&e

22os = . 4/

6"bstit"tin) 2/ and 3/ into / yields a "sef"l form of 2/, which is

02 22

2

=++ idt

di

dt

ido 5/

For an RLC Parallel Circuit

$f the circ"it is a parallel circ"it, then there is one &olta)e. +L at the top node yields

0,

=+++=+++ dtdv

CvdtLR

vIiiiI CLR .

Tain) the deri&ati&e

0,

2

2

=++dt

vdC

L

v

dt

dv

R.

!ewritin),

a)e 2 of 20

$&

i!

L !

iL i


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2,0

22

0

,,/0. AAeAeAti

t

ts

t

ts+=+==

==,

22,,0

222

0

,,,

0

sAsAesAesAdt

di

t

ts

t

ts

t

+=+====

.

Th"s, we ha&e two e("ations and two "nnowns. To sol&e, one m"st now /0. =ti and0=tdt

di

. 6ol&in), we )et

==== =

2

,,0

2, /0./0.s

sAdt

di

tiAtiA t ,

so

2

,

2

0

,

,

/0.

s

s

s

dt

di

ti

A

t

=

=

=

%ote see modification needed to incl"de final response/ =/

Then, find 2A from

,2 /0. AtiA == %ote see modification needed to incl"de final response/ >/

The key to finding 1and !is al"ays to kno" the inductor current and capacitor voltage

at t # $%& !emember that

*nless there is an infinite imp"lse of c"rrent thro")h a capacitor, the &olta)e across a

capacitor and the stored ener)y in the capacitor/ remains constant d"rin) a switchin)

transition from t ? 0-to t ? 0.

*nless there is an infinite imp"lse of &olta)e across an ind"ctor, the c"rrent thro")h an

ind"ctor and the stored ener)y in the ind"ctor/ remains constant d"rin) a switchin)

transition from t ? 0-to t ? 0.

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@or e9ample, consider the series circ"it at t ? 0. "rrent +== 0/0. LIti .

Ance $L0 and '0are nown, then we )et0=tdt

dias follows. @rom +'L

0000 =+++ +++ CLL VVRIV

th"s

+++ = 000 CLL VRIVV . 0/

6ince

00

=

+=

tL

dt

diLV

then we ha&e

L

V

dt

di L

t

+

== 0

0. /

6imilarly, for the parallel circ"it at t ? 0, the &olta)e across all elements +== 0/0. CVtv .


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th"s

+++ = 000 LRC IIII . 2/

6ince

00

=

+=

tC

dt

dvCI

then we ha&e

C

I

dt

dv C

t

+

== 0

0. 3/

Case !: 'nderdamped


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[ ] [ ]/sin./cos./sin./cos. 2, tjteAtjteA ddt

ddt

++= ,

[ ]/sin././cos./. 2,2, tAAjtAAe ddt

++= . 5/

6ince it/ is a real &al"e, then 5/ cannot ha&e an ima)inary component. This means that

/. 2, AA + is real, and /. 2, AAj is real which means that /. 2, AA is ima)inary/. Theseconditions are met if B,2 AA = . Th"s, we can write 5/ in the form of

[ ]/sin./cos./. 2, tBtBeti ddt

+= /

where Cand C2are real n"mbers.

To e&al"ate Cand C2, it follows from / that

[ ] ,2,0

/0sin./0cos./0. BBBeti =+==

,

so

/0., == tiB %ote see modification needed to incl"de final response/ 7/

To find 2B , tae the deri&ati&e of / and e&al"ate it at t ? 0,

[ ] [ ]/cos./sin./sin./cos. 2,2, tBtBetBtBedt

didddd

tdd

t

+++= ,

[ ] [ ] ddt

BBBBdt

di 2,2,

0

+=+=

=

,

so we can find 2B "sin)

d

t

Bdt

di

B

,0

2

+= = =/

Case ! Solution in Polar Form

The form in /, [ ]/sin./cos./. 2, tBtBeti ddt

+= is most "sef"l in e&al"atin)

coefficients Cand C2. C"t in practice, the answer is "s"ally con&erted to polar form.

roceedin), write / as

++

++= /sin./cos./.

22

2,

2

22

2,

,22

2, t

BB

Bt

BB

BeBBti dd

t

a)e 7 of 20


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[ ]/sin./sin./cos./cos./. 222, tteBBti dd

t

++=

The 2

2

2

,

,

BB

B

+and 2

2

2

,

2

BB

B

+

terms are the cosine and sine, respecti&ely, of the an)le shown

in the ri)ht trian)le. *nless C is ;ero, yo" can find D "sin)

=

,

2,tanB

B .

C"t be caref"l beca"se yo"r calc"lator will )i&e the wron) answer 50E of the time. The reason

is that /,=0tan./tan. = . 6o chec the ("adrant of yo"r calc"lator answer with the("adrant consistent with the ri)ht trian)le. $f the calc"lator ("adrant does not a)ree with the

fi)"re, then add or s"btract =0F from yo"r calc"lator an)le, and re-chec the ("adrant.

The polar form for it/ comes from a tri)onometric identity. The e9pression

[ ]/sin./sin./cos./cos./. 222, tteBBti dd

t

++=

becomes the damped sin"soid

/cos./. 2

22,

+= teBBti dt .

Case (: Critically )amped


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@or the case with o= , the e("ation becomes

02 22

2

=++ idt

di

dt

id .


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( ) tfinal eAtAIti

++= 2,/. ,

[ ] ( ) tfinal eAtAIti += 2,/. for critically damped/. 25/

Ho" can see that the presence of the final term finalI will affect the 1 and C coefficients

beca"se the initial &al"e of it/ now contains the finalI term. Take finalI into account "hen

you evaluate the +s and ,+s -y replacing /0. =ti in ./0 .20 .130 and .!10 "ith

finalIti = /0. &

Now do yo" )et the final &al"esG $f the problem has a # so"rce, then remember that after alon) time when the time deri&ati&es are ;ero, capacitors are Jopen circ"its,M and ind"ctors are

Jshort circ"its.M omp"te the Jfinal &al"esM of &olta)es and c"rrents accordin) to the Jopen

circ"itM and short circ"itM principles.

The 4eneral Second Order Case

6econd order circ"its are not necessarily simple series or parallel !L circ"its. 1ny two non-

combinable stora)e elements e.)., an L and a , two Ls, or two s/ yields a second order

circ"it and can be sol&ed as before, e9cept that the and o are different from the simple

series and parallel !L cases. 1n e9ample follows.


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0, =++

LCL ILVIR . 2/

%ow, write +L at the node P"st to the left of the capacitor,

02 =++

CC

L VCR

V

I ,

which yields

+= C

CL VC

R

VI

2. 27/

Tain) the deri&ati&e of 27/ yields

+= CC

L VCR

VI

2

. 2=/


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omparin) 2>/ to the standard form in 5/, we see that the circ"it is second-order and

+= ,

,

2

,2

R

R

LCo ,

+=

CRL

R

2

, ,2 , so

+=

CRL

R

2

, ,

2

, .

The sol"tion proced"re for the nat"ral response and total response of either /.tvC or /.tiL can

then proceed in the same way as for series and parallel !L circ"its, "sin) the and o

&al"es shown abo&e.

%otice in the abo&e two e("ations that when 2R ,

LCo

,2 , andL

R

2

, ,

which correspond to a series !L circ"it. #oes this mae sense for this circ"itG

%ote that when 0,R , then

LCo

,2 , andCR22

, ,

which correspond to a parallel !L circ"it. #oes this mae sense, tooG

%ow, consider irc"it O2.

T"rnin) off the independent so"rce and writin) +'L for the ri)ht-most mesh yields

02222,, =++

LLL ILIRIL . 30/

+'L for the center mesh is

[ ] 0,,2,, =++LLL ILIIR

,

yieldin)

a)e 3 of 20

$

eneral ase irc"it O2 rob. =.0/

L2

!2

$L2L! $L


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= ,

,

,,2 LLL I

R

LII . 3/

Tain) the deri&ati&e of 3/ yields

= ,,

,,2 LLL I

R

LII . 32/

6"bstit"tin) into 3/ and 32/ into 30/ yields

0,,

,

,2,,

,

,2,, =

+

+

LLLLL IR

L

ILIR

L

IRIL.

atherin) terms,

0,2,2

,

,2,,

,

2, =

+

LLL IRILR

LRLI

R

LL,

and p"ttin) into standard form yields

0,2,

2,,

,

,

2

2

2

,, =

+

+++

LLL ILL

RRI

L

R

L

R

L

RI . 33/

omparin) 33/ to 5/,

2,

2,2

LL

RRo = ,

++=,

,

2

2

2

,2L

R

L

R

L

R , so

++=

,

,

2

2

2

,

2

,

L

R

L

R

L

R . 34/

The sol"tion proced"re for the nat"ral response and total response of either /., tiL or /.2 tiL

can then proceed in the same way as for series and parallel !L circ"its, "sin) the and o

&al"es shown abo&e.

a)e 4 of 20


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*ormali6ed )amping Ratio

A"r pre&io"s e9pression for sol&in) for s was

02 22 =++ oss . 35/

8("ation 35/ is sometimes written in terms of a Jnormali;ed dampin) ratioM as follows:

02 22 =++ ooss . 3/

Th"s, the relationship between dampin) coefficient and normali;ed dampin) ratio is

o

= . 37/

%ormali;ed dampin) ratio has the con&enient feat"re of bein) .0 at the point of critical

dampin).


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a)e of 20

Response of Second Order System

(zeta = 0.99, 0.8, 0.6, 0.4, 0.2, 0.1)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 2 4 6 8 10

0.99

0.1

0.4

0.2


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Case !: 'nderdamped . o< so s1and s!are comple7 con8ugates0

The nat"ral i.e., so"rce free/ response for any c"rrent or &olta)e in the networ has the

rectan)"lar form

[ ]/sin./cos./. 2, tBtBeti ddt

+= /

22

, += ojs ,

22

2 = ojs .

#amped resonant fre("ency 22 = od , 4/

so that

djs +=, ,

djs =2

/0., == tiB %ote see modification needed to incl"de final response/ 7/

d

t

Bdt

di

B

,0

2

+= = =/

$n polar form,

/cos./. 222,

+= teBBti dt

=

,

2,tanB

B .

!e)ardin) the arctan)ent - be caref"l beca"se yo"r calc"lator will )i&e the wron)

answer 50E of the time. The reason is that /,=0tan./tan. = . 6o chec the("adrant of yo"r calc"lator answer with the ("adrant consistent with the ri)ht

trian)le. $f the calc"lator ("adrant does not a)ree with the fi)"re, then add or s"btract=0F from yo"r calc"lator an)le, and re-chec the ("adrant.

a)e = of 20

C2

C

D


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Case (: Critically )amped . o= , so == 2, ss /

( ) ttt eAtAeAteAti +=+= 2,2,/. . >/

/0.2 == tiA . %ote see modification needed to incl"de final response/ 2/

20

, Adt

diA

t

+==

. 22/

Total Response # *atural Response Plus Final Response

$f the circ"it has # so"rces, then the final response is also #. The total response is

tsts

final eAeAIti 2

2

,

,/. ++=

,

[ ] tstsfinal eAeAIti 22,,/. += 23/

The effect of $finalon determinin) 1and 12simply means that yo" "se 23/ at t ? 0 instead of

it ? 0/ by itself.

$f the circ"it has 1 so"rces, the final response is the phasor sol"tion.

a)e > of 20


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*ormali6ed )amping Ratio

A"r pre&io"s e9pression for sol&in) for s was

02 22 =++ oss . 35/

8("ation 35/ is sometimes written in terms of a Jnormali;ed dampin) ratioM as follows:

02 22 =++ ooss . 3/

Th"s, the relationship between dampin) coefficient and normali;ed dampin) ratio is

o

= . 37/

%ormali;ed dampin) ratio has the con&enient feat"re of bein) .0 at the point of critical

dampin).

Documents

35 EE394J Spring11 Notes on Second Order Systems