Chapter 4-1.doc

8/13/2019 Chapter 4-1.doc

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Chapter 4The Laplace transform and system response

The advantage of the Laplace transform is that it converts lineardifferential equations into algebraic relations. With properalgebraic manipulation of the resulting quantities, the solution ofthe differential equation can be recovered in an orderly fashion

by inverting the transformation process to obtain a function oftime.In addition to providing a systematic solution procedure fordifferential equations, the Laplace transform allows us todevelop a graphical representation of the system's dynamics interms of block diagrams. With the Laplace transform, thediagram concept can be e tended to include dynamic elementswith the use of transfer function concept.

4.1 The Laplace transformThe Laplace transform L[y(t)] of a function y(t) is defined to be

=!

"#"$#% dt et yt y L st

#&. ( "

The integration removes t as a variable, and the transform is thusa function of only the Laplace variable s, which may be acomple number.)n alternative notation is the use of the uppercase symbol torepresent the transform of the corresponding lowercase symbol*that is,

Y(s) = L[y(t)] #&. (+"

The variable y(t) is assumed to be ero for negative time t


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The transform of a derivative is of use. applying integration by parts to the definition of the transform, we obtain

=+

==

!!

"!#"$#%"#!

"# yt y sLdt et y set ydt edt

dy

dt

dy L st st st

#&. ( "

We have /ust seen how a multiplicative constant can be factoredout of the integral. )lso, the integral of a sum equals the sum ofthe integrals. These facts point out the linearity property of thetransform* namely, that

L[af 1(t) + bf 2(t)] = aL[f 1(t)] + bL[f 2(t)] #&. (&"

Table &. is a short table of transforms.

Application to differential equations

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The derivative and linearity properties can be used to solve thedifferential equation

bvrydt

dy +=#&. (0"

If we multiply both sides of #&. (0" by e p (-st " and thenintegrate over time from t 1! to t 1 , we obtain

L(dy/dt) = L(ry + bv) = rL(y) + bL(v)

2sing #&. ( " and the alternative transform notation, we obtain

sY(s) - y(0) = rY(s) + bV(s) #&. (3"

where V(s) is the transform of v. This equation is an algebraicequation for Y(s) in terms of V(s) and y(0). The solution is

"#"!#

"# sV r s

br s

y sY

+= #&. (4"

The inverse Laplace transform L-1[Y(s)] is that time function y(t) , whose transform is Y(s). The inverse operation is alsolinear, and when applied to #&. (4", it gives

+

= "#"!#"# '' sV

r sb

Lr s

y Lt y #&. (5"

-rom table &. , it is seen that

rt e yr s

y L "!#

"!#' = #&. (6"

which is the free response. The forced response must therefore be given by

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"#' sV

r sb

L #&. ( !"

This cannot be evaluated until 7#s" is specified.

Exa !"e #.18ompute the step response of #&. (0".9olution:If v is a step of magnitude M , V(s)=M/s . -or this case, the forcedresponse is

s M

r sb

L '

This transform can be converted into a sum of simple transforms by partial fraction e pansion

s

$

r s

$

sr sbM +'

"#+=

This is true only if $ 1 s + $ 2(s-r) = bM

-or arbitrary values of s. This implies that$ 1 + $ 2 =0-r$ 2 =bM

or $ 1 = -$ 2 = bM/r

Thus, the forced response is

r bM

er

bM rt

The addition of the forced and free responses gives the previousresults # .+(3" or # .+(&" for r 1 ( ; .

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The second-order caseIt is useful to see how the preceding concepts e tend to thesecond(order model

"#+

+

t f %xdt

dx&

dt

xd =++ #&. ( "

To transform this equation, we need a new property #5" fromTable &. . In terms of , this gives

"!#"!#"#$% +++

dt dx

sx s ' sdt

xd L = #&. ( +"

Transforming #&. ( " gives

"#"#"$!#"#%"!#"!#"#% + s ( s%' x s s' &dt dx

sx s ' s =++

or "#"!#"!#"!#"#"# + s ( &x

dt dx

sx s ' % &ss +++=++ #&. ( "

The free response is

++

++=

% &ss

&xdt dx

sx Lt x +

'"!#"!#"!#

"# #&. ( &"

The forced response is

++=

% &ss

s ( Lt x

+' "#"# #&. ( 0"

In order to evaluate either response, we must e pand thetransforms within the square brackets into a series of transformsthat appear in Table &. . To perform this e pansion, we need toknow the values of & , and % to determine whether the roots ofs 2+&s+% are # " real and distinct, #+" real and repeated, or # "comple , because the form of the e pansion is different for eachcase. In addition, we also need to know f(t) so that we can find

(s). Exa !"e #.2

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= "#"#

"# ' s / s -

Lt x #&. ( 5"

and the forced response is

= "#"#

"# ' s / s (

Lt x #&. ( 6"

where the characteristic polynomial is

+,

,,

, a sa sa sa s / ++++= ''' ..."# #&. (+!"

Thus, the Laplace transform provides a systematic way offinding the solution for any constant(coefficient lineardifferential equation.


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4.2 Transfer functions=quation #&. (4" contains much information about the system's

behavior, as we have seen.


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"#'

"#"#

"# s / s (

s ' s0 == #&.+( "

or

+,

,,

, a sa sa sa s0

++++= ''' ...'"# #&.+(&"

since L(d , x/dt , ) 1 s , '(s) for ero initial conditions.

-rom this, we see that the denominator of the transfer function isthe system's characteristic polynomial. Thus, if we are givenonly the transfer function, we could still make some assessment

of the system's behavior, because the roots of the characteristic polynomial give us stability information, time constants,damping ratios, and oscillation frequencies. In the precedinge amples, the numerator of the transfer function does not giveany particularly useful information, but we will soon seee amples where the numerator also contains information aboutthe system's response.

Transfer functions from the state-variable formIf the system model is in state(variable form, some algebra isrequired to obtain the transfer function. The general procedure isto transform the state equations using ero initial conditions.Then solve for the desired output variable in terms of the input.To do this, all the other variables must be eliminatedalgebraically. The following simple e ample illustrates this

procedure.

Exa !"e #.#The state(variable form of the model for the mass(spring(damper system is given by

vdt dx = #&.+(0"

%x&v f

dt

dv = #&.+(3"-ind the transfer function between the input f and the output .

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9olution:Transform the state equations with ero initial conditions toobtain

"#"# sV s s' = #&.+(4"

"#"#"#"# s%' s&V s ( ssV = #&.+(5"

=liminate V(s) algebraically by substituting V(s) = s'(s) into#&.+(5".

"#"#"#"#+ s%' s&s' s ( s ' s =

or "#"#"# + s ( s ' % &ss =++

9olve for '(s)/ (s) to obtain the same transfer function given by#&.+(+".

ultiple inputs


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Transform #&.+(6" and collect terms to obtain"#"#"#"# ++''+' s / a s / a s / aa sa +=++ #&.+( !"

To find the transfer function ?#s";? #s", set ?+#s" 1 ! and solve#&.+( !" to obtain

+'

'

' "#"#

aa sa

a

s / s /

++= #&.+( "

9imilarly, setting ? #s" 1!, we obtain the second transferfunction.

+'

+

+ "#

"#

aa sa

a

s /

s /

++=

#&.+( +"

@ote that the transfer functions in = ample &.0 have the samedenominator. )ll transfer functions of any given system willalways have the same denominator, because it is the system'scharacteristic polynomial.

ultiple outputs

) single input can generate more than one transfer function ifwe are interested in more than one variable as output. =achtransfer function is found by algebraically eliminating the othervariables, as shown in = ample &.3.

Exa !"e #. -ind the transfer functions for the speed and the current a. ofthe A8 motor, described by:

e

aa 3

dt d. L 4.v ++=

&. 3 dt

d - a0 =

The input is the voltage v.

9olution:Transforming and collecting terms, we obtain

"#"#"#"# sV s 3 s - 4 Ls ea =++ #&.+( 0"

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!"#"#"# =++ s& -s s - 3 a0 #&.+( 3"

These equations can be solved for (s) and "# s by standardalgebraic means. 8ramer's determinant for this system is

0 e

0 e0

e

3 3 4& s&L 4- L-s

3 3 & -s 4 Ls& -s 3

3 4 Ls s /

++++=

+++=++=

"#

""##"#

"#"#

+

#&.+( 4"

The solutions for (s) and "# s are

"#"#"#

"#!

"#

"# sV s /

& -s s /

& -s

3 sV

s -

e

+=+= #&.+( 5"

"#"#"#

!"#"#

"# sV s /

3 s /

3 sV 4 Ls

s 0 0 =+

= #&.+( 6"

The two transfer functions are the coefficients of V(s) in#&.+( 5" and #&.+( 6". @ote that 8ramer's determinant is thecharacteristic polynomial of the system.

!ecovering the differential equation modelThere will be applications where we are given a component'stransfer function, but the differential equation model is requiredfor analysis. The equation is easily recovered from the transferfunction by cross multiplying by the denominators, using thefollowing identity, which is valid for ero initial conditions.

"#"# s ' sdt

xd L ,,

,

= #&.+(+!"

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The general form of a transfer function (s) between an input (s) and output Y(s) is

"#

"#"#

"#

"#

s 5

s 6 s0

s (

sY ==#&.+(+ "

where 6(s) and 5(s) are the numerator and denominator of (s). This gives

5(s)Y(s) = 6(s) (s) #&.+(++"

=quation #&.+( " in this form is

(s-r)Y(s) = bV(s)

which implies

bvrydt dy =

To recover the differential equation for the current a in = ample&.3, use #&.+( 4" and #&.+( 5" to obtain

0 e

a

3 3 4&&Ls 4- L-s

& -s

sV

s -

+++++

="#"#

"#+ #&.+(+ "

or [L s 2 + (4 +&L)s + 4& + 3 e 3 ] a(s) = ( s+&)V(s)

This gives

&vdt

dv - . 3 3 4&

dt

d.&L 4-

dt

.d L-

a0 e

aa +=++++ "#"#++

#&.+(+&"

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"umerator #ynamics

The transfer function given by #&.+(+ " is the first e ample thusfar of a transfer function, whose numerator is a polynomial in s.)ll the numerators previously seen were constants. Thenumerator's s term is seen from #&.+(+&" to have the effect ofcomputing the time derivative of the input v.


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4.$ %requency response and transfer functions8onsider the linear model

bvrydt dy += #&. ( "

wheret 7t v sin"# = #&. (+"

with constant amplitude 7 and frequency . -or the stable casewith b a constant, the steady(state solution, with ;'=r ,

"sin#"# += t 8t y #&. ( "

where

++'

+= 7b 8 #&. (&"

"#tan ' = #&. (0"

with !+; for bB! and +; for bC!.

The frequency transfer functionThe solution given by #&( . "(#&. (0" points out a usefulapplication of the transfer function. )ssume that the system isstable and write the transfer function (s) = b/(s-r) in terms of

r ;'= . This is

'"#

+= sb

s0

#&. (3"

The transform of a sinusoidal steady(state output of frequency has the denominator ++ + s . The roots are . s = * that is, thevalues of s corresponding to sinusoidal oscillation of frequency are . s = . With this as a guide, let us see what the transferfunction reveals when s is replaced with . .

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'"#

+= .b

.0

#&. (4"

This comple number is a function of only for fi ed and b.

It can be e pressed in a more convenient form by multiplyingthrough by the comple con/ugate of the denominator.

++'

"'#''

'"#

+=

+=

.b..

.b

.0 #&. (5"

"# .0 can be thought of as a vector in the comple plane wherethe vector's components are the real and imaginary parts of

"# .0 . -igure #&. " shows the case for bB!. The vector rotates,and its length changes as varies from ! to . The locus ofvector's tip is shown, and it is a semicircle in the fourth quadrantfor positive .

-igure &. ?olar plot of a first(order linear system.The transfer function with s replaced by . is called thefrequency transfer function, and its plot in vector form is the

polar plot. The vector's magnitude M and angle relative to the positive real a is are easily found from trigonometry to be

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++'"#

t

b M

+= #&. (6"

"#tan"# ' = #&. ( !"

where M and are e plicitly written as functions of .8omparing these e pressions for D# " and "# with those of

8 and in #&. (&" and #&. (0" reveals the following useful fact#see -igure &.+". The magnitude D of the frequency transferfunction # . " is the ratio of the sinusoidal steady(state outputamplitude to the sinusoidal input amplitude. The phase shift ofthe output relative to the input is , the angle of # . " %the

argument of # . "$.

-igure &.+ 9teady(state sinusoidal response of a linear system.#a" Input(output relation. #b" ?olar representation of thefrequency transfer function # . ".


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"#"# .0 = #&. ( +"

"# 7M 8 = #&. ( "

where "# .0 and "# .0 denote the magnitude and angle ofthe comple number .


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This gives a straight line versus log . Its slope is(+!db;decade, where a decade is any !: frequency range. )t

;'= , #&. ( 0" gives

!'."log#+!+log'!"log#+!"# = bb

#&. ( 5"

Thus, at ;'= , "# is .! db below the low(frequencyasymptote given by #&. ( 3". The low(frequency and high(frequency asymptotes meet at ;'= , which is the breakpointfrequency. It is also called the corner frequency.The curve of versus is constructed as follows. -or ;'


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-igure &. Logarithmic frequency response plots for two first(order transfer functions. #a" Dagnitude ratio m in decibels. #b"?hase angle.&eneral formulationThe previous results are for first(order models withoutnumerator dynamics, but they can be easily e tended to a stable

251


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linear system of any order. The general form of a transferfunction is

"...#"#

"...#"#

"# +'+'

s 5 s 5

s 6 s 6

3 s0 = #&. ( 6"

where 3 is a constant real number. In general, if a complenumber T# . " consists of products and ratios of complefactors, such that

"...#"#"...#"#

"#+'

+'

. 5. 5. 6 . 6

3 .0 = #&. (+!"

where G is a real constant, then from the properties of complenumbers

..."#"#..."##

"#+'

+'

. 5. 5. 6 . 6

3 .0 = #&. (+ "

In decibel units, this implies that

..."#log+!"#log+!

..."#log+!"#log+!log+!"#log+!"#

+'

+'

+++==

. 5. 5

. 6 . 6 3 .0

#&. (++"

That is, when e pressed in logarithmic units, multiplicativefactors in the numerator are summed, while those in thedenominator are subtracted. We can use this principle

graphically to add or subtract the contribution of each term inthe transfer function to obtain the plot for the overall systemtransfer function.

)lso, for the form #&. (+!", the phase angle is

..."#"#

..."#"#"#"#

+'

+'

+++==

. 5. 5

. 6 . 6 3 .0 #&. (+ "

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The phase angles of multiplicative factors in the numerator aresummed, while those in the denominator are subtracted. Thisallows us to build the composite phase angle plot from the plotsfor each factor.

Common building bloc'sDost transfer functions occur in the form given by #&. ( 6". Inaddition, the >building blocks>, 6 9(s) and 5 9(s) , usually take theforms shown in Table &.+.

Table &.+ -actors commonly found in transfer functions of the

form "...#"#"...#"#

"#+'

+'

s 5 s 5

s 6 s 6 3 s0

=

-or (s) = 3 #&. (+&"

We have

3 log+!= #&. (+0"

==

!'5!

!!

3

3 3

+

+

#&. (+3"

The plots are shown in -igure &.&

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-igure &.& -requency response plots for (s) =3 .

-or

(s) = s , #&. (+4"

we have

log+!log+!log+! ,.,., === #&. (+5"

+, ,.,. 6!"#"# ===

#&. (+6"

The most common case is ,= , and these plots are given in-igure &.0.

254


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-igure &.0 -requency response plots for (s) =s .

Exa !"e #.: The presence of numerator dynamics in a first(order system cansignificantly alter the system's frequency response. 8onsider theseries F8 circuit, where the output voltage is now taken to beacross the resistor #-igure &.3". The two amplifiers serve toisolate the circuit from the loading effects of ad/acent elements.The input impedance between the voltage v and the current i isfound from the series law.

-igure &.3 Eigh(pass filter circuit.

255


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$s 4

s - sV '

"#"# +=

Thus, 4

$s 4

sV 4 s - sY '"#"#"#

+==

Aetermine the circuit's frequency response, and interpret itseffect on the input.9olution:The transfer function is

'"#"#

"#+== 4$s

4$s sV sY

s0 #&. ( !"

Let 4$ = to obtain

'"#

+= s s

s0

This is of the form #&. ( 6" with = 3 , 6 1(s) = s and 5 1(s) 1 '+ s . Therefore, from #&. (++",

'log+!log+!log+!"# ++= .. #&. ( "

-or the low(frequency asymptote, '


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!log+!log+!log+!log+!

log+!log+!log+!"#

=+=

=+

for ;'>> . )t ;'= , the term '+ s contributes ( db. Thecomposite curve "# is obtained by >blending> the low(

frequency and high(frequency asymptotes through this point, asshown in -igure &.4.

-igure &.4 -requency response of the high(pass filter. #a"Dagnitude plot. #b" ?hase plot.

) similar technique can be used to sketch "# . -rom #&. (+ ","#tan6!!"'#"# ' +=++= ++.

-or '> , the high(frequency asymptote is

++++ !6!6!"#tan6!"# ' == The result is sketched in -igure &.4b. The log magnitude plotshows that the circuit passes signals with frequencies above

257


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;'= with little attenuation, and thus it is called a high(passfilter.

)nother e ample of numerator dynamics is the transfer function

'

'"#

+

'

++=

s s

3 s0

#&. ( "

)n e ample of this form is the transfer function of the leadcompensator circuit.Thus,

'

'"#

+

'

++=

.

. 3 .0

#&. ( &"

-rom #&. (++",

'log+!'log+!log+!"# +' +++= .. 3 #&. ( 0"Thus, the plot of "# can be obtained by subtracting the plot of

'+ + s from that of '' + s . The scale is then ad/usted by 3 log+! .The term '' + s causes the curve to break upward at ';' = .The term '+ + s causes the curve to break downward at +;' = .

If +' ;';' > , the composite curve looks like -igure &.5a, and thesystem is a low(pass filter. If +' ;';' < , it is a high(pass filter#-igure &.5b".-rom #&. (+ ", the phase angle is

"#tan"#tan!

"'#"'#"#

+'

''

+'

+=

=+++=+

.. 3 #&. ( 3"

Its plot can also be found by combining the plots of"# for 3 ,'' + s , and '+ + s . The sketches shown in -igure &.5 can be

obtained by using the low(frequency, corner frequency, andhigh(frequency values of ! o, &0o, and 6! o for the term '+ s ,supplemented by evaluations of #&. ( 3" in the region betweenthe corner frequencies. This sketching technique is moreaccurate when the corner frequencies are far apart.

258


28/37

259


29/37

-igure &.5 appro imate sketches of the frequency response plots

for ''

"#+

'

++=

s s

3 s0

. -or the low(pass filter in #a", +' < . -or the

high(pass filter in #b", +' > .

Dore accurate plots can be obtained by using #&. ( 0" and #&. ( 3" with a calculator or computer. 9ome e amples areshown in -igure &.6 for various values of ' and + .

-igure &.6 representative plots for ''

"#+

'

++=

s s

3 s0

with 3 1 . -or

a , !+.!,+.! +' == . -or b, '.!,+.! +' == .

260


30/37

(econd-order systems

The denominator of a second(order transfer function with '


31/37

-igure &. ! -requency response plots for "'#"# += s s 3 s0

with

3 1 and +.!= .

262


32/37


33/37

-igure &. -requency response plots for the overdamped

system "'"#'#;

"#+' ++

= s s

% d s0

264


34/37

An underdamped systemIf the transfer function given by #&. ( 5" has comple con/ugateroots, it can be e pressed as form & in table &.+.

'+

;

'

;"# +

+

++

=++

=

,,

s s

% d

s% &

s%

% d s0

#&. (&+"

We have seen that the constant term d/% merely shifts themagnitude ratio up or down by a fi ed amount and adds either ! o

or ( 5! o to the phase angle plot. Therefore, for now, let us taked/% 1 and consider the following quadratic factor, obtained from#&. (&+" by replacing s with . .

...

.0

,,,,

+

'

'

'+

'"# ++

+

=++

=#&. (& "

The magnitude ratio is

+

=

+

=

+

=

++

+

+

++

+

+

+

+'log'!

+'log+!

+

'

'log+!"#

,,

,,

,, .

#&. (&&"

The asymptotic appro imations are as follows. -or ,


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++

+

+ +'

'"#

+

=

,,

M

#&. (&0"

This has a ma imum value when the denominator has aminimum. 9etting the derivative of the denominator with respectto equal to ero shows that the ma imum "# M occurs at

++' = , . This frequency is the resonant frequency r . The peak of "# M e ists only when the term under the radical is positive* that is, when 4!4.! . Thus,

4!4.!!+' + = ,r #&. (&3"

The value of the peak M ! is found by substituting r into "# M .This gives

4!4.!!'+

'"#

+

==

r ! M M #&. (&4"

If 4!4.!> , no peak e ists, and the ma imum value of M occursat != where M 1 . @ote that as ! , ,r , and ! M . -oran underdamped system, the resonant frequency is the naturalfrequency , .) plot of "# versus log is shown in -igure &. +a forseveral values of . @ote that the correction to the asymptoticappro imation in the vicinity of the corner frequency dependson the value of . The peak value in decibels is

"'+log#+!"# + == r ! #&. (&5"

)t , = ,

+log+!"# =, #&. (&6"The curve can be sketched more accurately by repeatedevaluation of #&. (&&" for values of near , .

266


36/37


37/37

Thus,

= +'

+

"#tan

,

,

#&. (0!"

where "# is in the third or fourth quadrant. -or , >>+'5!"#

)t the corner frequency,

+, 6!tan"# ' ==

This result is independent of . The rest of the plot can besketched by evaluating #&. (0!" at various values of . The plotis shown for several values of in -igure &. +b.)t the resonant frequency.

+' +'tan"#

= r #&. (0 "

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