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o e n go ont nuoust mesystemWe have two approaches for modeling the linearcontinuous time s stem:

1) Transfer function approach

Transfer function approach make use of the LaplaceTransform.

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Idea

the Laplace transform converts integral and differentialequations intoalgebraicequations

this is like phasors, but

applies to general signals, not just sinusoids

handles non-steady-state conditions

allows us to analyze

LCCODEs

complicated circuits with sources, Ls, Rs, and Cs

complicated systems with integrators, differentiators, gains

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The Laplace transform

well be interested in signals defined for t 0

the Laplace transformof a signal (function) f is the function F = L(f)defined by

F(s) = 0

f(t)est dt

for those s C for which the integral makes sense

F is a complex-valued function of complex numbers

s is called the (complex) frequency variable, with units sec1; t is calledthe time variable (in sec); st is unitless

for now, we assume fcontains no impulses at t= 0

common notation convention: lower case letter denotes signal; capitalletter denotes its Laplace transform, e.g., U denotes L(u), Vin denotes

L(vin), etc.

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Example

lets find Laplace transform off(t) =et:

F(s) =

0

et est dt=

0

e(1s)t dt= 1

1 se(1s)t

0

= 1

s 1

provided we can say e(1s)t 0 as t , which is true for s >1:e(1s)t

=

ej(s)t

=1e(1s)t

=e(1s)t

the integral defining Fmakes sense for all s C with s >1 (theregion of convergence ofF)

but the resulting formula forFmakes sense for all s C except s= 1

well ignore these (sometimes important) details and just say that

L(et) = 1

s 1

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More examples

constant: (or unit step) f(t) = 1 (for t 0)

F(s) =

0

est dt= 1

sest

0

=1

s

provided we can say est 0 as t , which is true for s >0 since

est= ej(s)t =1

e(s)t=e(s)t

the integral defining Fmakes sense for all s with s >0

but the resulting formula forFmakes sense for all s except s= 0

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sinusoid: first express f(t) = cos t as

f(t) = (1/2)ejt + (1/2)ejt

now we can find F as

F(s) =

0

est

(1/2)ejt + (1/2)ejt

dt

= (1/2)

0 e

(s+j)t

dt+ (1/2)

0 e

(sj)t

dt

= (1/2) 1

s j+ (1/2)

1

s+j

=

s

s2 +2

(valid for s >0; final formula OK for s = j)

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powers oft: f(t) =tn (n 1)

well integrate by parts, i.e., use ba

u(t)v(t)dt= u(t)v(t)

b

a

ba

v(t)u(t)dt

with u(t) =tn, v(t) =est, a= 0, b=

F(s) =

0

tnest dt = tnest

s

0

+n

s

0

tn1est dt

= nsL(tn1)

provided tnest 0 ift , which is true for s >0

applying the formula recusively, we obtain

F(s) = n!

sn+1

valid for s >0; final formula OK for all s = 0

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Impulses at t = 0

iffcontains impulses at t= 0 we choose to includethem in the integraldefiningF:

F(s) =

0

f(t)est dt

(you can also choose to not include them, but this changes some formulaswell see & use)

example: impulse function, f=

F(s) =

0

(t)est dt= estt=0

= 1

similarly for f=(k) we have

F(s) =

0

(k)(t)est dt= (1)kdk

dtkest

t=0

= skestt=0

=sk

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Linearity

the Laplace transform is linear: iff andg are any signals, and a is anyscalar, we have

L(af) =aF, L(f+g) =F+G

i.e., homogeneity & superposition hold

example:

L

3(t) 2et

= 3L((t)) 2L(et)

= 3

2

s 1

= 3s 5

s 1

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One-to-one property

the Laplace transform is one-to-one: ifL(f) = L(g) then f=g(well, almost; see below)

F determines f

inverse Laplace transform L1 is well defined(not easy to show)

example(previous page):

L13s 5

s 1

= 3(t) 2et

in other words, the only function f such that

F(s) =3s 5

s 1

is f(t) = 3(t) 2et

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what almost means: iff and g differ only at a finite number of points(where there arent impulses) then F =G

examples:

fdefined asf(t) =

1 t= 20 t = 2

has F = 0

fdefined as

f(t) =

1/2 t= 01 t >0

has F = 1/s (same as unit step)

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Inverse Laplace transform

in principle we can recover f from F via

f(t) = 1

2j

+jj

F(s)est ds

where is large enough that F(s) is defined for s

surprisingly, this formula isnt really useful!

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Time scaling

define signal g byg(t) =f(at), where a >0; then

G(s) = (1/a)F(s/a)

makes sense: times are scaled by a, frequencies by 1/a

lets check:

G(s) =

0f(at)e

st dt= (1/a)

0f()e

(s/a) d= (1/a)F(s/a)

where =at

example: L(et

) = 1/(s 1) so

L(eat) = (1/a) 1

(s/a) 1=

1

s a

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Exponential scaling

let fbe a signal and a a scalar, and define g(t) =eatf(t); then

G(s) =F(s a)

lets check:

G(s) =

0esteatf(t)dt=

0e(sa)tf(t)dt=F(s a)

example: L(cos t) =s/(s2 + 1), and hence

L(et cos t) = s+ 1

(s+ 1)2 + 1=

s+ 1

s2 + 2s+ 2

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Time delay

let fbe a signal and T >0; define the signal g as

g(t) = 0 0 t < T

f(t T) t T

(g is f, delayed by T seconds & zero-padded up to T)

ttt = T

f(t) g(t)

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then we have G(s) =esTF(s)

derivation:

G(s) =

0

estg(t)dt =

T

estf(t T)dt

= 0

es(+T)f()d

= esTF(s)

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example: lets find the Laplace transform of a rectangular pulse signal

f(t) = 1 ifa t b

0 otherwise

where 0< a < b

we can write f as f=f1 f2 where

f1(t) =

1 t a0 t < a

f2(t) =

1 t b0 t < b

i.e.,f is a unit step delayed aseconds, minus a unit step delayed bseconds

hence

F(s) = L(f1) L(f2)

= eas ebs

s

(can check by direct integration)

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Derivative

if signal f is continuous at t= 0, then

L(f) =sF(s) f(0)

time-domain differentiation becomes multiplication by frequencyvariable s (as with phasors)

plusa term that includes initial condition (i.e., f(0))

higher-order derivatives: applying derivative formula twice yields

L(f) = sL(f) f(0)

= s(sF(s) f(0)) f(0)

= s2F(s) sf(0) f(0)

similar formulas hold for L(f(k))

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Integral

let g be the running integral of a signal f, i.e.,

g(t) =

t0

f()d

then

G(s) =1

sF(s)

i.e., time-domain integral becomes division by frequency variables

example: f=, soF(s) = 1; g is the unit step function

G(s) = 1/s

example: f is unit step function, so F(s) = 1/s; g is the unit rampfunction (g(t) =t fort 0),

G(s) = 1/s2

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derivation of integral formula:

G(s) = t=0

t=0

f()d

est dt= t=0

t=0

f()est d dt

here we integrate horizontally first over the triangle 0 tt

lets switch the order, i.e., integrate vertically first:

G(s) =

=0

t=f()est

dt d =

=0 f()

t=est

dt d=

=0

f()(1/s)esd

= F(s)/sThe Laplace transform 326

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Multiplication by t

let fbe a signal and define

g(t) =tf(t)

then we haveG(s) = F(s)

to verify formula, just differentiate both sides of

F(s) =

0

estf(t)dt

with respect to s to get

F(s) =

0

(t)estf(t)dt

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examples

f(t) =et, g(t) =tet

L(tet) = d

ds

1

s+ 1=

1

(s+ 1)2

f(t) =tet, g(t) =t2et

L(t2et) = dds 1(s+ 1)2 = 2(s+ 1)3

in general,

L(tket) = (k 1)!(s+ 1)k+1

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Convolution

the convolution of signals f andg, denoted h=f g, is the signal

h(t) =

t0

f()g(t )d

same as h(t) =

t0

f(t )g()d; in other words,

f g=g f

(very great) importance will soon become clear

in terms of Laplace transforms:

H(s) =F(s)G(s)

Laplace transform turns convolution into multiplication

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lets show that L(f g) =F(s)G(s):

H(s) =

t=0est t

=0f()g(t )d dt

=

t=0

t=0

estf()g(t )d dt

where we integrate over the triangle 0 t

change order of integration: H(s) =

=0

t=

estf()g(t )dt d

change variable t to t=t ; dt=dt; region of integration becomes 0, t 0

H(s) =

=0

t=0

es(t+)f()g(t)dt d

=

=0

esf()d

t=0

estg(t)dt

= F(s)G(s)

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examples

f=, F(s) = 1, givesH(s) =G(s),

which is consistent with

t0

()g(t )d=g(t)

f(t) = 1, F(s) =esT/s, gives

H(s) =G(s)/s

which is consistent with

h(t) = t0

g()d

more interesting examples later in the course . . .

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Finding the Laplace transform

you should knowthe Laplace transforms of some basic signals, e.g.,

unit step (F(s) = 1/s), impulse function (F(s) = 1)

exponential: L(eat) = 1/(s a)

sinusoids L(cos t) =s/(s2 +2), L(sin t) =/(s2 +2)

these, combined with a table of Laplace transforms and the propertiesgiven above (linearity, scaling, . . . ) will get you pretty far

and of course you can always integrate, using the defining formula

F(s) =

0

f(t)est d t . . .

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Patterns

while the details differ, you can see some interesting symmetric patternsbetween

the time domain (i.e., signals), and the frequency domain (i.e., their Laplace transforms)

differentiation in one domain corresponds to multiplication by thevariable in the other

multiplication by an exponential in one domain corresponds to a shift

(or delay) in the other

well see these patterns (and others) throughout the course

The Laplace transform 333