(306557551) laplace (12) (1)

S. Boyd EE102

Lecture 3The Laplace transform

definition & examples•

properties & formulas

– linearity– the inverse Laplace transform– time scaling– exponential scaling– time delay– derivative– integral– multiplication by t– convolution

•

3–1

Idea

the Laplace transform converts integral and differentialalgebraic equations

equations into

this is like phasors, but

• applies to general signals, not just

sinusoids

• handles non-steady-state conditionsallows us to analyze

• LCCODEs

• complicated

• complicated

circuits with sources, Ls,

systems with integrators,

Rs, and Cs

differentiators, gains

The Laplace transform 3–2

Complex

complex number in Cartesian form: z

• x = <z, the real part of z

• y = =z, the imaginary part of z

• j = √−1 (engineering notation);

numbers

= x + jy

= √−1 is polite term in mixedicompany

rejφcomplex number in polar form: z =

• r is the modulus or magnitude of z

• φ is the angle or phase of z

• exp(jφ) = cos φ + j sin φ

complex exponential of z = x + jy:

ez ex+jy exejy ex(cos y= = = + j sin y)


The Laplace transform

defined for t ≥ 0

signal (function) f is the

we’ll be interested in signals

function F = L(f )the Laplace transform of adefined by Z ∞

f (t)e−st dtF (s) =0

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 sec−1; t is calledthe time variable (in sec); st is unitless

• for now, we assume f contains no impulses at t = 0

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


Vin denotes

3–4

Example

let’s find Laplace transform of f (t) = et:¯Z Z ∞¯0

1 e(1−s)t 1 ∞ ∞

et e−st e(1−s)tF (s) = dt = = =dt1 − s

which

s − 1¯0 0

e(1−s)t → 0 as t → ∞, is true for <s > 1:provided we can say¯

(1−s)t¯ ¯

−j(=s)t¯ ¯

(1−<s)t¯

(1−<s)t= ¯e =¯¯e

¯ ¯ ¯ ¯ ¯e¯

e¯ ¯ ¯|

=z

1

• the integral defining F makes sense for all s ∈ C with <s > 1 (the‘region of convergence’ of F )

• but the resulting formula for F makes sense for all s ∈ C except s = 1

we’ll ignore these (sometimes important) details and just say that

1L(et) =

s − 1


More examples

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

¯∞Z1 1∞ ¯

e−st e−stF (s) = dt = − =¯s s¯

00

e−st → 0 as t → ∞, which is true for <s > 0 sinceprovided we can say¯

−j(=s)t¯ ¯

−(<s)t¯¯ ¯

¯e−st¯ −(<s)t= ¯e = e¯ ¯ ¯ ¯e

¯¯ ¯

|=z

1

0• the integral defining F makes sense for all s with <s >

• but the resulting formula for F makes sense for all s except s = 0


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

f (t) = (1/2)ejωt + (1/2)e−jωt

now we can find F asZ ∞

e−st ¡(1/2)ejωt + (1/2)e−jωt¢ dtF (s) =

0 Z Z∞ ∞

e(−s+jω)t e(−s−jω)t= (1/2) dt + (1/2) dt0 0

1 1= (1/2) + (1/2)

s + jωs − jω s

=s2 ω2+

> 0; OK for s = ±jω)(valid for <s final formula


tnpowers of t: f (t) = (n ≥ 1)

we’ll integrate by parts, i.e., use¯bb bZ Z

u(t)v0(t) dt = u(t)v(t)¯

v(t)u0(t) dt¯ −¯aa a

with u(t) = tn, v0(t) = e−st, a = 0, b = ∞µ −e−st ¶¯∞

Z Z ∞ n ∞¯tne−st tn

n s

tn−1e−stF (s) = dt = + dt¯sn−1

s¯00 0

= L(t )

tne−st → 0 <s > 0provided if t → ∞, which is truewe obtain

forapplying the formula recusively,

n! F (s) =

sn+1

valid for <s > 0;


final formula OK for all s = 0

3–8

Impulses at t = 0

if f contains impulses at t = 0 we choose to include them in the integral defining F :

Z ∞

f (t)e−st dtF (s) =0−

(you can also choose to not include them, but this changes we’ll see & use)

example: impulse function, f = δZ

some formulas

∞

e−st¯δ(t)e−st dt =F (s) = = 1¯t=0

0−

δ(k)similarly for f = we have¯Z

k∞

= (−1)k d e−st¯ ske−st¯δ(k)(t)e−st skF (s) = = =dt ¯ ¯

t=0kdt ¯t=00−


Linearity

the Laplace transform is linear : if f and g arescalar, we have

any signals, and a is any

L(af ) = aF,

i.e., homogeneity & superposition

L(f + g)

hold

= F + G

example:

L ¡3δ(t) − 2et¢ 3L(δ(t)) − 2L(et)=

2= 3 −

s − 13s − 5=s − 1


One-to-one property

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

f =then g

• F determines f

• inverse Laplace transform(not easy to show)

L−1 is well defined

example (previous page):µ

3s − 5¶L−1 t= 3δ(t) − 2e

s − 1

in other words, the only function f such that

3s − 5F (s) =s − 1

is f (t) = 3δ(t) − 2et


what ‘almost’ means: if f and g differ only at a finite number of pointsimpulses) then F = G(where there aren’t

examples:

• f defined as ½10

t = 2t = 2f (t) =

has F = 0

• f defined as ½1/21

=>

00

ttf (t) =

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


Inverse Laplace transform

in principle we can recover f from F via

σ+j∞Z1 stf (t) = F (s)e ds

2πj σ−j∞

enough that F (s) is definedwhere σ is large for <s ≥ σ

surprisingly, this formula isn’t really useful!


Time scaling

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

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

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

let’s check:Z Z∞ ∞

f (at)e−st f (τ )e−(s/a)τG(s) = dt = (1/a) = (1/a)F (s/a)dτ0 0

where τ = at

example: L(et) = 1/(s − 1) so

L(eat) = (1/a)1 1

=(s/a) − 1 s − a


Exponential scaling

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

G(s) = F (s − a)

let’s check:Z Z∞ ∞

e−steatf (t) e−(s−a)tf (t)G(s) = dt = dt = F (s − a)0 0

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

s + 1 + 1sL(e−t cos t) = =(s + 1)2 + 1 s2 + 2s + 2


Time delay

let f be a signal and T > 0; define the signal g as½

0f (t − T )

0 ≤ t < Tt ≥ Tg(t) =

(g is f , delayed by T seconds & ‘zero-padded’ up to T )

f (t) g(t)

t tt = T


then we have G(s) = e−sT F (s)

derivation:Z Z∞ ∞

e−stg(t) e−stf (t − T ) dt

e−s(τ +T )f (τ )

G(s) = =dt0 T

∞Z

= dτ0

e−sT F (s)=


example: let’s find the Laplace transform of a rectangular pulse signal½

10

if a ≤ t ≤otherwise

bf (t) =

where 0 < a < b

we can write f as f = f1 − f2 where½ ½

10

10

t ≥ at < a

tt

≥ b< bf1(t) = f2(t) =

i.e., f is a unit step delayed

hence

a seconds, minus a unit step delayed b seconds

F (s) = L(f1) − L(f2)

e− − e as −bs=

s

(can check by direct integration)


Derivative

if signal f is continuous at t = 0, then

L(f 0) = sF (s) − f (0)

• time-domain differentiation becomes multiplicationvariable s (as with phasors)

by frequency

• plus a term that includes initial condition (i.e., −f (0))

higher-order derivatives: applying derivative formula twice yields

L(f 00) ==

=

sL(f 0) − f 0(0)

s(sF (s) − f (0)) − f 0(0)

s2F (s) − sf (0) − f 0(0)

similar formulas hold for L(f (k))


examples

• f (t) = et, etso f 0(t) = and

1L(f ) = L(f 0) =

1

s − 1

which isusing the formula, L(f 0) = s(s

) − 1,1 the same−

1 d−• sin ωt = cos ωt, soω dtµ ¶

− 1

=

1 s ωL(sin ωt) = −ω

ss2 ω2 s2 ω2+ +

• f is unit ramp, so f 0 is unit stepµ ¶

1L(f 0) = s 0 = 1/s−s2


derivation of derivative formula: start from the defining integralZ ∞

f 0(t)e−stdtG(s) =0

integration by parts yieldsZ ∞

e−stf (t)¯∞ − f (t)(−se−st)G(s) = dt¯

00

lim f (t)e−st − f (0) + sF (s)=t→∞

for <s large enough the limit is zero, and we recover the formula

G(s) = sF (s) − f (0)


derivative formula for discontinuous functions

if signal f is discontinuous at t = 0, then

L(f 0) = sF (s) − f (0−)

so f 0(t) = δ(t)example: f is unit step,

µ1

¶

L(f 0) = − 0 = 1ss


Example: RC circuit

1Ω

yu 1F

t = 0, i.e., y(0) = 0• capacitor is uncharged at

• u(t) is a unit step

from last lecture,y0(t) + y(t) = u(t)

take Laplace transform, term by term:

sY (s) + Y (s) = 1/s

(using y(0) = 0 and U (s) = 1/s)


solve for Y (s) (just algebra!) to get

1/s 1Y (s) = =

s + 1 s(s + 1)

to find y, we first express Y as

1 1Y (s) = − s + 1s

(check!)

therefore we have

y(t) = L−1(1/s) − L−1(1/(s + 1)) = 1 − e−t

Laplace transform turned a differential equation into an algebraic equation(more on this later)


Integral

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

tZ

g(t) = f (τ ) dτ0

then1

G(s) = F (s)s

integral becomes division by frequency variable s

i.e., time-domain

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

G(s) = 1/s

example: f is unit step function, so F (s) =function (g(t) = t for t ≥ 0),

G(s) = 1/s2

1/s; g is the unit ramp


derivation of integral formula:

t tµZ ¶Z Z Z∞ ∞

e−st f (τ )e−st dτG(s) = f (τ ) dτ dt = dtt=0 τ =0 t=0 τ =0

over the triangle 0here we integrate horizontally firstt

≤ τ ≤ t

τ

switch the order, i.e., integratelet’s vertically first:Z µZ ¶

dt dτZ Z∞ ∞ ∞ ∞

f (τ )e−st e−stG(s) = = f (τ )dt dττ =0 t=τ τ =0

∞t=τZ

f (τ )(1/s)e−sτ dτ=τ =0

F (s)/s

=


Multiplication by t

let f be a signal and define

g(t) = tf (t)

then we haveG(s) = −F 0(s)

just differentiate both sides ofto verify formula,Z ∞

e−stf (t) dtF (s) =0

with respect to s to getZ ∞

(−t)e−stf (t) dtF 0(s) =0


examples

e−t, te−t• f (t) = g(t) =

1 1 d− L(te−t) = =(s + 1)2s + 1ds

= te−t, t2e−t• f (t) g(t) =

1 2 d− L(t2e−t) = =(s + 1)2 (s + 1)3ds

• in general, (k − 1)!

L(tke−t) =(s + 1)k+1


Convolution

signals f and g, denoted h = f ∗ g, isthe convolution of the signal

tZ

h(t) = f (τ )g(t − τ ) dτ0

tZ

as h(t) = f (t − τ )g(τ ) dτ ; insame other words,•0

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


let’s show that L(f

H (s)

∗ g)

=

=Z

F (s)G(s):t

µZ ¶∞

e−st

t

f (τ )g(t − τ ) dτ dtt=0

∞τ =0

Z Z

e−stf (τ )g(t − τ ) dτ dt=t=0 τ =0

the triangle 0 ≤ τ ≤ twhere we integrate overZ Z∞ ∞

e−stf (τ )g(t − τ ) dt dτH (s) =change order of integration:•τ =0 t=τ

t = t − τ ; dt = dt; region of integration becomeschange variable t to•≥ 0, ≥ 0τ t

Z Z∞ ∞

e−s(t+τ )f (τ )g(t) dtH (s) = dττ =0

µZt=0

¶ µZ ¶∞ ∞

e−sτ f (τ ) e−stg(t)= dτ dtτ =0 t=0

= F (s)G(s)


examples

f = δ, F (s) = 1, gives•H (s) = G(s),

which is consistent withZ t

δ(τ )g(t − τ )dτ = g(t)0

f (t) = 1, F (s) = e−sT /s, gives•

H (s) = G(s)/s

which is consistent withtZ

h(t) = g(τ ) dτ0

more interesting examples later in the course . . .•


Finding the Laplace transform

you should know the 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 ωt) = s/(s2 + ω2), L(sin ωt) = ω/(s2 + ω2)

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 formulaZ ∞

f (t)e−st dtF (s) = . . .0


Patterns

while the details differ, you can see some interesting symmetricbetween

patterns

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

•

•

we’ll see these patterns (and others) throughout the course


(306557551) laplace (12) (1)

Documents