laplace transforms - creating web pages in your...

Laplace Transforms

• Definition

• Region of convergence

• Useful properties

• Inverse & partial fraction expansion

• Distinct, complex, & repeated poles

• Applied to linear constant-coefficient ODE’s

J. McNames Portland State University ECE 222 Laplace Transform Ver. 1.73 1

Laplace Transform Motivation

t

LinearCircuit

vs(t)

vo

-

+

vs(t)

• In ECE 221, you learned

– DC circuit analysis

– Transient response (limited to simple RL & RC circuits)

– Sinusoidal steady-state response (Phasors)

• We did not learn how to find the total response (transient andsteady-state) to an arbitrary waveform

• The Laplace transform enables us to do this

• Circuit elements limited to resistors, capacitors, inductors,transformers, op amps, and ideal sources until ECE 321


Laplace Transform Motivation Continued

Why are we studying the Laplace transform?

• Makes analysis of circuits

– Easier than working with multiple differential equations

– More general than the types of analysis we discussed inECE 221

• Used extensively in

– Controls (ECE 311)

– Communications

– Signal Processing

– Analog circuits (ECE 32X sequence)

• Expected to know for interviews

• Gives you insight in circuit analysis and design


Laplace Transform Analysis Illustration

t = 0

vo

-

+1 kΩ

sin(1000t) 1 μF

Given vo(0) = 0, solve for vo(t) for t ≥ 0.

vo(t) = 12e−t/0.001 + 1√

2sin(1000t − 45◦)

= vtr(t) + vss(t)

vtr(t) = 12e−t/0.001

vss(t) = 1√2

sin(1000t − 45◦)


Laplace Transform Analysis Illustration Continued

0 5 10 15 20 25−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1 TotalTransientSteady State

Time (ms)

vo(t

)(V

)

vo(t) = 12e−t/0.001 + 1√

2sin(1000t − 45◦)

= vtr(t) + vss(t)


Laplace Transform for ODE’s

x(t)LinearCircuit

y(t)

-

+N∑

k=0

akdky(t)dtk

=M∑

k=0

bkdkx(t)

dtk

• Relationship of a voltage (or current) in a linear circuit to anyother voltage (or current) is defined by a linear, time-invariantconstant-coefficient ordinary differential equation (ODE)

• Describes the behavior of many types of systems: Electrical,Mechanical, Chemical, Biological, etc.

• Laplace transform is an easier approach than applying standardtechniques of differential equations or convolution


Approach

• We will begin with a thorough discussion of the Laplace transform

• The elegance and simplicity of using this approach for circuitanalysis will not become apparent for several lectures

• We will spend a lot of time on this topic

• Bear with me


Laplace Transform Definition

L{x(t)} = X(s) �∫ ∞

0−x(t)e−st dt

• Transform will be written with an upper-case letter

• Defined from 0− to include impulses at t = 0

• s = σ + jω is a complex variable

• s has units of inverse seconds (s−1)

• Known as the one-sided (unilateral) Laplace transform

• There is also a two-sided (bilateral) version:

X(s) =∫ +∞−∞ x(t)e−st dt

• We will only work with the one-sided version

+ Easier to obtain the transient response

+ Consistent with common practice

– Ignores x(t) for t < 0


Laplace Transform Convergence

• The Laplace transform does not converge to a finite value for allsignals and all values of s

• Does converge for all signals we will be interested in

– Sinusoids

– e−atu(t) for any real |a| < ∞– δ(t)

• The values of s for which the Laplace transform converges iscalled the region of convergence (ROC)

• Will not discuss in detail this term, but may see this in otherclasses on linear systems

• See Signals and Systems chapter for more information


Example 1: Laplace Transform of x(t)

Find the Laplace transform of x(t) = u(t). What is the region ofconvergence? What is the transform of x(t) = 1?



Find the Laplace transform of x(t) = e−atu(t). What is the region ofconvergence? What is the transform of x(t) = e−at?



Find the Laplace transform of x(t) = δ(t). What is the region ofconvergence?



Find the Laplace transform of x(t) = cos(ωt)u(t). What is the regionof convergence? What is the Laplace transform of cos(ωt)?


Example 4: Workspace


Laplace Transform Properties

• The Laplace transform has many important properties

• We need to know these for at least three reasons

– Improves our understanding of the transform

– Enables us to find the transform more easily

– Enables us to find the inverse transform more easily

• Will use the following notation for Laplace transform pairs

x(t) u(t) L⇐⇒ X(s)X(s) = L{x(t)}

x(t) u(t) = L−1 {X(s)}


Linearity

X1(s) = L{x1(t)}X2(s) = L{x2(t)}

then you should be able to show that

[a1x1(t) + a2x2(t)]u(t) L⇐⇒ a1X1(s) + a2X2(s)

Example: Find the Laplace transform of x(t) = 5δ(t) − 2 cos 5t.

L{δ(t)} = 1

L{cos ωt} =s

s2 + ω2

L{5δ(t) − 2 cos 5t} = 5(1) − 2(

s

s2 + 52

)

= 5 − 2s

s2 + 25


Scaling

Given X(s) = L{x(t)}, what is L{x(at)} for a > 0?

L{x(at)} =∫ ∞

0−x(at)e−st dt

τ = at t =τ

a

dτ = a dt dt =1a

dτ

L{x(at)} =1a

∫ ∞

0−x(τ)e−s

τa dτ

=1a

∫ ∞

0−x(τ)e−( s

a )τ dτ

x(at) u(t) L⇐⇒ 1aX

( s

a

)


Translation in Time

Given X(s) = L{x(t)}, what is L{x(t − t0) u(t − t0)} for t0 > 0?

L{x(t − t0)u(t − t0)} =∫ ∞

0−x(t − t0) u(t − t0)e−st dt

τ = t − t0

dτ = dt

t = τ + t0

L{x(t − t0) u(t − t0)} =∫ ∞

−t0

x(τ) u(τ)e−s(τ+t0) dτ

=∫ ∞

0−x(τ) u(τ)e−s(τ+t0) dτ

= e−st0

∫ ∞

0−x(τ)e−sτ dτ

x(t − t0) u(t − t0)L⇐⇒ e−st0X(s)


Translation in Frequency

Given X(s) = L{x(t)}, what is the inverse Laplace transform ofX(s + s0)?

X(s + s0) =∫ ∞

0−x(t) e−(s+s0)t dt

=∫ ∞

0−

(x(t) e−s0t

)e−st dt

= L{e−s0tx(t)

}L−1 {X(s + s0)} = e−s0tx(t) u(t)

e−s0tx(t) u(t) L⇐⇒ X(s + s0)


Time Differentiation

Given X(s) = L{x(t)}, what is the Laplace transform of x(t)?

L{

dx(t)dt

}=

∫ ∞

0−

dx(t)dt

e−st dt

u = e−st du = −se−st dt

dv =dx(t)

dtdt v = x(t)

L{

dx(t)dt

}=

∫ ∞

0−u dv = uv|∞0− −

∫ ∞

0−v du

= e−stx(t)∣∣∞0− −

∫ ∞

0−x(t)

(−se−st)

dt

=(0 − x(0−)

)+ s

∫ ∞

0−x(t) e−st dt

dx(t)dt

u(t) L⇐⇒ sX(s) − x(0−)


Time Differentiation Continued

In general

dnx(t)dtn

u(t) L⇐⇒ snX(s) − sn−1x(0−) − · · · − s0 dn−1x(t)dtn−1

∣∣∣∣t=0−

If all of the initial conditions are zero,

dnx(t)dt

u(t) L⇐⇒ snX(s)


Time Integration

Given X(s) = L{x(t)}, what is the Laplace transform of∫ t

0− x(τ) dτ?

L{∫ t

0

x(τ) dτ

}=

∫ ∞

0−

(∫ t

0

x(τ) dτ

)e−st dt

u =∫ t

0

x(τ) dτ du = x(t) dt

dv = e−stdt v =−1s

e−st

L{∫ t

0

x(τ) dτ

}=

(∫ t

0

x(τ) dτ

) −1s

e−st

∣∣∣∣∞

0−−

∫ ∞

0−

−1s

e−stx(t) dt

= (0 − 0) +1s

∫ ∞

0−x(t)e−st dt

∫ t

0

x(τ) dτL⇐⇒ 1

sX(s)


Other Properties

• Other properties of the Laplace transform are derived in the text

• See Table 15.1 (page 687) of the electric circuits text

• Common Laplace transform pairs are listed in Table 15.2 (Page687)

• You should put copies of these tables on your notes that you bringto the exams


Example 5: Laplace Transform Properties

Find the Laplace transform of t u(t).



Find the Laplace transform of

r(t) � dr(t)dt



Find the Laplace transform of e−at cos(ωt) u(t). Hint:L{cos(ωt)} = s

s2+ω2


Inverse Laplace Transform Overview

L−1 {X(s)} =1

j2π

∫ σ1+j∞

σ1−j∞X(s) est ds

• The inverse Laplace transform is given above

• σ1 is such that the integral is taken over a line in the region ofconvergence

• Very difficult to apply directly

• We will use a different approach

• Convert X(s) to a form such that we can easily find the inverse


Inverse Laplace Transform Example

X(s) =s + 8

s(s + 2)=

4s− 3

s + 2Since we know

u(t) L⇐⇒ 1s

e−atu(t) L⇐⇒ 1s + a

a1x1(t)u(t) + a2x2(t)u(t) L⇐⇒ a1X1(s) + a2X2(s)

we know that the inverse Laplace transform of X(s) is

L−1 {X(s)} = 4u(t) − 3e−2tu(t)


Partial Fraction Expansion

A critical step in the previous example was finding following equation:

X(s) =s + 8

s(s + 2)=

4s− 3

s + 2

• In general, this can be done by partial fraction expansion

• In practice, we can do this using your calculators, MATLAB, or asimilar tool


PFE: Overview

In general, most functions will have a general form

X(s) =N(s)D(s)

=∑N

k=0 aksk∑Mk=0 bksk

• N(s) and D(s) are a polynomials in s.

• The roots of N(s) = 0 are called zeros of X(s)

• The roots of D(s) = 0 are called poles of X(s)

• To find x(t) = L−1 {X(s)}, we need to

1. Find the poles of X(s)2. Apply partial fraction expansion (via MATLAB)

3. Find the inverse of each term by table lookup


PFE: Distinct Poles

X(s) =N(s)D(s)

=N(s)

(s + p1)(s + p2) . . . (s + pn)

=k1

s + p1+

k2

s + p2+ · · · + kn

s + pn

ki = (s + pi)X(s)|s=−pi

• A distinct pole is a unique root of D(s) = 0

• The coefficients ki are called the residues of X(s)

• To find ki multiply both sides by (s + pi) and evaluate at s = −pi


Example 8: Partial Fraction Expansion

Given X(s) = L{x(t)}, find x(t).

X(s) =5s + 29

s3 + 8s2 + 19s + 12

� [r,p,k] = residue([5 29],[1 8 19 12])

r = 3.0000, -7.0000, 4.0000

p = -4.0000, -3.0000, -1.0000

k = []


PFE: Distinct Complex Poles Method 1

X(s) =N(s)D(s)

=N(s)

(s + α − jβ)(s + α + jβ)

=k1

s + α − jβ+

k∗1

s + α + jβ

k1 = (s + α − jβ)X(s)|s=−α+jβ

• There are two methods for handling complex poles

• Often the residues of X(s) will be complex

• In this case, the complex roots of X(s) will be in complexconjugate pairs

• The residues will also be complex conjugate pairs


PFE: Distinct Complex Poles Method 2

X(s) =N(s)D(s)

=k1s + k2

s2 + as + b+ R(s)

=k1s + k1α + k2 − k1α

(s + α)2 + β2+ R(s)

=k1(s + α) + k2 − k1α

(s + α)2 + β2+ R(s)

=c1(s + α)

(s + α)2 + β2+

c2β

(s + α)2 + β2+ R(s)

c1 = k1

c2 =k2 − k1α

β

x(t) = c1e−αt cos(βt)u(t) + c2e−αt sin(βt)u(t) + L−1 {R(s)}


Example 9: Distinct Complex Poles

Given X(s) = L{x(t)}, find x(t) using both methods of handlingcomplex poles.

X(s) =2

s3 + 2s2 + 2s

� [r,p,k] = residue([2],[1 2 2 0])

r = -0.5000 + 0.5000i, -0.5000 - 0.5000i, 1.0000

p = -1.0000 + 1.0000i, -1.0000 - 1.0000i, 0

k = []


PFE: Useful Transforms

k

s + a⇔ ke−atu(t)

k

(s + a)2⇔ kte−atu(t)

k

s + α − jβ+

k∗

s + α + jβ⇔ 2|k|e−αt cos(βt + θk)u(t)

k

(s + α − jβ)2+

k∗

(s + α + jβ)2⇔ 2t|k|e−αt cos(βt + θk)u(t)

where k = |k|∠θk

• You solve for k just as for a single pole (residue)

• Can also complete the square, as described in the textbooks


PFE: Repeated Poles

X(s) =N(s)D(s)

=N(s)

(s + p)n

=k1

s + p+

k2

(s + p)2+ · · · + kn

(s + p)n

kn−m =1m!

d(m)

dsm(s + p)nX(s)|s=−p

L−1

{1

(s + p)n

}=

1(n − 1)!

tn−1e−ptu(t)

• You can apply the equations above to handle repeated poles

• The algebraic method is usually easier


Example 10: Repeated Poles

Given X(s) = L{x(t)}, find x(t).

X(s) =s − 2

s(s + 1)3

� [r,p,k] = residue([1 -2],poly([0 -1 -1 -1]))

r = 2.0000, 2.0000, 3.0000, -2.0000

p = -1.0000, -1.0000, -1.0000, 0

k = []


PFE: Improper Rational Functions

X(s) =N(s)D(s)

= A(s) +B(s)C(s)

• If X(s) is an improper rational function, you must convert it to anexpression that contains a proper rational function

• Will not explain conversion

• Key point: if the order of N(s) is greater than or equal to theorder of D(s), you cannot apply PFE directly

• MATLAB’s residue will find the coefficients of A(s) as part of thepartial fraction expansion


Solving Ordinary Differential Equations

N∑k=0

akdky(t)dtk

= x(t)

N∑k=0

akskY (s) −N∑

k=0

k∑�=1

sk−�y(�−1)(0−) = X(s)

Y (s) =X(s) +

∑Nk=0

∑k�=1 sk−�y(�−1)(0−)∑N

k=0 aksk

If L{x(t)} is a rational function of s, then the linear ordinarydifferential equation shown above can be solved (more easily) usingthe Laplace Transform


Example 11: Solving ODE’s

Solve the following ODE for y(t) given that y(0−) = y(0−) = 0

d2y(t)dt2

+ 3dy(t)dt

+ 2y(t) = 20 cos(2t)u(t) + 12 sin(2t)u(t)

Hint:� [r,p,k] = residue([20 1],[conv([1 3 2],[1 0 4])])

r = 4.8750, -0.5375 - 1.4875i, -0.5375 + 1.4875i, -3.8000

p = -2.0000, -0.0000 + 2.0000i, -0.0000 - 2.0000i, -1.0000

k = []

� [abs(r) angle(r)*180/pi]

ans = 4.8750 0, 1.5816 -109.8670, 1.5816 109.8670, 3.8000 180.0000,


Example 11: Workspace 1


Example 11: Workspace 2


Summary

• The Laplace transform can be used to solve ordinary differentialequations

• This includes circuits and many other linear time-invariant systems

• We use the one-sided Laplace transform

• The inverse of this transform is always 0 for t < 0

• We usually solve for the inverse by using known transform pairsand the properties of the transform

• This topic is covered in both books


laplace transforms - creating web pages in your...

Documents