Laplace Transforms Definition Region of convergence Useful - - PowerPoint PPT Presentation

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 Linear Circuit 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 and

steady-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

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

2

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 in ECE 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
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

3

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) =

1 2e−t/0.001 + 1 √ 2 sin(1000t − 45◦)

= vtr(t) + vss(t) vtr(t) =

1 2e−t/0.001

vss(t) =

1 √ 2 sin(1000t − 45◦)

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

4

Laplace Transform Analysis Illustration Continued

5 10 15 20 25 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 Total Transient Steady State

Time (ms) vo(t) (V)

vo(t) =

1 2e−t/0.001 + 1 √ 2 sin(1000t − 45◦)

= vtr(t) + vss(t)

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

5

Laplace Transform for ODE’s

x(t) Linear Circuit

y(t)

• +

N

• k=0

ak dky(t) dtk =

M

• k=0

bk dkx(t) dtk

• Relationship of a voltage (or current) in a linear circuit to any
• ther voltage (or current) is defined by a linear, time-invariant

constant-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 standard

techniques of differential equations or convolution

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

6

Approach

• We will begin with a thorough discussion of the Laplace transform
• The elegance and simplicity of using this approach for circuit

analysis will not become apparent for several lectures

• We will spend a lot of time on this topic
• Bear with me
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

7

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

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

8

Laplace Transform Convergence

• The Laplace transform does not converge to a finite value for all

signals 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 is

called the region of convergence (ROC)

• Will not discuss in detail this term, but may see this in other

classes on linear systems

• See Signals and Systems chapter for more information
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

9

Example 1: Laplace Transform of x(t) Find the Laplace transform of x(t) = u(t). What is the region of convergence? What is the transform of x(t) = 1?

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

10

Example 2: Laplace Transform of x(t) Find the Laplace transform of x(t) = e−atu(t). What is the region of convergence? What is the transform of x(t) = e−at?

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

11

Example 3: Laplace Transform of x(t) Find the Laplace transform of x(t) = δ(t). What is the region of convergence?

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

12

Example 4: Laplace Transform of x(t) Find the Laplace transform of x(t) = cos(ωt)u(t). What is the region

• f convergence? What is the Laplace transform of cos(ωt)?
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

13

Example 4: Workspace

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

14

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)}

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

15

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

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

16

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 = 1 a dτ L {x(at)} = 1 a ∞

0− x(τ)e−s τ a dτ

= 1 a ∞

0− x(τ)e−( s a)τ dτ

x(at) u(t)

L

⇐ ⇒ 1 aX s a

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

17

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)

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

18

Translation in Frequency Given X(s) = L {x(t)}, what is the inverse Laplace transform of X(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)

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

19

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) dt dt 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−)

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

20

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)

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

21

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

0− x(τ) dτ?

L t x(τ) dτ

• =

∞

0−

t x(τ) dτ

• e−st dt

u = t x(τ) dτ du = x(t) dt dv = e−stdt v = −1 s e−st L t x(τ) dτ

• =

t x(τ) dτ −1 s e−st

• ∞

0−

− ∞

0−

−1 s e−stx(t) dt = (0 − 0) + 1 s ∞

0− x(t)e−st dt

t x(τ) dτ

L

⇐ ⇒ 1 sX(s)

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

22

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 (Page

687)

• You should put copies of these tables on your notes that you bring

to the exams

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

23

Example 5: Laplace Transform Properties Find the Laplace transform of t u(t).

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

24

Example 6: Laplace Transform Properties Find the Laplace transform of ˙ r(t) dr(t) dt

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

25

Example 7: Laplace Transform Properties Find the Laplace transform of e−at cos(ωt) u(t). Hint: L {cos(ωt)} =

s s2+ω2

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

26

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 of

convergence

• 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
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

27

Inverse Laplace Transform Example X(s) = s + 8 s(s + 2) = 4 s − 3 s + 2 Since we know u(t)

L

⇐ ⇒ 1 s e−atu(t)

L

⇐ ⇒ 1 s + 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)

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

28

Partial Fraction Expansion A critical step in the previous example was finding following equation: X(s) = s + 8 s(s + 2) = 4 s − 3 s + 2

• In general, this can be done by partial fraction expansion
• In practice, we can do this using your calculators, MATLAB, or a

similar tool

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

29

PFE: Overview In general, most functions will have a general form X(s) = N(s) D(s) = N

k=0 aksk

M

k=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
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

30

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
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

31

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 = []

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

32

Example 8: Workspace

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

33

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 complex

conjugate pairs

• The residues will also be complex conjugate pairs
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

34

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)}

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

35

Example 9: Distinct Complex Poles Given X(s) = L {x(t)}, find x(t) using both methods of handling complex 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 = []

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

36

Example 9: Workspace

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

37

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
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

38

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 = 1 m! 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
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

39

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 = []

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

40

Example 10: Workspace

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

41

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 an

expression that contains a proper rational function

• Will not explain conversion
• Key point: if the order of N(s) is greater than or equal to the
• rder of D(s), you cannot apply PFE directly
• MATLAB’s residue will find the coefficients of A(s) as part of the

partial fraction expansion

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

42

Solving Ordinary Differential Equations

N

• k=0

ak dky(t) dtk = x(t)

N

• k=0

akskY (s) −

N

• k=0

k

• ℓ=1

sk−ℓy(ℓ−1)(0−) = X(s) Y (s) = X(s) + N

k=0

k

ℓ=1 sk−ℓy(ℓ−1)(0−)

N

k=0 aksk

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

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

43

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) + 1

2 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,

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

44

Example 11: Workspace 1

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

45

Example 11: Workspace 2

• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

46

Summary

• The Laplace transform can be used to solve ordinary differential

equations

• 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 pairs

and the properties of the transform

• This topic is covered in both books
• J. McNames

Portland State University ECE 222 Laplace Transform

• Ver. 1.73

47