EECE 301 Discussion 10 Laplace Transform Examples_2.pdf

8/17/2019 EECE 301 Discussion 10 Laplace Transform Examples_2.pdf

1/19


2/19

2/19

Examples of Solving Differential Equations using LT

Notice how easy this is!

-LT converts the differential equation into an algebraic equation

-We can easily solve an algebraic equation for an output Y (s)

-We can do partial fraction expansion

-We can use the LT tableBut, this is where the hard part lies…although it is easy for certain inputs.


3/19

3/19

Example 6.29: 2nd Order Differential Equation

Given )(2)(8)(6)(2

2

t xt ydt t dy

dt t yd =++

Find the output y(t ) for t ≥ 0 when the input is x(t ) = u(t )

2)0(1)0( ==

−−

yand y &

1

)(t x

t

Input “starts” att = 0

Assume that the system has ICs given by:


4/194/19

Gives: [ ] [ ] )(2)(8)0()(6)0()0()(2 s X sY yssY ys ysY s =+−+−− −−− &

Solving for Y (s) algebraically gives:

)(

86

2

86

)0(6)0()0()(

22 s X

ssss

y ys ysY

++

+

++

++=

−−−&

Using the specific IC’s gives:

)(86

286

8)(22 s X

ssss

ssY ⎥⎦⎤

⎢⎣⎡

+++⎥⎦

⎤⎢⎣⎡

+++=

IC part H (s) “Transfer Function” (notice howthis comes directly from the D.E.)

)(2)(8)(

6)(

2

2

t xt ydt

t dy

dt

t yd =++Applying the LT to this D.E.


5/195/19

Notice how the IC part can be easily found using PFE and an LT table (Forlinear, constant coefficient differential equations it will always be like that!)

)(

86

2

86

8)(

22 s X

ssss

ssY ⎥

⎦

⎤⎢

⎣

⎡

++

+⎥

⎦

⎤⎢

⎣

⎡

++

+=

IC part H (s) “Transfer Function” (notice howthis comes directly from the D.E.)

Notice that the input part may not be easy to do PFE/ILT if theinput X (s) is complicated.

But in control systems and sometimes in electronics we are often interested in howthe system responds to a step function → Called the “step response”

ss X t ut x 1)()()( : Note =↔=


6/196/19

⎥⎦

⎤⎢⎣

⎡

+++⎥⎦

⎤⎢⎣

⎡

++

+=

)86(

2

86

8)(

22sssss

ssY

IC part Input part

( )[ ]( )

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−=

+−

−=

+−

−

−

n

n

n

n

t

n

n

A

t ut Ae

ζω α

ζ ω φ

ζ

ζω β

φ ζ ω

ω α

ζω

21

2

2

2

1tan

11

:where

)(1sin

22

2 nnss

s

ω ζω

α β

++

+

First… compare to this:

Is… 0< | | < 1?

06.124/62/662

228

182

===⇒=

=⇒=

==

nn

nn

ω ζ ζω

ω ω

β α And identify:

No!!!So… factor!!


7/197/19

⎥

⎦

⎤⎢

⎣

⎡

+++⎥

⎦

⎤⎢

⎣

⎡

++

+=⎥

⎦

⎤⎢

⎣

⎡

+++⎥

⎦

⎤⎢

⎣

⎡

++

+=

)2)(4(

2

)2)(4(

8

)86(

2

86

8)(

22

sssss

s

sssss

ssY

Doing PFE:

>> [R,P,K]=residue([1 8],[1 6 8])R =

-23

P =-4-2

K =[]

>> [R,P,K]=residue(2,[1 6 8 0])R =0.2500

-0.50000.2500

P =-4-20

K =[]

⎟

⎠

⎞⎜

⎝

⎛

⎥⎦

⎤

⎢⎣

⎡+

⎥⎦

⎤

⎢⎣

⎡

+

−+

⎥⎦

⎤

⎢⎣

⎡

+

+⎟

⎠

⎞⎜

⎝

⎛

⎥⎦

⎤

⎢⎣

⎡

+

+

⎥⎦

⎤

⎢⎣

⎡

+

−=

sssss

sY 25.0

2

5.0

4

25.0

2

3

4

2)(

IC part Input part IC part Input part

Factored!!!

IC part SS partTransient part


8/19

8/19

0,4/12/14/132)( 2424 ≥+−++−= −−−− t eeeet y t t t t

4/14/1 0 =t e

Specifyingthis means

we can leaveoff the u(t )’s

Using the LT table:

⎟ ⎠

⎞⎜⎝

⎛ ⎥⎦

⎤⎢⎣

⎡+⎥⎦

⎤⎢⎣

⎡

+

−+⎥⎦

⎤⎢⎣

⎡

++⎟

⎠

⎞⎜⎝

⎛ ⎥⎦

⎤⎢⎣

⎡

++⎥⎦

⎤⎢⎣

⎡

+

−=

ssssssY

25.0

2

5.0

4

25.0

2

3

4

2)(

complexorreal),( bt ue bt − complexorreal,1

bbs +




9/19

9/19

Note: IC part & Transient part have the same kind of decaying parts:These come from the system’s characteristic polynomial!

t t ee24 & −−

Note: The book combined Y (s) into one big thing and found y(t ) from that.Same answer…but we cannot see the impact of three parts!

0,4/12/14/132)( 2424 ≥+−++−= −−−− t eeeet y t t t t



10/19

10/19

Plots for ex. 6.29

Zero state system takes time to respond to step input

Zero state “transient”

Eventually the IC’s decay away, the ZS transient diesand the system reaches its “steady-state” response


11/19

11/19

Now Modify Example 6.29: 2nd Order Differential Equation

)(2)(8)(

6)(

2

2

t xt ydt

t dy

dt

t yd =++

)(2)(8)()(

2

2

t xt ydt

t dy

dt

t yd =++

18.024/12/112

228

18

2

===⇒=

=⇒=

==

nn

nn

ω ζ ζω

ω ω

β α

And identify:

Was…

Now…

⎥⎦

⎤⎢⎣

⎡

+++⎥⎦

⎤⎢⎣

⎡

++

+=

)86(

2

86

8)(

22sssss

ssY

IC part Input part

⎥⎦

⎤⎢⎣

⎡

+++⎥⎦

⎤⎢⎣

⎡

++

+=

)8(

2

8

8)(

22 sssss

ssY

IC part Input part

Is… 0< |

| < 1? Yes… Complex Roots… so use oneof the 2nd order LT Pairs


12/19


13/19

13/19

⎥⎦

⎤⎢⎣

⎡

++

+−+⎥⎦

⎤⎢⎣

⎡

++

+=

8

125.0

25.0

8

8)(

22ss

s

sss

ssY

)4.178.2sin(25.025.0)36.078.2sin(87.2)( 5.05.0 +−++= −− t et et y t t

LTTable

Zero-InputResponse

Zero-StateResponse

TotalResponse


14/19

14/19

Compare the Two Cases:

)2)(4(

2

86

2)(

2 ++=

++=

sssss H

)78.25.0)(78.25.0(

2

8

2)(

2 js jsss

s H ++−+

=++

=

σ

jω

σ

jω


15/19

15/19

Ex. 6.38 “RLCLC” Circuit

In analyzing circuits we often

-Have zero IC’s

-Have no specific input signal in mind

We only want to find the transfer function and/or the frequency response⇒

Suppose you must analyze the following circuit to find the current through C 1

(You might need to do that because in the physical circuit there is a device being

driven by that current and that device has an input impedance that is modeled as acapacitor)

)(t x device )(t y

⇒

C2

C1

Re-Drawn Versionof Fig. 6.13… but

equivalent!!


16/19

16/19

Now, replace “device” by its capacitor model and find the transfer function between:

input = voltage x(t )output = current y(t )&

Use s-domain impedances Ls

1/Cs

And analyze the circuit as if X (s) is a DC voltage source.

[ ] 1)()()(

212

222113

2214

2121

13

221

+++++++

+=

sC C RsC LC C Ls LC RC s L LC C

sC s LC C s H

Note that there are 6 “adjustable” coefficients but only 5 adjustable componentvalues

⇒ Can’t achieve all possible coefficient combinations

This is the wholeidea behind the LT

approach!!

Can find that (see book for details):


17/19

17/19

( )[ ] 2121212121

2212122211

31

422

21

/1/)(/)()/(

/1)/1()(

L LC C s L LC C C C Rs L LC C C LC C Ls L Rs

LC ss Ls H

+++++++

+=

A zero “at the origin”forces low frequencies

to be attenuated

A purely-imaginary-rootsterm… puts conjugate zeroson the jω axis… nulls one

specific frequency

T h hi i i d l k i f Thi i lid


18/19

18/19

To see what this circuit can do we look at its frequency response. This is valid because all poles are in the region of convergence. (True for any passive RLCcircuit)

Get H (ω ) by replacing s→ jω

The following plot shows | H (ω )| for two sets of component values

similarities…

but differences!! Note

How do we choose the component values to do what we want?!

STAY TUNED!!!


19/19

19/19

Due to a pole

near the jω axis

Due to zeros on

the jω axis

157 Hz1570 Hz 157 kHz

Documents

EECE 301 Discussion 10 Laplace Transform Examples_2.pdf