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# Laplace Example

Summary: Example of a Laplace system

## Example 1

y0-=-1 y 0- -1
(1)
ddty0-=2 t1 y 0- 2
(2)
d2dt2y0-=-4 t2 y 0- -4
(3)

Find the step response for the system above, when utut is the input and ytyt is the output (i.e. find ytyt for ut=stept u t step t ).

(4)
Admittance=1Impedance=1+1s+1s+11+2s=2s3+3s2+4s+2s3+2s2+s+2 Admittance 1 Impedance 1 1 s 1 s 1 1 2 s 2 s 3 3 s 2 4 s 2 s 3 2 s 2 s 2
(5)
Ys=(2s3+3s2+4s+2s3+2s2+s+2)Us Y s 2 s 3 3 s 2 4 s 2 s 3 2 s 2 s 2 U s
(6)

With our previous definition of qdtyt=pdtut q d t y t p d t u t we can define the Laplace domain equivalents of qq and pp as:

qs=s3+2s2+s+2 q s s 3 2 s 2 s 2
(7)
ps=2s3+3s2+4s+2 p s 2 s 3 3 s 2 4 s 2
(8)

When we multiply qsqs times YsYs, we have to remember to include terms relating to the initial conditions of ytyt. We normally think of the Laplace transform of ddtyt tyt as sYs sYs. However, in reality, the general transform is as follows:

dnytdtn=snYssn1y0-sn2d1y0-dt1sdn2y0-dtn2dn1dtn1y0- tn y t s n Y s s n 1 y 0- s n 2 t1 y 0- s tn2 y 0- tn1 y 0-
(9)

Therefore, using the initial conditions stated above, we can find the Laplace transforms of the first three derivatives of ytyt.

d3ytdt3=s3Ys+s22s+4 t3 y t s 3 Y s s 2 2 s 4
(10)
d2ytdt2=s2Ys+s2 t2 y t s 2 Y s s 2
(11)
d1ytdt1=sYs+1 t1 y t s Y s 1
(12)

We can now get a complete ss-domain equation relating the output to the input by taking the Laplace transform of qdtyt=pdtut q d t y t p d t u t . The transform of the right-hand side of this equation is simple as the initial conditions of ytyt do not come into play here. The result is just the product of psps and the transform of the step function (1s)1s.

The left-hand side is somewhat more complicated because we have to make certain that the initial conditions are accounted for. To accomplish this, we take a linear combination of Laplace transform of the third derivative of y(t), Laplace transform of the second derivative of y(t), and Laplace transform of the first derivative of y(t) according to the polynomial qsqs. That is to say, we use the coefficients of the s terms in qsqs to determine how to combine these three equations. We take 1 of Laplace transform of the third derivative of y(t) plus 2 of Laplace transform of the second derivative of y(t) plus 1 of Laplace transform of the first derivative of y(t) plus 2.

When we sum these components, collect the YsYs terms, and set it equal to the right-hand side, we have:

(s3+2s2+s+2)Ys+s2+1=(2s3+3s2+4s+2)(1s) s 3 2 s 2 s 2 Y s s 2 1 2 s 3 3 s 2 4 s 2 1 s
(13)

Rearranging, we can find the solution to YsYs:

Ys=(2s3+3s2+4s+2s3+2s2+s+2)(1s)(s2+1s3+2s2+s+2) Y s 2 s 3 3 s 2 4 s 2 s 3 2 s 2 s 2 1 s s 2 1 s 3 2 s 2 s 2
(14)

This solution can be looked at in two parts. The first term on the right-hand side is the particular (or forced) solution. You can see how it depends on psps and usus. The second term is the homogeneous (or natural) solution. The numerator of this term describes how the initial conditions of the system affect the solution (recall that s2+1 s 2 1 was the part of the result of the linear combination of Laplace transform of the third derivative of y(t), Laplace transform of the second derivative of y(t), Laplace transform of the first derivative of y(t)). The denominator of the second term is the qsqs polynomial; it serves to describe the system in general.

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##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

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##### What are tags?

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