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
).
Ys=AdmittanceUs
Y
s
Admittance
U
s
(4)
Admittance=1−Impedance=1+1−s+1−s+1−1+2−s=2s3+3s2+4s+2−s3+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+2−s3+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
qd−ⅆtyt=pd−ⅆtut
q
d
ⅆ
t
y
t
p
d
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:
ℒdnytdtn=snYs−sn−1y0-−sn−2d1y0-dt1−…−sdn−2y0-dtn−2−dn−1dtn−1y0-
ℒ
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.
ℒd3ytdt3=s3Ys+s2−2s+4
ℒ
t3
y
t
s
3
Y
s
s
2
2
s
4
(10)
ℒd2ytdt2=s2Ys+s−2
ℒ
t2
y
t
s
2
Y
s
s
2
(11)
ℒd1ytdt1=sYs+1
ℒ
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
qd−ⅆtyt=pd−ⅆtut
q
d
ⅆ
t
y
t
p
d
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 (1−s)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)(1−s)
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+2−s3+2s2+s+2)(1−s)−(s2+1−s3+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.
