Assign. #10 - Gene and Linda Voiland School of Chemical ...

BE 440/ChE 441
Unified Systems Bioengineering II/
Process Control
Fall, 2009
Washington State University
Voiland School of Chemical Engineering and
Bioengineering
Richard L. Zollars
Homework Assignment #10
Due: October 2, 2009
You are working with a system described by the following transfer function
G(
s)
K
a
s 1
s 1
s 1
1 2
where K = 1, τ 1 = 4 and τ 2 = 1. This system will be subjected to a unit step change at t = 0.
Enter this transfer function in to Simulink and plot the response of the system for the
following values of τ a : -4, -1, 1, 4, 8 and 16. (Plot the values out to t = 20). Using a partial
fraction expansion determine equation representing the time domain response of this system
for three values of τ a : τ a = -4, 1, and 8. In the time domain equation, what has changed in the
equation for the various values of τ a ?
SOLUTION
The Simulink file used to generate the data for this problem is shown below
Clock
time
To Workspace1
Step
-4s+1
4s 2+5s+1
Transfer Fcn
Out
To Workspace
The plot of the results is shown on the following page. Substituting -4 for τ a into the transfer
function and using a unit step function for the input gives the following expression for the
output (Y(s))

3
Assignement #10
2.5
2
Taua = -4
Taua = -1
Taua = 1
Taua = 4
Taua = 8
Taua = 16
1.5
Output
1
0.5
0
-0.5
0 2 4 6 8 10 12 14 16 18 20
Time
Y ( s)
s
4 s 1
4
s 1s
1
1
s
4
1
s s s
1
4
The form for a partial fraction expansion is
Y ( s)
1
s
4
1
s s s
1
4
A
s
B
1
s
4
C
s 1
Multiply both sides by s and let s = 0 to get A = 1. Multiply both sides by (s + ¼) then let
s = -1/4 to get B = -8/3. Finally multiply both sides by (s + 1) then let s = -1 to get C = 5/3.
So for τ a = -4.0 the time domain response (y(t)) is
t
8
4
( t)
1 e
y
3
5
e
3
t

For τ a = 1 the LaPlace domain expression is
Y ( s)
s
s
1
4
s 1s
1
1
4s
s
1
4
A B
s 1
s
4
Multiply by s and let s = 0 to get A = 1. Multiply by (s + ¼) and let s = -1/4 to get B = -1.
This gives a time domain response of
y(
t)
1 e
t
4
For τ a = 4 the LaPlace domain expression is
Y ( s)
s
8s
1
4
s 1s
1
8s
1
1
4s
s s
1
4
A B
s 1
s
4
C
s 1
Multiply by s and let s = 0 to get A = 1. Multiply by (s + ¼) and let s = -1/4 to get B = 4/3.
Multiply by (s + 1) and let s = -1 to get C = -7/3. This gives a time domain response of
t
4
4
( ) 1
y t
e
3
7
e
3
t
There are two time dependent portions of all of the solutions; e -t/4 and e -t . Since the time
constant for the first of these is 4 it is the slower of the two solutions. For τ a = -4 the
coefficient on this slower term is larger in magnitude than the coefficient on the faster term.
In addition the coefficient on the slower term is negative while that on the faster term is
positive. Thus at small times, when both solutions influence the final result, the system
response first goes in the wrong direction (lower for a positive step change). For τ a = 8 the
situation is reversed with the coefficient on the slower solution being positive and small in
magnitude. Thus at short times the faster solution plays a major role resulting in an
overshoot. As the faster solution decays away the system response approaches the new
steady-state from the side opposite the initial state. For τ a = 1 we get a cancellation between
a term in the numerator and the denominator. As a result there is only one time dependent
solution, he slower of the two responses in the other two cases. Thus the system response is
a monotonic increase to the new steady-state.