f(t)
co
sin bt
b /( s 2 + b 2 )
cos bt
s /( s 2 + b 2 )
e at
e at sin bt
1 /( s a )
b /[(s a ) 2 + b 2 ]
e at cos bt
( s a ) /[(s a ) 2 + b 2 ]
LT of derivatives:
L[f(t)]
co/s
L[ f ' (t )] = sF ( s ) f (0)
(t )] = s n F ( s ) s n 1 f (0) s n 2 f ' (0) . f

PTFE 4761
Homework #4 Solution
1) Problem 4.1 from the textbook
0.83sec -1
In problem 3.4, we found that the transfer function can be modeled as Gp(s) =
s + 0.83 sec 1
(For simplicity, let us drop the sec units. The book is not very consistent when it spe

PTFE 4761
Homework #3 Solution
1) Problem 3.1 parts a) and d) in textbook.
Transfer function G(s) is defined as G = Y/C:
a) dy/dt + 4y = c
Laplace: sY + 4Y = C
Re-arrange: Y = C/(s+4)
So the transfer function G = 1/(s+4). Plotting the response in Simulink

PTFE 4761
Homework #2 Solution
1)
With iv = KP (L-y), we get: dy/dt = fin = iv2 = KP2(L-y)2.
So we have the ODE: dy/dt = KP2(L-y)2
Re-arranging:
dy/(L-y)2 = KP2 dt
Integrating from t = 0 (y = 0) to some time t (and level y):
dy/(L-y)2 = KP2 dt
which give

PTFE 4761
Homework #1 Solution
1) Problem 1.7 from the textbook
a)
Generally, if something listed as the condition for the controller (e.g. If heater is on.), it is an
input. If something is stated as the action (e.g. ., then the controller turns on the h

Question 1 (10 of 100 points)
a) In a feedback closed loop, for what purpose do we set the poles of the closed loop transfer function?
We set the poles to set the rate of the response of the closed loop system. The poles usually translate to
reciprocals o

PTFE 4761
Homework #5 Solution
1) Problem 5.3 from textbook
a) We are asked to design an analog PI controller to get a response of the form:
y(nT) = ro + c1(e-2T)n + c2(e-4T)n
This is equivalent to getting the poles of Gcl to be -2 and -4
Gp(s) =
2
s+5
.

PTFE 4761
Homework #7 Solution
1) Problem 7.4 in textbook
a) From the diagram:
I1 = 0 and X2 = 0
X1 = 0
X1 = 1
(I1 = 0 and X2 = 0)
or (I2 = 1 and X2 = 0)
(I3 = 0 and X1 = 0)
or (I1 = 0 and X1 = 1)
X2 = 0
X2 = 1
I3 = 1 and X1 = 0
b)
X1 : Finding the sets a

1. a) Taking Laplace transform on both sides:
s 2Y ( s ) + 3sY ( s ) + 2Y ( s ) = C ( s )
1
Y (s) = 2
C ( s)
s + 3s + 2
1
G (s) = 2
s + 3s + 2
1
1
1
b) lim y (t ) = lim sY ( s ) = lim sC ( s )G ( s ) = lim s G ( s ) = lim G ( s ) = lim 2
=
t
s0
s0
s 0
s0

f(t)
co
L[f(t)]
co/s
tn
n! / s n +1
sin bt
b /( s 2 + b 2 )
cos bt
s /( s 2 + b 2 )
e at
1 /( s a )
t n e at
n! /( s a ) n +1
sinh at
a /( s 2 a 2 )
cosh at
s /( s 2 a 2 )
e at sin bt
b /[(s a ) 2 + b 2 ]
e at cos bt
( s a ) /[(s a ) 2 + b 2 ]
1 for t t 0

PTFE 4761
Laboratory assignment #1
due: Wed, Jan 30
Solve the ODEs below using Simulink. The initial conditions are zeros: y(0) = 0, y(0) = 0, etc.
Also provide a printout of your Simulink diagram. For each result, provide a plot of y(t)
from t = 0 to t =

PTFE 4761
Homework #6 Solution
1)
Performing regression / fitting on the given data, we can find the best fit expression:
y(n+1) = 25.07 c(n-1) + 456.24 c(n) + 6.17 y(n-2) + 24.49 y(n-1) - 44.13 y(n)
Plotting the result from the regression:
35
Data
30
The

PTFE 4761
Lab Assignment #4
due: Wed, Apr 9
Posted on t-square is an Excel file that contains the data from experiment we did on the coupled tank
system. The data are not exactly the ones you obtained but they were done with exactly the same
method (just

PTFE 4761 Final Practice Problems
(problems from Spring 2007 Final Exam)
1) (note that here I leave T to be whatever value; its also okay to take a value of T, e.g. T = 1 to
make the notation simpler)
a) The difference equation for a discrete time step dt

Question 1 (25 of 100 points)
A process is described by the differential equation:
a) Find Y(s)
dy
+ y = e 2t .
dt
At t = 0, y = dy/dt = 0.
(function of s only)
Laplace transform of both sides: sY(s) + Y(s) =
1
( s + 2)
(1+s) Y(s) =
1
( s + 2)
Y (s) =
1
(

PTFE 4761 Final Practice Problems
(problems from Spring 2007 Final Exam)
1. Given a process governed by
dy (t )
dc(t )
+ 2 y (t ) =
+ c(t ) with all initial conditions setting to
dt
dt
zero.
a) Find the corresponding difference equation.
b) Obtain the dig