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EEL 3105 Fall 2011 Final Exam
- Consider the differential equation 2 2 0.
d y (^) a dy by dt dt
Suppose y t ( ) e t sin(3 ) t is a solution to this differential equation. Find a and b. [4 pts]
Answer: Based on the form of y(t), 1 j 3 are roots of the polynomial corresponding to the differential
equation: 2 a b .Thus,
2 a b ( 1 j 3)( 1 j 3) 2 2 10.
Therefore, a=2 and b=10.
An alternative approach would be to plug the expression for y(t) into the differential equation:
2 2
sin(3 ) 3 cos(3 )
sin(3 ) 6 cos(3 ) 9 sin(3 ) , [ 8sin 3 6 cos(3 ) sin(3 ) 3 cos(3 ) sin(3 )] 0 8 0 6 3 0 2, 10.
t t
t t t
t
dy (^) e t e t dt d y (^) e t e t e t dt Thus e t t a t a t b t a b a a b
- Find Laplace transform of
( ) ( ) sin 5 , 0. 3
f t tU t ^ t t
Here U(t) is the unit step function. [4 pts]
Answer: We will use superposition. Laplace transform of the first term is
2 LT tU t ( ( )) 1. s
We can use basic trigonometry to write
2 2
2
sin 5 sin(5 ) cos( / 3) cos(5 ) sin( / 3) 3 (^3) sin(5 ) 1 cos(5 ). 2 2 sin 5 3 5 1 3 2 25 2 25 1 5 3 2 25
t t t
t t
LT t s s s s s
Thus, the answer is
2 2
s s s
- Consider the differential equation 2 2 ( ).
d y (^) u t dt
Suppose the input u(t) is the unit step function and the initial conditions are y(0)=1,
dy (0) 0. dt
Find the solution y(t). [4 pts]
Answer: The easiest way to do this problem is by using Laplace transforms. Taking LT of both sides, we get
2
3 2
( )^1
s Y s sy s Y s s s y t invLT Y s t U t
^
- Find Laplace transform of
f t ( ) sin ( ),^2 t t 0. [4 pts]
Answer: We first need to use basic trigonometric identity to simplify f(t):
Please find suitable choices for u(t) and initial conditions y(0) and^ dy (0) dt
. [6 pts]
Answer: Since y(t) is the solution, we can first calculate the required initial conditions:
(0) 15, (0) 25 10 35.
y dy dt
To find u(t), we plug the given y(t) into the differential equation and calculate the LHS:
2 2 5 5 5
( ) ( )^7 ( ) 10 ( )
t t t t t t t
u t d y t^ dy t y t dt dt e e e e e e e
- Consider the differential equation 3 3 10 ( ).
d y (^) y u t dt
Find A, B, C for a state‐space representation of the form
( ) ( ) ( ) ( ).
dx (^) Ax t Bu t dt y t Cx t
Answer: This is an easy one ‐‐‐ no du/dt here to worry about! Following the examples worked out in the homework and practice sets:
A B C
^ ^
