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EEL 3105 Fall 2011
Homework 7 ‐ Solution
1. Consider the 2 nd^ order differential equation
2
2 16 0,^ (0)^ 1,^ (0)^0
d y dy dy
k y y
dt dt dt
Find solution y(t) for the following values of k: k=12, k=6, k=5.6, k=2, k=0, k=‐1. Choose a suitable time interval and plot these solutions. You can use MATLAB or some other computational math tool for plotting y(t). Comment on your answers. Answer: First, let us calculate the roots of the polynomial corresponding to the differential equation: 2
k 16
The roots are
k k
There are four distinct cases: k >8 in which case we have two distinct real roots, k = 8 in which case we have two roots at ‐ 4 and ‐ 8 < k < 8 in which case we have a pair of complex conjugate roots and k < ‐ 8 when we again have two real roots. We will show solution for three values of k: k=12, k = 6, and k = ‐ 1 as these exhibit different behaviors. a. k=12: the roots are ‐1.52 and ‐10.48. The general form of solution is
y t ( ) Ae ^ 1.52^ t^ Be 10.48 t
We need to find coefficients A and B. Applying initial conditions gives us:
y A B
dy
A B
dt
We can solve for A and B to get A = 1.171 B = ‐0.
t=0:.1:2pi; a=1.171exp(‐1.52t)‐0.171exp(‐10.48*t); plot(t,a)
b. k=6: the roots are ‐3+j2.65 and ‐ 3 ‐j2.65. There are two ways to proceed here. We can use either of the following general forms of solutions: ( 3 2.65) ( 3 2.65) 3
( ) ( cos(2.65 ) sin(2.65 ))
j t j t t
y t Ae Be
or
y t e P t Q t
In either case, we have to use the initial conditions to find the unknown constants A, B or P, Q. We will use the second from as it avoids complex calculations. The initial conditions lead to the following simultaneous equations:
(0) [ cos(2.65 ) sin(2.65 ) [ 2.65 sin(2.65 ) 2.65 cos(2.65 ) (0)
t t
y P
dy
e P t Q t e P t Q t
dt
P Q
^
Then P = 1 and Q = 1.13 which gives the solution
ି݁ൌ ሻݐሺݕ ଷ௧^ ሺcosሺ2.65ݐሻ 1.13 sinሺ2.65ݐሻሻ
t=0:.1:2pi; d=exp(‐3t).(cos(2.65t)+1.13sin(2.65t)); plot(t,d) (^00 1 2 3 4 5 6 )
1
2
2 5.6^16 ( )
d y dy
y r t
dt dt
Find impulse response of this system. In other words, find y(t) for r(t) = unit impulse function and initial
conditions y(0)=0 and (0) 0.
dy
dt
Answer: We first recall from class that the unit impulse response function with zero initial conditions can be found by solving the differential equation with zero input but with a specific different set of initial conditions. In our case, the differential equation is second order. So, the initial condition on the first derivative should be set to one. Thus, we need to solve 2
2 5.6^16
d y dy
y
dt dt
dy
y
dt
We can follow the solution technique from Problem 1 above. The roots of the polynomial 2
are ‐2.8+j2.86 and ‐2.8‐j2.86. The general solution is of the form
y t ( ) e 2.8 t P cos(2.86 ) t Q sin(2.86 ) t
Applying the initial conditions gives:
y P
dy
P Q
dt
So, Q=1/2.86 giving the unit impulse response
( ) 0.35 sin(2.86 )
t
h t e t
As you can see, this is a decaying exponential. It is zero at t=0. Its derivative at t=0 is 1. The effect of the impulse at t=0 is to change the value of the first derivative from zero to one instantaneously.
3. Consider the differential equation
3 2
d y d y dy
y u t
dt dt dt
Set this up as a state‐space linear system of the form
dx t
Ax t Bu t
dt
y t Cx t
Find suitable matrices A, B, and C. What are the eigenvalues of A? Is this system stable? Solution: We start by defining the state vector 1 2 3 2 2
y
x
dy
x t x t t
dt
x
d y
dt
^ ^ ^
Now, we get
1 2 2 2 2 3 3 1 2 3 3 3 1 2 3
dx^ dy
dt dt
x
dx dx d y
t t x t
dt dt dt
x x x u
dx d y
dt dt
x
x t u t
x
^
^
^ ^ ^
^ ^ ^
^ ^ ^
^ ^ ^ ^ ^
^
^
Thus, if we set
A
B
and C =[0 0 1], we will get the needed state‐space model. Characteristic polynomial of A is
s^3^ 5 s^2 7 s 10
