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EEL 3105 Fall 2011
Homework 8
1. Solve the following problems for both of these the differential equations
2 2 2 2
d y dy
y u t
dt dt
d y dy
y u t
dt dt
To get ready to solve the problems below, let us first calculate the roots of the corresponding polynomials: 2 2
and
The roots are 3 j 7 [labeled as 1 , 2 ] and ‐10.47,‐1.53 [labeled as 3 , 4 ] respectively. Thus, the
solution the homogeneous part of the first differential equation [I will label it (i)] has the form: ( 3 2.65) ( 3 2.65)
3 [ cos(2.65 ) sin(2.65 )]
j t j t t
Ae Be
or
e M t N t
where A,B or M,N are unknown constants that depend on the initial conditions. Note that this function is a decaying sinusoid. Please plot it using MATLAB to understand this important point. Similarly, the solution to the homogeneous part of the second differential equation [labeled as (ii)] has the form
Ae ^ 10.47^ t^ Be 1.53 t
This function, by contrast, has no oscillations but simply decays exponentially to zero. The slower time constant 1.53 determines, to a dominant extent, the rate of decay to zero. Again, please plot using MATLAB to get a sense of this function. We are now ready to attack the problems below.
A. Suppose y(0)=0, (0)^0
dy
dt
and u=U(t) where U(t) is the unit step function. Find y(t).
Solution: I will do this first the Equation (i). Since initial conditions are zero, we can directly try to calculate the solution. Using the general form of solution for step functions, the solution has the form 3
( ) 1 ( ) [ cos(2.65 ) sin(2.65 )]
t
y At C U t e M t N t
I am using subscript A to denote that this is solution to Problem A. We will reuse this solution to solve other parts below. In the sequel, I will use subscript to denote solutions to various parts. We need to find the three constants C 1 , M, N. This will be done using two facts: y must satisfy the differential equation and it must satisfy the initial conditions on y and dy/dt. Let’s plug y into the differential equation (i). Note that since U(t) is a constant for all t> 0, its derivatives are all zero for t>0. Moreover, the second term in the solution form for y(t) will render the left hand side zero since it is a solution to the homogeneous equation. Therefore, when we plug y into the Equation (i), we get
16 C 1 1.
Therefore C 1 =1/16. Let’s now apply the initial conditions:
y M
dy
M N
dt
Thus, M=‐1/16 and N=
. Thus, the solution is
( ) 1 1 3 [cos(2.65 ) 1.13sin(2.65 )] ( )
t
yA t e t t U t
^
We can repeat this process for Equation (ii). In this case, the general form of solution is 10.47 1.
1 ( )^ ,^ 0.
t t
y A C U t Ae Be t
^
Plugging it into the differential equation yields
16 C 1 1.
Application of initial conditions yields
C A B
A B
Solving for A and B gives rise to A=0.0107, B=‐.073. Thus, the solution for differential equation (ii) is
t t
y A e e U t
^ ^
1 2 3
[cos(2.65 ) 1.51sin(2.65 )]
B B B t
y t y t y t
e t t
Same method can be applied to solve Problem B for Equation (ii). I will not repeat all the arguments but simply show the key equations. In this case, the general form of the solution for u=0 and the given initial conditions is: 10.47 1.
1 ,^ 0.
t t
y B Ae Be t
To calculate A and B we apply the initial conditions: 1 1
B B
y A B
dy
A B
dt
We can solve for A and B to get A=‐0.17,B=1.17. Thus, we get 10.47 1.
1 0.17^ 1.17^ ,^ 0.
t t
y B e e t
We can now solve for the impulse response with zero initial conditions by setting up a new initial condition problem. Following the approach as above, we get 10.47 1.
2 ,^ 0.
t t
y B Ae Be t
^
We will now apply the initial conditions for the calculation of impulse response:
A B
A B
Solving for A and B gives: A=‐0.112, B=0.112. 10.47 1.
2 0.112^ 0.112^ ,^ 0.
t t
y B e e t
^
Thus, the solution to Problem B for the differential equation (ii) is 10.47 1.
t t
yB t y B t y B t e e t
C. Suppose y(0)=0, (0) 0
dy
dt
and u(t)=5sin(20t). Find y(t).
Solution: Let’s do this for Equation (i). The general form of solution is
( ) sin(20 ) cos(20 ) 3 t [ cos(2.65 ) sin(2.65 )]
yC t A t B t e M t N t
We solve for A, B, M, N by plugging it into the differential equation and applying the initial conditions. Here are the equations that arise from the initial conditions:
C C
y B M
dy
A M N
dt
Now plugging y c into the differential equation (i) leads to:
[ 400 2 A 120 B 16 A ]sin(20 t ) [ 400 2 B 120 A 16 B ]cos(20 t ) 5sin(20 t )
This leads to two additional equations: 2 2
[ 400 120 16 ] 5
[ 400 120 16 ] 0
A B A
B A B
We can now solve A,B,M, and N. The values are: A=‐0.00126; B=‐0.00012; M=‐0.00012; N=0.03001. Thus, 3
( ) 0.00126sin(20 ) 0.00012 cos(20 ) [ 0.00012 cos(2.65 ) .03001sin(2.65 )], 0.
t
yC t t t e t t t
Following the same approach for equation (ii) leads to the general solution
( ) sin(20 ) cos(20 ) 10.47^ t^ 1.53 t.
yC t A t B t Ce De
^
We can again apply initial conditions to get the equations:
C C
y B C D
dy
A C D
dt
Plugging yc into the differential equation (ii) yields the following equation: 2 2
[ 400 A 240 B 16 A ]sin(20 t ) [ 400 B 240 A 16 B ]cos(20 t ) 5sin(20 t )
Solution: In this case, we again apply superposition. Using reasoning as in the solution to Part D, we get solution to problem E, denoted as y (^) E is given by combining y (^) 1B and 5 times y (^) 1A , i.e.,
yE y 1 B 5 y 1 A.
We can use the formulae for y (^) 1B and y (^) 1A to get yE for differential equation (i) and (ii).
- Solve Problem 1 using the Laplace transform method. A. Let’s start with differential equation 1. Taking LT of both sides gives
dy
s Y s sy sY s y Y s
dt s
Solving for Y(s) gives 2
Y s
s s s
I will now write a partial fraction expansion for Y(s). In this case, the denominator has two complex roots. I will avoid having to deal with complex numbers and use the completion of squares and keep the quadratic factor:
A Bs C A Bs C
Y s
s s s s s
We can use many methods to find A, B, C. A simple method is to find A first:
A sY s ( ) | s 0 1/16.
Then 2 2
Bs C A s
Y s Y s
s s s s
Thus, 2 2 2
s s
Y s
s s s s s
^
We can now read off the inverse Laplace transform to get
( ) ( ) cos( 7 ) sin( 7 ) , 0.
t t
y At U t e t e t t
^ ^
Let’s now turn to differential equation (ii). Again, taking Laplace transforms of both sides gives
2
Y s
s s s s s s
Now, we can calculate the partial fraction expansion:
A B C
Y s
s s s
A, B, and C can be computed by many different techniques: A=1/16, B=.0107, C=‐0.073. Taking inverse Laplace transform gives:
( ) 1 ( ) 0.0107 10.47^ 0.073 1.53, 0.
t t
yA t U t e e t
^
B. Again, we start with differential equation (i). Taking LT of both sides gives 2 2
dy
s Y s sy sY s y Y s
dt
s s Y s s
Solving for Y(s) gives 2 2 2 2
s s s
Y s
s s s s s
We can now read off the inverse Laplace transform:
( ) cos( 7 ) sin( 7 ), 0.
t t
y Bt e t e t t
We now turn to differential equation (ii). Again, taking Laplace transform of both sides gives: 2 2
s s Y s s
s s
Y s
s s s s
Now, we can calculate the partial fraction expansion:
A B
Y s
s s s s
Finally, taking inverse Laplace transform leads to
( ) 1.283 1.53^ t^ 0.283 10.47 t , 0.
yB t e e t
^
We can now solve differential equation (ii) using a similar approach. Again, taking Laplace
transforms, we get
2 2 2
s s Y s
s
Solving for Y(s), we get 2 2 2 2 2
Y s
s s s
s s s
We can write a partial fraction expansion of the form:
A B Cs D
Y s
s s s
We can calculate A and B easily: 2 2
2 2
s s
A
s s
B
s s
Now, we can solve for C and D using
A B Cs D
Y s
s s s
Plugging values of A and B, we can find C and D: C=‐0.00003; D=0. Now, we can read off the inverse Laplace transform: 10.47 1.
cos(20 ) sin(20 ), 0.
t t C
D
y Ae Be C t t t
Parts D and E can be done by following LT methods in parts A, B, and C or by superposition as in solution to Problem 1.
