Dynamic Systems Biology Modeling and Simulation (2015) – Answers to Exercises – DiStefano

ALL 17 CHAPTERS COVERED

, Dynamic Systems Biology Modeling & Simulation 1-26-2014
2014 by JJ DiStefano III All rights reserved.



SOLUTIONS or ANSWERS to SELECTED EXERCISES in
CHAPTERS 1 – 17
Chapter 1
Has no Exercises.

Chapter 2
2.1 is a discussion question.
S2.2: Substitute et for y, differentiate four times and substitute in the homogeneous
equation:

( 4   3  3 2  5  2)e  t  0
Factoring:
(  1)3 (  2)e t  0

Two distinct solutions are λ = 1 and λ = −2, or y(t) = et and e−2t, which are linearly
independent because the only c1 and c2 that satisfy c1et  c2 e2t  0 for all t are
c1  c2  0. Since one of the roots has multiplicity 3, the overall solution of the ODE is

y (t )  c1et  c2 e 2t  c3tet  c4t 2 et  (c1  c3t  c4t 2 )et  c2 e 2 t

Since the only c1, …, c4 that satisfy this equation for all t are c1 = c2 = c3 = c4 = 0, the four
solutions are independent.

S2.3: Let
x1  y
x2  y  x1
x3  y  x 2
x4  
y  x3

y  x 4

From the equation, x4  2 x1  5 x2  3 x3  x4  u. Therefore, we have:

 x1   0 1 0 0 0 
 x   0 0 1 0  0 
x   2     x u
 x3   0 0 0 1 0 
     
 x 4   2 5 3 1 1 




1

, Dynamic Systems Biology Modeling & Simulation 1-26-2014
2014 by JJ DiStefano III All rights reserved.

S2.4: Following the hints, let x1 (k )  y2 (k ), the output. Then

x1 (k  1)  y2 (k  1)  x2 (k )
x2 (k  1)  y2 (k  2)  x3 (k )
x3 (k  1)  y2 (k  3)  x4 (k )

Now substitute xi into the second equation:

y1 (k  2)
x4 (k )   x2 (k )  0
2
and solve for y1:

y1 (k  2)  2  x2 (k )  x4 (k ) 

Now substitute y1 into the first equation:

5  x2 (k  1)  x4 (k  1)   x2 (k  2)  x4 (k  2)   u (k  1)
2  x2 (k )  x4 (k )   
3 3

Then simplify and substitute x1 (k )  x2 (k  1), etc.

5 5 1 1
2 x2 (k )  2 x4 (k )  x1 (k )  x3 (k )  x1 (k  1)  x2 (k )  u (k  1)
3 3 3 3

Which is equivalent to

5 5 1 1
2 x2 (k  1)  2 x4 (k  1)  x1 (k  1)  x3 (k  1)  x1 (k )  x2 (k  1)  u (k )
3 3 3 3
and
5 5 1 1
2 x3 (k )  2 x4 (k  1)  x2 (k )  x4 (k )  x1 (k )  x3 (k )  u (k )
3 3 3 3

Therefore, the difference equation for x4 is:

1 5 7 5 1
x4 (k  1)   x1 (k )  x2 (k )  x3 (k )  x4 (k )  u (k )
6 6 6 6 2

which, in matrix form, is:




2

, Dynamic Systems Biology Modeling & Simulation 1-26-2014
2014 by JJ DiStefano III All rights reserved.

 0 1 0 0   0 
 x1 (k  1)     
 x (k  1)   0 0 1 0
  0 
x (k  1)   2  0 0 0 1  x (k )   0  u (k )
 x3 (k  1)     
  1 5 7 5  1 
 x4 (k  1)       
 6 6 6 6   2 


S2.5: (a) The question implies finding the impulse response h(t), the inverse transform
of H(s).

(b) One way is to solve by using the fact that a step is the integral of an impulse, etc.
Another is to convert into state variable form and find the Transfer Function (TF) for the
2x2 system of ODEs (which is what is done in Example 2.32). Another approach is to
convert the equation directly into the Laplace domain and find the TF that way, i.e.

L  
y  3 y  2 y   L 5  1(t ) 
L  
y   3L  y   2L  y   5L 1(t ) 
s ( sY ( s )  y (0))  y (0)  3( sY ( s )  y (0))  2Y ( s )  5 / s


Substituting in the ICs and solving for Y(s):
5
Y (s) 
s ( s  3s  2)
2


Now we do PFE:
5 c c c 
Y (s)   5  11  21  31 
s ( s  1)( s  2)  s s 1 s  2 
1 1
c11  sY ( s ) s  0  
(0  1)(0  2) 2
1
c21  ( s  1)Y ( s ) s 1   1
(1)(1  2)
1 1
c31  ( s  2)Y ( s ) s 2  
(2)(2  1) 2
Thus
 1 1 1 
Y (s)  5    
 2 s s  1 2( s  2) 
Taking the inverse Laplace transform:



3