Biochemical Reactions and Enzyme Kinetics - Introduction to Biomedical Engineering - page 468

Step Input
We now draw our attention to the case in which the input to the system in Eq. (8.51) is a
step input,

f ð t Þ¼

z

u ð t Þ:

With initial condition

q S ð

0 Þ , we have for

t

0

q S ¼ V

ð z V Þ q S þ z K M

max
q S þ K M

max

Þ q S þ

z ¼ ð 8 : 54 Þ

ð ð q S þ K M Þ

Separating q S and t in Eq. (8.54) gives

K M dq S q S dq S

dt ¼

K M þ ð z V Þ q S þ z ð z V Þ q S þ z K M

max max

0
@

1
A

0
@

1
A

1

K M dq S q S dq S

¼

K M þ

z K M

ð z V max Þ q S þ z ð z V max Þ

q S þ

ð 8 : 55 Þ

ð z V Þ

max

0 1

z K M

0
@

1
A 1

@ A

ð z V max Þ

K M dq S 1

¼

K M þ

dq S

z K M

ð z V Þ q S þ z ð z V Þ

max max

q S þ

ð z V max Þ

Then we integrate Eq. (8.55)

0
@

0
@

1
A

1
A dq S

z K M

1

Z t
0 dt ¼

Z q S

K M dq S 1 ð z V max Þ

K M þ

ð 8 : 56 Þ

z

ð z V Þ q S þ z z V K M

q S ð 0 Þ max max

q S þ

ð z V max Þ

which gives

0
@

1
A ln

0
@

1
A þ

0
@

1
A

0
@

1
A

K M ð z V Þ q S þ z K M q S q S ð 0 Þ

max

ð z V Þ ð z V Þ q S ðÞþ 0 z K M z V

max max max

0
@

1
A

z K M

t ¼

q S þ

ð z V Þ

z K M

max

2 ln

ð 8 : 57 Þ

z

K M

ð z V max Þ

q S ð 0 Þþ

ð z V max Þ

0
@

1
A

¼ q S q S ð

0 Þ V K M ð z V Þ q S þ z K M

max max

V max

2 ln

z ð z V max Þ q S ð 0 Þþ z K M

ð z V max Þ