Chemical Engineering - Process Control and Instrumentation - Discussion
Discussion Forum : Process Control and Instrumentation - Section 2 (Q.No. 23)
23.
The initial value (t=0) of the unit step response of the transfer function [(s + 1)/(2s + 1)] is
Discussion:
6 comments Page 1 of 1.
Hitesh Pawade said:
8 years ago
@Deepti right.
y(s)={(1/s)* [(s + 1)/(2s + 1)] }.
=1/s*[s/(2s+1) +1/(2s+1)].
=1/(2s+1) + 1/s*(2s+1).
=1/(2s+1) + 1/s -2/(2s+1) ...(Partial Fraction).
y(s) = 1/s - 1/(2s+1).
taking inverse laplace.
y(t)=u(t) - e^(t/2)/2.
t=0.
Y(0) = u(0) - e^0/2.
Y(0) = 1-1/2.
y(0) = 1/2.
[B].
y(s)={(1/s)* [(s + 1)/(2s + 1)] }.
=1/s*[s/(2s+1) +1/(2s+1)].
=1/(2s+1) + 1/s*(2s+1).
=1/(2s+1) + 1/s -2/(2s+1) ...(Partial Fraction).
y(s) = 1/s - 1/(2s+1).
taking inverse laplace.
y(t)=u(t) - e^(t/2)/2.
t=0.
Y(0) = u(0) - e^0/2.
Y(0) = 1-1/2.
y(0) = 1/2.
[B].
(2)
Bhavin trada said:
8 years ago
@Udaykiran.
Why you have applied final value theorem? we are asked for an initial value.
Why you have applied final value theorem? we are asked for an initial value.
Kel said:
9 years ago
Great answer @Udaykiran. This helps greatly to simplify the solution, especially in time constrained situations.
Deepti said:
1 decade ago
One can also take inverse Laplace transform.
Babar munir said:
1 decade ago
I have not understand what are you saying please tell me the whole procedure?
Udaykiran said:
1 decade ago
Option B.
For a unit step response X(s) = 1/s.
Y(s)/X(s)= [(s + 1)/(2s + 1)] becomes,
Y(s)={(1/s)* [(s + 1)/(2s + 1)] }.
Now apply final value value theorem i.e.,
lt y(t) = lt sY(s)
t->0 s->(1/0).
On RHS,
Yes will get cancelled now apply limits. So the answer is 1/2.
Therefore the opt is B.
For a unit step response X(s) = 1/s.
Y(s)/X(s)= [(s + 1)/(2s + 1)] becomes,
Y(s)={(1/s)* [(s + 1)/(2s + 1)] }.
Now apply final value value theorem i.e.,
lt y(t) = lt sY(s)
t->0 s->(1/0).
On RHS,
Yes will get cancelled now apply limits. So the answer is 1/2.
Therefore the opt is B.
(1)
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