Chemical Engineering - Process Control and Instrumentation - Discussion
Discussion Forum : Process Control and Instrumentation - Section 3 (Q.No. 21)
21.
A process is initially at steady state with its output y=l for an input u=l. The input is suddenly changed to 2 at time t=0. The output response is y(t)=1+2t. The transfer function of the process is
2/s
Discussion:
5 comments Page 1 of 1.
Albaz said:
4 years ago
The Correct answer is A.
Dev said:
6 years ago
The Correct answer is option C.
(2)
JERRY145 said:
1 decade ago
Answer A.
Y(t) = 2t.
U(t) = 1.
Y(s) = 2/s2.
U(s) = 1/s.
Y(s)/U(s) = 2/s.
Y(t) = 2t.
U(t) = 1.
Y(s) = 2/s2.
U(s) = 1/s.
Y(s)/U(s) = 2/s.
Jarat said:
1 decade ago
y(t=0) = 1, u(t=0) = 1.
u(t) = u(t)-u(t=0) = 2-1 = 1.
y(t) = 2t+1.
Now, y(t) = y(t)-y(t=0) = 2t+1-1 = 2t = 2t*u(t).
y(s) = 2/s2*U(s).
y(s)/u(s) = 2/s2.
u(t) = u(t)-u(t=0) = 2-1 = 1.
y(t) = 2t+1.
Now, y(t) = y(t)-y(t=0) = 2t+1-1 = 2t = 2t*u(t).
y(s) = 2/s2*U(s).
y(s)/u(s) = 2/s2.
Patan ameer khan said:
1 decade ago
The answer is 1+2/s not 1/s(1+2/s).
Because y(s) = 1/s(1+2/s).
u(t) = 2- 1 = 1.
u(s) = 1/s.
y(s)/u(s) = 1+2/s
Because y(s) = 1/s(1+2/s).
u(t) = 2- 1 = 1.
u(s) = 1/s.
y(s)/u(s) = 1+2/s
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