Electronics and Communication Engineering - Automatic Control Systems - Discussion
Discussion Forum : Automatic Control Systems - Section 1 (Q.No. 3)
3.
For the system in the given figure. The transfer function C(s)/R(s) is
Answer: Option
Explanation:
Transfer function = (G1 + 1) G2 + 1 = G1 G2 + G2 + 1.
Discussion:
20 comments Page 1 of 2.
Ammar KHAN said:
6 years ago
Use simple block reduction. The first block G1 and the parallel path below that add just after G1 add up to give you G1+1. Replace this portion with G1+1.
Now you also see that there is another parallel path from before G1+1 that adds up after G2. This parallel path will again add up to the gain of the forward path i.e (G1+1)'- G2. So we get (G1+1)'- G2 + 1 = G1G2 + G2 + 1 which is the answer.
Now you also see that there is another parallel path from before G1+1 that adds up after G2. This parallel path will again add up to the gain of the forward path i.e (G1+1)'- G2. So we get (G1+1)'- G2 + 1 = G1G2 + G2 + 1 which is the answer.
Arnab Bhattacharya said:
9 years ago
Using block diagram reduction technique, shift the summing point from the left of G2 to its right. It will place a block of TF G2. Now the lower paths are just summed up to produce (G2+1). The upper path is G1G2. Both the paths are in parallel. So, the ultimate TF is.
C(s) / R(s) = G1G2 + G2 + 1. So, answer (C) is correct.
C(s) / R(s) = G1G2 + G2 + 1. So, answer (C) is correct.
Anurag said:
9 years ago
Suppose the middle summing point is considered T(s).
T(s) = R(s)G1+R(s).
C(s) = C(s)T(s)+G2T(s).
Now replace T(s) with R(s)G1+R(s).
C(s) = C(s)R(s)G1+C(s)R(s)+G2R(s)G1+G2R(s).
C(s)-C(s)R(s)G1-C(s)R(s) = (G2G1+G2)R(s).
C(s)/R(s)-C(s)(G1+1) = G2+G1G2.
C(s)/R(s) = C(s)(1+G1 G2+G1G2.
Please correct me.
T(s) = R(s)G1+R(s).
C(s) = C(s)T(s)+G2T(s).
Now replace T(s) with R(s)G1+R(s).
C(s) = C(s)R(s)G1+C(s)R(s)+G2R(s)G1+G2R(s).
C(s)-C(s)R(s)G1-C(s)R(s) = (G2G1+G2)R(s).
C(s)/R(s)-C(s)(G1+1) = G2+G1G2.
C(s)/R(s) = C(s)(1+G1 G2+G1G2.
Please correct me.
(2)
Sinchana Shetty said:
5 years ago
When the first summing point is shifted beyond G2 the block G2 will come in parallel. As G1 and G2 are in series they will result in G1G2 which are now parallel with G2.
So, far we got G1G2 + G2.
Then unity gain=1 is in parallel with (G1G2+G2) resulting in G1G2+G2+1=C (s)/R (s).
So, far we got G1G2 + G2.
Then unity gain=1 is in parallel with (G1G2+G2) resulting in G1G2+G2+1=C (s)/R (s).
Lordm said:
8 years ago
Answer is (G1G2+G2)/(1-G2). The summer could be split into two summers; the left side being an unity gain signal and signal with gain G2 being added. The right side is positive unity feedback loop. These are then in cascade.
Bitra hanumantharao said:
1 decade ago
Number of forward paths are 2.
Number of loops are 2.
That too no loop are not repeated.
The transfer function of the system is: G1G2/ (1-G1G2-G2).
Number of loops are 2.
That too no loop are not repeated.
The transfer function of the system is: G1G2/ (1-G1G2-G2).
ANISH KUMAR said:
1 decade ago
Answer (G2+G1G2)/(1-G2).
Explanation:
Forward path = 2 (G2 & G1G2).
Loop = 1(G2).
Using Masson formula:
We get (G2+G1G2)/(1-G2);
Explanation:
Forward path = 2 (G2 & G1G2).
Loop = 1(G2).
Using Masson formula:
We get (G2+G1G2)/(1-G2);
(1)
Anish said:
1 decade ago
If anyone knows any answer which is different from my answer please explain it clearly so that everyone will understand.
Hamed said:
9 years ago
Hello, I'm very sorry to the last comment because after searching the answer is correct.
C(s) / R(s) = G1G2 + G2 + 1.
C(s) / R(s) = G1G2 + G2 + 1.
Rajarajan said:
1 decade ago
The above answer is not correct. Number of forward path = 2.
Number of loops = 1.
Answer is [G1G2+G2]/[1-G2].
Number of loops = 1.
Answer is [G1G2+G2]/[1-G2].
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