Discussion :: Automatic Control Systems - Section 1 (Q.No.3)
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For the system in the given figure. The transfer function C(s)/R(s) is
Answer: Option C
Transfer function = (G1 + 1) G2 + 1 = G1 G2 + G2 + 1.
|Beu said: (Apr 25, 2014)|
|No. of fwd path = 3.
No. of loop = 0.
|Rithika said: (Jun 27, 2014)|
|Can anyone please solve it briefly ?|
|Bitra Hanumantharao said: (Jul 9, 2014)|
|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).
|Rajarajan said: (Oct 31, 2014)|
|The above answer is not correct. Number of forward path = 2.
Number of loops = 1.
Answer is [G1G2+G2]/[1-G2].
|Roshan said: (May 5, 2015)|
|There is only one loop.
Number of +ve loop = 1.
So, (G1G2 + G2) / (1 - G2).
|Anish Kumar said: (May 27, 2015)|
Forward path = 2 (G2 & G1G2).
Loop = 1(G2).
Using Masson formula:
We get (G2+G1G2)/(1-G2);
|Anish said: (May 27, 2015)|
|If anyone knows any answer which is different from my answer please explain it clearly so that everyone will understand.|
|R C Pandey said: (Jun 9, 2015)|
You are calculating TF between C and are. So there is only one forward path and that is G1G2.
|Praveen said: (Jan 10, 2016)|
|Can anyone say the correct answer please as soon as possible?|
|Fizza said: (Jan 10, 2016)|
|What is the difference between transfer function and characteristic equation?|
|Ayushri said: (Jan 18, 2016)|
|Answer is wrong.
|Khetendra Vashishtha said: (Jun 10, 2016)|
|Forward path = 2 (G2 & G1G2).
Loop = 1(G2).
Using Masons formula:
then TF = (G2 + G1.G2)/(1 - G2).
|Arnab Bhattacharya said: (Jun 27, 2016)|
|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.
|Hamed said: (Jul 24, 2016)|
The answer is [(G1 + 1) G2]/[1 - G2].
|Hamed said: (Aug 2, 2016)|
|Hello, I'm very sorry to the last comment because after searching the answer is correct.
C(s) / R(s) = G1G2 + G2 + 1.
|Anurag said: (Feb 13, 2017)|
|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.
|Lordm said: (Sep 17, 2017)|
|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.|
|Ammar Khan said: (Jun 10, 2019)|
|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.
|Yogendra said: (Jul 6, 2019)|
|No loop is there, 3 forward path is g2 and g1g2 and 1 only.|
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