### Discussion :: Automatic Control Systems - Section 1 (Q.No.3)

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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) | |

Answer (G2+G1G2)/(1-G2). Explanation: 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) | |

@Anish. 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. Answer: (G1G2/((1+G1)(1+G2))). |

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) | |

Not correct. 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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