# 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*) isAnswer: Option

Explanation:

Transfer function = (G_{1} + 1) G_{2} + 1 = G_{1} G_{2} + G_{2} + 1.

Discussion:

20 comments Page 1 of 2.
Sinchana Shetty said:
4 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).

Yogendra said:
5 years ago

No loop is there, 3 forward path is g2 and g1g2 and 1 only.

Ammar KHAN said:
5 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.

Lordm said:
7 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.

Anurag said:
7 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.

(1)

Hamed said:
8 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.

Hamed said:
8 years ago

Not correct.

The answer is [(G1 + 1) G2]/[1 - G2].

The answer is [(G1 + 1) G2]/[1 - G2].

Arnab Bhattacharya said:
8 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.

Khetendra Vashishtha said:
8 years ago

Forward path = 2 (G2 & G1G2).

Loop = 1(G2).

Using Masons formula:

then TF = (G2 + G1.G2)/(1 - G2).

Loop = 1(G2).

Using Masons formula:

then TF = (G2 + G1.G2)/(1 - G2).

(1)

Ayushri said:
8 years ago

Answer is wrong.

Answer: (G1G2/((1+G1)(1+G2))).

Answer: (G1G2/((1+G1)(1+G2))).

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