Signal Flow Graphs (Solved Problem 3)

Control Systems: Signal Flow Graphs (Solved Problem 3)
Topics discussed:
1. Solved GATE Problem (EC 2010, IIT Guwahati) based on calculation of overall gain of a Block Diagram by using Mason’s Gain Rule.
2. Homework problem (GATE EE 1992, IIT Delhi) based on calculation of overall gain of a Block Diagram by using Mason’s Gain Rule.
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#ControlSystemByNeso #ControlSystems #MasonsGainFormula #SignalFlowGraphs

[@shraddharaju4862]

Request you to please complete the series. Awaiting R-H criteria, Root locus, bode plot and Nyquist classes soon!

[@sanjusingha1714]

Yes please complete it

[@Rising_sun_408]

C/R= 2G/(1-GH)

[@sameermeshram3062]

Homework prob.
Answer :- Option (B) is correct
(B) = 2*G / 1 - G*H

[@khookengsuen1428]

how bro

[@khookengsuen1428]

4 fwd paths right? I got 2G + 2G^2H / 1 - 2G^2H^2

[@ginnasomeshwar5068]

@@khookengsuen1428 yeah bro I also got the same eq

[@mohamedaitelcaid5969]

@@khookengsuen1428
Hey bro!
Here's the final step!
2[G+(G^2)H]/ [1-(G^2)(H^2)]
=
2G(1+GH)/[(1-GH)(1+GH)]
1+GH gets canceled out!
The final value is 2G/(1-GH)

[@khookengsuen1428]

@@mohamedaitelcaid5969 oh I see now, there is only one loop, thus 1-G²H², I initially thought there are two loops but it is actually the same loop 🤦 good tricky question, thanks for the answer btw ✨✨

[@mayanksoni2399]

for the homework problem, if we swap the take-off points for both H, there will be no change in the situation.
then this gives us two positive feedbacks, hence option B ( 2 * positive feedback gain )

[@mayanksoni2399]

Also for discussed problem, if we shift node b to node c, we get a feedback with zero gain. so our system reduces to open loop with gain of 1/1+s

[@Knowledge_oriented]

Sir, please make all control system course videos regularly other wise we will back in this topic

[@mounikareddymgucet6648]

pls complete this series as soon as possible. we are preparing for competative exams.

[@sameermeshram3062]

Nice lecture sir...

[@samikshabagde20]

Sir, is there any video regarding RLC Circuit using Signal Flow graph?

[@user-haritha22]

Thank you so much sir

[@chandruloki6900]

can i please
get the explanation for the homework problem sir ?

[@162-gcet-tahamehraj5]

More please

[@srikanthnanda5723]

B 4 forward paths, 1loop, no associated path factor, hence Y/R = 2G/(1-GH)

[@khalidismailshaikh2661]

2 loops are there

[@LIVE_LIVELY]

10:38 Nice point

[@vivek4604]

Thanks a lot 🙏

[@shubhanshugupta9754]

Sir can u upload all lecture
We have to wait too much for 1 lecture

[@PunmasterSTP]

Maybe it takes him so long to upload because he makes the quality so good? I'm also thinking that he has lots of other stuff to do.

[@verissakissiedu1302]

Hello sir, thank you for your wonderful lectures. Please I want you to help me know the answer to the assignment, I had 2G/1-GHGH. Which is close to 2G/1-GH that is answer b. I'm thinking it should be 2G/ 1-G^2.H^2

[@syedsalarjung202]

I am afraid u r wrong . Becuz u have to consider the delta k for the four forward parts. By isolating the for forward paths one by one, you see that there are no isolated loops. So u obtain delta 1, delta 2 and delta 3 as "1".
No. Of forward paths 4
No. of individual loops 1
Therefore U/Y = [(G)(1)+(G)(1)+(G^2 H)(1)+(G^2 H)(1)] / 1-G^2 H^2
Solving it further u get the 2nd option.
Hope this answered ur question :⁠-⁠)

[@Ijlalwasiq-tg7gk]

​@@syedsalarjung202 why individual loop is 1 ??

[@sanskarkumar6484]

thanks Sir🙏❤

[@praneethkumarreddy7674]

Option B

[@rishav.2305]

2G/(1-GH)

[@g-fourgaming8657]

Don't we have another individual loop
Which is b-c-d-a-b ???

[@malekzitouni1779]

i think also like you

[@praneshsrikanth2364]

2G/(1-GH) is the answer to hw problem sir

[@fxcapes]

Is it correct?
Answer to Homework Q :- Option (B) = 2*G / (1 - G*H)

[@HemanthMaddila-n8m]

G/1-GH

[@soumyadeeppradhan]

nice question

[@tejassrivastavaK_EE_]

Ans opt B for the HW question

[@abd5184]

How to solve it by shortcut method?

[@vepaguntaps8722]

B

[@ravi-321]

B-C-D-A also a loop i think sir

[@ravinchauhan7947]

No

[@annaveia2417]

No

[@rizwanakhan0610]

@@ravinchauhan7947 I think so b-c-d-b is also another loop

[@vivek4584]

Shift Node b to right,so that bottom T.F becomes 1 ( Node d summer adds same Y(s) but with different polarity, do feed back =0). Only left TF is 1/s+1

[@PunmasterSTP]

Yes my thoughts exactly!

[@63_tanishkgupta55]

yes i got the answer as 0 initially

[@gsivarani6747]

Complete database management system

[@agent_spidey]

b-c-d-b is not an Individual loop???

[@g-fourgaming8657]

It's b-c-d-a-b
Yes I have this question as well

[@annaveia2417]

No

[@rizwanakhan0610]

@@annaveia2417 Why?

[@SachinKumar-bb1iy]

Because arrow is from b-->d not from d-->b

[@PunmasterSTP]

Signal flow? More like "I gotta know"...so keep on teachin' me!

[@saniakouser3832]

nhi maine direction dekh kr judge kiya

[@adityaaryam]

i did that mistake.

[@1hourmusic855]

sorry

[@HemanthMaddila-n8m]

G/1-GH

[@arindamdutta2797]

B