@@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)
@@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 ✨✨
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 )
A faster method solve the h.w problem would be to use symmetry actually. if u see it's symmetric about the horizontal, so we can just exchange the end branches and use the feedback formula to solve the block, no need to use SFG and MGF.
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
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 :-)
Observe the right side summing point...we are subtracting two same functions..which means we get zero...this zero is then fed back to the left side summing point...R(s)-0=R(s).... That means our block diagram can simply reduced to.. R(s)-->1/(s+1)-->Y(s)..simply here the transfer function is 1/s+1
Request you to please complete the series. Awaiting R-H criteria, Root locus, bode plot and Nyquist classes soon!
Yes please complete it
C/R= 2G/(1-GH)
Homework prob.
Answer :- Option (B) is correct
(B) = 2*G / 1 - G*H
how bro
4 fwd paths right? I got 2G + 2G^2H / 1 - 2G^2H^2
@@khookengsuen1428 yeah bro I also got the same eq
@@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)
@@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 ✨✨
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 )
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
Sir, please make all control system course videos regularly other wise we will back in this topic
pls complete this series as soon as possible. we are preparing for competative exams.
Nice lecture sir...
10:38 Nice point
Thank you so much sir
A faster method solve the h.w problem would be to use symmetry actually. if u see it's symmetric about the horizontal, so we can just exchange the end branches and use the feedback formula to solve the block, no need to use SFG and MGF.
Sir, is there any video regarding RLC Circuit using Signal Flow graph?
2G/(1-GH)
Thanks a lot 🙏
B 4 forward paths, 1loop, no associated path factor, hence Y/R = 2G/(1-GH)
2 loops are there
thanks Sir🙏❤
Ans opt B for the HW question
can i please
get the explanation for the homework problem sir ?
More please
Option B
2G/(1-GH) is the answer to hw problem sir
Sir can u upload all lecture
We have to wait too much for 1 lecture
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.
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
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 :-)
@@syedsalarjung202 why individual loop is 1 ??
nice question
Don't we have another individual loop
Which is b-c-d-a-b ???
i think also like you
G/1-GH
Its 2G/1-GH
There are four forward paths
yes i got the answer as 0 initially
Is it correct?
Answer to Homework Q :- Option (B) = 2*G / (1 - G*H)
How to solve it by shortcut method?
B-C-D-A also a loop i think sir
No
No
@@ravinchauhan7947 I think so b-c-d-b is also another loop
How can we solve with the method of block diagram reduction ?
Observe the right side summing point...we are subtracting two same functions..which means we get zero...this zero is then fed back to the left side summing point...R(s)-0=R(s)....
That means our block diagram can simply reduced to..
R(s)-->1/(s+1)-->Y(s)..simply here the transfer function is 1/s+1
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
Yes my thoughts exactly!
Complete database management system
B
b-c-d-b is not an Individual loop???
It's b-c-d-a-b
Yes I have this question as well
No
@@annaveia2417 Why?
Because arrow is from b-->d not from d-->b
I did that same mistake too😂
nhi maine direction dekh kr judge kiya
i did that mistake.
Signal flow? More like "I gotta know"...so keep on teachin' me!
sorry
G/1-GH
B