Thanks for making and sharing another great video! I and some of the other commenters were curious about why you disregarded the lower branch for the first part of the problem (finding C1/R1) but you didn't disregard the upper branch for the second part (finding C2/R2). I understand the reasoning that 3:45 that because nothing from R1 can get to the lower branch, you can safely disregard it. However, for the second part, since nothing can get from R2 to the upper branch *and then come back down*, I think we can safely disregard the upper branch. Would you agree?
at 2:12, it was calculating the transfer function of C1/R1 keeping R2,C2 equal to zero; and at 10:08, it was calculating the transfer function of C2/R2 keeping R1,C1 equal to zero. Based on the direction of arrows, it can only go from R2 to R1 but not from R1 to R2, that's why we should consider the loop of R1,C1 when calculating TF of C2,R2 coment form @bingxuli2368
at 2:12 you have written the possible transfer function for c2/R2 keeping R1 and c1 equal to zero. But at 10:08 why you are counting the loop of R1 and C1 for finding transfer function of C2/R2?
at 2:12, it was calculating the transfer function of C1/R1 keeping R2,C2 equal to zero; and at 10:08, it was calculating the transfer function of C2/R2 keeping R1,C1 equal to zero. Based on the direction of arrows, it can only go from R2 to R1 but not from R1 to R2, that's why we should consider the loop of R1,C1 when calculating TF of C2,R2.
IN part 1 of the question why didn't you consider the lower part of the graph while considering the isolated loops. Because in part 2 you considered the loops in the upper part of the graph. This confused me a lot. COuld you please explain this again ?
I was confused at that part too, the part where the upper loops were also considered for C2/R2 Why is it still considered if the signal won't even return to the forward path R2 - p - q - r -C2 (logically speaking)
In the first part, you will get the same same answer which is 2/5 either considering the lower part or removing it, try it by yourself! He said that the lower part is redundant, removing it will save us time and effort! If you keep it then C1/R1=(2*3)/15=(2/5), if you remove it, you will get C1/R1 = 2/5 directly with less time and calculations.
@@mohamedaitelcaid5969 but in second part, its the same thing. if we do not consider upper part then still the answer is 2/3. then why do we have to consider upper part this time?
@@mayanksoni2399 I was confused on the second part as well. My best guesses are that he either didn't realize considering the upper part of the network was unnecessary, or he wanted to demonstrate the algorithm more. Perhaps he made and/or edited the video a bit out of order as well. Either way, I think we can safely argue that causality would show that the upper part would have no effect on C2, and so we can disregard it entirely.
Yeah I and others were confused about that as well. My best guess is that he either didn't realize that branches could be discarded in other parts of the question, or he just wanted to demonstrate the algorithm more. Perhaps he also made or edited the video out of order as well...
3:49 If you eliminate on the basis of signal flow from upper to lower then, one must consider the contribution of signal from lower to upper also. This will affect the calculation of associated co factor of forward path (∆¹) & determinant of S.G.F. (∆)!. I Strongly recommend this is the wrong way of calculating the Transfer Function. ❌
You guys are doing a great job 👏
VERY VERY IMPORTANT QUESTION SIR
REALLY IT WAS VERY USEFUL SIR
THANK YOU SO MUCH FOR YOUR EFFORT SIR🙏🙏🙏🙏🙏
Amazing lecture, beautifully explained
Can you upload lectures regarding different forms of signal flow graphs based on IIR and FIR systems? This would be a great help.
That is under different subject which is digital signal processing
That will not be covered here.. Are you from India or somewhere else?
@@princechoudhary5772 Yes i know that.
Thanks for making and sharing another great video!
I and some of the other commenters were curious about why you disregarded the lower branch for the first part of the problem (finding C1/R1) but you didn't disregard the upper branch for the second part (finding C2/R2). I understand the reasoning that 3:45 that because nothing from R1 can get to the lower branch, you can safely disregard it. However, for the second part, since nothing can get from R2 to the upper branch *and then come back down*, I think we can safely disregard the upper branch. Would you agree?
at 2:12, it was calculating the transfer function of C1/R1 keeping R2,C2 equal to zero; and at 10:08, it was calculating the transfer function of C2/R2 keeping R1,C1 equal to zero. Based on the direction of arrows, it can only go from R2 to R1 but not from R1 to R2, that's why we should consider the loop of R1,C1 when calculating TF of C2,R2 coment form @bingxuli2368
Can u pls upload a lecture regarding "mathematical models"-differential equations of translational and rotational mechanical systems
Thanks sir for this work 😀😃
C2/R2 = 2 / (1+2)
Superb teaching sir.... please complete the syllabus sir
Bhai , control engineering portion is completed or not
I mean only 60 videos are there... I don't know that's why I'm asking
at 2:12 you have written the possible transfer function for c2/R2 keeping R1 and c1 equal to zero. But at 10:08 why you are counting the loop of R1 and C1 for finding transfer function of C2/R2?
at 2:12, it was calculating the transfer function of C1/R1 keeping R2,C2 equal to zero; and at 10:08, it was calculating the transfer function of C2/R2 keeping R1,C1 equal to zero. Based on the direction of arrows, it can only go from R2 to R1 but not from R1 to R2, that's why we should consider the loop of R1,C1 when calculating TF of C2,R2.
@@bingxuli2368 Thank u bro
keep it up sir
Thank you very much :)
thanks Sir🙏❤
IN part 1 of the question why didn't you consider the lower part of the graph while considering the isolated loops. Because in part 2 you considered the loops in the upper part of the graph. This confused me a lot. COuld you please explain this again ?
I was confused at that part too, the part where the upper loops were also considered for C2/R2
Why is it still considered if the signal won't even return to the forward path R2 - p - q - r -C2 (logically speaking)
In the first part, you will get the same same answer which is 2/5 either considering the lower part or removing it, try it by yourself!
He said that the lower part is redundant, removing it will save us time and effort!
If you keep it then C1/R1=(2*3)/15=(2/5), if you remove it, you will get C1/R1 = 2/5 directly with less time and calculations.
@@mohamedaitelcaid5969 but in second part, its the same thing. if we do not consider upper part then still the answer is 2/3. then why do we have to consider upper part this time?
Also in last part why are we considering loop in bottom part. I applied the mason formula without considering pqrp loop and got the same result.
@@mayanksoni2399 I was confused on the second part as well. My best guesses are that he either didn't realize considering the upper part of the network was unnecessary, or he wanted to demonstrate the algorithm more. Perhaps he made and/or edited the video a bit out of order as well. Either way, I think we can safely argue that causality would show that the upper part would have no effect on C2, and so we can disregard it entirely.
Thank you so much sir
u choosed all 3 loops in every output/input beets of Question but why u discarded pqrp loop in the 1st beet of Question 🙋?
Yeah I and others were confused about that as well. My best guess is that he either didn't realize that branches could be discarded in other parts of the question, or he just wanted to demonstrate the algorithm more. Perhaps he also made or edited the video out of order as well...
pls complete the syllabus soon.
3:49 If you eliminate on the basis of signal flow from upper to lower then, one must consider the contribution of signal from lower to upper also. This will affect the calculation of associated co factor of forward path (∆¹) & determinant of S.G.F. (∆)!. I Strongly recommend this is the wrong way of calculating the Transfer Function. ❌
Sir i completed solved the problem answer write or worng how to find that sir
sir why don't we consider the loops x-y-z-y-x
We don't have to repeat nodes
conceptual problem