I just went through the problem again starting with moving the right-most summing junction to the right of G4 putting a G4 into the upper right branch. I got the same answer as in the video. The steps I followed were as follows: • move right most summing junction to right of G4 and make upper right branch contain G4 • multiply original G4 with G3 since they are now in series • move takeoff point to right of G2 upstream similar to how I started in the video which makes the upper right branch G2G4 • combine G1 and original G2 in parallel to get G1-G2 • move summing junction to the left of G3G4 downstream of G3G4 making the bottom branch G3G4G5 • multiply (G1-G2) with G3G4 since they are in series • combine left side of diagram in parallel to get (G2G4 + G1G3G4 - G2G3G4) • use feedback rule on right side of diagram (before G6) to get 1/(1-G3G4G5) • multiply together remaining three blocks in parallel to get final transfer function I hope this helps. There are many different ways you can do this kind of problem, but they should all get you to the same answer
The rules are shown on a sheet of paper I use throughout the video. Please pause the video and copy the rules down if you would like to have them for your notes
the most easiest explanation I've heard on anything till date, . thanks mate
Best with clear details, thanks sir, I'm seeing your lecture from Bangladesh 🇧🇩❤️
Very good explanations!
you are holding the pen like you were suppose to be left handed
great video, thank you so much for making this!
Very well explained
Very nice. Thank you!
That sheet you have is so so good
Well explained sir
Спасибо! ❤❤❤❤
Thanks for your video mate
that rules paper helps a lot
Well done pro
perfect
Bro where are you from?
I don't get the same answer when I move the junction between G3 and G4 in front of G4 , Why?
I just went through the problem again starting with moving the right-most summing junction to the right of G4 putting a G4 into the upper right branch. I got the same answer as in the video. The steps I followed were as follows:
• move right most summing junction to right of G4 and make upper right branch contain G4
• multiply original G4 with G3 since they are now in series
• move takeoff point to right of G2 upstream similar to how I started in the video which makes the upper right branch G2G4
• combine G1 and original G2 in parallel to get G1-G2
• move summing junction to the left of G3G4 downstream of G3G4 making the bottom branch G3G4G5
• multiply (G1-G2) with G3G4 since they are in series
• combine left side of diagram in parallel to get (G2G4 + G1G3G4 - G2G3G4)
• use feedback rule on right side of diagram (before G6) to get 1/(1-G3G4G5)
• multiply together remaining three blocks in parallel to get final transfer function
I hope this helps. There are many different ways you can do this kind of problem, but they should all get you to the same answer
thx
Perfect
Can I get the rules please
The rules are shown on a sheet of paper I use throughout the video. Please pause the video and copy the rules down if you would like to have them for your notes
Can someone one take the junction downstream and get the same answer?...
Yes! There are many ways to solve this kind of problem, and I just showed one
I didn't moving g2 after g4 and not before g3. I got a different answer
godsent
Use masons gain formula! It always works
❤
+rep
the G5 is not a feedback, just evaluate it by parallel lol
Yes it is lol, look at the direction of the arrows
How the hell are you holding that pen!!🫥
😂😂