Thanks for your adviceHi,
Your 1/G4 idea looks good, but your G2+G5 does not look right because the input to G5 is not what you changed it to, the input to G5 comes directly from the input in the original flow chart.
If you wanted to put G5 there you would have to find a way to invert everything before so you get the equivalent of the direct input.
Thanks for your response and i think i finally understand my mistakes i really apreciate itI'm not sure you're right on try #2 either.
In your first try, your initial reduction is correct, though the step with 1/G4 wasn't really needed, I would have just duplicated G3, its less confusing, since that's what you actually did in the following step.
Where you went wrong however was in step 2 when you pulled G5 into the G2 block. G5 is a gain on the original input but G2 is a gain on the output of the second summation so you'd need to divide G5 by the overall function across the two summations, which you don't yet know.
Then I think you tried the same trick again in step 4 where you pulled -G3G4 into the G2 block. But again G2 and -G3G4 are acting on different values.
Here's my take on this... its hidden, just highlight it to reveal all... once you've tried again
The output of the second summation x2 = (G1 + G3) . x1, where x1 is the output of the first summation. Therefore
x1 = V - x2
= V - (G1 + G3) . x1
x1 = V / (1 + G1 + G3)
x2 = V . (G1 + G3)/(1 + G1 + G3)
Y = x2 . G2 - x1 . G3 . G4 + G5 . V
= V . (G2 . (G1 + G3)/(1 + G1 + G3) - (G3 . G4)/(1 + G1 + G3) + G5 )
= V . ((G2 . (G1 + G3) - G3 . G4)/(1 + G1 + G3) + G5)
What kind of mistake? Because I already run out of ideas I think I tried everything I could but still can't make it to work I just strted to losse hope that I will solve this in near future because I really dont't know what else I can do.Sadly no... but closer..
I think you made a mistake in Step 3 with your treatment of the new G3G4 block...
Hello again,What kind of mistake? Because I already run out of ideas I think I tried everything I could but still can't make it to work I just strted to losse hope that I will solve this in near future because I really dont't know what else I can do.
So I tried your method and made some calculations and made few changes, and know If I hadn't made any mistakes in my calculations the output should be 20.6 so approximately 21 which should be correct.Hello again,
Yes it looks like there is still a mistake. What you need is a sanity test to get an idea if your result is correct or not, that way you can test yourself with a future problem if you need to also.
Note that if you multiply your result out you get a G3^2 factor which is probably indicative that something is wrong right away.
The attachment shows a test of the technique i used to solve it and as you can see if you start at ANY node in the circuit using the values i assigned to G1 to G5 and a=5, you always get the correct result for the node that depends on any of the other nodes. For example, e=4 and G2=2 so node g MUST come out to 4*2=8. If it doesnt, then something is wrong, but really all the nodes must calculate out to the correct values when we assign arbitrary gains to G1 to G5 and an arbitrary input in this case a=5.
The method i used and you can use to test your results is algebraic block diagram reduction. A block diagram like this one represents a set of simultaneous equations that once solved, gives you the solutions to every node not just the output. You can then test every node in the manner shown above.
To start, we label each node in just about any order, i used 'a' through 'i' to label each node.
Then we start writing one equation for each node. The first one i wrote was:
and the second one could be:
but you can do them in any order as long as you include them all.
Once you have one equation for each node (except the input) you can then solve them simultaneously and you get a solution for each and every node.
In the end, assign values for the gains and calculate the numerical values of the nodes and see if they all work out to the correct values. To be more sure, use unusual numbers for the gains like 1.123, 4.654, etc. I used 1,2,3,4,5 which is not the best choice but i wanted it to be simple to illustrate. You might use prime numbers if you like. Even better would be to do two tests, one with arbitrary numbers then a second test with those same numbers all incremented by 1, and both tests must come out correct of course. So if we used 2,3,5,7,11 then the second test would use 3,4,6,8,12, or maybe just the next set of primes 3,5,7,11,13.
Hi,So I tried your method and made some calculations and made few changes, and know If I hadn't made any mistakes in my calculations the output should be 20.6 so approximately 21 which should be correct.
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by Steve Arar
by Steve Arar
by Luke James