I am not sure if that is viable way to determine stability. I have linked another excellent video that explains:th-cam.com/video/WBCZBOB3LCA/w-d-xo.html
sorry but you had a mistake when you made the second matrix marked with green it is (8*2-6*0)/2 while it should be only the first with third raw (4*2-6*0)/2=2 1 2 4 0 but the answer is still the same :)
yes ,u will get the same answer but the correct approach is different u will use: 1 2 4 0 for the second product and not: 6 2 8 0 try it for q(s)=[2 5 6 4 3] the two approaches will have a different answer lu chen
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31-8/4=6 Genius 😎
By the way Mam Your Voice❤️😌
correct me if i'm wrong but we can determine the system is stable due to no variable K in the characteristic equation
I am not sure if that is viable way to determine stability. I have linked another excellent video that explains:th-cam.com/video/WBCZBOB3LCA/w-d-xo.html
sorry but you had a mistake when you made the second matrix marked with green
it is (8*2-6*0)/2 while it should be only the first with third raw (4*2-6*0)/2=2
1 2
4 0
but the answer is still the same :)
Hamad, yeah we're both right. Thanks for the check.
hahahah It is Emad
Good luck :)
Sorry I didn't get it. I have the same answer as Raiya's. Can you put you full matrix here?Thanks.
yes ,u will get the same answer but the correct approach is different
u will use:
1 2
4 0
for the second product and not:
6 2
8 0
try it for
q(s)=[2 5 6 4 3]
the two approaches will have a different answer
lu chen
it will be stable because u avoid to multiply the big number to a negative u always put the negative to the small value!
you are wrong 4*2- 1*0/4 must be
8*2 can never be 16.
do a better practice on multiplication.....