Root Locus Special Points - Break in and Break away

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@asdfasdf6587 7 жыл бұрын

This video has an error, the when he set dk/ds=0 and solved for the roots of s, those are the locations of the breakin/away points. When he subbed the values back in to the eqn and solved for k, those are the corresponding gains of those points. when he solved s=+/-2 THAT is the time were you look on the root locus plot and see if they both actually lie on the root locus, in this case only -2 does. So he was correct in subbing back -2 to find the gain of that point. However, when he switched k from +3/2 to -3/2, that is incorrect, the answer is k= +3/2 for that point as the equation predicted. K only goes from 0 to infinite for points on the root locus, so he is clearly incorrect. If he were to sub in s=2 back in and solve for k, it would give a negative k value, which would imply that the s being solved for IS NOT on the root locus plot, as predicted when you originally solved for the roots of dk/ds=0 and compared them to the graph. Hope this clears things up for ppl.

6 жыл бұрын

Exactly ! I was gonna punch the monitor ...

@aliberkdemir7080 6 жыл бұрын

Let this comment rise in the ranks. You are right sir.

@amierulasyraf3956 4 жыл бұрын

i'm glad, i read this comment 1st

@MrMattsquatch 10 жыл бұрын

A 7:38, should it be: -1/(4sigma) + 1/2 - 1/sigma?

@TheSalvagehamilcar 7 жыл бұрын

great explanation of root locus! when you solve for K=-3/2 is that the break away point? or is that the gain associated with sigma=-2 which is the break away point. I was a little unclear because i thought that sigma was the real axis and since we found the real axis value to be -2 this is the break away point. K would just be the gain associated with that sigma found. Not 100 percent sure, appreciate if you would clarify thanks you.