The first thing to say is that the final bit of the video is certainly the least important. So it certainly doesn't matter if that bit isn't too clear to you - you understand all the important parts of the circle criterion! But to fill in the details, at the end we are just trying to show that for the transfer function G(s)=1/(s^2+s+1), the Nyquist plot does not intersect or encircle the circle that passes through the -2 and -3/2 point, since this is what we would need to do to prove stability using the circle criterion. The best way to do this is to just plot the Nyquist diagram, and look at it! In practice this is what I would recommend that you do. But suppose that we don't have a computer to make the plot or we can't be bothered to make the Nyquist plot by hand - can we prove this algebraically? Because G(s) is so simple, and actually misses the circle by quite a lot, this is easy to do. One way that is sufficient is to show that the Nyquist plot of G(s) lies inside a circle centred on the origin with radius less that 3/2. After all, if this is the case, the Nyquist plot cannot intersect the other circle, and cannot encircle it. The rest of the video is essentially just doing the algebra to show that the Nyquist plot does in fact lie within this circle. In fact, the calculation shows that the smallest circle that contains the Nyquist has a radius 4/3 (this is what gamma is). Note though, this method of finding the smallest circle containing the Nyquist plot is only a sufficient condition for showing that the circle criterion holds. It gave a convenient way to check things algebraically that worked for this simple G(s), but it will not work every time. You can probably think of much better ways to capture the actual circle criterion requirement algebraically. In fact, if G(s) is the ratio of two polynomials, you can always turn the circle criterion requirement into an equivalent algebraic requirement (not just a sufficient one). If you want a challenging problem, try to figure this out!
sir thank you. that was a perfect explanation.
Happy to hear!
Thank you for your excellent presentation. I only do not understand the gamma inequality like 2/sqrt(3) and 1.5 .
The first thing to say is that the final bit of the video is certainly the least important. So it certainly doesn't matter if that bit isn't too clear to you - you understand all the important parts of the circle criterion! But to fill in the details, at the end we are just trying to show that for the transfer function G(s)=1/(s^2+s+1), the Nyquist plot does not intersect or encircle the circle that passes through the -2 and -3/2 point, since this is what we would need to do to prove stability using the circle criterion. The best way to do this is to just plot the Nyquist diagram, and look at it! In practice this is what I would recommend that you do. But suppose that we don't have a computer to make the plot or we can't be bothered to make the Nyquist plot by hand - can we prove this algebraically? Because G(s) is so simple, and actually misses the circle by quite a lot, this is easy to do. One way that is sufficient is to show that the Nyquist plot of G(s) lies inside a circle centred on the origin with radius less that 3/2. After all, if this is the case, the Nyquist plot cannot intersect the other circle, and cannot encircle it. The rest of the video is essentially just doing the algebra to show that the Nyquist plot does in fact lie within this circle. In fact, the calculation shows that the smallest circle that contains the Nyquist has a radius 4/3 (this is what gamma is).
Note though, this method of finding the smallest circle containing the Nyquist plot is only a sufficient condition for showing that the circle criterion holds. It gave a convenient way to check things algebraically that worked for this simple G(s), but it will not work every time. You can probably think of much better ways to capture the actual circle criterion requirement algebraically. In fact, if G(s) is the ratio of two polynomials, you can always turn the circle criterion requirement into an equivalent algebraic requirement (not just a sufficient one). If you want a challenging problem, try to figure this out!
Nyquist plot should start at -1 and and in 0.