John, thank you so much. I was absolutely struggling with this topic. I watched a dozen videos before finding yours. The multiple examples at the end made it click!
At 7:50 you say, its clear that w=sqrt(2) at -90°. How do you see that so fast without calculating? Thank you very much for the videos, they are very helpful. Greetings from Germany.
+keenwasp You can show that atan(w)+atan(w/2)=90 if (w*w/2 = 1). Use a right angle triangle and this should be obvious (2 other angles add to 90).Anthony
At 4:09, where did you get the 5 and the 20 when you calculated the magnitude of the gain? The denominator is s^2+5s+4 so the magnitude would be 1/sqrt((s^2+4)^2+(5s)^2))? Nice videos by the way, I wished I watched these before my midterm.
You're not consistent in your labeling of quadrants. In the previous video (nyquist-2) at 2:26 you label quadrant 2 as being -180 to -270 whereas here at 16:32 you label quadrant 2 to be 0 to -90 whereas 0 to -90 is actually quadrant 4.
John, thank you so much. I was absolutely struggling with this topic. I watched a dozen videos before finding yours. The multiple examples at the end made it click!
You're just amazing professor. Many thanks from Br!
I can't thank you enough. You don't know how much time and frustration you saved me. Thank you soooo much
Thank you very much, I wasn't exactly sure how to sketch these diagrams before checking your video. Great job !
Your videos make things easier and clear.... Thanks.
Superb description
Thanks, really good and clear explanation!
Helped me a lot, thanks!
You are just amazing/; very clear teaching, nice diction. Many thanks from BR|
it's 3 hours prior to my final, thank you
Thank you for help.
First of all, thank you very much for the video. I'd like to ask you where do you get the -26 degrees at 12:37.
atan(1/2)
At 7:50 you say, its clear that w=sqrt(2) at -90°.
How do you see that so fast without calculating?
Thank you very much for the videos, they are very helpful. Greetings from Germany.
+keenwasp You can show that atan(w)+atan(w/2)=90 if (w*w/2 = 1). Use a right angle triangle and this should be obvious (2 other angles add to 90).Anthony
Hii is it correct that for a closed loop it should be GH/(1+GH) instead of G/(1+GH)?
At 4:09, where did you get the 5 and the 20 when you calculated the magnitude of the gain? The denominator is s^2+5s+4 so the magnitude would be 1/sqrt((s^2+4)^2+(5s)^2))?
Nice videos by the way, I wished I watched these before my midterm.
+smyleylily Put jw=j2 which is the geometric mean of the 2 corner frequencies (1 and 4). Then |jw+1|=sqrt(5) and |jw+4|=sqrt(20)Anthony
I'm assuming when you say Quad2 later in the video you mean Quad4.
Jee Charns thats at 16:38 and i think you are right
At 18:80 , you said all functions start at 1.5 (3/2).
But I think the third function starts at 1.7 (5/3).?
Can you check this?
Thanks
+keenwasp Sorry not sure where you get this from. 0.3*10/2 = 1.5
How do you know that the phase equals 90 degrees when w=2, without finding the arctan of Im{G(jw)} / Re{G(jw) } ?
*****
I think I got it:
We need to solve the equation: Re{denominator} = 0 and then get the "w" which sets the whole Phase of G(jw) to be -90 degrees.
You're not consistent in your labeling of quadrants. In the previous video (nyquist-2) at 2:26 you label quadrant 2 as being -180 to -270 whereas here at 16:32 you label quadrant 2 to be 0 to -90 whereas 0 to -90 is actually quadrant 4.