Callisto
  • Callisto
How can we know if the system is oscillating and if it is decaying with only a quadratic equation? PS: <1> Do not involve any calculus <2> Haven't learnt damping.
Engineering
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Callisto
  • Callisto
@ash2326
ash2326
  • ash2326
First of all find the closed loop transfer function of the system
Callisto
  • Callisto
Suppose it is \[Y= \frac{1}{1-kz^{-1} + bz^{-2}}\]where k is an unknown constant and b is a known constant.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Callisto
  • Callisto
I meant \[\frac{Y}{X}=...\]
Callisto
  • Callisto
Solving the denominator =0, \[1-kz^{-1}+bz^{-2}=0\]\[z^2 - kz + b =0\]\[z = \frac{k \pm \sqrt{k^2 - 4b}}{2}\]
ash2326
  • ash2326
You'd need to find the poles of the system, sorry I was dealing in the s- domain
Callisto
  • Callisto
Poles are at \[\frac{k\pm\sqrt{k^2-4b}}{2}\]
ash2326
  • ash2326
There will be many cases, \[ K=2\sqrt b, K>2\sqrt b\ and\ K<2\sqrt b\] for first case, \[z= \sqrt b\] if b<1 then system will decay if b=1 system will be constant if b>1 then system will be unbounded.
ash2326
  • ash2326
@Callisto I found a good document on this, please see it. You'll understand better http://www.eng.ox.ac.uk/~conmrc/dcs/dcs-lec4.pdf
ash2326
  • ash2326
Let me know if you have doubt anywhere
Callisto
  • Callisto
Now, I see why I can never get the answer. The reason why I have been stressing that no calculus is involved is because when we learnt this topic, our lecturer didn't not teach us using calculus, i.e. solving D.E., using expressions in exponential forms, nor mentioning those fancy terms like damping. Anyway, thanks for trying to help!
ash2326
  • ash2326
so you undestood now ?
Callisto
  • Callisto
No.
ash2326
  • ash2326
oops, did you ask this doubt to your teacher?
Callisto
  • Callisto
Ha! I don't even have to ask as in the lecture, he has written "you will find out the reason if you take EEE"
Callisto
  • Callisto
*in the lecture notes
ash2326
  • ash2326
umm, where did calculus was used?
Callisto
  • Callisto
He has NEVER used calculus in this course.
ash2326
  • ash2326
But the solution requires just solving the quadratics. then depending on the poles, we classify the system
Callisto
  • Callisto
p = pole p = 1 => remains (unchanged) |p| >1 => diverge |p| < 1 => converge
ash2326
  • ash2326
now you need to check which one lies in, out or on the unit circle. then you can classify the system
ash2326
  • ash2326
@Callisto are you here?
Callisto
  • Callisto
checking the magnitude? I did it. But the problem is how I can identify if the system oscillates. Sorry, I was on other page.
ash2326
  • ash2326
if the poles are on the unit circle, system will oscillate if they are inside, it'll decay if they are outside, oscillations will grow unboundedly
ash2326
  • ash2326
???
Callisto
  • Callisto
Hmm... I think the magnitude of the pole only tell us if the system is converging/diverging/remaining unchanged?!
Callisto
  • Callisto
I think I understand how to analyze the system now, thanks :)
ash2326
  • ash2326
welcome :P

Looking for something else?

Not the answer you are looking for? Search for more explanations.