@@drpeyam It's Heaviside cover-up method!!! e.g: if I have 1/(s(s+1)(s+2)) and want to use PF to separate it into A/s + B/(s+1) + C/(s+2), we say that A = 1/((s+1)(s+2)) @ s = 0, thus A = 1/2 B = 1/(s(s+2)) @ s + 1 = 0 (or s = -1), thus B = -1 C = 1/(s(s+1)) @ s + 2 = 0 (or s = -2), thus C = 1/2 it's a quicker way to find A, B and C without having to go through finding a common denominator and doing a comparison... etc. N.B: it's quickest when the fraction is in terms of s only (like this example) or s² only (in the video), but not a mix of both. It's a great hack for the students :D... well.. with limitations to its speed of course :D
Dr. Peyam Thank you for covering this! I didn’t learn this topic in university but have been interested in learning about them. Much appreciated as always 😊
Have you tried using modulation theorem?-I guess in mathematician jargon this will be a multiplication by an exponential in time domain vs a frequency shift in s-domain or Frequency Domain respectivel. Just passed by my mind… that’s all I should try and see too for the middle term (the φ may need to be converted to an exponential)..looks has if it’s unnecessary..but I just mentioned it with a possibility of finding a new way so…
At 5:55 you could do the cover method by treating the s^2 as x for example
What’s the cover method?
@@drpeyam
It's Heaviside cover-up method!!!
e.g: if I have 1/(s(s+1)(s+2)) and want to use PF to separate it into A/s + B/(s+1) + C/(s+2), we say that
A = 1/((s+1)(s+2)) @ s = 0, thus A = 1/2
B = 1/(s(s+2)) @ s + 1 = 0 (or s = -1), thus B = -1
C = 1/(s(s+1)) @ s + 2 = 0 (or s = -2), thus C = 1/2
it's a quicker way to find A, B and C without having to go through finding a common denominator and doing a comparison... etc.
N.B: it's quickest when the fraction is in terms of s only (like this example) or s² only (in the video), but not a mix of both.
It's a great hack for the students :D... well.. with limitations to its speed of course :D
Yeah it is very powerful. Nice explanation BTW@@ELKCreativist
If one does not know Laplace transform one probably would be lost. With L-transforms it is rather straighforward.
What are L-transforms?
That's only he short form for Laplace-Transforms
@@renesperb I see. The reply of the other person left me confused.
Dr. Peyam Thank you for covering this! I didn’t learn this topic in university but have been interested in learning about them. Much appreciated as always 😊
Glad it was helpful!
Mistake in the video: You didn't make your usual intro.
Is there a way to transform a problem like this into a differential equation?
Just differentiate it.
Twice
@@vascomanteigas9433 Why twice? The unknown function is convolved with a degree 1 polynomial.
Have you tried using modulation theorem?-I guess in mathematician jargon this will be a multiplication by an exponential in time domain vs a frequency shift in s-domain or Frequency Domain respectivel. Just passed by my mind… that’s all I should try and see too for the middle term (the φ may need to be converted to an exponential)..looks has if it’s unnecessary..but I just mentioned it with a possibility of finding a new way so…
Oh yeah that’s a convolution integral 😅
are You From tabriz ?