Laplace Transform Electric Circuit Example
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- Опубліковано 5 лис 2018
- Shows an example of using the Laplace Transform to analyse a basic electric circuit.
* Note that I made a small typo in the video. I should have made the value of R = 1.4 ohms. Then the middle term on the denominator of H(s) would be 7s (not 5s) and then the rest of the maths would be correct.
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For a full list of Videos and accompanying Summary Sheets, see: www.iaincollings.com
Thank you so much for your videos. I never understood this in college. Now many years ago, it’s so much clearer thanks to your explanations!
I'm really glad to hear that my videos have helped clarify things.
You forgot to point out the full Laplace transform:
dy(t)/dt => sY(s) - s(0).
And:
dy^2(t)/dt^2 => (s^2)*Y(s) - sY(s)- s'(0).
This is important as the system could have initial conditions, for example if the capacitor has charge in t=0.
Good point. Yes, I should make a video to explain where the initial conditions come into it.
3:05 - 3:20 Is what I am looking for ... please suggest a reference for more information on the relationship between ROC and the existence of fourier transform to stability
I can see from your other comments that you've already found the answer in the following video, but in case others have the same question, then here's the link: "Why are Poles Negative in Laplace Transform of Real Stable LTI Systems?" ua-cam.com/video/apdh8ZXW3a0/v-deo.html
1:09 Dude got a spider nearby 👀
Thanks. By the way the quadratic formula roots should be complex conjugate numbers with negative real parts -2.5, so it’s still stable and causal system.
Oh yes. Thanks for picking this up. That's annoying. I must have had a brain fade. I should have made the value of R = 1.4 ohms, to end up with those nice whole numbers in the denominators of the final two terms. But as you say, either way, the main point of the video is still the same.
@@iain_explains Thanks
hi teacher,what is difference between unileteral and bileteral Laplace Transform? why you use df(t)/dt=sF(s) instead of df(t)/dt=sF(s)-f(0) ?
The difference is in the lower limit of the integration in the definition of the transform. The Unilateral transform integrates over positive time (only). It takes into account the initial conditions, but is limited to only considering causal systems.
Thanks! Very clear explanation
Glad it was helpful!
This is one fantastic video. Thanks alot.
Glad you liked it!
You didn't consider the initial conditions i mean the transient analysis ? Is there any video on that ?
Good point. Thanks for the suggestion. I'll add it to my "to do" list. I don't have anything on that yet.
Thank you very much for your nice explanation
Glad it was helpful!
Why don't we consider the current given by the inductor?
I'm not sure I understand what you're asking. The inductor doesn't "give" current. The equation I have written is for the voltage drops around the circuit.
@@iain_explains For both the resistor and the inductor voltage drop you include the term C*dy/dt as the current. This current comes from the capacitor in the circuit, i assume. My question is if the inductor is also capable of giving current why don't we also use it as well? I would think, for example, the voltage drop for the resistor would be something like R*(Current from capacitor + Current from inductor). This would be because the current going through the circuit would be the sum of the two sources, that being the inductor and the capacitor.
how did you get (x+2), (x+5)
i meant (S+2), (S+5)
Yes, it's a "typo". In the Description under the video it says: "* Note that I made a small typo in the video. I should have made the value of R = 1.4 ohms. Then the middle term on the denominator of H(s) would be 7s (not 5s) and then the rest of the maths would be correct."
Thank you full mister
You are welcome
Am i dumb, or does s^2 + 5s + 10 ≠ (s+2)(s+5)
If you look at the description under the video, you'll see that I already added a comment about that "typo". Nobody's perfect. You don't need to be rude about it.
Your math is incorrect for the s2+5s+10
It seems right to me. R/L = 1/0.2 = 5 and 1/(LC) = 1/(0.2*0.5) = 1/0.1= 10.
@@iain_explains I think he means that you can't factorise s^2 + 5s + 10 to equal (s+2)(s+5). Because (s+2)(s+5) = s^2 +7s +10.
Ahh, ooops, now I see. Yes, silly mistake. I should have made the resistance R=1.4 ohms, then it would give (s^2+7s+10), and everything else holds. Thanks for pointing it out.
Hi Sir, how do you solve x(t)->X(s) for dy(t)/dt?
s^2+5s+10=(s+2)(s+5)??? In what world? lol. I'm pretty sure it's a complex conjugate solution. Either way, I get what you're saying.
Yes. I already explained that in the Details below the video.
@@iain_explains Gotcha. lol. Thank you for making these videos. They are very helpful.
You're welcome. I'm glad you're finding them to be helpful.