I'm a working analog Design engineer. I have no shame in admitting that initially I made a mistake in the solution when I tried to quickly think by myself by using steady state concept. Now I learned I have to be more careful. Your answer has definitely taught me something. Thanks for your work. Keep going. Your vids are useful and enjoyable.
Maybe this was a failure in my background, but I never thought about solving it in the time domain. Usually I converted it to frequency domain and then converted back to time domain. Solving and understanding it in the time domain was a novel experience for me LOL Thank you very much.
would be a good move to solve it by the transfer function? and i don't quite understand how you assume Vin - Vout is constant in the integral and the result is still correct though
Hello Will the output voltage be a decaying amplitude ringing signal over time, as in the ringing amplitude reduces until it settles down? (since this is a step input and the edge transition from low to high happens only once according to the diagram)
The output will not be decaying in this case as inductor and capacitor are ideal (they have no resistance components). In a practical circuit however due to parasitic resistance, output voltage will be decaying and settles to 1V at steady state. This can also be easily proven by solving above circuit in frequency domain and applying inverse laplace to get time domain response
6:20 No this assumption is wrong. You don't know what L and C are. If L is really small and C very large then, then its possible no ringing occurs at all. But that's not really what I take issue with, in your explanation. I don't think you can say what voltage the capacitor will be at. Now that I'm thinking about it, a more correct answer would also consider cases where C>>L and L>>C. All that aside keep it up! I enjoy watching your videos. I think a few have stumped me as well, and I've only watched like 10 so far.
Hi! Thank you for your constructive feedback, we very much appreciate it! However, we'd like to point the following: The result holds true independent of the values of L and C (assuming they are ideal). This is because this LC filter is a double pole system and varying the values of L and/or C will only shift the frequency at which they resonate. Let us know if you agree. Cheers!
Thank you so much for contributing to the discussion! You're absolutely right. Common follow up questions are: 1. Imax 2. Frequency of Oscillation 3. Bode Plot of transfer function 4. What happens if there's a resistor in parallel/series with the capacitor
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I'm a working analog Design engineer. I have no shame in admitting that initially I made a mistake in the solution when I tried to quickly think by myself by using steady state concept. Now I learned I have to be more careful. Your answer has definitely taught me something. Thanks for your work. Keep going. Your vids are useful and enjoyable.
Maybe this was a failure in my background, but I never thought about solving it in the time domain. Usually I converted it to frequency domain and then converted back to time domain.
Solving and understanding it in the time domain was a novel experience for me LOL
Thank you very much.
same! I was using laplace method, in the end I was figuring out how to now apply inverse laplace now :)
The solution is vc(t) = 1-cos(w0t) where w0 is the natural frequency given by the expression 1/sqrt(1LC) .
would be a good move to solve it by the transfer function?
and i don't quite understand how you assume Vin - Vout is constant in the integral and the result is still correct though
Thanks for these, I have a job interview in a couple of days. I need all the help i can get :D
Happy to help! Please do let us know how it goes!
Thank you for this Video..!!!
Nice video,
Thank you for this video... keep going ;)
Thank you for the kind words, please share with your colleagues!
Hello
Will the output voltage be a decaying amplitude ringing signal over time, as in the ringing amplitude reduces until it settles down? (since this is a step input and the edge transition from low to high happens only once according to the diagram)
The output will not be decaying in this case as inductor and capacitor are ideal (they have no resistance components). In a practical circuit however due to parasitic resistance, output voltage will be decaying and settles to 1V at steady state.
This can also be easily proven by solving above circuit in frequency domain and applying inverse laplace to get time domain response
6:20 No this assumption is wrong. You don't know what L and C are. If L is really small and C very large then, then its possible no ringing occurs at all.
But that's not really what I take issue with, in your explanation. I don't think you can say what voltage the capacitor will be at.
Now that I'm thinking about it, a more correct answer would also consider cases where C>>L and L>>C.
All that aside keep it up! I enjoy watching your videos. I think a few have stumped me as well, and I've only watched like 10 so far.
Hi!
Thank you for your constructive feedback, we very much appreciate it! However, we'd like to point the following: The result holds true independent of the values of L and C (assuming they are ideal). This is because this LC filter is a double pole system and varying the values of L and/or C will only shift the frequency at which they resonate. Let us know if you agree.
Cheers!
@@HardwareNinja Agreed.
add question: what is Imax?
Thank you so much for contributing to the discussion! You're absolutely right. Common follow up questions are:
1. Imax
2. Frequency of Oscillation
3. Bode Plot of transfer function
4. What happens if there's a resistor in parallel/series with the capacitor
@@HardwareNinja Can we get an explanation for those too?😜
Great work !