Laplace Domain Circuit Analysis

[@EfieldHfield_377]

Helping my daughter with some HW. I never cease to be amazed how little school taught me vs how much I had to learn on my own. I've learned way more on KZbin University then from "most" of my professors in school. The one thing I have learned is not everyone was meant to teach and most teachers should do something else. Teaching is the highest form of communicating an idea to someone else. 13 years ago and your video has relevance to me - Thank you sir.

[@michaelt126]

Doing some circuit analysis in school this semester, your video is still helping students out a decade later!

[@ebarbie5016]

There are two S-domain circuit models for the capacitor and inductor (voltage source and current source models). Voltage source models are ideal for applying KVL and current source models are ideal for applying KCL. I encourage my students to use both.

[@DarrylMorrell]

The sigma is the real part of s. When taking inverse Laplace transforms, sigma tends to show up as multipliers of real exponentials: e^{-sigma t}. w shows up as frequencies of sines and cosines. The Laplace variable is used in conjunction with impedance to find voltages and currents in the circuit, but I am not aware of a more specific relationship.

[@xMrJanuaryx]

Well my request wont be of any use to me now because my circuits test is tomorrow but for future students it might help. Could you make a video expansion on this topic that covers the derivation for these circuits in more detail? For instance, you can also represent the s-domain equivalent circuits for a capacitor and inductor as a circuit which contains a parallel current sources rather than a series voltage source representing the initial conditions. I managed to figure out how it works based on your video but a more in-depth explanation as to why you can do both would be helpful to future students I am sure. Otherwise, thanks for the helpful video!

[@moeinmaleki4084]

Did what a 60 yr old professor couldn't. Forever grateful

[@mnada72]

First of all I like your videos and I am following from the beginning of the course. I wish I have the same style of explanation in all the fields of my interest. I have a comment regarding the laplace transform of a capacitor voltage, I was not comfortable to the explanation of 1/s V(0) because of the 1/s yet it's true, I believe if we start for the capacitor with I(s) = s C V(s) - C v(0), it will make more sense. Thank you very much for the nice videos

[@Fisheatpho]

Such a simple and clear way of explaining!! Thank you!

[@parinita1031]

This just cleared all my doubts and concept as well . Thanks

[@DarrylMorrell]

The frequency domain and the Laplace domain are closely related; the frequency domain is the part of the Laplace domain that has only imaginary values (zero real values). The Laplace domain, however, allows you to take into account initial conditions of your circuit elements; you can't easily do this in the frequency domain.

[@hippiechickie18]

I could listen to your voice all day

[@DarrylMorrell]

Yes, in theory that would be the case. An ideal voltage source supplies whatever current is necessary to maintain the voltage. To charge the capacitor instantly, the current would be infinite. In reality, capacitors cannot change voltage instantaneously, and real voltage sources cannot supply infinite current-there will always be some internal resistance in the voltage source.

[@ChristianKrause89]

Cool! Now I get the extra voltage thing and where the polarity comes from. Thanks!

[@RedHeaded411]

It's been awhile since I've done these. This is a great review - thanks.

[@daltonortega]

Good explanation. I'm taking a signal processing class and was wondering the difference between Fourier and Laplace. Having the real part of s, i can see the advantages of using Laplace for circuit analysis. My question is this, what would be a situation that one would favor Fourier?.

[@shawalmbalire]

I'm watching in 2022. 11 years later.

[@francisfigueroaiii5383]

your inverse laplace is wrong @ 13:05 you supposed to use the method completing the square for dinominator and and the answer would be e^-3sin(-3t)

[@mjeezyca]

this is awesome dr. morrell

[@amandafalke7670]

Great job as usual, Darryl. Thank you so much!

[@DeviL-yw6ud]

Where are you from

[@ahelleso]

Nice video. I was wondering if you could make one more, with an R, L and C, to se how that is done? Would have been very helpful! And one more question; if V(t) is 10V, would still V(s) be equal to 1/s? I didn't quit get that part? Thank you...

[@KazKylheku]

Ideal capacitors charge instantly because they have no parasitic resistance or inductance to limit the current. What is the RC time constant when R is zero?

[@mghinto]

This is great! Do you have a more complex example with initial conditions?

[@sami-ullahkhaliq3388]

if vs(t) = 10t, R=1M ohms, C=1micro Farad. how would this be done? im sorry I still cant understand how to do it after watching your video.

[@n7.allameh704]

What is the meaning of the s-domain in the Laplacian function? and what is the inverse function of that?

[@shabinanaaz7581]

But for DC sources the capacitor acts as open circuit... so the voltage across capacitor should directly be equal to supply voltage

[@Vivenk88]

Is the Laplace variable s = jw + sigma related to the impedance Z = R + j(Xl - Xc)? Also the Laplace variable has frequency as imaginary part, but what is the "sigma" in the s = jw + sigma. How can we get a clearer understanding of this relation: s = jw + sigma?

[@swatinegi8941]

If the current entering terminal is negative in both inductor and capacitor then euation will be???...make a video on that

[@ayakhateeb6399]

hi have some questions about laplace how can i contact you??

[@vbidawat93]

The initial condition polarity for inductor, can someone elaborate on it for me? Like why it's opposite to the actual polarity across the inductor?

[@simpzic772]

i got quiz tomorrow, thanks alot

[@souhardyapal1731]

This means in s domain all capacitive and inductive properties are just converted to resistive properties?

[@kamzy98]

Thanks Sir, this was helpful and you explained it very well

[@gabrielgodbout3206]

great video

[@photon2724]

am i missing something. dont remember doing step response for lapance transform...

[@photon2724]

i dont understand how inductance and capacitance in the s domain convert to that equation....

[@flutterwind7686]

Laplace transforms and then inverse laplace transforms

[@amrmoneer5881]

i understood everything except the initial condition part. but thanks for the video god bless you

[@MyIntestinesLikeFood]

The initial condition is there because we cannot always assume that a capacity/inductors initial value will be zero. A capacity may still be discharging and therefore have Voltage/current still in the capacitor. For most easy question, the initial conditions will be zero, but for some harder questions you may be presented with a question which states "take into consideration the capacitor is not fully discharged and has this Voltage at time = 0. Hope this makes sense.

[@alexplastow9496]

Dozed off in lecture, thanks for bailing me out

[@adityasahu96]

thank you sir

[@hippiechickie18]

Also I fucking love you for this video. Thank you.

[@aneguitar]

Thank you very much! It was great!

[@Ugenetic]

I failed to make connection. forgot too much preliminaries. Besides, Phasors are way better tools for solving a particular circuit, Laplace is to analyze the circuit responses to different functions (of inputs)

[@nashs.4206]

IIRC, phasor analysis only works if your source is sinusoidal. If it is sinusoidal, then the source can be represented as a phasor (a complex number with a magnitude and phase), and performing a phasor analysis will give you a steady state result. This is intuitive because, since the input is steady (i.e. it is sinusoidal, but it remains sinusoidal), the output will also be in steady-state. Fourier series works if the source is periodic, but not necessarily sinusoidal. Of course, due to Fourier's theorem, it is possible to represent this periodic source as a sum of sinusoids, so that phasor analysis can once again be performed, along with the superposition principle. Once again, the output will be in steady-state. So, phasor and fourier series methods only gives us a steady-state solution. This is because the input itself is in steady-state. However, if the input is not in steady-state, then the above methods can't be used. Instead, the laplace and fourier transform methods must be used. Whether to use laplace transform or fourier transform depends on whether the laplace integral or fourier integral converges. It turns out that for certain functions, the laplace transform exists, but the fourier transform doesn't. Likewise, there are certain functions for which the fourier transform exists, but the laplace transform doesn't. In this case, we would use the laplace or fourier transform as appropriate.

[@ECOMMUSK]

excellent!

[@timothdev]

Thank you so much ;)

[@coolwinder]

Thank you so much!

[@ssssssssssurvey]

powerful method

[@tharunishadi3980]

Thank you very much

[@borabrothers9011]

hey darryl morrell cant able to understand ....

[@GrimmNinetyThr33]

youre a god

[@VFXCommander]

very helpful!

[@DaShogun]

nice

[@MasterTuttoo]

why does it sound like making this video is giving u physical pain

[@alejandroalonsosanchez2418]

💀💀💀

[@omegaK-v5j]

dude you seem drunk .....

[@saichaithrik7134]

Thank you so much