Relationship between the Laplace transform and impedance

[PauloConstantino]

Howdy,

I've been thinking about the relation between the Laplace transform / s-domain / frequency domain and impedance of reactive circuits.

Could anyone tell me what is the relation between the laplace transform and dealing with impedances? basically in an LCR circuit driven by Vi * sin(wt) if you take the voltage across the capacitor, the value will be Vi * (1/jwC) / (R + jwL + 1/jwC), but why does this correspond to the laplace transform of the differential equation ? jw = s in the laplace domain but why ? The laplace transform will be (1/sC) / (R + sL + 1/sC), but why ?

I wont write the diff eq of the LCR here, but we know that taking the laplace transform of it will yield the simple s domain equation.

I am a mathematician by trade so you can spare some of the math, I want to really know why this is done in electronics.

Who was the first to find out that impedances correspond to laplace transforms? And lastly.... why does it make sense to just deal with impedances in ractive circuit as thought you were dealing with DC voltage dividers ? i.e. the s-domain.

[tggzzz]

Could anyone tell me what is the relation between the laplace transform and dealing with impedances?
...
Who was the first to find out that impedances correspond to laplace transforms?

None whatsoever, and they don't. (Unless, of course, you are taking the Laplace transform of an impedance.)

[PauloConstantino]

Could anyone tell me what is the relation between the laplace transform and dealing with impedances?
...
Who was the first to find out that impedances correspond to laplace transforms?

None whatsoever, and they don't. (Unless, of course, you are taking the Laplace transform of an impedance.)

I'm talking about taking the laplace transform of a differential equation, not of "an impedance". You tried being "clever" but you failed.

And don't use your time replying if you can't contribute to the issue.

[Mattjd]

Howdy,

I've been thinking about the relation between the Laplace transform / s-domain / frequency domain and impedance of reactive circuits.

Could anyone tell me what is the relation between the laplace transform and dealing with impedances? basically in an LCR circuit driven by Vi * sin(wt) if you take the voltage across the capacitor, the value will be Vi * (1/jwC) / (R + jwL + 1/jwC), but why does this correspond to the laplace transform of the differential equation ? jw = s in the laplace domain but why ? The laplace transform will be (1/sC) / (R + sL + 1/sC), but why ?

I wont write the diff eq of the LCR here, but we know that taking the laplace transform of it will yield the simple s domain equation.

I am a mathematician by trade so you can spare some of the math, I want to really know why this is done in electronics.

Who was the first to find out that impedances correspond to laplace transforms? And lastly.... why does it make sense to just deal with impedances in ractive circuit as thought you were dealing with DC voltage dividers ? i.e. the s-domain.

I have a hard time believing you're a mathematician by trade by the way you're loosely throwing around terms here.

Anyway...

s = sigma + j*omega

by definition, being a math guy you'd think you know this. This isn't an electronics thing, its a math thing. Now why is sigma assumed to be zero in the realm of EE? Go figure it the hell out. Hint, it has to do with Fourier, and it has to do with the different frequencies that both sigma and omega represent.

edit: Another FYI, Z (impedance) = R + j*X, R is the resistance, X is the reactance. Resistance for inductors and capacitors is zero, reactance of a resistor is zero. 

Think of when Laplace Analysis is used vs when Phasor Analysis is used.

tggzzz may have worded it poorly but sure are being a dick. Of course Laplace Transform of an "impedance" means nothing. Laplace Transform is used to solve certain types of differential equations.

[PauloConstantino]

Howdy,

I've been thinking about the relation between the Laplace transform / s-domain / frequency domain and impedance of reactive circuits.

Could anyone tell me what is the relation between the laplace transform and dealing with impedances? basically in an LCR circuit driven by Vi * sin(wt) if you take the voltage across the capacitor, the value will be Vi * (1/jwC) / (R + jwL + 1/jwC), but why does this correspond to the laplace transform of the differential equation ? jw = s in the laplace domain but why ? The laplace transform will be (1/sC) / (R + sL + 1/sC), but why ?

I wont write the diff eq of the LCR here, but we know that taking the laplace transform of it will yield the simple s domain equation.

I am a mathematician by trade so you can spare some of the math, I want to really know why this is done in electronics.

Who was the first to find out that impedances correspond to laplace transforms? And lastly.... why does it make sense to just deal with impedances in ractive circuit as thought you were dealing with DC voltage dividers ? i.e. the s-domain.

I have a hard time believing you're a mathematician by trade by the way you're loosely throwing around terms here.

Anyway...

s = sigma + j*omega

by definition, being a math guy you'd think you know this. This isn't an electronics thing, its a math thing. Now why is sigma assumed to be zero in the realm of EE? Go figure it the hell out. Hint, it has to do with Fourier, and it has to do with the different frequencies that both sigma and omega represent.

edit: Another FYI, Z (impedance) = R + j*X, R is the resistance, X is the reactance. Resistance for inductors and capacitors is zero, reactance of a resistor is zero. 

Think of when Laplace Analysis is used vs when Phasor Analysis is used.

tggzzz may have worded it poorly but sure are being a dick. Of course Laplace Transform of an "impedance" means nothing. Laplace Transform is used to solve certain types of differential equations.

I don't think you have answered even 1% of the question.......


P.s I am graduated in pure mathematics. I can understand the maths of transforms, the question posed is the why of it being used in electronics. I throw away loose terms because this is a forum and not a maths textbook.

[Benta]

You've basically answered yourself in your question.
It's an absolute pain to work with differential equations. By using the Laplace transform, the differential eqs are transformed to linear eqs.
In practice as an engineer, you'd use R, 1/sC and sL for impedances, or G, sC and 1/sL for admittances (depending on the circuit, it's sometimes much easier using admittance).

What's normally interesting for synthesis and analysis is not the impedances, but the transfer function, which expresses Uout/Uin or Iout/Iin or even Pout/Pin.

By working in the s-domain, you are not limited to sinusoidal signals, but can work with arbitrary waveforms. The step response is an example of a very often made calculation.

At the end, jw is normally substituted into the transfer function to get amplitude, phase and delay response.

[ironmonkey]

I'll try to answer your question in a very simple way, so don't bash me too hard if it is not what you have asking for.

The Laplace transform is only a domain change system, for example, the voltage over a ideal inductor is directly proportional to the rate of change of current over time ( v(t)=L di(t)/dt ), in a Laplace domain it will be v(s)=L*1/s), so, so you can write the impedance also in time domain and after a little bit more painful demonstration get the same result with Laplace, however you are not changing any fundamental law, is just a Homomorphism (I think its how its called after all this years outside a linear algebra class, please don' be too hard with me)

How this beautiful simple and convenient space change of integrals/derivatives for products/quotients was deducted (or invented?), I don't now, I'm not mathematician

[tggzzz]

Howdy,

I've been thinking about the relation between the Laplace transform / s-domain / frequency domain and impedance of reactive circuits.

Could anyone tell me what is the relation between the laplace transform and dealing with impedances? basically in an LCR circuit driven by Vi * sin(wt) if you take the voltage across the capacitor, the value will be Vi * (1/jwC) / (R + jwL + 1/jwC), but why does this correspond to the laplace transform of the differential equation ? jw = s in the laplace domain but why ? The laplace transform will be (1/sC) / (R + sL + 1/sC), but why ?

I wont write the diff eq of the LCR here, but we know that taking the laplace transform of it will yield the simple s domain equation.

I am a mathematician by trade so you can spare some of the math, I want to really know why this is done in electronics.

Who was the first to find out that impedances correspond to laplace transforms? And lastly.... why does it make sense to just deal with impedances in ractive circuit as thought you were dealing with DC voltage dividers ? i.e. the s-domain.

I have a hard time believing you're a mathematician by trade by the way you're loosely throwing around terms here.

Just so.

Anyway...

s = sigma + j*omega

by definition, being a math guy you'd think you know this. This isn't an electronics thing, its a math thing. Now why is sigma assumed to be zero in the realm of EE? Go figure it the hell out. Hint, it has to do with Fourier, and it has to do with the different frequencies that both sigma and omega represent.

edit: Another FYI, Z (impedance) = R + j*X, R is the resistance, X is the reactance. Resistance for inductors and capacitors is zero, reactance of a resistor is zero. 

Think of when Laplace Analysis is used vs when Phasor Analysis is used.

tggzzz may have worded it poorly but sure are being a dick. Of course Laplace Transform of an "impedance" means nothing. Laplace Transform is used to solve certain types of differential equations.

I don't think you have answered even 1% of the question.......


P.s I am graduated in pure mathematics. I can understand the maths of transforms, the question posed is the why of it being used in electronics. I throw away loose terms because this is a forum and not a maths textbook.

In which case you shouldn't be surprised if you get loose responses.

OT here - but relevant to one of your other topics - is that employers value people that can not only do their own research, but also can express themselves concisely and precisely.

[PauloConstantino]

You've basically answered yourself in your question.
It's an absolute pain to work with differential equations. By using the Laplace transform, the differential eqs are transformed to linear eqs.
In practice as an engineer, you'd use R, 1/sC and sL for impedances, or G, sC and 1/sL for admittances (depending on the circuit, it's sometimes much easier using admittance).

What's normally interesting for synthesis and analysis is not the impedances, but the transfer function, which expresses Uout/Uin or Iout/Iin or even Pout/Pin.

By working in the s-domain, you are not limited to sinusoidal signals, but can work with arbitrary waveforms. The step response is an example of a very often made calculation.

At the end, jw is normally substituted into the transfer function to get amplitude, phase and delay response.

^ This is the correct answer.  The attached drawing helps illustrate the point. 



Differential equations are a pain.  The Laplace transform allows us to bypass differential equations and do algebra instead.  We can then translate back to the time domain or frequency domain.

It's not the correct answer to my question. I know transforms make it easier.

I am asking here the relationship between impedances and transforms.

If anyone here actually has studied by the text book and actually knows an answer please go ahead.


For those like tgzgzz who are here to waste time and offend others, please don't use your time this well.

[Simon]

ok the half time gong has rung and you have all had an equal pop at each other so it's a draw and thread closed.

[EEVblog]

FYI, telling people to go kill themselves gets you banned from the forum.