Meaning of "Quasi Static"

quasistatic meaning

Hi all,

We often encounter the term "Quasi Static" analysis in EM. What exactly does it mean ? Can anyone clarify on this ?

-svarun

meaning of quasi static analysis

hi svarun
i think the "Quasi Static" means that the transverse component of the electramagnetic field is dominating and the longitudinal component is neglectable,so the wave equation can be reduced to the Possion's equation,and the 3-D EM(FM) problem can be tansformed to 2-D problem.In this condition,the EM(FM) is similar to the static electromagnetic field.
regards

meaning of quasistatic

Hi Svarun

there are two other terms which are in the same box one is "stationary" and the other is "steady"

quasi-static analysis definition

hi svarun
the quasi-static is neglect the displacement currents item in the Maxwell equation

displacement currents quasi static approximation

Lets say we have the following function F, which is a function of position, frequency and time. The question is, how would we represent this equation in the quasi-static limit?

\[F=F(x,\omega,t)\]

In my opinion, we would make a Taylor expansion and only retain the orders up to ω in the expansion. That is because the zero-th order would actually mean we have only retained the static value.

Well, a more physical meaning is the following:
" "quasi-static" refers to the regime for which the finite speed of light can be neglected and fields treated as if they propagated instantaneously."

This would mean that we ignore the contribution from the displacement current in the Ampere's Law.

quasi static meaning

irfan1 said:

Lets say we have the following function F, which is a function of position, frequency and time. The question is, how would we represent this equation in the quasi-static limit?

\[F=F(x,\omega,t)\]

Click to expand...

Irfan has asked the question very clearly, I think, an answer to this question will certainly clarify the "quasi-static" concept.

quasi-static manner means

I also would like to add the following bit. The term "quasi-static" is quite relative. For instance, if you are considering the effect of radio-waves on a small piece of semiconductor, you can safely use the "quasi-static" approximation. On the other hand, if you are investigating the propagation of radio waves between two stars, obviously you can not use such an approximation.

quasi static concept

So does that mean that for quasi static analyses, you do not need to take into account, the finite speed of light ?

-svarun

meaning of quasi arabic

yes. indeed, we would treat the fields as if they were instantaneous.

Hi Svarun -- Quasi means "as if" (in Latin). Static means no chage with respect to time. You can have "electrostatic" when charge is constant with time, and "magnetostatic" when current is constant with time.

If you have a lossless transmission line and homogenous dielectric, you can solve for the capacitance per unit length using an electrostatic analysis. This is easy to do for, for example, coax. See any book on EM theory. Since the dielectric is homogenous, we know the velocity of propagation exactly. Once you have the capacitance per unit length and the velocity of propagation, you can calculate the inductance per unit length. From the inductance and capacitance per unit length you can get the characterisitic impedance.

You can also start with a magnetostatic analysis and proceed in a manner very similar to the above and arrive at exactly the same result.

All the above is exact, provided the line is lossless and the dielectric is homogenous.

If you add a little bit of loss (metal resistance, or dielectric loss tangent), or the dielectic is not homogenous (e.g., microstrip), the above analysis is now approximate. In this case, we call it "quasi-static". Since we never have the exact static case (except for electrostatic at zero frequency!), most transmission lines will be quasi-static, provided we don't go too high in loss or too high in frequency.

Static means simplifying Maxwell's equations such that there is no coupling at all between E and B fields:

ε div(E)= ρ
curl(B)=µ J

With quasi-static, one form of coupling between B and E is considered: the E field generates conventional current in conductive materials (Ohms Law) and then this conventional current adds to the external J stimulus and generates B in the normal (Biot-Savart) way.

curl(B)=µ(Jexternal+σE)

In both static and quasi-static, the time derivative terms in Maxwell's equations are set to zero. In other words, the displacement current term that Maxwell added to Ampere's Law is set back to zero and Faraday's Law becomes:

curl(E) = -dB/dt = 0

In contrast, full wave solvers consider all the time derivative coupling terms in Maxwell's equations to be finite.

Best regards,

-- Colin Warwick
High Speed Digital and Signal Integrity Design Flow Manager, Agilent EEsof EDA
Blog: **broken link removed**