Active power dissipated by a resistor from DC to RF

[aabbaa11]

TL;DR

My question is what is the connection between equations on the power dissipated by resistor in DC and high frequency?

Hello all PF members!

I was wondering about the power dissipated by resistor at high frequency which is:
$P=Re\left \{ U\cdot I^{*} \right \}=Re\left \{ \frac{\left |V^{+} \right |^{^{2}}}{2\left |Z_{c} \right |}\cdot (1-\left |\Gamma \right |^{2} )\right \}$, (1)
where:
$Z_{c}$ - characteristic impedance of the waveguide (transmission line) = $\sqrt{\frac{R_{c}+ j\omega L_{c}}{G_{c}+ j\omega C_{c}}}$
$V^{+}$ - amplitude of incident wave (to the resistor)
$\Gamma$ - reflection coefficient = $\frac{Z_{L}-Z_{c}}{Z_{L}+Z_{c}}$, where $Z_{L}$ is load impedance.

These equations work fine at high frequency. The problem (and my question) is about the case of lower frequency and DC. When the frequency is going down the $\left | Z_{c} \right |$ is going to +inf, as well as $1-\left |\Gamma \right |^{2}$ which is going to 0. Thus, the total active power is equal to 0 at DC. I can assume, that equation (1) represents the active power delivered only by wave, so for DC = 0. But as we know, the active power at DC is still transferred to the resistor:
$P=\frac{U^{2}}{R_{L}}$ (2).
My question is: Which is the relation between (1) and (2)? Is equation (1) simplified somehow?

Thank you in advance!

 

[sophiecentaur]

For DC the equation for Z₀ is indeterminate so it won't deliver.

 

[aabbaa11]

For DC the equation for Z₀ is indeterminate so it won't deliver.

Sorry, there was my mistake, there should be + instead of *, now is correct. But my question is still valid. Which is the relation between (1) and (2)?

 

[tech99]

At radio frequencies the Zo of the line is involved and at DC it is not, apart from the intial switch-on step function.
Further, if the load impedance has series capacitance, no current flows at DC, apart from switch-on.

 

[aabbaa11]

Thank you for response! I understand your point of view. But when the impedance consists only resistance, the current flows for DC as well. I still don't see the mathematical relation between (1) and (2). There should be "smooth" switch from (1) to (2).

 

[sophiecentaur]

There should be "smooth" switch from (1) to (2).

Ii see what you are getting at but, seriously, could you expect to be working with reactances at 0.001Hz? (Add some more zeros, if you want). There is no sharp switch between AC and DC in practice. The wavelength involved with a very low frequency is so great that any transmission line connecting a source and a load is negligibly short so Transmission Line Equations become meaningless. If you want to reconcile the two regimes then you would need to take the limit of the AC equations as f approaches Zero and that would make the values of some of the parameters impossible to calculate.
As with all of Science, the Maths is the slave of Science and not the other way round. Eventually, the Maths doesn't follow reality.