time average

Average power for circuit elements

February 9, 2016 ece1236 No comments , , , , , , ,

[Click here for a PDF of this post with nicer formatting] or [Click here for my notes compilation for this class]

In [2] section 2.2 is a comparison of field energy expressions with their circuit equivalents. It’s clearly been too long since I’ve worked with circuits, because I’d forgotten all the circuit energy expressions:

\begin{equation}\label{eqn:averagePowerCircuitElements:20}
\begin{aligned}
W_{\textrm{R}} &= \frac{R}{2} \Abs{I}^2 \\
W_{\textrm{C}} &= \frac{C}{4} \Abs{V}^2 \\
W_{\textrm{L}} &= \frac{L}{4} \Abs{I}^2 \\
W_{\textrm{G}} &= \frac{G}{2} \Abs{V}^2 \\
\end{aligned}
\end{equation}

Here’s a recap of where these come from

Energy lost to resistance

Given
\begin{equation}\label{eqn:averagePowerCircuitElements:40}
v(t) = R i(t)
\end{equation}

the average power lost to a resistor is

\begin{equation}\label{eqn:averagePowerCircuitElements:60}
\begin{aligned}
p_{\textrm{R}}
&= \inv{T} \int_0^T v(t) i(t) dt \\
&= \inv{T} \int_0^T \textrm{Re}( V e^{j \omega t} ) \Real( I e^{j \omega t} ) dt \\
&= \inv{4 T} \int_0^T
\lr{V e^{j \omega t} + V^\conj e^{-j \omega t} }
\lr{I e^{j \omega t} + I^\conj e^{-j \omega t} }
dt \\
&= \inv{4 T} \int_0^T
\lr{
V I e^{2 j \omega t} + V^\conj I^\conj e^{-2 j \omega t}
+ V I^\conj + V^\conj I
}
dt \\
&= \inv{2} \textrm{Re}( V I^\conj ) \\
&= \inv{2} \textrm{Re}( I R I^\conj ) \\
&= \frac{R}{2} \Abs{I}^2.
\end{aligned}
\end{equation}

Here it is assumed that the averaging is done over some integer multiple of the period, which kills off all the exponentials.

Energy stored in a capacitor

I tried the same sort of analysis for a capacitor in phasor form, but everything cancelled out. Referring to [1], the approach used to figure this out is to operate first strictly in the time domain. Specifically, for the capacitor where \( i = C dv/dt \) the power supplied up to a time \( t \) is

\begin{equation}\label{eqn:averagePowerCircuitElements:80}
\begin{aligned}
p_{\textrm{C}}(t)
&= \int_{-\infty}^t C \frac{dv}{dt} v(t) dt \\
&= \inv{2} C v^2(t).
\end{aligned}
\end{equation}

The \( v^2(t) \) term can now be expanded in terms of phasors and averaged for

\begin{equation}\label{eqn:averagePowerCircuitElements:100}
\begin{aligned}
\overline{{p}}_{\textrm{C}}
&= \frac{C}{2T} \int_0^T \inv{4}
\lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} }
\lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} } dt \\
&= \frac{C}{2T} \int_0^T \inv{4}
2 \Abs{V}^2 dt \\
&= \frac{C}{4} \Abs{V}^2.
\end{aligned}
\end{equation}

Energy stored in an inductor

The inductor energy is found the same way, with

\begin{equation}\label{eqn:averagePowerCircuitElements:120}
\begin{aligned}
p_{\textrm{L}}(t)
&= \int_{-\infty}^t L \frac{di}{dt} i(t) dt \\
&= \inv{2} L i^2(t),
\end{aligned}
\end{equation}

which leads to

\begin{equation}\label{eqn:averagePowerCircuitElements:140}
\overline{{p}}_{\textrm{L}}
= \frac{L}{4} \Abs{I}^2.
\end{equation}

Energy lost due to conductance

Finally, we have conductance. In phasor space that is defined by

\begin{equation}\label{eqn:averagePowerCircuitElements:160}
G = \frac{I}{V} = \inv{R},
\end{equation}

so power lost due to conductance follows from power lost due to resistance. In the average we have

\begin{equation}\label{eqn:averagePowerCircuitElements:180}
\begin{aligned}
p_{\textrm{G}}
&= \inv{2 G} \Abs{I}^2 \\
&= \inv{2 G} \Abs{V G}^2 \\
&= \frac{G}{2} \Abs{V}^2
\end{aligned}
\end{equation}

References

[1] J.D. Irwin. Basic Engineering Circuit Analysis. MacMillian, 1993.

[2] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.

Updated notes for ece1229 antenna theory

March 16, 2015 ece1229 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

I’ve now posted a first update of my notes for the antenna theory course that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides which go by faster than I can easily take notes for (and some of which match the textbook closely). In class I have annotated my copy of textbook with little details instead. This set of notes contains musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book), as well as some notes Geometric Algebra formalism for Maxwell’s equations with magnetic sources (something I’ve encountered for the first time in any real detail in this class).

The notes compilation linked above includes all of the following separate notes, some of which have been posted separately on this blog:

Energy momentum conservation with magnetic sources. Comparison to frequency domain and reciprocity theorem.

February 20, 2015 ece1229 No comments , , ,

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In the frequency domain

In the frequency domain with \( \boldsymbol{\mathcal{E}} = \textrm{Re} \BE e^{j \omega t}, \boldsymbol{\mathcal{H}} = \textrm{Re} \BH e^{j \omega t} \). Using the electric field dot product as an example, note that we can write

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:480}
\boldsymbol{\mathcal{E}} = \inv{2} \lr{ \BE e^{j \omega t} + \BE^\conj e^{-j \omega t} },
\end{equation}

so

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:500}
\begin{aligned}
\boldsymbol{\mathcal{E}}^2
&=
\inv{2} \lr{ \BE e^{j \omega t} + \BE^\conj e^{-j \omega t} }
\cdot
\inv{2} \lr{ \BE e^{j \omega t} + \BE^\conj e^{-j \omega t} } \\
&=
\inv{4} \lr{
\BE^2 e^{2 j \omega t}
+ \BE \cdot \BE^\conj + \BE^\conj \cdot \BE
+\lr{\BE^\conj}^2 e^{-2 j \omega t}
} \\
&=
\inv{2} \textrm{Re}
\lr{
\BE \cdot \BE^\conj
+
\BE^2 e^{2 j \omega t}
}.
\end{aligned}
\end{equation}

Similarly, for the cross product

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:540}
\begin{aligned}
\boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}}
&=
\inv{4}
\lr{
\BE \cross \BH e^{2 j \omega t}
+ \BE \cross \BH^\conj + \BE^\conj \cross \BH
+ \lr{ \BE^\conj \cross \BH^\conj } e^{-2 j \omega t}
} \\
&=
\inv{2}
\textrm{Re}
\lr{
\BE \cross \BH^\conj
+
\BE \cross \BH e^{2 j \omega t}
}.
\end{aligned}
\end{equation}

Given phasor representations of the sources \( \boldsymbol{\mathcal{M}} = \BM e^{j \omega t}, \boldsymbol{\mathcal{J}} = \BJ e^{j \omega t} \), \ref{eqn:energyMomentumWithMagneticSources:40} can be recast into (a messy) phasor form

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:560}
\begin{aligned}
\inv{2} &\textrm{Re} \inv{2} \PD{t}{} \lr{
\epsilon_0 \BE \cdot \BE^\conj
+ \mu_0 \BH \cdot \BH^\conj
+ \epsilon_0 \BE^2 e^{ 2 j \omega t}
+ \mu_0 \BH^2 e^{ 2 j \omega t}
} \\
&+
\inv{2} \textrm{Re} \spacegrad \cdot \lr{
\BE \cross \BH^\conj
+\BE \cross \BH e^{ 2 j \omega t}
} \\
&=
\inv{2} \textrm{Re}
\lr{
– \BH \cdot \BM^\conj
– \BE \cdot \BJ^\conj
– \BH \cdot \BM e^{2 j \omega t}
– \BE \cdot \BJ e^{2 j \omega t}
}.
\end{aligned}
\end{equation}

In particular, when averaged over one period, the oscillatory terms vanish. The time averaged equivalent of the Poynting theorem is thus

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:580}
0 =
{\left[
\textrm{Re}
\lr{
\inv{2} \PD{t}{} \lr{
\epsilon_0 \BE \cdot \BE^\conj
+ \mu_0 \BH \cdot \BH^\conj
}
+
\spacegrad \cdot \lr{
\BE \cross \BH^\conj
}
+
\BH \cdot \BM^\conj
+
\BE \cdot \BJ^\conj
}
\right]
}_{\textrm{av}}.
\end{equation}

Comparison to the reciprocity theorem result

The reciprocity theorem had a striking similarity to the Poynting theorem above, which isn’t suprising since both were derived by calculating the divergence of a Poynting like quantity.

Here’s a repetition of the reciprocity divergence calculation without the single frequency (phasor) assumption

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:600}
\begin{aligned}
\spacegrad \cdot &\lr{
\boldsymbol{\mathcal{E}}^{(a)} \cross \boldsymbol{\mathcal{H}}^{(b)}
-\boldsymbol{\mathcal{E}}^{(b)} \cross \boldsymbol{\mathcal{H}}^{(a)}
} \\
&=
\boldsymbol{\mathcal{H}}^{(b)} \cdot \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}}^{(a)} } -\boldsymbol{\mathcal{E}}^{(a)} \cdot \lr{ \spacegrad \cross \boldsymbol{\mathcal{H}}^{(b)} } \\
&\quad
-\boldsymbol{\mathcal{H}}^{(a)} \cdot \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}}^{(b)} } +\boldsymbol{\mathcal{E}}^{(b)} \cdot \lr{ \spacegrad \cross \boldsymbol{\mathcal{H}}^{(a)} } \\
&=
-\boldsymbol{\mathcal{H}}^{(b)} \cdot \lr{ \mu_0 \partial_t \boldsymbol{\mathcal{H}}^{(a)} + \boldsymbol{\mathcal{M}}^{(a)} }
-\boldsymbol{\mathcal{E}}^{(a)} \cdot \lr{ \boldsymbol{\mathcal{J}}^{(b)} + \epsilon_0 \partial_t \boldsymbol{\mathcal{E}}^{(b)} } \\
&\quad
+\boldsymbol{\mathcal{H}}^{(a)} \cdot \lr{ \mu_0 \partial_t \boldsymbol{\mathcal{H}}^{(b)} + \boldsymbol{\mathcal{M}}^{(b)} }
+\boldsymbol{\mathcal{E}}^{(b)} \cdot \lr{ \boldsymbol{\mathcal{J}}^{(a)} + \epsilon_0 \partial_t \boldsymbol{\mathcal{E}}^{(a)} } \\
&=
\epsilon_0
\lr{
\boldsymbol{\mathcal{E}}^{(b)} \cdot \partial_t \boldsymbol{\mathcal{E}}^{(a)}
-\boldsymbol{\mathcal{E}}^{(a)} \cdot \partial_t \boldsymbol{\mathcal{E}}^{(b)}
}
+
\mu_0
\lr{
\boldsymbol{\mathcal{H}}^{(a)} \cdot \partial_t \boldsymbol{\mathcal{H}}^{(b)}
-\boldsymbol{\mathcal{H}}^{(b)} \cdot \partial_t \boldsymbol{\mathcal{H}}^{(a)}
} \\
&+\boldsymbol{\mathcal{H}}^{(a)} \cdot \boldsymbol{\mathcal{M}}^{(b)}
-\boldsymbol{\mathcal{H}}^{(b)} \cdot \boldsymbol{\mathcal{M}}^{(a)}
+\boldsymbol{\mathcal{E}}^{(b)} \cdot \boldsymbol{\mathcal{J}}^{(a)}
-\boldsymbol{\mathcal{E}}^{(a)} \cdot \boldsymbol{\mathcal{J}}^{(b)}
\end{aligned}
\end{equation}

What do these time derivative terms look like in the frequency domain?

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:620}
\begin{aligned}
\boldsymbol{\mathcal{E}}^{(b)} \cdot \partial_t \boldsymbol{\mathcal{E}}^{(a)}
&=
\inv{4}
\lr{
\BE^{(b)} e^{j \omega t}
+
{\BE^{(b)}}^\conj e^{-j \omega t}
}
\cdot
\partial_t
\lr{
\BE^{(a)} e^{j \omega t}
+
{\BE^{(a)}}^\conj e^{-j \omega t}
} \\
&=
\frac{j \omega}{4}
\lr{
\BE^{(b)} e^{j \omega t}
+
{\BE^{(b)}}^\conj e^{-j \omega t}
}
\cdot
\lr{
\BE^{(a)} e^{j \omega t}

{\BE^{(a)}}^\conj e^{-j \omega t}
} \\
&=
\frac{\omega}{4}
\lr{
j \BE^{(a)} \cdot { \BE^{(b)} }^\conj
-j \BE^{(b)} \cdot { \BE^{(a)} }^\conj
+j \BE^{(a)} \cdot \BE^{(b)} e^{ 2 j \omega t }
-j { \BE^{(a)}}^\conj \cdot { \BE^{(b)} }^\conj e^{ -2 j \omega t }
} \\
&=
\inv{2} \textrm{Re}
\lr{
j \omega \BE^{(a)} \cdot { \BE^{(b)} }^\conj
+ j \omega \BE^{(a)} \cdot \BE^{(b)} e^{ 2 j \omega t }
}
\end{aligned}
\end{equation}

Taking the difference,

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:640}
\begin{aligned}
\boldsymbol{\mathcal{E}}^{(b)} \cdot \partial_t \boldsymbol{\mathcal{E}}^{(a)}
-\boldsymbol{\mathcal{E}}^{(a)} \cdot \partial_t \boldsymbol{\mathcal{E}}^{(b)}
&=
\inv{2} \textrm{Re}
\lr{
j \omega \BE^{(a)} \cdot { \BE^{(b)} }^\conj
– j \omega \BE^{(b)} \cdot { \BE^{(a)} }^\conj
+ j \omega \BE^{(a)} \cdot \BE^{(b)} e^{ 2 j \omega t }
– j \omega \BE^{(b)} \cdot \BE^{(a)} e^{ 2 j \omega t }
} \\
&=
– \omega \textrm{Im}
\lr{
\BE^{(a)} \cdot { \BE^{(b)} }^\conj
+ \BE^{(a)} \cdot \BE^{(b)} e^{ 2 j \omega t }
},
\end{aligned}
\end{equation}

so we have

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:660}
0
=
{
\left[
\spacegrad \cdot \textrm{Re} \lr{
\BE^{(a)} \cross {\BH^{(b)}}^\conj
-\BE^{(b)} \cross {\BH^{(a)}}^\conj
}
+
\omega \textrm{Im}
\lr{
\epsilon_0
\BE^{(a)} \cdot { \BE^{(b)} }^\conj
+
\mu_0
\BH^{(a)} \cdot { \BH^{(b)} }^\conj
}
+ \textrm{Re}
\lr{
-\BH^{(a)} \cdot { \BM^{(b)} }^\conj
+\BH^{(b)} \cdot { \BM^{(a)} }^\conj
-\BE^{(b)} \cdot { \BJ^{(a)} }^\conj
+\BE^{(a)} \cdot { \BJ^{(b)} }^\conj
}
\right]
}_{\textrm{av}}.
\end{equation}

Observe that the perfect cancellation of the time derivative terms only occurs when the cross product differences were those of the phasors. When those cross differences are those of the actual fields, like those in the Poynting theorem, there is a frequency dependent term is that expansion.