Hewlet Packard's appnote A-95 is a classic text on this subject. Since the related divisions of HP were spun off to form Agilent in the mid 90's, this text is now hosted at http://cp.literature.agilent.com/litweb/pdf/5952-1130.pdf.

Check pages 2 and 3 for the derivations of the S parameters. To summarize, given a two-port network as shown:

with impedance \$Z_0\$, the normalized incident voltages \$a_1\$ and \$a_2\$ are calculated as follows:

\$a_1=\frac{V_1+I_1*Z_0}{2\sqrt{Z_0}}\$

\$a_2=\frac{V_2+I_2*Z_0}{2\sqrt{Z_0}}\$

The reflected voltages \$b_1\$ and \$b_2\$ are calculated as follows. Note the similar form:

\$b_1=\frac{V_1-I_1*Z_0}{2\sqrt{Z_0}}\$

\$b_2=\frac{V_2-I_2*Z_0}{2\sqrt{Z_0}}\$

These values are derived from the voltages and currents. They're simply normalized so that the following equations make more sense and are more consistent and compact. The S parameters can be calculated as

\$s_{11} = \left.{\frac{b_1}{a_1}} \right|_{a_2=0}\$

\$s_{22} = \left.{\frac{b_2}{a_2}} \right|_{a_1=0}\$

\$s_{21} = \left.{\frac{b_2}{a_1}} \right|_{a_2=0}\$

\$s_{12} = \left.{\frac{b_1}{a_2}} \right|_{a_1=0}\$

The text goes into a good deal more detail on the meanings of each of these values and their derivations, but those equations should allow you to make progress on your problem.

The error is in that you're using the amplitude instead of the RMS value:

$$P = V_{RMS} \cdot I_{RMS} \cdot cos(\theta) $$

but for sinusoidal signals, \$V_{RMS}=\dfrac{V_{M}}{\sqrt{2}}\$ and \$I_{RMS}=\dfrac{I_{M}}{\sqrt{2}}\$ so multiplying:

$$P = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} V_{M} \cdot I_{M} \cdot cos(\theta)
= \frac{1}{2} V_{M} \cdot I_{M} \cdot cos(\theta) =\\
= \sqrt{\frac{1}{T} \int_T [p(t)]^2 \:\mathrm{d}t}
$$

## Best Answer

Let

VandIbe the instantaneous voltage and current on a load. From thedefinitionof power, voltage and current, we have the relation for instantaneous power:\$ p(t) = v(t) \cdot i(t) \$

Which means that the power on a given instant \$ t \$ is equal to the product of the voltage and the current exactly on that instant.

I'll assume you're familiar with what the phasor representation actually means. Just to state that shortly: a phasor is a mathematical shorthand for representing a sinusoid at a given unknown frequency.

So, \$ V=V_{M} ∠ \phi _{V} \$ is a shorthand for \$ v(t) = V_{M} \cdot cos(\omega t+ \phi _{V})\$. Similarly: \$ I=I_{M} ∠ \phi _{I} \$ means \$ i(t) = I_{M} \cdot cos(\omega t+ \phi _{I})\$.

Multiplying \$ v(t) \cdot i(t) \$ for all \$ t \$, gives us the waveform of the

instantaneous powerfor every \$t\$. Working on that multiplication:\$s(t) = v(t) \cdot i(t) = V_{M} \cdot cos(\omega t+ \phi _{V}) \cdot I_{M} \cdot cos(\omega t+ \phi _{I})\$

As \$ cos(u) \cdot cos(v) = \cfrac{1}{2} \cdot [cos(u-v)+cos(u+v) ]\$, with \$ u = \omega t+ \phi _{V} \$ and \$ v = \omega t+ \phi _{I} \$, we can simplify the equation above to:

\$ s(t) = v(t) \cdot i(t) = \cfrac{V_{M}I_{M}}{2} \cdot [cos(\phi _{V} - \phi _{I}) + cos(2\omega t+ \phi _{V} + \phi _{I})] \$

This waveform is pretty interesting for itself: it is a constant value \$\cfrac{V_{M}I_{M}}{2} \cdot cos(\phi _{V} - \phi _{I}) \$ summed by a sinusoid \$ \cfrac{V_{M}I_{M}}{2} cos(2\omega t+ \phi _{V} + \phi _{I})]\$.

This clearly shows that the instantaneous power

is notconstant with time.Based on that result, we can see that the

meanpower is equal to the non-varying component of \$ s(t) \$ (it's pretty straightforward to prove that mathematically, one just have to solve the integral \$ \cfrac{1}{T}\int_{t}^{t+T}{s(t)dt}\$ )Motivated by this result, and by the pretty sweet geometrical interpretation of \$ VIcos(\phi _{V} - \phi _{I}) \$, that value has been defined as the

real power, that is, the power that is actually delivered to the load. Now you know that this so calledreal poweris nothing more than the mean power at the load.Diving into this concept a little bit (it's a pitty I can't draw here, but I'll try):

Let

vbe a vector with magnitude ||v|| and phase \$ \phi_v \$, andibe a vector with magnitude ||i|| and phase \$ \phi_i \$ If you multiply ||i|| by \$cos(\phi_v-\phi_i)\$ you have theprojection of i over v. On the other hand, \$||i||sin(\phi_v-\phi_i)\$ is said to be the component ofiinquadraturewithv.Now you can understand why the mean power has a cool geometric interpretation: the mean power is the voltage multiplied by the projection of the current over the voltage, on the phasor space.

This motivated the creation of the complex power

Sas:With this definition, the real part of the vector is exactly the mean power delivered to the load, and the complex part is the power said to be in

quadrature, called reactive power (google for Power Triangle to see the geometrical interpretation of this result).Ok, now going back to the \$ s(t) \$ definition, we see that \$ P = \cfrac{V_M I_M}{2} \cdot cos(\phi_v - \phi_i) \$ and \$ Q \$, by definition, and to comply with the definition of S, is equal to \$ \cfrac{V_M I_M}{2} \cdot sin(\phi_v - \phi_i) \$

So, as we wanted to prove at the begining:

\$ S = P + jQ = \cfrac{V_M I_M}{2} \cdot cos(\phi_v - \phi_i) + j\cfrac{V_M I_M}{2} \cdot sin(\phi_v - \phi_i)\$

\$ S = \cfrac{V_M I_M}{2} \cdot [cos(\phi_v - \phi_i) + jsin(\phi_v - \phi_i) ]\$

\$ S = \dfrac{V_{M} ∠ \phi _{V} \cdot I_{M} ∠ -\phi _{I}}{2} \$

\$ S = \cfrac{V \cdot I*}{2} \$

So, there you go, what you wanted to see ;)

edit: What's the physical interpretation of Q?I've shown above what's the physical interpretation of the real part of the complex power, P, that is, the mean power delivered to the load. But what's exactly Q, how can one visualize it? It's based on the fact that cos and sin are

orthogonal, and the principle of superposition can be applied to power if the two waveforms involved in the calculation are orthogonal. Let's go into the math, because that's really what matters.Using the result obtained above: \$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [cos(\phi _{V} - \phi _{I}) + cos(2\omega t+ \phi _{V} + \phi _{I})] \$

\$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [1 + cos(2(\omega t+ \phi _{V}))] \$

That is a sinusoid centered on \$ \cfrac{V_{M}I_{M}}{2} \$ with that same amplitude (its minimum value is 0 and its maximum value is \$ V_{M}I_{M} \$ ). Let's call it

P\$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [0 - cos(2(\omega t+ \phi _{V}) - \cfrac{\pi}{2} )]\$

\$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [sin(2(\omega t+ \phi _{V}))] \$

That is a purely oscillatory waveform with mean value equal to 0. Let's call this result

Q.In this case, s(t) is exactly the general equation we found on the discussion above. But we can rewrite that to make use of the result of the two previous cases, like this:

First, we rewrite the equation in terms of \$ \theta \$ (notice that \$ \phi_V + \phi_I = \phi_V -\phi_V + \phi_V + \phi_I = 2\phi_V - \theta \$): \$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [cos(\theta) + cos(2(\omega t+ \phi _{V}) - \theta)] \$ Knowing that: \$ cos(x-y) = cos(x)cos(y) + sin(x)sin(y)\$, letting \$ x = 2(\omega t+ \phi _{V}) \$ and \$ y = \theta \$

\$ s(t) = \cfrac{V_{M}I_{M}}{2} \cdot [cos(\theta) + cos(\theta)cos(2(\omega t + \phi_V)) + sin(\theta)sin(2(\omega t + \phi_V))]\$

Rearranging the terms:

\$ s(t) = cos(\theta) \cdot \cfrac{V_{M}I_{M}}{2} \cdot [1 + cos(2(\omega t + \phi_V))] + sin(\theta) \cdot \cfrac{V_{M}I_{M}}{2} sin(2(\omega t + \phi_V))\$

Using the result of the two first cases above:

\$ s(t) = cos(\theta)P + sin(\theta)Q \$

An amazing result, right? What does that mean?

Let's go back to what we are doing: calculating the power for the generic case where \$ \phi _{V} - \phi _{I} = \theta \$, that is, solvig the equation:

\$ s(t) = V_{M}cos(\omega t + \phi_V) \cdot I_{M}cos(\omega t + \phi_I) \$

Can we rewrite \$ i(t) = I_{M}cos(\omega t + \phi_I) \$ in the form of \$ i(t) = K_1 cos(\omega t + \phi_V) + K_2 sin(\omega t + \phi_V) \$?

Let's try:

\$ \phi_I = \phi_V - \theta \$ \$ i(t) = I_{M}cos(\omega t + \phi_V - \theta \$) \$

Letting \$ \omega t + \phi_V = u \$ and \$ \theta = v \$

With the relation:

\$ cos(u-v) = cos(u)cos(v) + sin(u)sin(v) \$

We have:

\$ i(t) = I_{M}cos(\theta)cos(\omega t + \phi_V) + I_{M}sin(\theta)sin(\omega t + \phi_V) \$

Just what we wanted, to rewrite i(t) as a sum of two components: one in phase with v(t), and one in quadrature with v(t)!

Now the result of the case 3 can be explained: i(t) can be decomposed in two components, as shown above, and the power generated by i(t) is equal to the power generated by each one of these components

individually. Whoa, just like superposition but for power! (Remember that this is only true, and it was proven above, because cos and sin are orthogonal)So

Qis the amount of power generated by the component of i(t) that's in quadrature with v(t). It is purely oscillatory and has no mean value.Pis the amount of power generated by the component of i(t) that's in phase with v(t). It is oscillatory but has a mean value that's equal the mean power delivered to the load.And the complex power

S, the total power, is exactly the sum of these two components