# S parameters of a capacitor

I am struggling to understand S parameters. As an example, I am considering the S matrix of a capacitor in series with a transmission line. It has two ports, so must be represented by 2x2 matrix. But the form of this matrix eludes me. I thought I could deduce it from the reflection coefficient, giving

$$S=\left( \begin{array}{cc} \frac{Z-Z_0}{Z+Z_0} & 1-\frac{Z-Z_0}{Z+Z_0}\\ 1-\frac{Z-Z_0}{Z+Z_0} & \frac{Z-Z_0}{Z+Z_0}\\ \end{array} \right)$$

where $Z=1/i \omega C$, but this doesn't seem to give the right answers. Is this correct? Or have I misunderstood the concept?

• Is your capacitor connected in parallel (from the trace to ground) or in series (connecting the input trace to the output trace)? – The Photon Sep 29 '14 at 19:54
• In series. I shall clarify... – NLambert Sep 29 '14 at 19:58
• No time for a full answer now, but for calculating S11 you need to consider the input impedance when the output is properly terminated. This means the Z in your equation should be $1/(j\omega{}C) + Z_0$. – The Photon Sep 29 '14 at 20:03

When calculating the S-parameters, you should terminate all of the ports that don't have stimulus applied. So, in your situation, to calculate $S_{11}$ and $S_{21}$, you'd be working with this circuit:

simulate this circuit – Schematic created using CircuitLab

Notice that the current passing through the capacitor from port 1 to port 2 is $i_1$ and $-i_2$ at the same time.

Now you can easily calculate $i_1$ and $i_2$ using the series impedance rule:

$$i_1 = -i_2 = \frac{v_s}{2Z_0+1/j\omega{}C}$$

And you can also calculate $v_1$ and $v_2$ using the voltage divider rule,

$$v_1 = \frac{Z_0+1/j\omega{}C}{2Z_0+1/j\omega{}C}v_s$$

$$v_2 = \frac{Z_0}{2Z_0+1/j\omega{}C}v_s$$

Then you can convert these voltages to the incident and reflected wave variables at each port by

$$a_n=\frac{1}{2}\frac{v_n+Z_0i_n}{\sqrt{\Re\left({Z_0}\right)}}$$

$$b_n=\frac{1}{2}\frac{v_n-Z_0^\star{}i_n}{\sqrt{\Re\left({Z_0}\right)}}$$

(which get a lot less hairy when you start plugging in numbers, since your $Z_0$ is purely real)

And then you have

$$S_{11}=\frac{b_1}{a_1}$$

and

$$S_{21}=\frac{b_2}{a_1}$$

I'm pretty sure this will show you that your equations for the off-diagonal parameters aren't quite right. You might be thinking that $S_{11} + S_{21} = 1$ because of conservation of energy, however this isn't correct because the travelling wave parameters are not proportional to the signal power but to its square root. To get the correct conversions from Z-parameters to S-parameters, see the Wikpedia page on Z-parameters.

• Doesn't a_2, and thus S21, always reduce to zero here? As v_2 = -Z0 i_2 – NLambert Sep 30 '14 at 9:19
• @NLambert, you're right --- I was thinking in terms of forward and reverse waves, but the formulas are really for incident and scattered waves. I've edited to fix it. – The Photon Sep 30 '14 at 15:31
• +1. Small q-n from a student, if I may. For S11=b1/a1 to be true, a2 must be zero. Right? a2=0 assumes ideal termination of port 2. But how to properly terminate a capacitor? It seems Z0 (50 Ohm) would not work here or I may be wrong. – Sergei Gorbikov Nov 9 '16 at 15:34
• @SergeiGorbikov, yes, $a_2$ is zero, because there is no incoming wave from the right. Proper termination means matching $Z_0$, not matching the device being tested (DUT). If the DUT generates a reflection when properly matched, that is considered part of the reflection produced by the device, the $S_{11}$ you are trying to measure, so you don't want to supress it. – The Photon Nov 9 '16 at 16:47
• @ThePhoton Now clear. 10x!! – Sergei Gorbikov Nov 10 '16 at 8:25

ThePhoton has given a very good answer. However, below is an expanded version with complete derivation of the formula and check in LTSpice, as well as power transfer analysis.

I. The S-matrix for the 2-port system is

$$\begin{pmatrix} {\frac{Z}{{2{Z_0} + Z}}}&{\frac{{2{Z_0}}}{{2{Z_0} + Z}}}\\ {\frac{{2{Z_0}}}{{2{Z_0} + Z}}}&{\frac{Z}{{2{Z_0} + Z}}} \end{pmatrix},$$

where $Z \equiv {1 \over {j\omega C}}$, ${Z_0}$ is the real characteristic impedance of the system. The matrix is true for any Z, both with resistive and reactive components.

II. Derivation

For real ${Z_0}$ $${a_i} = {{{V_i} + {I_i}{Z_0}} \over {2\sqrt {{Z_0}} }},\:{b_i} = {{{V_i} - {I_i}{Z_0}} \over {2\sqrt {{Z_0}} }}$$ Source: "Power Waves and the Scattering Matrix", K. Kurokawa, IEEE, 1965, URL.

From voltage divider rule
$${V_1} = {V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}},\:{V_2} = {V_s}{{{Z_0}} \over {2{Z_0} + Z}}$$

From Ohm's law
$${I_1} = {{{{V_s}} \over {2{Z_0} + Z}}}$$

Then the value for the input reflection coefficient is

$${S_{11}} = {\left. {{{{b_1}} \over {{a_1}}}} \right|_{{a_2} = 0}} = {{{V_1} - {I_1}{Z_0}} \over {{V_1} + {I_1}{Z_0}}} = {{{V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}} - {{{V_s}} \over {2{Z_0} + Z}}{Z_0}} \over {{V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}} + {{{V_s}} \over {2{Z_0} + Z}}{Z_0}}} = {Z \over {2{Z_0} + Z}}$$

Applying ${I_2} = - {I_1}$, the value for the forward gain is
$${S_{21}} = {\left. {{{{b_2}} \over {{a_1}}}} \right|_{{a_2} = 0}} = {{{V_2} - {I_2}{Z_0}} \over {{V_1} + {I_1}{Z_0}}} = {{{V_2} + {I_1}{Z_0}} \over {{V_1} + {I_1}{Z_0}}} = {{{V_s}{{{Z_0}} \over {2{Z_0} + Z}} + {{{V_s}} \over {2{Z_0} + Z}}{Z_0}} \over {{V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}} + {{{V_s}} \over {2{Z_0} + Z}}{Z_0}}} = {{2{Z_0}} \over {2{Z_0} + Z}}$$

Note that ${S_{11}} + {S_{21}} = 1$, what was expected if no incoming wave at the second port (${a_{2}}=0$).

III. Conservation of energy

As for the law of conservation of energy
$${\left| {{a_1}} \right|^2} - {\left| {{b_1}} \right|^2} = {\left| {{b_2}} \right|^2} + \Delta,$$
where the ${\left| {{a_1}} \right|^2} - {\left| {{b_1}} \right|^2}$ is the power supplied by the generator, ${\left| {{b_2}} \right|^2}$ is the power consumed by the load with the delta being the power consumed by the DUT.

Pls, refer to the same paper by Kurokawa if not clear why these particular expressions are used to calculate power transfers.

Let's show that, in case of a capacitor, $\Delta$ is zero (no power is consumed by the DUT). Remembering $Z \equiv {1 \over {j\omega C}}$

$${\left| {{a_1}} \right|^2} - {\left| {{b_1}} \right|^2} = {\left| {{{{V_1} + {I_1}{Z_0}} \over {2\sqrt {{Z_0}} }}} \right|^2} - {\left| {{{{V_1} - {I_1}{Z_0}} \over {2\sqrt {{Z_0}} }}} \right|^2} = {1 \over {4{Z_0}}}\left( {{{\left| {{V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}} + {{{V_s}} \over {2{Z_0} + Z}}{Z_0}} \right|}^2} - {{\left| {{V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}} - {{{V_s}} \over {2{Z_0} + Z}}{Z_0}} \right|}^2}} \right) = {{{V_s}^2} \over {4{Z_0}}}\left( {{{\left| {{{2{Z_0} + Z} \over {2{Z_0} + Z}}} \right|}^2} - {{\left| {{Z \over {2{Z_0} + Z}}} \right|}^2}} \right) = {{{V_s}^2} \over {4{Z_0}}}\left( {1 - {{{{\left| Z \right|}^2}} \over {{{\left| {2{Z_0} + Z} \right|}^2}}}} \right) = {{{V_s}^2} \over {4{Z_0}}}\left( {1 - {{{1 \over {{C^2}{\omega ^2}}}} \over {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}}}} \right) = {{{V_s}^2} \over {4{Z_0}}}{{4Z_0^2} \over {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}}} = {V_s}^2{{{Z_0}} \over {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}}}$$

Now, the same for the transmitted wave: $${\left| {{b_2}} \right|^2} = {\left| {{{{V_2} - {I_2}{Z_0}} \over {2\sqrt {{Z_0}} }}} \right|^2} = {\left| {{{{V_2} + {I_1}{Z_0}} \over {2\sqrt {{Z_0}} }}} \right|^2} = {1 \over {4{Z_0}}}\left( {{{\left| {{V_s}{{{Z_0}} \over {2{Z_0} + Z}} + {{{V_s}} \over {2{Z_0} + Z}}{Z_0}} \right|}^2}} \right) = {{{V_s}^2} \over {4{Z_0}}}{\left| {{{2{Z_0}} \over {2{Z_0} + Z}}} \right|^2} = {{{V_s}^2} \over {4{Z_0}}}{{4Z_0^2} \over {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}}} = {V_s}^2{{{Z_0}} \over {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}}} = {\left| {{a_1}} \right|^2} - {\left| {{b_1}} \right|^2},\, \Rightarrow \Delta = 0.$$

IV. Extraction of s-parameters in LTSpice and check of the analytical formula

Finally, let's extract an s-parameter with LTSpice and compare it with our analytical formula. Let's do it for, say, ${S_{21}}$:

$$\left| {{S_{21}}} \right| = \left| {{{2{Z_0}} \over {2{Z_0} + Z}}} \right| = {{2{Z_0}} \over {\sqrt {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}} }} = 0.30,\,$$
at ${Z_{0}}$=50 Ohm, $\omega$=10 MHz and C=50 pF.

Extraction in LTSpice:

What this chart says is that
1) at lower frequencies forward gain is low (nothing breaks through to the load), as the capacitor behaves as open,
2) at higher frequencies forward gain is close to one: incoming power wave is fully transmitted to the load as the capacitor behaves as short.