# What are S-parameters, anyway?

Figure 1. You can evaluate three-terminal networks like this one by solving three simultaneous equations having three unknowns. |

Elementary circuit theory provides many methods for describing electronic networks. Those methods, however, best describe DC and low-frequency circuits. They fall short when the wavelengths of the signals of interest shrink to become comparable to the physical dimensions of the circuit of interest. To characterize high-frequency circuits, you can employ S-parameters (or scattering parameters) in place of the impedance or admittance parameters that describe low-frequency circuits.

To give you a basis for understanding S-parameters, I will first review low-frequency analysis techniques. Most college texts present circuit analysis in terms of equations describing node voltages and loop currents. For a three-terminal circuit, such as the one shown in **Figure 1**, you can write three simultaneous equations in six variables; for the node voltages and current directions shown in Figure 1, these equations suffice:

*i _{1}* +

*i*=

_{3}*i*(1)

_{2}*i _{1}* = (

*v*–

_{1}*v*)/

_{3}*R*(2)

_{1}*i _{2}* = (

*v*–

_{3}*v*)/

_{2}*R*(3)

_{2}If you specify any three variables, you can calculate the rest.

Of course, without having to solve any equations, you recognize that if the current into node 3 of the Figure 1 circuit is zero (*i _{3}* = 0), the voltage at node 3 is 80% of the difference between the node 1 and node 2 voltages.

A problem

But address this question: for *v _{1}* –

*v*= 10 V (assume that

_{2}*v*= 0), what voltage at node 3 will sustain a 1-A current into node 3? To get the answer, substitute the specified values into Equations 1 through 3 to obtain Equations 4 through 6:

_{2}(4)

(5)

(6)

Substituting Equations 5 and 6 into Equation 4 yields

(7)

If you multiply both sides of Equation 7 by 8 W, Equation 7 reduces to 40 V – 4*v _{3}* + 8 V =

*v*, or 5

_{3}*v*= 48 V, so

_{3}*v*= 9.6 V.

_{3}That’s not too tough a calculation, but you probably would need paper and pencil to solve it. Further, the algebra increases dramatically with circuit node count. Although you can use a computer to solve sets of algebraic equations, you might find it difficult to conveniently load your equations into a computer. (The easiest way, in fact, is probably to enter your circuit graphically using a schematic-capture program and then use a simulator such as Spice to develop and solve the equations for you.)

But whether you intend to calculate circuit values by hand or with computer assistance, you can simplify the computational and data-management aspects of the problem if you can group your circuit nodes into appropriate pairs, which leads to the concepts of ports and matrix representations of circuit characteristics (see “Matrix math methods”). In fact, you needn’t know anything about internal circuit topology to make use of the port concept. Given a black box, you can make lab measurements that let you develop a simple matrix representation of the internal circuitry.

Figure 2. (a) You can completely describe the external functionality of a two-port network by means of a 2x2 matrix. (b) The voltage divider of Figure 1 constitutes a two-port network. |

So, what is a port?

**Figure 2a** shows a two-port network. You can treat any circuit as a two-port network if you can select two pairs of nodes (for example, an input pair and an output pair) such that the current into the positive node of a pair equals the current out of the negative node of the same pair—that is, *i _{1}* must equal

*i*and

_{A}*i*must equal

_{2}*i*in Figure 2a. Figure 2b shows the Figure 1 circuit rearranged to emphasize that it is indeed a two-port network. (See clarification, below)

_{B}A two-port network can be represented by a 2-by-2 matrix. (An *n*-port network, having *n* pairs of nodes, can be represented by an *n*-by-*n* matrix). In Figure 2a, the *x _{ij}* terms (where

*x*represents the value at row

_{ij}*i*column

*j*) stand in for impedance parameters (Z-parameters), admittance parameters (Y-parameters), hybrid parameters (h-parameters), chain parameters (A-, B-, C-, and D-parameters), or S-parameters. I will briefly review Z-parameter representations to illustrate how matrix representations work and how you can derive matrix parameters from laboratory measurements. Then, I will show you how you can apply similar matrix representations to characterize high-frequency circuits using S-parameters.

In a Z-parameter representation, the matrix elements take on the values that satisfy this matrix equation:

(8)

To measure *z _{11}*, you leave port 2 open-circuited, apply a test voltage

*v*to port 1, and divide that voltage by the resulting current

_{1}*i*into port 1:

_{1}*z _{11}* =

*v*/

_{1}*i*for

_{1}*i*= 0 (9)

_{2}Similarly, to measure *z _{22}*, you leave port 1 open-circuited, apply a test voltage

*v*to port 2, and divide that voltage by the resulting current

_{2}*i*into port 2:

_{2}*z _{22}* =

*v*/

_{2}*i*for

_{2}*i*= 0 (10)

_{1}Parameters *z _{11}* and

*z*are called the open-circuit driving-point impedances (Ref. 1).

_{22}Visualizing the measurements of the remaining two matrix entries—the open-circuit transfer impedances—with respect to the instrumentation you would use is slightly more difficult. Mathematically, the upper-right matrix parameter is

*z _{12}* =

*v*/

_{1}*i*for

_{2}*i*= 0 (11)

_{1}To make the measurement, you can use two voltage sources: use one to apply test voltage *v _{1}* to port 1; then monitor current

*i*into port 1 and adjust the second voltage source, connected to port 2, until

_{1}*i*= 0. Then, divide

_{1}*v*by the resulting current

_{1}*i*into port 2. Similarly, to measure the lower-left matrix parameter, you apply

_{2}*v*to port 2, adjust the

_{2}*v*voltage until

_{1}*i*goes to zero, and divide

_{2}*v*by the resulting

_{2}*i*:

_{1}*z _{21}* =

*v*/

_{2}*i*for

_{1}*i*= 0 (12)

_{2}You can try these out for the circuit values shown in Figure 2b, either on the bench or with some quick calculations. Calculating the *z* terms on the matrix major diagonal (top left to bottom right) is simple:

With *i _{2}* = 0,

*z*=

_{11}*v*/

_{1}*i*where

_{1},*i*=

_{1}*v*/(2 W + 8 W ). So

_{1}*z*= 10 V.

_{11}With *i _{1}* = 0,

*z*=

_{22}*v*/

_{2}*i*where

_{2},*i*=

_{2}*v*/(8 W ). So

_{2}*z*= 8 W.

_{22}Calculating the remaining two terms is a tad more difficult. To determine *z _{12}* =

*v*/

_{1}*i*for

_{2}*i*= 0, note that if

_{1}*i*= 0, then

_{1}*v*=

_{2}*v*and

_{1},*i*=

_{2}*v*/8 W, so

_{1}*z*= 8 W.

_{12}Finally, to determine *z _{21}* =

*v*/

_{2}*i*for

_{1}*i*= 0, note that

_{2}*i*=

_{1}*v*/10 W, so

_{1}*z*=

_{21}*v*/(

_{2}*v*/10 W), or (10 W)

_{1}*v*/

_{2}*v*Note also that if

_{1}.*i*= 0, then

_{2}*v*= 0.8

_{2}*v*, so

_{1}*z*= 8 W. These calculations yield the following matrix equation:

_{12}(13)

You can test out this representation on the Figure 2 circuit problem that was solved above with three node equations: if *v _{1}* = 10 V, what value of

*v*will sustain a 1-A current into port 2? With these values, Equation 13 becomes

_{2}(14)

which yields

10 V = (10 V) *i _{1}*+(8 V)(1 A) = (10 V)

*i*+ 8 V (15)

_{1}and

*v _{2}* = (8 V)

*i*+ (8 V)(1 A) (16)

_{1}Therefore, from Equation 15, *i _{1}* = (2 V)/10 W, or 0.2 A, which, substituted into Equation 16, yields

*v*= (8 W)(0.2 A) + 8 V = 9.6 V, which is the same result obtained from solving the node equations.

_{2}Active networks

Figure 3. Matrix parameters can describe active as well as passive networks. This amplifier includes a transistor having a beta of 1000 and a 1-MW input resistance. |

You can apply matrix parameters to active networks as well as to passive ones like the Figure 2 voltage divider. **Figure 3** models a transistor amplifier using two resistors and a dependent current source. Using equations 9 and 10, you can determine that *z _{11}* =

*v*/

_{1}*i*for

_{1}*i*= 0 is

_{2}*R*and that

_{b}*z*=

_{22}*v*/

_{2}*i*for

_{2}*i*= 0 is

_{1}*R*.

_{L}To determine *z _{12}* =

*v*/

_{1}*i*for

_{2}*i*= 0 (Equation 11), imagine applying a test current (1 A, for instance) to port 2. Then, while holding

_{1}*i*to zero, measure

_{1}*v*. If

_{1}*i*= 0, then

_{1}*v*must equal zero, so

_{1}*z*= 0/(1 A) = 0. This result helps to illustrate the physical meaning of Z- and other matrix parameters. The subscript

_{12}*ij*indicates the effect on port i of a test input applied to port j. The fact that

*z*= 0 for the transistor amplifier simply means that nothing you do to the amplifier’s output will change its input.

_{12}Conversely, parameter *z _{21}* does have a nonzero value, meaning that something done to the input will affect the output (as you would expect for an amplifier). To calculate

*z*=

_{21}*v*/

_{2}*i*for

_{1}*i*= 0 (Equation 12), first note that for

_{2}*i*= 0, 1000

_{2}*i*must equal

_{b}R_{L}*v*. Therefore,

_{2}*z*= (1000

_{21}*i*)/

_{b}R_{L}*i*, and since

_{1}*i*equals

_{1}*i*(representing transistor base current),

_{b}*z*= 1000

_{21}*R*This matrix equation therefore represents the transistor amplifier:

_{L}.(17)

You can use this equation to calculate the no-load output voltage in response to a 1-V input with the resistance values as shown in Figure 3:

(18)

Therefore, 1 V = (1 MW) *i _{1}*, so

*i*= 1 µA, and

_{1}*v*= (10 MW)(1 µA) = 10 V.

_{2}Figure 4. You can cue up an incident voltage waveform and direct it toward an n-port network. An S-parameter matrix describes the relative strengths of reflected and transmitted signals at each port. The plus superscript represents an incident wave (moving toward the network). The minus superscript indicates a wave moving away from the network (whether transmitted or reflected). |

**Z-parameters for microwaves?**

The impedance matrix is a general analytical tool applicable in theory to any multiport network. Practically, though, it is difficult to apply to microwave networks. You can’t conveniently apply a test signal and simultaneously monitor the response. Current and voltage ratios aren’t constant throughout a microwave network; they depend on position relative to wavelength. You can choose to make network measurements at a location that you define as the network’s reference plane (or calibration plane), but even then your instrumentation leads will introduce loading and other errors. To accurately measure millimeter-wavelength voltage and current signals simultaneously at a reference plane, you would need miniature sources and meters with submillimeter-length leads.

What you can do to measure a microwave network is apply incident waveforms and measure the resulting waveforms that your network reflects and transmits (**Figure 4**). Just as Z-parameters relate port voltages and currents, S-parameters relate incident waves to transmitted or reflected ones.

Note that S-parameters are not independent of Z-parameters. Z-parameters are not unique to low-frequency circuits, and S-parameters are not unique to microwave networks. You can describe any network in terms of either, and if you have one set of parameters, you can derive another. Here, for example, is *s _{11}* in terms of the Z-parameters:

(19)

where *z _{01}* and

*z*are the characteristic impedances of ports 1 and 2. Ref. 2 provides a table of the algebraic relationships among all the S-parameters, Y-parameters, and chain parameters for two-port networks. As Equation 19 suggests, deriving S-parameter descriptions based on network topology would be a nightmare. Because S-parameters directly relate incident and reflected or transmitted waves, however, they are ideal for characterizing microwave devices in a laboratory.

_{02}Consider only ports 1 and 2 in Figure 4. For source and measurement instruments having the same characteristic impedance (typically 50 W ), parameter *s _{11}* describes the relationship between port 1’s incident signal

*V*

_{1}^{+}and port 1’s resulting reflected signal

*V*

_{1}^{–}:

*s*=

_{11}*V*

_{1}^{–}/

*V*

_{1}^{+}. Similarly,

*s*=

_{22}*V*

_{2}^{–}/

*V*

_{2}^{+}

_{,}relating incident and reflected waves on port 2. (In commonly used terminology, the plus superscript always indicates a waveform propagating toward a network, and the minus superscript represents a waveform, either reflected or transmitted, moving away from the network. Some instrument vendors use the variable

*a*to represent incident waves and

_{ij}*b*to represent transmitted and reflected waves.) The remaining two S-parameters for two-port networks relate incident waveforms at one port to transmitted waves at the other:

_{ij}*s*=

_{12}*V*

_{1}^{–}/

*V*

_{2}^{+}(that is, the transmitted signal at port 1 divided by the incident signal at port 2) and

*s*=

_{21}*V*

_{2}^{–}/

*V*

_{1}^{+}(that is, the transmitted signal at port 2 divided by the incident signal at port 1). The complete matrix representation follows:

(20)

Figure 5. Vector network analyzers include the source and receiver instruments needed to measure S-parameters. You can choose S-parameter test-set options to conveniently route signals to and from your DUT. |

Figure 6. These measurement examples and corresponding S-parameter matrices are based on instrumentation having a 50-W characteristic impedance. (a) For a 25-W resistor connected to port 1, the S-parameter matrix reduces to a single parameter, s_{11} = 0.333 with a 180° phase angle. (b) A 3-in. length of cable represents about a quarter wavelength at 1 GHz, so s_{12} = s_{21} = 1 at 90°. (c) A matched amplifier exhibits no reflections on input or output, and a signal applied to the output won’t affect the input, so s_{11} = s_{12} = s_{22} = 0. The nonzero s_{21} parameter represents the amplifiers gain and phase shift. |

To make S-parameter measurements, you need a signal source and receivers capable of measuring your source signal as well as the response signals reflected from or transmitted through your DUT. Vector network analyzers (which measure signal magnitude and phase) include such instrumentation and are readily adaptable to S-parameter measurements. Vendors offer their vector network analyzers with S-parameter test-set options (**Figure 5**), which include the power splitters, switches, and couplers necessary to route signals to and from your DUT.

The measurement examples and corresponding S-parameter matrices shown in **Figure 6** are based on a vector network analyzer and S-parameter test set having 50-W characteristic impedance. If you connect a 25-W resistor to port 1 and leave port 2 unused (Figure 6a), the S-parameter matrix reduces to a single parameter (the resistor is a one-port network requiring a 1x1 matrix), which represents the reflection coefficient. The following equation (Ref. 3) describes the reflection coefficient due to a load impedance *Z _{L}* with respect to characteristic impedance

*Z*:

_{0}r = (*Z _{L}* –

*Z*)/(

_{0}*Z*+

_{L}*Z*) (21)

_{0}For the values shown in Figure 6a, r = *s _{11}* = (25 W-50 W)/(25 W+50 W) = –0.333, or 0.333 with a 180° phase angle.

Figure 6b shows the application of a 1-GHz signal to a 3-in. length of lossless 50-V cable. The lossless cable contributes no amplification or attenuation, but the 3-in. length represents about a quarter wavelength at 1 GHz and therefore contributes a 90° phase shift, so *s _{12}* = 1 at 90° (that is, a signal applied to port 2 produces a signal delayed by 90° at port 1) and

*s*= 1 at 90° (that is, a signal applied to port 1 produces a signal delayed by 90° at port 2).

_{21}Figure 6c shows a matched 15-dB amplifier. It exhibits no reflections on input or output, and, as for the Figure 3 circuit, a signal applied to the output won’t affect the input, so *s _{11}* =

*s*=

_{12}*s*= 0. The nonzero

_{22}*s*parameter represents the amplifier’s gain and phase shift. Note that the gain in decibels relates to the S-parameter as follows: 15 dB = 20log

_{21}_{10}(

*V*

_{2}^{–}/

*V*

_{1}^{+}) where

*s*=

_{21}*V*

_{2}^{–}/

*V*

_{1}^{+}, so 0.75 = log

_{10}(

*s*). Therefore,

_{21}*s*= 10

_{21}^{0.75}= 5.62 (at an unspecified phase angle).

These examples illustrate how easy it is to relate S-parameters to the performance of microwave circuits. The April issue of *T&MW* will provide more information on the vector network analyzers you can use to make S-parameter measurements. *T&MW*

**References**

1. Murdoch, Joseph B., *Network Theory,* McGraw-Hill, New York, NY, 1970. p. 69.

2. Mongia, Rajesh, Inder Bahl, and Prakash Bhartia, *RF and Microwave Coupled-Line Circuits*, Artech House, Boston, MA, 1999. p. 52.

3. Nelson, Rick, “High Speeds and Fine Precision Knock PCB Traces Off Pedestal,”*Test & Measurement World,* January 2000. p. 22.

**Rick Nelson***received a BSEE degree from Penn State University. He has six years experience designing electronic industrial-control systems. A member of the IEEE, he has served as the managing editor of* EDN*, and he became a senior technical editor at* T&MW *in 1998. E-mail: rnelson@tmworld.com.*

Matrix math methods
Matrix notation offers a convenient way of manipulating the sets of simultaneous algebraic equations often used to represent circuit performance. For example, the set of n equations which might relate an n-node circuit’s voltages, currents, and impedances, is equivalent to this matrix representation: which you can also write as V = ZINote that this representation lets you directly calculate values for V as a function of Z and I. You can manipulate this equation algebraically to provide I in terms of V and Z:I = Z^{-1}VWhere Z is the inverse of matrix ^{-1}Z.Obtaining the inverse of a matrix is tedious and difficult for a human (it involves calculation of matrix cofactors and determinants as well as transposition, not to mention a trip to the library to find a textbook that explains those terms). But matrix inversion is easy for a computer or calculator ( Figure).Furthermore, matrix representations provide a consistent way to represent circuit properties. If you and a colleague sit down to develop sets of loop or node equations for a moderately complex circuit, your sets will probably differ—they may both be right, but either or both may contain errors that wouldn’t be readily apparent. If you solve the equations and get different numerical answers, you’ll have no easy way of telling who, if either of you, is right. If you agree on port definitions ahead of time, however, the matrix representations you and your colleague derive should be identical. If not, you can rework your derivations until you both agree. Then, let a computer or calculator do the heavy computational lifting.— Rick Nelson |

**Clarification**

** A reader wrote in:** In Fig. 2(b) shouldn't the voltage at the right node be V sub 3 and the current be i sub 3?

** Rick Nelson clarifies:** What I was trying to suggest here is that the arbitrary node assignments made to the three-terminal circuit in Figure 1 should give way to port-specific nomenclature when that circuit is represented as a two-port network. Although it's perfectly reasonable to assign v-sub-1, v-sub-2, and v-sub-3 (with corresponding i values) in any order to the nodes of the 3-terminal circuit, the subscripts for the 2-port descriptions are most meaningful if they are arranged to correspond to the port designations: v-sub-1 and i-sub-1 for port 1, and v-sub-2 and i-sub-2 for port 2. That way, for example, in a matrix description of the network, a matrix parameter in row 2 column 1 would reflect an effect on v-sub-2 (at port 2) of an applied i-sub-1 (at port 1).

I made another attempt at emphasizing the significance of node assignments in the sidebar, saying "If you and a colleague sit down to develop sets of loop or node equations for a moderately complex circuit, your sets will probably differ...If you agree on port definitions ahead of time, however, the matrix representations you and your colleague derive should be identical."

It is jarring to see "v-sub-3=9.6 V" on p. 24 and then "v-sub-2=9.6 V" on p. 26. In retrospect, I wish I had:

• Commented explicitly on the change in subscript numbering,

• Swapped i-sub-2 and v-sub-2 with i-sub-3 and v-sub-3 in Figure 1 (although the quiet disappearance of v-sub-3 and i-sub-3 in Figure 2b would have de-emphasized one of the points I was trying to make), or

• Perhaps the best approach, used sub-A, sub-B, and sub-C to describe the voltages and currents in Figure 1.