# [SOLVED]SmithChart Fundamentals

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#### seackone

##### Junior Member level 1
Hi,
I got a question about simple smith chart basic:

Szenario: I got the s-parameters of a network at only one frequency. The s-parameters are splitted into real -and imaginary part and measured with a network analyzer (50 Ohm termination). The parameter of interest is only S11.

I want to plot S11 in a smith chart, so i have to calculate the impedance:

Z = (1+(real(S11)+j*imag(S11))) / (1-(real(S11)+j*imag(S11)))

An other parameter could be the absolute reflection coefficient (and the angle..):

|r| = sqrt(real(S11)^2 + imag(S11)^2)

With Z, is possible to plot the smith chart with a reference to 50 ohm. But what is, if I change my reference to 60 ohm for example? Are there any changes of the position and reflection coefficient of my measured network in the smith chart or is it just a change if Im looking at my denormalized impedance:

impedance = Z0 * Z

If im looking at the last formular, there is no change in position.. or should the formula be more like:

impedance = Z0_NWA * Z / Z0 = 50*Z / 60

than there is a change in position or the value of the reflection coefficient.

#### FvM

##### Super Moderator
Staff member
I'm stumbling at the very first statement
I want to plot S11 in a smith chart, so i have to calculate the impedance.

Smith chart is a linear plot of S11 in complex coordinates, so why calculate impedance?

#### seackone

##### Junior Member level 1
Good morning FvM,

I got only the real -and imaginary part of the s-parameter. They could be in a range of -1 and 1 and if I got no reflections at the output of my network (50 ohm termination), S11 and the reflection coefficient is the same. How is it possible to plot S11 in a smith chart (with a range of 0 to infinity), directly with S11? I only know to plot with the absolute value of the reflection coefficient and the angle, or calculate the normalized impedances before..

- - - Updated - - -

Okay, sorry. I found my failure..

But the result is: if i change the reference to something else like 50 ohm, there are no changes with S11 in the smithchart?

#### FvM

##### Super Moderator
Staff member
If you plot S11 of a given impedance, changing the reference surely changes S11. E.g. 50 ohm load gives S11=0 for 50 ohm reference, but different value for 60 ohm.

seackone

### seackone

Points: 2

#### biff44

I got only the real -and imaginary part of the s-parameter. ?

the smith chard is a polar plot of S11. IF you have real and imaginary values for S11 at one frequency....you simply take out a pencil, and put a dot on the polar plot where it belongs. End of story.

Like if you have an S11 = .707 +j.707, you put a dot at +45 degrees at the edge of the smith chart (i.e. radius=1)

#### seackone

##### Junior Member level 1
But that what I say: you need the absolute value of S11 and the angle to draw it in the smith chart. it is no problem to calculate them with real and imaginary part of S11.

|r| = sqrt(.707^2+.707^2) = 1
phi = arctan(.707/.707) * 180/pi = 45

An other possibility is to convert the real and imaginary part into normalized impedances:

z = (1+(.707+j*.707))/(1-(.707+j*.707)) = 0 + j*2.4142

The result is the same.. But my question wasn't about the drawing. I want to know what is the mathematic background if i change the normalized impedance

#### FvM

##### Super Moderator
Staff member
The math background is completely given in the S11 definition:

S11 = (Z11-Z0)/(Z11+Z0)

seackone

### seackone

Points: 2

#### seackone

##### Junior Member level 1
Thank you very much! That was what im looking for!

#### biff44

The math background is completely given in the S11 definition:

S11 = (Z11-Z0)/(Z11+Z0)

exactly. The smith chart is a bilinear transform of the impedance plane.

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