Editing Smith chart (section)

==Using the Smith chart to solve conjugate matching problems with distributed components==
Distributed matching becomes feasible and is sometimes required when the physical size of the matching components is more than about 5% of a wavelength at the operating frequency. Here the electrical behaviour of many lumped components becomes rather unpredictable. This occurs in microwave circuits and when high power requires large components in shortwave, FM and TV broadcasting.

For distributed components the effects on reflection coefficient and impedance of moving along the transmission line must be allowed for using the outer circumferential scale of the Smith chart which is calibrated in wavelengths.

The following example shows how a transmission line, terminated with an arbitrary load, may be matched at one frequency either with a series or parallel reactive component in each case connected at precise positions.

[[Image:SmithEx4.png|thumbnail|Smith chart construction for some distributed transmission-line matching.]]
Supposing a loss-free air-spaced transmission line of characteristic impedance <math>Z_0 = 50 \ \Omega</math>, operating at a frequency of 800&nbsp;MHz, is terminated with a circuit comprising a 17.5 <math>\Omega</math> resistor in series with a 6.5 nanohenry (6.5 nH) inductor. How may the line be matched?

From the table above, the reactance of the inductor forming part of the termination at 800&nbsp;MHz is
:<math>Z_L = j \omega L = j 2 \pi f L = j32.7 \ \Omega\,</math>
so the impedance of the combination (<math>Z_T</math>) is given by
:<math>Z_T = 17.5 + j32.7 \ \Omega\,</math>
and the normalised impedance (<math>z_T</math>) is
:<math>z_T = \frac{Z_T}{Z_0} = 0.35 + j0.65\,</math>
This is plotted on the Z Smith chart at point P<sub>20</sub>. The line OP<sub>20</sub> is extended through to the wavelength scale where it intersects at the point <math>L_1 = 0.098 \lambda\,</math>. As the transmission line is loss free, a circle centred at the centre of the Smith chart is drawn through the point P<sub>20</sub> to represent the path of the constant magnitude reflection coefficient due to the termination. At point P<sub>21</sub> the circle intersects with the unity circle of constant normalised resistance at 
:<math>z_{P21} = 1.00 + j1.52\,</math>.
The extension of the line OP<sub>21</sub> intersects the wavelength scale at <math>L_2 = 0.177 \lambda\,</math>, therefore the distance from the termination to this point on the line is given by
:<math>L_2 - L_1  = 0.177\lambda - 0.098\lambda = 0.079\lambda\,</math>
Since the transmission line is air-spaced, the wavelength at 800&nbsp;MHz in the line is the same as that in free space and is given by
:<math>\lambda = \frac{c}{f}\,</math>
where <math>c\,</math> is the velocity of [[speed of light|electromagnetic radiation in free space]] and <math>f\,</math> is the frequency in hertz. The result gives <math>\lambda = 375 \ \mathrm{mm}\,</math>, making the position of the matching component 29.6&nbsp;mm from the load.

The conjugate match for the impedance at P<sub>21</sub> (<math>z_{match}\,</math>) is 
:<math>z_{match} = - j (1.52),\!</math>
As the Smith chart is still in the normalised impedance plane, from the table above a series capacitor <math>C_m\,</math> is required where
:<math>z_{match} = - j 1.52 = \frac{-j}{\omega C_m Z_0} = \frac{-j}{2 \pi f C_m Z_0}\,</math>
Rearranging, we obtain
:<math>C_m=\frac{1}{(1.52) \omega Z_0} = \frac{1}{(1.52)(2 \pi f) Z_0}</math>.
Substitution of known values gives
:<math>C_m = 2.6 \ \mathrm{pF}\,</math>
To match the termination at 800&nbsp;MHz, a series capacitor of 2.6 pF must be placed in series with the transmission line at a distance of 29.6&nbsp;mm from the termination.

An alternative shunt match could be calculated after performing a Smith chart transformation from normalised impedance to normalised admittance. Point Q<sub>20</sub> is the equivalent of P<sub>20</sub> but expressed as a normalised admittance. Reading from the Smith chart scaling, remembering that this is now a normalised admittance gives
:<math>y_{Q20} = 0.65 - j1.20\,</math>
(In fact this value is not actually used). However, the extension of the line OQ<sub>20</sub> through to the wavelength scale gives <math>L_3 = 0.152 \lambda\,</math>. The earliest point at which a shunt conjugate match could be introduced, moving towards the generator, would be at Q<sub>21</sub>, the same position as the previous P<sub>21</sub>, but this time representing a normalised admittance given by 
:<math>y_{Q21} = 1.00 + j1.52\,</math>.
The distance along the transmission line is in this case 
:<math>L_2 + L_3  = 0.177\lambda + 0.152\lambda = 0.329\lambda\,</math>
which converts to 123&nbsp;mm.

The conjugate matching component is required to have a normalised admittance (<math>y_{match}</math>) of
:<math>y_{match} = - j1.52\,</math>.
From the table it can be seen that a negative admittance would require an inductor, connected in parallel with the transmission line. If its value is <math>L_m\,</math>, then
:<math>-j1.52 = \frac{-j}{\omega L_m Y_0}= \frac{-jZ_0}{2\pi f L_m}\,</math>
This gives the result 
:<math>L_m = 6.5 \ \mathrm{nH}\,</math>
A suitable inductive shunt matching would therefore be a 6.5 nH inductor in parallel with the line positioned at 123&nbsp;mm from the load.