GeezerVison

A transmission line with characteristic impedance $Z_0$ has a characteristic admittance $Y_0=1/Z_0$. Any load impedance $Z$ or admittance $Y$ is normalized by dividing it by the corresponding characteristic value:

$z=\frac{Z}{Z_0},y=\frac{Y}{Y_0}=\frac{1}{z}.$

While impedance is measured in ohms (Ω) and admittance in siemens (S), dividing by $Z_0$ or $Y_0$ cancels these units. The resulting normalized values are dimensionless quantities, which are the only values plotted directly on a Smith chart.

When a transmission line is terminated in a load impedance $Z_L$ that differs from its characteristic impedance $Z_0$, the incident (forward) wave $V_{\text{F}}$ and reflected wave $V_{\text{R}}$ superimpose to form a standing wave.

In complex phasor notation, measuring distance $\ell$ along the line from the load toward the generator gives:

$\begin{array}{cc}{V}_{\text{F}}\left(\ell \right)& =A{e}^{+\gamma \ell },\\ V_{\text{R}}\left(\ell \right)& =B{e}^{-\gamma \ell },\end{array}$

where $A$ and $B$ are the complex voltage amplitudes at the load ($\ell =0$). The common time factor $e^{j\omega t}$ is suppressed assuming steady-state single-frequency operation.

The complex propagation constant is:

$\gamma =\alpha +j\beta ,$

where $\alpha$ is the attenuation constant and $\beta$ is the phase constant.

Variation along a line

The complex voltage reflection coefficient $\Gamma$ is the ratio of the reflected wave to the incident wave:

$\Gamma =\frac{V_{\text{R}}}{V_{\text{F}}}=\frac{B{e}^{-\gamma \ell }}{A{e}^{+\gamma \ell }}=C{e}^{-2\gamma \ell },$

where $C$ is a constant.

For a uniform line with constant propagation constant $\gamma$, $\Gamma$ varies with position. On a lossy line ($\alpha >0$), this variation forms a spiral toward the center. On a lossless line ($\alpha =0$), attenuation is zero, simplifying the expression to:

$\Gamma ={\Gamma }_{\text{L}}{e}^{-2j\beta \ell }={\Gamma }_{\text{L}}\exp (-j\frac{4\pi \ell }{\lambda }),$

where ${\Gamma }_{\text{L}}$ is the reflection coefficient at the load, $\ell$ is line length from the load, and $\lambda =2\pi /\beta$ is the line wavelength.

Because the angle changes by $2\pi$ radians over a distance of $\lambda /2$, the reflection coefficient and impedance repeat every half-wavelength. The outer scale of the Smith chart measures line length from $0$ to $0.5\lambda$.

Solving matching networks often requires converting between series elements (easiest in impedance $Z$) and parallel elements (easiest in admittance $Y$).

Converting a point between normalized impedance $z$ and normalized admittance $y=1/z$ corresponds to a 180° rotation around the origin at the same radius. For example, point ${\text{P}}_1$ ($z=0.80+j1.40$) rotates to diametrically opposite point ${\text{Q}}_1$ ($y=0.30-j0.54$).

Engineers choose between Z and Y charts based on circuit topology. Series components add directly as impedances ($Z_{\text{series}}=Z_1+Z_2+\dots$), while parallel components add directly as admittances ($Y_{\text{parallel}}=Y_1+Y_2+\dots$).

The normalized impedance and admittance for basic passive elements are:

$\begin{array}{ccc}\text{Resistor:}& z=\frac{R}{Z_0},& y=\frac{G}{Y_0}=\frac{Z_0}{R}\\ \text{Inductor:}& z=j\frac{\omega L}{Z_0},& y=-j\frac{1}{\omega L{Y}_0}\\ \text{Capacitor:}& z=-j\frac{1}{\omega C{Z}_0},& y=j\frac{\omega C}{Y_0}\end{array}$

where $Z_0=1/Y_0$ is the characteristic impedance.

A reflection coefficient of $0.63\angle 60^{\circ }$ plots at point ${\text{P}}_1$: 60° on the outer angle scale and at a radius of 0.63 from the center.

For distributed components, the effect of moving along the transmission line on the reflection coefficient and impedance is accounted 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 can be matched at a single frequency using either a series or a parallel reactive component placed at a precise location.

Suppose a lossless air-spaced transmission line of characteristic impedance $Z_0=50\Omega$, operating at 800 MHz, is terminated with a circuit comprising a 17.5 Ω resistor in series with a 6.5 nH inductor. The reactance of the inductor at 800 MHz is:

$j\omega L=j2\pi fL=j32.7\Omega ,$

giving a total load impedance of:

$Z_{\text{T}}=17.5+j32.7\Omega ,$

and a normalized impedance of:

$z_{\text{T}}=\frac{Z_{\text{T}}}{Z_0}=0.35+j\mathrm{0.65.}$

This point is plotted on the Z Smith chart as ${\text{P}}_{20}$. Extending the line from the origin $\text{O}$ through ${\text{P}}_{20}$ to the outer wavelength scale intersects it at $L_1=0.098\lambda$. Because the line is lossless, a circle centered at the origin through ${\text{P}}_{20}$ represents the path of constant-magnitude reflection coefficient. At point ${\text{P}}_{21}$, this circle intersects the $r=1$ circle of constant normalized resistance at:

$z_{{\text{P}}_{21}}=1.00+j\mathrm{1.52.}$

Extending the line ${\text{OP}}_{21}$ to the wavelength scale gives an intersection at $L_2=0.177\lambda$, making the distance from the termination:

$L_2-L_1=0.177\lambda -0.098\lambda =0.079\lambda .$

For an air-spaced transmission line, the wavelength at 800 MHz equals its free-space value:

$\lambda =\frac{c}{f},$

where $c$ is the speed of light and $f$ is frequency. This yields $\lambda =375\text{mm}$, placing the matching component 29.6 mm from the load.

The conjugate match for the impedance at ${\text{P}}_{21}$ is:

$z_{\text{match}}=-j\mathrm{1.52.}$

Since the chart is in the normalized impedance plane, a series capacitor $C_{\text{m}}$ is required:

$z_{\text{match}}=-j1.52=-\frac{j}{\omega C_{\text{m}}{Z}_0}=-\frac{j}{2\pi f{C}_{\text{m}}{Z}_0}.$

Rearranging gives:

$C_{\text{m}}=\frac{1}{1.52\omega Z_0}=\frac{1}{1.52\times 2\pi f{Z}_0},$

which yields $C_{\text{m}}=2.6\text{pF}$. To match the load at 800 MHz, this capacitor is placed in series with the line 29.6 mm from the termination.

An alternative shunt match can be found by transforming the chart from normalized impedance to normalized admittance. Point ${\text{Q}}_{20}$ is the admittance equivalent of ${\text{P}}_{20}$; extending ${\text{OQ}}_{20}$ to the wavelength scale gives $L_3=0.152\lambda$. The nearest position toward the generator for a shunt conjugate match is ${\text{Q}}_{21}$ (at the same location as ${\text{P}}_{21}$, but representing admittance):

$y_{{\text{Q}}_{21}}=1.00+j\mathrm{1.52.}$

The distance along the line for this shunt match is:

$L_2+L_3=0.177\lambda +0.152\lambda =0.329\lambda ,$

or 123 mm.

The conjugate matching component requires a normalized admittance of:

$y_{\text{match}}=-j\mathrm{1.52.}$

A negative susceptance requires an inductor connected in parallel with the line. For an inductance $L_{\text{m}}$:

$-j1.52=-\frac{j}{\omega L_{\text{m}}{Y}_0}=-\frac{j{Z}_0}{2\pi f{L}_{\text{m}}},$

yielding $L_{\text{m}}=6.5\text{nH}$. A suitable shunt match is therefore a 6.5 nH inductor in parallel with the line, 123 mm from the load.

For lumped-element circuits, movements on the chart are generated by the normalized impedances and admittances of individual components at the operating frequency. The outer wavelength scale is not used.

The circuit shown below is analyzed at an operating frequency of 100 MHz (wavelength $\lambda =3\text{m}$). It consists of two series capacitors and two shunt elements between the input and a load resistor. Because the component dimensions are on the order of millimeters, the lumped-element assumption holds. A reference impedance is still required for normalization; $Z_0=50\Omega$ is chosen because it equals the load resistance $R_1$.^[c]

The analysis begins at the load and proceeds toward the input. Looking into $R_1$ alone, the normalized impedance is 1, represented by point $\text{O}$ at the chart center. Points in the Z plane are labeled $\text{P}$ and points in the Y plane are labeled $\text{Q}$; series elements are added in the Z plane and shunt elements in the Y plane. The table records the complete path around the chart, returning to the center and confirming a matched 50 Ω condition at the input.

The admittance (Y) Smith chart expresses the voltage reflection coefficient in terms of normalized admittance $y$ instead of impedance $z$. Because $y=1/z$:

$y=\frac{1-\Gamma }{1+\Gamma },\Gamma =\frac{1-y}{1+y}.$

The Y chart is identical to the Z chart, but rotated 180°. Writing $y=g+jb$ for conductance $g$ and susceptance $b$ gives:

$\Gamma =\frac{1-g^2-b^2}{(1+g{)}^2+b^2}-j\frac{2b}{(1+g{)}^2+b^2}.$

The region above the horizontal axis represents capacitive admittances ($b>0$), and the region below represents inductive admittances ($b<0$). A matched load ($\Gamma =0$) sits at the center, while pure reactances lie on the bounding unit circle ($\left|\Gamma \right|=1$).

The Smith chart can be extended for negative resistance ($\text{Re}\left(z\right)<0$). In this region, $\left|\Gamma \right|>1$, meaning reflected power exceeds incident power. The extra power comes from an external DC source, as in an active amplifier.

Smith chart projected onto a hemisphere^[e]

Constant resistance ($\text{Re}\left(z\right)$)

Constant reactance ($\text{Im}\left(z\right)$)

Combined grid lines

In 2011, Muller et al. proposed stereographically mapping the 2D reflection coefficient plane onto a 3D Riemann sphere.^[f] The resulting sphere has the following features:^[24]