How to Read a Smith Chart from One Impedance

A Smith chart does not plot ohms on a rectangular grid. It plots the reflection coefficient, and it bends the grid so that constant resistance and constant reactance are circles.

Take $25 + j50\,\Omega$ on a $50\,\Omega$ chart. Divide by 50 first. The point is $z = 0.5 + j1$, in the upper half of the chart, on the $r = 0.5$ circle and the $x = +1$ arc. Its distance from the center is $|\Gamma| = 0.620$ at $82.9^\circ$. That is $4.15\,\mathrm{dB}$ return loss and a VSWR of $4.27$. The center is $50\,\Omega$, not zero, and not automatically the conjugate of a complex source.

Search interest in the chart picked up after Veritasium’s July 2026 video “The Scariest Chart in Electrical Engineering.” The drawing is old. P. H. Smith published the form in 1939. The rule for reading one point has not changed.

What is drawn on a Smith chart?

The reflection coefficient of an impedance $Z$ against a reference $Z_{\mathrm{ref}}$ is

$$\Gamma = \frac{Z - Z_{\mathrm{ref}}}{Z + Z_{\mathrm{ref}}}$$

$\Gamma$ is a complex number inside the unit disk for a passive impedance. The chart is that disk. Real $\Gamma$ runs left to right. Imaginary $\Gamma$ runs bottom to top, so the upper half of the chart is the upper half of the complex plane.

Two families of curves make the disk useful:

• Circles of constant normalized resistance. Every one of them passes through the open-circuit point on the right.
• Arcs of constant normalized reactance. Their centers sit on the vertical line through that same open-circuit point.

You read a point as the intersection of one resistance circle and one reactance arc. You do not interpolate $X$ the way you would on graph paper.

The number printed as the center, $Z_{\mathrm{ref}}$, is a choice. Most RF work uses $50\,\Omega$. A chart drawn at $75\,\Omega$ or $100\,\Omega$ is the same picture with a different scale. IndieRF Match keeps $Z_{\mathrm{ref}}$ separate from the source impedance. The Smith chart calculator defaults to $50\,\Omega$.

Where are the short, the open, and 50 Ω?

Three landmarks are enough to orient the disk.

Point	Impedance	$\Gamma$
Center	$Z_{\mathrm{ref}}$, here $50\,\Omega$	$0$
Left rim	short, $0\,\Omega$	$-1$
Right rim	open circuit	$+1$

Anything on the horizontal diameter is a real impedance. Moving right from the center raises the resistance. Moving left lowers it. The rim is $|\Gamma| = 1$, a pure reactance, or a short, or an open. A passive load cannot fall outside the rim. A negative resistance would, and this calculator rejects it.

Wavelength scales are sometimes printed around the rim. Toward the generator is clockwise on the usual chart: adding a length of line rotates $\Gamma$ clockwise by twice the electrical angle. The notes on matching a chip antenna use that rotation for a $50\,\mathrm{ps}$ port extension. The chart itself does not know your cable.

How do you read 25 + j50 Ω?

Normalize before you look for circles. With $Z_{\mathrm{ref}} = 50\,\Omega$,

$$z = \frac{25 + j50}{50} = 0.5 + j1$$

Find the circle marked $0.5$ (or $25\,\Omega$ on a chart that is already scaled in ohms). Find the arc marked $+1$ (or $+j50\,\Omega$). They cross once in the upper half. That crossing is the load.

The same point, computed rather than read off a printed chart:

Quantity	Value
Normalized $z$	$0.5 + j1$
$\|\Gamma\|$	$0.620$
Angle of $\Gamma$	$82.9^\circ$
Return loss	$4.15\,\mathrm{dB}$
VSWR	$4.27$
Mismatch loss into $50\,\Omega$	$2.11\,\mathrm{dB}$

Return loss is $-20\log_{10}|\Gamma|$. VSWR is $(1+|\Gamma|)/(1-|\Gamma|)$. Mismatch loss is the power that reflection keeps from the load, $-10\log_{10}(1-|\Gamma|^2)$, and it is not the return loss. The conversion, including this point, is in return loss, VSWR, and mismatch loss.

The dashed segment on the figure is $|\Gamma|$. A longer segment is a worse match. Angle is measured from the positive real axis, counterclockwise, the same way you read a complex number. $82.9^\circ$ is almost straight up, which is why the dot sits near the top of the chart and only slightly to the right of the vertical diameter. The small real part of $\Gamma$ is what “slightly to the right” means. Most of this reflection is imaginary.

A printed chart is coarse around the rim and fine near the center. That is the point of the mapping. A $1.1{:}1$ VSWR and a $3{:}1$ VSWR are both easy to see, which they are not on a rectangular plot of $R$ and $X$.

What do the upper and lower halves mean?

Positive reactance is the upper half. Negative reactance is the lower half. A series inductor moves a load upward along its resistance circle, because a series element cannot change $R$. A series capacitor moves it downward along the same circle.

A shunt element does not follow those circles. It follows a constant-conductance circle, which you can see by reading the chart as an admittance chart: the same disk, rotated $180^\circ$, or the same point read as $y = 1/z$. The conductance circles pass through the short on the left. A shunt capacitor adds positive susceptance. A shunt inductor adds negative susceptance.

That is the whole geometry an L-network uses. One element rides a resistance circle. The other rides a conductance circle. Their job is to land on the match, which for a real $50\,\Omega$ source is the center. The step-by-step choice among the two or four networks that can do it is which L-network to build.

When is the center not the match?

The center is $Z_{\mathrm{ref}}$. The conjugate match of a source $Z_s$ is $Z_s^*$, the impedance that absorbs the available power. If $Z_s$ is $50\,\Omega$ and the chart is $50\,\Omega$, those are the same dot. If the source is complex, they are not.

A source of $40 + j30\,\Omega$ on a $50\,\Omega$ chart is matched at $40 - j30\,\Omega$, down and to the right of center. Building a network that lands on the center matches the load to $50\,\Omega$, not to that source. The worked case is on the L-network note. The chart reference stays $50\,\Omega$ either way, so a shared screenshot still means something.

What this picture is not

The chart does not include loss in the parts, a fixture, or the line you forgot to port-extend. A dot is one frequency. A load that moves with frequency is a track, not a dot, which is why a match that looks perfect at one marker can be narrow. That track is the subject of why an L-match is narrow.

A quarter-wave transformer and a single-stub tuner are motions on this same disk: the transformer slides along a constant-$|\Gamma|$ circle, and a stub adds a susceptance. Neither is calculated here yet.

The numbers above are the closed-form $\Gamma$, not a reading from a paper chart. A paper chart is fine for seeing the region. Use the calculator when the third digit matters.

References

• P. H. Smith, “Transmission Line Calculator,” Electronics, vol. 12, January 1939.
• D. M. Pozar, Microwave Engineering, the chapter on impedance matching and tuning.
• K. Kurokawa, “Power Waves and the Scattering Matrix,” IEEE Transactions on Microwave Theory and Techniques, 1965. The power-wave $\Gamma$ used when the source is complex is the one in that paper. For a real $Z_{\mathrm{ref}}$ it reduces to the formula above.

Related

• Which L-network to build
• Return loss, VSWR, and mismatch loss
• Smith chart calculator

FAQ

What is a Smith chart?

A Smith chart is the reflection-coefficient plane. The center is a perfect match to the chart reference, usually 50 Ω. The right-hand rim is an open circuit and the left-hand rim is a short. Circles of constant resistance and arcs of constant reactance are drawn so a normalized impedance can be read without converting Γ by hand.

How do you read 25 + j50 ohms on a Smith chart?

Divide by 50 Ω first. The point is z = 0.5 + j1, on the r = 0.5 circle and the x = +1 arc, in the upper half. For a 50 Ω reference that point is |Γ| = 0.620 at 82.9°, return loss 4.15 dB, and VSWR 4.27.

Is the center of the Smith chart always the match?

The center is Zref, the impedance the chart was normalized to. A conjugate match to a complex source ends at Zs*, which is not the center unless the source is real and equal to Zref.

Which half of the chart is inductive?

The upper half is positive reactance, inductive in the series sense. The lower half is negative reactance, capacitive. A shunt inductor is a negative susceptance, so it does not live at the same point as a series inductor of the same magnitude.