According to "Smith Chart Supplement - Corrected Jan 2012" when f.e. calculating input impedance (from load impedance) it is necessary to use WLTG (wave lengths towards generator).

WLTG range is from 0 to 0.5 of wave length so input impedance will be same if length of line is multiplies of 0.5 wave length. But if f.e. transmission line length is 0.20WL impedance will be different.

Also if load impedance is matched to characteristic impedance of line f.e. 50 ohms. In this case impedance is same regardless of length of transmission line (so parts different than 0.5x wave length doesn't affect input impedance it is always 50).

I don't understand how length (as parts of 0.5 wavelengths) of transmission line affects input impedance but it doesn't affect when length is multiple of 0.5?

  • \$\begingroup\$ Do you understand why reflections occur in a t-line that is not terminated in the correct impedance? \$\endgroup\$
    – Andy aka
    Commented Sep 9, 2021 at 7:40

1 Answer 1


For a lossless transmission line, the input reflection coefficient \$\Gamma_{in}\$ can be calculated from load reflection coefficient \$\Gamma_{L}\$ using this formula:

$$\Gamma_{in} = \Gamma_{L}*e^{-j2\beta d}$$

Because \$-j2\beta d\$ is pure imaginary, this represents a vector rotation.

Basically on Smith chart, you rotates the vector \$\Gamma_{L}\$ toward generator \$k\$ wavelengths to find \$\Gamma_{in}\$, which you can find \$Z_{in}\$ from. After each full rotation on the Smith chart (\$ 0.5 \lambda\$ WTG), \$\Gamma_{in}\$ repeats and is equal to \$\Gamma_{L}\$ and thus \$ Z_{in} = Z_{L} \$.

Let me proof this mathematically. Given \$d = k\lambda\$, the formula can be rewritten:

$$\Gamma_{in} = \Gamma_{L}*e^{-j4k \pi}$$

When \$k\$ equals to multiples of 0.5, \$e^{-j4k \pi} = 1\$, so \$\Gamma_{in} = \Gamma_{L}\$ and \$ Z_{in} = Z_{L} \$.


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