This article explains some very useful equivalences of very short very mismatched transmission lines. They can be very useful in:
- understanding / explaining /anticipating some measurement errors; and
- applying port extension corrections to VNA measurements where the fixture can be reasonably be approximated as a uniform transmission line.
Port extension commonly applies a measurement correction assuming a section of lossless 50Ω transmission line specified by the resulting propagation time, e-delay, but as explained below, can be used to correct other approximately lossless uniform transmission line sections of other characteristic impedance.
In the following, Z0 is the characteristic impedance of the transmission line, and Zl is the load at the end of that transmission line.
The line section can usually be considered lossless because of its very short length, but whilst its loss might be insignificant, the phase change along the line may be significant and is considered in the following analysis.
The expression for Zin of a lossless transmission line terminated in Zl is:
\(Z_{in}=Z_0 \frac{Z_l+\jmath Z_0 tan(\beta l)}{Z_0+\jmath Z_l tan(\beta l)}\\\)where:
- Z0 is the characteristic impedance of the line;
- β is the phase velocity of the wave on the transmission line, the imaginary part of the complex wave propagation constant γ (the real part α is zero by virtual of it being assumed lossless);
- l is the length of the line section.
βl is the electrical length or phase length in radians.
β can be calculated:
\(\beta= \frac{2 \pi f}{c_0 v_f}\) which allows the equivalent shunt capacitance to be calculated (an exercise for the reader).
- where f is the frequency;
- c0 is the speed of an EM wave in a vacuum, 299792458m/s; and
- vf is the applicable velocity factor.
The transmission line has a propagation time t:
\(t=\frac{l}{c_0 v_f}\) or rearranged \(l=t c_0 v_f\), or \(t=\frac{3.336}{v_f} \text{ps/mm}\).
Two cases are discussed:
- Zl>>Zo; and
- Zl<<Zo.
Zl>>Zo
Recalling that the expression for Zin of a lossless transmission line terminated in Zl is:
\(Z_{in}=Z_0 \frac{Z_l+\jmath Z_0 tan(\beta l)}{Z_0+\jmath Z_l tan(\beta l)}\\\)Inverting both sides:
\(Y_{in}=\frac{1}{Z_0} \frac{Z_0+\jmath Z_l tan(\beta l)}{Z_l+\jmath Z_0 tan(\beta l)}\\\)For \(Z_l \gg Z_0 \text{, } Z_0 tan(\beta l) \ll Z_l\) and can be ignored.
\(Y_{in}= \frac{1+\jmath \frac{Z_l}{Z_0} tan(\beta l)}{Z_l}\\\) \(Y_{in}=Y_l +\jmath \frac{tan(\beta l)}{Z_0}\)Where \(\beta l \lt 0.1\) applies:
For short electrical length, \(\beta l<0.1\), \(tan(\beta l) \approx \beta l\):
\(Y_{in}=Y_l +\jmath \frac{\beta l}{Z_0}\\\)So, the effect of this transmission line is to add a small +ve (capacitive) shunt susceptance to Yl.
\(\frac{\beta l}{Z_0}\) is interesting:
\(\frac{\beta l}{Z_0}= \frac{2 \pi f}{Z_0 c_0 v_f} t c_0 v_f=2 \pi f \frac{t}{Z_0}\\\)Where \(\beta l\frac{50}{Z_0} \lt 0.1\) also applies:
At a given frequency, we can say that \(\frac{\beta l}{Z_0} \propto \frac{t}{Z_0}\) and that \(t_{50} \equiv t_{Z_0}\frac{50}{Z_0}\) and since \(l \propto t \text{, } l_{50} \equiv l_{Z_0}\frac{50}{Z_0}\). These equivalences allow transforming a length or propagation time at actual Z0 into an equivalent length or time at Z0=50. For example if the propagation time of a fixture of 6mm length of air spaced line with Z0=200Ω was 20ps, an equivalent one way e-delay at Z0=50Ω (the instrument reference) of \(t_{50} =20\frac{50}{200}=5 \text{ ps}\) would approximately correct the transmission line effects of the 20ps of 200Ω line. Note that for an s11 correction, the two way e-delay is needed, 10ps in this example.
An important thing to remember is that port extension using e-delay assumes a lossless port extension using transmission line where phase length is proportional to f.
Above is a SimNEC simulation of Zl=10000+j0Ω with 0.025rad 200Ω lossless line backed out by -0.1rad of lossless 50Ω line (comparable to e-delay). The green reversal path lies almost exactly over the magenta path of the original transmission line transformation.
Summary for Zl>>Zo
\(Y_{in}=Y_l +\jmath \frac{\beta l}{Z_0}\\\) \(t_{50} =t_{Z_0}\frac{50}{Z_0}\\\) \(l_{50} =l_{Z_0}\frac{50}{Z_0}\)Zl<<Zo
Recalling that the expression for Zin of a lossless transmission line terminated in Zl is:
\(Z_{in}=Z_0 \frac{Z_l+\jmath Z_0 tan(\beta l)}{Z_0+\jmath Z_l tan(\beta l)}\\\)For \(Z_l \ll Z_0 \text{, } Z_l tan(\beta l) \ll Z_0\) and can be ignored.
\(Z_{in}=Z_0 \frac{Z_l+\jmath Z_0 tan(\beta l)}{Z_0}\\\) \(Z_{in}=Z_l+\jmath Z_0 tan(\beta l)\)Where \(\beta l \lt 0.1\) applies:
For short electrical length, \(\beta l<0.1\), \(tan(\beta l) \approx \beta l\):
\(Z_{in}=Z_l+\jmath Z_0 \beta l\\\)So, the effect of this transmission line is to add a small +ve (inductive) series reactance to Zl.
\(\beta l Z_0\) is interesting:
\(\beta l Z_0= \frac{2 \pi f Z_0}{c_0 v_f} t c_0 v_f=2 \pi f Z_0 t\\\)Where \(\beta l\frac{Z_0}{50} \lt 0.1\) also applies:
At a given frequency, we can say that \(\beta l Z_0 \propto t Z_0\) and that \(t_{50} \equiv t_{Z_0}\frac{Z_0}{50}\) and since \(l \propto t \text{, } l_{50} \equiv l_{Z_0}\frac{Z_0}{50}\). These equivalences allow transforming a length or propagation time at actual Z0 into an equivalent length or time at Z0=50. For example if the propagation time of a fixture of 6mm length of air spaced line with Z0=200Ω was 20ps, an equivalent one way e-delay at Z0=50Ω (the instrument reference) of \(t_{50} =20\frac{200}{50}=80 \text{ ps}\) would approximately correct the transmission line effects of the 20ps of 200Ω line. Note that for an s11 correction, the two way e-delay is needed, 160ps in this example.
An important thing to remember is that port extension using e-delay assumes a lossless port extension using transmission line where phase length is proportional to f.
Above is a SimNEC simulation of Zl=1+j0Ω with 0.1rad lossless 200Ω line backed out by -0.025rad of lossless 50Ω line (comparable to e-delay). The green reversal path lies almost exactly over the magenta path of the original transmission line transformation.
Summary for Zl<<Zo
\(Z_{in}=Z_l+\jmath Z_0 \beta l\\\) \(t_{50} =t_{Z_0}\frac{Z_0}{50}\\\) \(l_{50} =l_{Z_0}\frac{Z_0}{50}\)