# ZL1AN's transmission line challenge

Gary, ZL1AN, posed the following problem to stimulate a little thinking about transmission lines. I would have labeled the diagram a little differently. I would remove the arrow adjacent to the generator, and indicate its polarity with a + symbol. It turns out that Gary had in mind that the 40V is RMS, and it ought to be labeled as such... but that approach taken in this article treats it as the amplitude of a sinusoid with phase angle zero to demonstrate the technique with a more complex problem.

The first step is to solve the right hand TL section.

Fig 2:

# RF Arbitrary Transmission Line Loss Calculator

## Right hand TL section

 Parameters Ro 50.00000 vf 1.000 k1 0.000000e+0 k2 0.000000e+0 Frequency 1.000 MHz Length 0.250 wl Zload 10.00+j0.00 Ω Yload 0.100000+j0.000000 S Results Zo 50.00+j0.00 Ω Velocity Factor, VF -2 1.000, 1.000 Length 90.00 °, 0.250 λ, 74.948 m Line Loss (matched) 0.00e+0 dB Line Loss 0.00e+0 dB Efficiency 100.00 % Zin 250.00+j0.00 Ω Yin 0.004000-j0.000000 S VSWR(50)in 5.00 Γ, ρ∠θ, RL, VSWR, MismatchLoss (source end) 6.667e-1+j8.164e-17, 0.667∠0.0°, 3.5 dB, 5.00, 2.55 dB Γ, ρ∠θ, RL, VSWR, MismatchLoss (load end) -6.667e-1+j0.000e+0, 0.667∠180.0°, 3.5 dB, 5.00, 2.55 dB Vout/Vin 2.449e-18-j2.000e-1, 2.000e-1∠-90.0° Iout/Iin 1.531e-15-j5.000e+0, 5.000e+0∠-90.0° S11, S21 0.000e+0+j0.000e+0, 6.123e-17-j1.000e+0 Y11, Y21 -7.498e-35-j1.225e-18, 0.000e+0+j2.000e-2 NEC NT NT t s t s -7.498e-35 -1.225e-18 0.000e+0 2.000e-2 -7.498e-35 -1.225e-18 'ATLLC, 0.250 wl, 1.000 MHz k1, k2 0.000e+0, 0.000e+0 C1, C2 0.000e+0, 0.000e+0 γ 0.000e+0+j6.283e+0

Using ATLLC ticking the long output box, the results in Fig 2 were obtained. Make k1 and k2 zero to model a lossless line. The key results are Zin=250+j0Ω and Vout/Vin=2.000e-1∠-90.0°.

Step 2 is to solve the left hand TL section with the calculated load.

Fig 3:

# RF Arbitrary Transmission Line Loss Calculator

## Left hand TL section

 Parameters Ro 100.00000 vf 1.000 k1 0.000000e+0 k2 0.000000e+0 Frequency 1.000 MHz Length 0.250 wl Zload 250.00+j0.00 Ω Yload 0.004000+j0.000000 S Results Zo 100.00+j0.00 Ω Velocity Factor, VF -2 1.000, 1.000 Length 90.00 °, 0.250 λ, 74.948 m Line Loss (matched) 0.00e+0 dB Line Loss 0.00e+0 dB Efficiency 100.00 % Zin 40.00-j0.00 Ω Yin 0.025000+j0.000000 S VSWR(50)in 1.25 Γ, ρ∠θ, RL, VSWR, MismatchLoss (source end) -4.286e-1-j5.248e-17, 0.429∠-180.0°, 7.4 dB, 2.50, 0.88 dB Γ, ρ∠θ, RL, VSWR, MismatchLoss (load end) 4.286e-1+j0.000e+0, 0.429∠0.0°, 7.4 dB, 2.50, 0.88 dB Vout/Vin 3.827e-16-j2.500e+0, 2.500e+0∠-90.0° Iout/Iin 9.797e-18-j4.000e-1, 4.000e-1∠-90.0° S11, S21 6.000e-1+j2.939e-17, 3.919e-17-j8.000e-1 Y11, Y21 -2.082e-18-j7.103e-19, 4.898e-20+j1.000e-2 NEC NT NT t s t s -2.082e-18 -7.103e-19 4.898e-20 1.000e-2 -2.082e-18 -7.103e-19 'ATLLC, 0.250 wl, 1.000 MHz k1, k2 0.000e+0, 0.000e+0 C1, C2 0.000e+0, 0.000e+0 γ 0.000e+0+j6.283e+0

Fig 3 shows the solution of the left hand TL section. The key results are Zin=40+j0Ω and Vout/Vin=2.500e+0∠-90.0°.

Step 3 is to find the voltage at the input to the left hand TL section. Since Zin=40+j0Ω, the voltage is exactly half the source voltage by simple voltage division with the 40Ω resistor, so Vin=20∠-90.0° (treating the input as having a phase dimension).

Step 4 is to find the load voltage, and that will be the input voltage multiplied by the Vout/Vin quantities for each line section, so Vload=20*2.000e-1∠-90.0°*2.500e+0∠-90.0°=10∠-180.0°V or -10+j0V.

Of course, if you treat the source as an RMS value, the phase angle is perhaps not relevant, depending on the application.

Though this problem is a trivial one and can be solved by mental arithmetic thanks to Gary's choice of values, more complex problems in the real world can be solved using the same techniques. Practical calculating tools include:

# Mental arithmetic solution

As the circuit values in the challenge are purely real numbers, the problem can be solved by mental arithmetic.

Step 1: solve the RH TL. The complex reflection coefficient Γ=(Zl-Zo)/(Zl+Zo)=-40/60=-2/3. For a lossless line of 90°, Γin=-Γout. So, Vout/Vin=(1+Γout)∠90/(1+Γin)=(1/3)∠90/(5/3)=(1/5)∠90. Zin=Zo^2/Zl=2500/10=250.

Step 2: solve the LH TL. The complex reflection coefficient Γ=(Zl-Zo)/(Zl+Zo)=150/350=3/7. For a lossless line of 90°, Γin=-Γout. So, Vin/Vout=(1+Γout)∠90/(1+Γin)=(10/7)∠90/(4/7)=(10/4)∠90. Zin=Zo^2/Zl=10000/250=40.

Step 3: by voltage division, LH Vin is half of 40 which equals 20V. RH Vout=LH(Vin) * LH(Vout/Vin) * RH(Vout/Vin) = 20*(1/5)∠90*(10/4)∠90=-10V.

# Smith chart solution

The problem can be solved graphically using a Smith chart. Such a simple problem is difficult to illustrate clearly because most of the construction lines lie along the real axis and would take a dozen graphics to show the development clearly, perhaps a movie one day!