Problems One

Question 1

A certain coaxial cable has characteristic impedance 50 ohms and velocity factor 0.63. Calculate the wave velocity on this cable in cm/nanosecond; also the inductance and capacitance per unit length of the cable.

• The wave velocity in vacuum times the velocity factor is equal to the wave velocity on the cable. The wave velocity in vacuum is 30 cm per nanosecond, or 3 times 10^8 metres per second. Thus the wave velocity is 30*0.63 = 18.90 cm per nanosecond or in SI units, 1.89 times 10^8 metres per second.
• The inductance per unit length L and capacitance per unit length C, in H/m or F/m respectively, are related to the velocity u in SI units and the characteristic impedance Zo in ohms by the relationships Zo = sqrt(L/C) and u = 1/sqrt(LC). Thus we see that L = Zo/u and C = 1/(Zo*u). Using these we find L = 50/(1.89*10^8) = 264.55 nH/m and C = 1/(50*1.89*10^8) = 105.8 pF/m.

Let us suppose that this cable feeds a load impedance 75 + j150 ohms. Calculate the complex reflection coefficient gamma, and express it both as real and imaginary parts, and also as modulus and phase angle.

• gamma = (ZL-Zo)/(ZL+Zo) = (75+j150-50)/(75+j150+50) = (1+j6)/(5+j6) = (honestly, by calculator) 0.6721+j0.3934 = 0.7788*exp(j30.34deg)

Determine the normalised load impedance in this case and plot it on the Smith chart. Measure gamma from the Smith chart plot and show that it agrees with the value you calculated.

• Normalised load impedance = ZL/Zo = 1.5+j3

• The return loss = -20 log10(|gamma|) = -20 log10(.7788) = 2.17dB

• at 1.2 GHz the wavelength in free space or vacuum is 30/1.2 cm or 25 cm. On the cable the wavelength is smaller than this by the velocity factor 0.63. On the cable, the wavelength is 25*0.63 = 15.75 cm.
• For a 1/4 wavelength section we need 15.75/4 = 3.9375cm = 39.375mm of cable.

An 80 cm length of this cable is connected to the load. Calculate the input impedance (to the combination of load and cable) from the formula for the transformation of impedance along the line. Repeat graphically using the Smith chart. Compare results.

• How many wavelengths of 15.75 cm are there in the 80 cm of cable? well, there are 80/15.75 = 5.0794 wavelengths. The phase shift is 360 degrees per wavelength, or 1828.5714 degrees. But because exp(j theta) is cyclic in theta every 360 degrees, this is entirely equivalent to a phase shift of .0794*360 = 28.58 degrees.
• The reflection coefficient at the input differs in phase from the reflection coefficient at the load by -2*28.58 degrees, or -57.17 degrees. The modulus of the reflection coefficient is unchanged along the line in a lossless line. At the input therefore, gamma(in) = gamma*exp(-j57.17) = 0.7788 exp(j[30.34-57.17]) = 0.6950-j0.3515
• Now Zin/Zo = (1+gamma(in))/(1-gamma(in)) = (1.695-j0.3515)/(0.3050+j0.3515) = 1.8165-j3.246
• So Zin = 90.83-j162.3 ohms.

For this example, determine the two values of entirely real impedance which can be found along the line. What is the VSWR on this line for this load impedance?

• The maximum value of impedance occurs at a VSWR maximum where the voltages add and the currents subtract, but all quantities are in phase. At this point the voltage is (1+|gamma|)*(V+) and the current is (1-|gamma|)*(I+). The impedance is real and is equal to Zo*(1+|gamma|)/(1-|gamma|) = 50*(1.7788/0.2212) = 402.1 ohms = 8.04*Zo = VSWR*Zo
• By a similar argument, the minimum value of impedance = Zo/VSWR = 50/8.04 = 6.218 ohms.
• Clearly, the max voltage is (1+|gamma|) and the min voltage (1-|gamma|) times V+ in each case, so the VSWR = (1+|gamma|)/(1-|gamma|) = 8.04 as above.

Question 2.

A certain 75 ohm coax cable feeds an antenna of (assumed) radiation impedance 80 + j15 ohms. Calculate the return loss. Calculate the proportion of forward wave power which is radiated.

• As in Qn 1, gamma = (80+j15-75)/(80+j15+75) = (5+j15)/(155+j15) = 0.0412+j0.0928 = 0.1015*exp(j66.04deg) so the return loss is -20*log10(0.1015) = 19.87dB.
• The reflected power proportion is |gamma|*|gamma| = 0.1015^2 = 0.0103 or 1.03 percent. Thus 98.97 percent is radiated.
• Comment. This isn't bad. However, suppose your antenna is presented with transmitter power of 100 kWatt. The 1.03 percent of reflected power represents 1030 watts which is quite a lot of heat to absorb somewhere in the transmitter.

The antenna is to be matched using a single series shorted stub. Calculate the position of attachment and length of this stub in terms of numbers of wavelengths. Calculate the physical stub dimensions for signals at 400MHz on cable of velocity factor 0.70.

• This is where the SMITH chart really comes in useful. The normalised antenna load impedance is 80/75+j15/75 = 1.067+j0.2000
• We plot this on the SMITH chart, place a ruler from the chart centre via the plotted point to intersect the "wavelengths towards generator" scale and read off an angle 0.156 wavelengths.
• We now transform the impedance towards the generator at constant radius (|gamma|) until we hit the R=Zo circle at X = -j0.20 and plot this second point. The angle for this point is 0.366 wavelengths towards the generator.
• We have moved through an angle 0.366-0.156 wavelengths or 0.21 wavelengths towards the generator, to transform the impedance from 1.067+j0.200 to 1.000-j0.20 (within the accuracy of our plot).
• We cut the transmission line here and insert a pure reactive shorted stub having normalised reactance 0.20j. On the generator side of the series stub we see 1.000-j0.20+j0.20 = 1.000+j0 normalised impedance, and we have a match.
• The stub length is, from the SMITH chart plots of short point and +j0.20 point, 0.0315 wavelengths.
• In free space the wavelength at 400 MHz is 30/0.4 = 75 cm. The velocity factor is 0.7 so on the transmission line the wavelength at 400 MHz is 0.7*75= 52.5 cm. The short series stub is placed 52.5*0.21 cm from the load, or 11.03 cm from the load. The stub length is 52.5*0.0315 = 1.65 cm.

Calculate the return loss for this arrangement at 390MHz, assuming the antenna impedance has not changed.

• At 390 MHz the wavelength is 52.5*(400/390) = 53.85 cm and the stub attachment point is 11.03/53.85 wavelengths from the load. This is 0.2048 wavelengths from the load. The stub length of 1.65 cm is 0.0306 wavelengths so its (normalised) reactance (from the SMITH chart) is j0.195
• Transforming the antenna load impedance 0.2048 wavelengths towards the generator we find the point 1.02-j0.21 normalised impedance, so adding the j0.195 of the stub we find the point 1.02-j0.015 for the impedance.
• As before, the return loss is found from |gamma| for this impedance which is modulus[(1.02-j0.015-1)/(1.02-j0.015+1)] which my calculator makes 0.0124, so the return loss is -20*log10(0.0124) = about 38dB.
• Comment. This matching network isn't really necessary in this particular scenario; we would be quite happy with a 20 dB return loss from the bare antenna. But if we put a simple matching network in we can improve the matching which gives 20 dB reduction in reflected power over a range 390-410 MHz. This is a bandwidth of 20 MHz, good enough for a TV channel and more, and is a "fractional bandwidth" of 20/400 = 5%. Over this range of frequencies the matched antenna radiates at least 99.99% of the incident power.