Transmission line impedance help

Discussion in 'Homework Help' started by geft, Dec 21, 2011.

  1. geft

    Thread Starter New Member

    Dec 8, 2011
    19
    0
    A lossless 50 ohm transmission line is terminated in 35 + j65 ohm. Find the shortest length of line for which impedance is purely resistive.

    How do I go about solving this? I know the formula to calculate impedance from distance x, but I'm not sure I see any way to remove all imaginary components.
     
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    coax has more capacitance per foot than inductance, so it can be measured as pF/foot.

    Also This might help, unsure

    The attached PDF may be of more help for your question.
     
    geft likes this.
  3. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    You may not be able to see it, but if you know the formula, then just apply it and work it forward. Eventually, you will see that you can find a relation that allows you to solve for lengths that make the impedance look real. If memory serves me, I think you may end up with a quadratic formula for tan(beta*x). But, don't worry about my memory or your ability to visualize it. Just work it forward and let the math take care of the details for you.

    If you can't work it out, at least post your attempt and then we can help you more.
     
    geft likes this.
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If you are familiar with its use the Smith Chart makes this a relatively simple task.
     
  5. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Good point. It certainly would help in the visualization of why it is possible to have points that "see" real impedance, and provide a way to double check the calculated value.

    It's good advice to learn them, even if one is not familiar with them.

    Back in the 1980's when I learned field theory, Smith charts were already falling out of favor, but my old-time professor saw fit to include one Smith chart problem on every exam he gave in undergrad fields I and II and the grad level class, as well. He correctly viewed them as more than just a calculation aid.
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Thanks Steve,

    I was playing around with the Smith Chart in the late 60's when all we had were pencils, erasers, graph paper, rulers, compass & slide rule.

    Interestingly, one can find a range of software which includes the Smith Chart as an optional aid to analysis & understanding.

    BTW the OP didn't tell us the operating frequency at which the analysis is to be done - unless they simply require the answer in wavelengths.
     
  7. geft

    Thread Starter New Member

    Dec 8, 2011
    19
    0
    We haven't learned the Smith Chart yet. Anyway, here's my working:

    Using this formula:

    Z_x = Z_0 \frac{Z_L + jZ_0 tan \beta x}{Z_0 + jZ_L tan \beta x}

    Then plugging in the numbers and simplifying:

    Z_x = \frac{35 + j(65 + 50tan\beta x)}{1 + j0.7tan\beta x - 1.3tan\beta x}

    What do I do next? Do I set the imaginary number to 0 or something? Because for the numerator, I got x = -0.1627\lambda whereas in the denominator, I got 0 if tan^{-1} \beta x = 0 and 1 if tan^{-1} \beta x = 2\pi.

    The given answer is 0.091\lambda.
     
    Last edited: Dec 21, 2011
  8. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    Here, now that you have the answer, play around with the smith chart.

    It gives you GREAT insight as to where you should be roughly with the numbers. It's a log scale, so remember that.

    Seriously, give it a shot with the problem you just solved. Print it out or scribble on it in a graphics program on the screen.

    Couple versions, one is for 2 way (color), the other is for one way (B&W) Click on them to see, I apparently messed up the attachment system.

    View attachment imped_admit_smithchart.pdf

    View attachment Smith_Chart.pdf
     
    geft likes this.
  9. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Yes, you have it correct to this point. This should lead to the given answer. First of all, you can check the answer just by plugging the given answer in, and making sure that the answer has an imaginary part that is small compared to the real part. But, that is not proof.

    To be formal, you just need to rearrange the equations and break it into real and imaginary parts, and then set the imaginary part to zero, as you said.

    In general, the number (a+jb)/(c+jd) can be separated into real and imaginary parts by multiplying the numerator and denominator by the complex conjugate of the denominator.

    (a+jb)/(c+jd)=(a+jb)*(c-jd)/(c^2+d^2)=(ac+bd)/(c^2+d^2)+j(cb-ad)/(c^2+d^2)

    You then have a choice to number crunch the rest on computer or continue with equations. The latter involves writing a quadratic equation to solve for tan(beta*x). Then you choose one of the two roots and then take the inverse tangent of that etc. ...
     
    geft likes this.
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hi geft,

    To solve your problem you simply need to know two things - given this is a loss-less line example.

    1. The magnitude of the complex reflection coefficient is the same anywhere on the line. Only the phase changes as the wave travels along the line.
    2. If the line input impedance seen at the source is purely resistive then the complex reflection coefficient is purely real. The coefficient phase angle must be zero at the source.
    So you have ...


    \Gamma=\frac{\frac{Z_L}{Z_0}-1}{\frac{Z_L}{Z_0}+1}

    You know ZL=35+j65

    So you find the reflection coefficient at the load.

    \Gamma=\frac{\frac{35+j65}{50}-1}{\frac{35+j65}{50}+1}=0.6234\angle{65.59^o}

    What is the phase angle?

    In terms of phase relationship the incident wave must travel from the source to the load and the reflected wave must travel back to the source. If the reflected wave is in phase with the input voltage then the total return phase change is 65.59° or 32.8° one way. A wavelength corresponds to 360° so that's a line length of 32.8°/360° or 0.091 wavelengths.
     
    geft likes this.
  11. geft

    Thread Starter New Member

    Dec 8, 2011
    19
    0
    Thank you all for the help!
     
  12. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    It looks like you solved it. However, since we've made the recommendation to consider Smith charts, we would be remiss if we didn't make some attempt to let you see their value.

    t_n_k gave a nice intuitive description on a method to solve this. How did he find this insight? I'm guessing from a familiarity with the theory and particularly with the visualization of that theory as enabled by using a Smith chart.

    I've attached a drawing of the Smith chart that goes along with t_n_k's description. The Smith chart may look complicated, but it is simple in concept. The plot is a polar plot in which the radius represents the magnitude of the reflection coefficient and the angle represents the angle of the reflection coefficient. Overlayed on top of this is a mapping of the impedance that goes with that reflection coefficient (impedance is normalized relative to Zo). One starts by marking the terminating impedance on the plot, and then one can move away from the load (toward the generator) by moving on a circle clockwise. As t_n_k said, the magnitude of the reflection coefficient does not change, which is why the circle is the correct shape. Once you hit the x-axis, you know the impedance at that distance is real. There you can read off the resistance value (a little over 200 Ohms in this case), and you can note the distance of traversal either in degrees in the reflection coefficient or in units of lambda. The outer rim of the Smith chart provides scales for both measures.
     
    Last edited: Dec 22, 2011
    geft likes this.
  13. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,286
    331
    What does it mean to say "coax has more capacitance per foot than inductance"?

    I cut off a foot of some small coax I have on hand and measured 55.7 pF and 58.4 nH. Is 55.7 pF greater than 58.4 nH? How does one compare picofarads to nanohenries?
     
  14. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    I meant that the capacitance of coax is a bigger factor in transmission than the inductance. When looking at different types of coax, capacitance per length is always listed, while inductance isn't.
     
  15. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,286
    331
    In what sense is capacitance of a coax a "bigger factor in transmission than the inductance"? It's not clear what you mean by that.

    Why do you suppose inductance isn't listed?
     
  16. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Both are important, but we seem to have a little more capability to control or change capacitance since the dielectric used has an impact on permittivity/capacitance. Typically, we don't use magnetic materials in transmission lines. This is probably why the capacitance is specified more often. That is, the permeability of free space is well-known, but the relative permittivity of the dielectric is likely to change in different waveguides. (note, the known geometry does affect inductance and capacitance)

    This gets to the one sense in which you might be justified to say that capacitance is bigger. Relatively speaking, in comparison to free space, a coax line might be viewed as having a large capacitance, due to the use of a dielectric with relative permeability greater than one. Relative permeability is typical about one in typical low-loss transmission lines.

    The characteristic impedance says a lot here because Zo=sqrt(L/C) and higher capacitance/permittivity will decrease the line impedance. In the old days, our TV antennas used 300 Ohm cable which is not too far off from the free space value of 377 Ohms because those waveguides allowed most of the wave to travel in the free space around the cable, and the geometry was less restrictive. Modern day coax line, with values in the 25 to 100 Ohm range (50 Ohm most common, and 75 Ohm for TV) confines the wave almost entirely in the dielectric, due to the shielding property of coax. Note geometry is relavent also, as one can see in the capacitance formula for a coax geometry.

    It's interesting to compare the formula for propagation speed v=1/sqrt(LC) with that for Zo. In the speed formula, the geometrical effects on L and C cancel out and the only thing which slows the speed to a value let than c=299792458 m/s is the relative permittivity of the dielectric. However, for Zo, the geometrical effects have a double impact in that they change both L and C inversely.
     
    Last edited: Dec 23, 2011
  17. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Re the earlier solution -

    For those who prefer a mathematical representation it's possibly worthwhile noting that for two points separated by distance 'd' on a loss-less line, the relationship between the complex reflection coefficients is given by

    \Gamma_1=\Gamma_2 e^{-2 j \beta d}

    where β is the wave number and

    \beta=\frac{2\pi}{\lambda} \ and \ \lambda=wavelength

    So one can formally find the solution given

    \Gamma_1=0.6234\angle{0^o} \ and \ \Gamma_2=0.6234\angle{65.59^o} \ for \ Zo=50\Omega

    where location '2' is at the load [(35+j65)Ω] and location '1' is the nearest point at which the line impedance is purely resistive.
     
    geft likes this.
Loading...