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need help in microwave oscillators

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Jun 3, 2002
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I'm studying microwave oscillators (two port) and I'm confused about the synthesis of the terminating (input) network. one example synthesized Zs = 28+(j1.9) as 90ohm resistor in series with (0.238lembda) transmission line. The same example with different book synthesized the same Zs as 28ohm resistor in series with (0.01lembda) transmission line.
the 0.238 and 0.01 are the lengths of the transmission lines.

I would like to know how they got these numbers.



Are the two examples the same topology? That is, is the resistor at the transistor end of the line with the other end of the line shorted?

I can see the second example being this type. The line looks like an inductance which is in series with the resistance to produce the complex impedance.

The first example looks like putting the resistor on the other end of the line relative to the end against the transistor. It is almost a quarter wave line which would transform the resistor to another resistance if it was a quarter wave. Since it is slightly off, the impedance transformed to has a slight reactance.

In both cases, the line impedance is an important factor.

I tried to get the second solution but i don't know how to get the (0.01lembda) transmission line.
I got like this:
1.9/50 = 0.038 then I find the location of 0.038 on smith chart but it does not give (0.01lembda)


do it with math

I think that you will have to do it with the math equations. Itterate the Zo until it works.

Another thing is to start with the square root of 90 times 38 to get a first guess for the line impedance. (58.5 ohms) Then do your smith chart graphs based on this line impeance normalizing factor or your math equation based on this line impedance.

After thinking about this, I suspect that 50 ohms will end up as the line impedance that works. You should try that one first.

Actually I don't understand.
I would like to know how to convert Zs = 28+(j1.9) to a resistor and a series transmission line.
I know how to convert Zs to a resistor in series with an inductor.


tried it

I have used your parameters in the WinSmith program. You can get 28 +j1.9 two ways.

1. short one end of a 50 ohm transmission line 0.005 wavelengths long to get the +j1.9 at the other end. Put the 28 ohm resistor in series withthe other end to get the 28 +j1.9.

2. Take a 0.238 wavelength long 50 ohm transmission line and put a 90 ohm rsostor across one end. The impeance at the other end is 28 +j1.9.

thanks for your effort.
Still I don't know how to convert (j1.9) to (0.238*lembda) using smith chart.
If you put Zs = 28+(j1.9) on smith chart and draw the VSWR circle, it will cross the center line at 1.8, so 1.8*50 = 90, and the distance from Zs location to 1.8 (0.25*lembda) = 0.25-0.008 = 0.242lembda and not 0.238lembda.
this what I don't understand.


how to use smith chart

All impedances are to be divided by the line impedance (50 in this case). Therefore you start on the chart at 90/50 (1.8) with no reactance part..

Draw a circle using the center of the chart as the circle center and the distance to the 1.8 point as the radius. Follow the circle from the 1.4 point in the towards generator direction (clockwise) for an angle of 0.238 wavelengths on the outside scale or for a rotation of 360 degrees times 0.238/0.5 or 171.4 mechanical rotation degrees on the paper. This should bring you to 28/50 +j1.9/50 or 0.56 +j0.038 on the impedance coordinate lines. (or close enough for demonstration purposes) [edited part: it actually brings you to -j1.9/50 which is not what we want. See further down the page for the correction of 0.258 for the right length.]

If you do some trial and error with the line length and load resistance you can get closer.

thanks again.
I cannot start from 90/50 = 1.8 because I assume I still don't know this point. All what I know is 28/50+j1.8/50. If you use the VSWR circle you will get the 90 but not the 0.238lembda. The problem here is that if you draw a straight line that goes from the center of smith chart and passes through the point where the j0.038 is, and then read the wavelegth scale, you will get 0.008lembda. Since the VSWR cross the center line to get 1.8, this is at 0.25lembda. Now the total distance travelled is:
0.25lembda-0.008lemda = 0.242lembda. This what makes me crazy.
The problem I will get similar thing in the exam and I don't like to lose the mark.
I have attached a file that contains the same problem in one of the books I have. Look at the example (11.8). This is the same Zs and the auther claims that Zs (Zg)can be synthesized by 28ohm resistor in series with 0.01lembda transmission line. this is the second solution I mentioned in my first post. The 90ohm solution is in pozar's text.


go backwards

If you want to go the other direction, you first specify the transmission line impedance. Then you divide the impedance you want to end up with by this line impedance. Then you find that "normalized" point on the chart. Then you draw the circle from the center with that radius. Where the line crosses the real axis is the resistor you use. The rotation around the circle is the line length.

After taking a look at the chart again, it looks like the 0.238 line is in error. It should be 0.258 of a wavelength. (I should get more sleep each day.)

I will edit the previous posts to the correct values. Also the 0.01 line is really about 0.008.

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