| Home |
| Search |
| Today's Posts |
|
#19
September 21st 09, 03:48 AM
posted to rec.radio.amateur.antenna
|
|||
|
|||
Resaonance and minimum SWR
On Sun, 20 Sep 2009 22:41:29 -0400, Walter Maxwell
wrote: My post below is not exactly on target for the thread, but I believe useful. It's Sec 11.3 from Chapter 11 of Reflections, the whole of which is available on my web page at www. w2du.com. The title of the Sec is "A Reader Self-test and Minimum-SWR Resistance." Sec 11.3 A Reader Self-Test and Minimum-SWR Resistance " Everyone knows that when a 50-ohm transmission line is terminated with a pure resistance of 50 ohms, the magnitude of the reflection coefficient,, rho , is 0, and the SWR is 1:1. Right? Of course! With that in mind, here is a little exercise to test your intuitive skill. If we insert a reactance of 50 ohm in series with the 50-ohm resistance, the load becomes Z = 50 + j50. The SWR will be 2.618:1. Now for the question. With this 50-ohm reactance in the load, is the SWR already at its minimum value with the 50-ohm resistance, or will some other value of resistance in the load reduce the SWR below 2.618:1? You say the SWR is already the lowest with the 50-ohm resistance, because, after all, the line impedance, ZC, is 50 ohms? Sorry, wrong. With reactance in the load, the minimum SWR always occurs when the resistance component of the load is greater than ZC. In fact, the more the reactance, the higher the resistance required for to obtain minimum SWR. For any specific value of reactance in the load there is one specific value of resistance that produces the lowest SWR. I call this resistance the "minimum-SWR resistance." Finding the value of this resistance is easy. First you normalize the reactance, X, by dividing it by the line impedance, ZC. The normalized value of X is represented by the lower case x. Thus x = XC / ZC. Then we solve for the normalized value of resistance r, from Eq 5-1, which is repeated here. r = sqrt (x^2 + 1) Eq 5-1 Let's try it on the example above. The normalized value of 50 ohms of reactance X, is x = 1. Substituting in Eq 5-1, r = sqrt 2 = 1.414. So the true value of the minimum-SWR resistance is 1.414 x 50 = 70.7ohms. While the 50-ohm resistance yields a 2.618:1 SWR, the 70.7-ohm resistance in series with the 50-ohm reactance yields an SWR of 2.414:1. Not a great deal smaller, but still smaller than with the 50-ohm resistance. So let's try a more dramatic example, this time with a 100-ohm reactance, which has a normalized value x = 2.0. With a 50-ohm resistance, the SWR is now 5.828:1. However, with the normalized minimum-SWR resistance, r = sqrt 5 = 2.236. Multiplying by 50, we get R = 111.8 ohms. With this larger resistance in series with the 100-ohm reactance, the SWR is reduced from 5.828:1 to 4.236:1. The results of this exercise didn't turn out quite the way you expected, did it?" For further proof of this concept I suggest reviewing the remainder of this Sec using the Smith Chart, available from my web page. Walt, W2DU |
| Thread Tools | Search this Thread |
| Display Modes | |
|
|
Similar Threads
|
||||
| Thread | Forum | |||
| Minimum gauge for groud... | Shortwave | |||
| Minimum gauge for groud... | Shortwave | |||
| 75 to 50 ohm minimum loss pad | Homebrew | |||
| Solar Minimum in 2006? | Shortwave | |||
| FA: Swan 350 $15 minimum bid! | Boatanchors | |||
Threaded Mode
