Wavelengths along a Smith Chart

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The discussion focuses on calculating input impedance and return loss using a Smith Chart, specifically at a distance of l = 5λ/8 from the load. It emphasizes the periodic nature of input impedance as a function of line length, necessitating the deduction of multiples of λ/2 to simplify analysis. The participants clarify that deducting 4/8 (or λ/2) from 5/8 results in a remainder of 1/8, allowing for effective use of the Smith Chart without unnecessary rotations.

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Hello,

This may be something that is more mathematical than electrical, but I figured here was the best place.

I am going through a worked example, and there is something I don't understand. The question is as follows

Find the imput impedance and return loss (in dB) of the loaded line in the last question as seen at the distance l = 5λ/8 from the load

The solution states that "because the input impedance is a periodic function of the line lenth, we need to decuct all multiples of λ/2 and analyse only the remainder (λ/8)..."

Now, I understand that λ/2 is a full rotation of the Smith chart, but why is that deducted? And, if you remove λ\2, how are you left with λ\8 ...from an origianl length of 5λ/8?

The last part may just be some mathematics that is escaping me at the moment, but I just cannot get past this part.

Thanks for any advice.

-S
 
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5/8 - 4/8 = 1/8.

You deduct it so you don't spin around the chart 4 times for nothing.
 
Hello.

Ok, that makes sense ...but, where is the 4 coming from? I mean, why not 6 times?

-S
 
1/2 Lambda = 4/8 Lambda.
 
Ahh, okay ...I am with you now!

That's fantastic. Thanks for your help.

-S
 

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