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Understanding S21 when there is an impedance mismatch (Read 1065 times)
iVenky
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Understanding S21 when there is an impedance mismatch
May 11th, 2020, 3:41pm
 
I have a very basic question regarding S21 when there is an impedance mismatch with the characteristic impedance.

I have attached a figure of the test circuit.

Assume that the Transmission line is loseless and has an effective length of λ/4 at Frequency=200GHz. When I plot the S11 at DC, it's simply

S11= (1M-10)/(1M+10)~0.99998 (pretty close to 0dB, it's complete reflection at the input).
S21 obtained is around -44 dB (based on simulation on ADS/Cadence). Is this because, |S11|2+|S21|2=1 for a lossless transmission line?
What I don't get is the following.

At DC, the gain of the above network is 1M/(1M+10) ~ 1, on the other hand, S21 is much below (i.e. -44 dB). What am I missing? I thought S21 is an indication of insertion loss but this result is counter intuitive. I am so confused...

*Note that I am not interested in transmission line s-parameters because in that case I have to use 50 ohm load. I am more interested in reflection coefficient and insertion loss when there is an impedance discontinuity and understand what s21 means in that case.
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Frank Wiedmann
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Re: Understanding S21 when there is an impedance mismatch
Reply #1 - May 13th, 2020, 5:52am
 
You need to normalize by the respective port impedances, see for example page 28 of http://rfic.eecs.berkeley.edu/~niknejad/ee242/pdf/eecs242_lect5_sparam.pdf.
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iVenky
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Re: Understanding S21 when there is an impedance mismatch
Reply #2 - May 13th, 2020, 2:55pm
 
Hi Frank Wiedmann,

Thank you very much for the reply. This was what I was looking for. It looks like a normalizing factor of 10 log10(1M/10) = 50 dB. So the V2-/V1+ is actually 50-44= 6 dB (which means I am missing  2x factor somewhere). What's that I am doing wrong here.
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Frank Wiedmann
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Re: Understanding S21 when there is an impedance mismatch
Reply #3 - May 14th, 2020, 3:23am
 
In order to determine S21, you must use a voltage source with amplitude 2. This will give you an amplitude of 1 at the output when both impedances are equal.
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