How to calculate an attenuation of a pi pad?

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Hi guys,

I have 10 Ohms series and 470 Ohms shunt resistors for a pi attenuator.
When I calculated using below equation,

Vin/Vout=(470//50)/((470//50)+10)
attenuation in dB = 10*log((Vin/Vout)^2)

I get 1.736 dB of attenuation

However when I simulate the exact same circuit using ADS I get 1.79dB of attenuation
Can anyone explain where I made a mistake?

Thanks in advance.
 

You ignored the mismatching caused by an input resistance slightly below 50 ohms. Need to calculate the full circuit including 50 ohm source impedance.
 

and you probably also forgot about the tolerance variations of your resistors

1.736 compared to 1.79 is not a huge difference

ADS will be assuming perfect (ideal) 10 Ohm and 470 Ohm resistors

Dave
 

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