How to calculate an attenuation of a pi pad?

[sp28212821]

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.

 

[FvM]

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

 

[davenn]

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