I've been trying to simulate the asymmetric pi-attenuator example on this page with different source and load impedances: http://www.electronics-tutorials.ws/attenuators/pi-pad-attenuator.html
I simulated it with an input voltage of 1V.
http://postimg.org/image/a25pao46z/
The output I get, however, is 0.41V which corresponds to an attenuation of 7.75dB. Why does this happen?
I also tried simulation the 10dB balanced pi attenuator shown in the page and get near perfect result from it. Would appreciate it if anyone could explain the discrepancy to me.
Best Answer
The pad is giving 6dB of power attenuation. Assuming a source resistance of 75 ohms and 1V open-circuit source voltage, the voltage across the pad input is 0.5V across 75 ohms for a power of
$$p_{in} = \frac{(0.5)^2}{75} = 3.33mW$$
The output voltage is .2047V and this is across 50 ohms so the output power is
$$p_{out} = \frac{(0.2047)^2}{50} = 0.838mW$$
The power is thus attenuated by
$$-10 \cdot \log{\frac{0.838}{3.33}} = 6\mathrm{dB}$$
If the source and load impedances are equal, the power and voltage attenuation are equal.
$$20 \log \frac{V_{out}}{V_{in}} = 10 \log \frac{V^2_{out}}{V^2_{in}} = 10 \log \frac{V^2_{out}}{Z_{out}}\frac{Z_{in}}{V^2_{in}} = 10 \log \frac{p_{out}}{p_{in}}, \, Z_{out} = Z_{in}$$
If the source and load impedances are unequal, the power and voltage attenuation will not be equal.
$$20 \log \frac{V_{out}}{V_{in}} = 10 \log \frac{p_{out}}{p_{in}} - 10 \log \frac{Z_{in}}{Z_{out}}$$
So, in your example, we should have
$$20 \log \frac{V_{out}}{V_{in}} = -6\mathrm{dB} - 10 \log \frac{75}{50} = -6\mathrm{dB} - -1.76 \mathrm{dB} = -7.76 \mathrm{dB}$$
which agrees with your results.