I'm trying to measure the bandwidth of a circuit that I've built. It's an analogue-to-digital converter (ADC) with some input protection and biasing. The ADC samples at a constant/fixed 50 kSa/s.
My approach is to inject a 2 V sine wave of variable frequency \$f_{in}\$ and take the maximum (\$Vmax\$) and minimum (\$Vmin\$) samples. That gives me \$Vpp = Vmax – Vmin\$, and I sweep \$fin\$ until \$ Vpp = 2 \sqrt{\frac{1}{2}}\ V \approx 1.41\ V \$.
I've read the Wikipedia articles Bandwidth (signal processing) and Half-power point, but I remain confused.
When I find the -3 dB point, what is the corresponding official bandwidth I've found? Is it \$fin\$ or \$ \frac{fin}{2} \$ with consideration of the Nyquist rate?
Best Answer
The bandwidth is the frequency for which the output is down 3 dB. It it a function of the analog portion of your circuit and does not depend on the sample rate nor the Nyquist rate which are basically digital concepts. Thus, in your experiment, the bandwith will be the frequency, fin, for which the output has dropped by 3 dB.