Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
I have tried the 820 kHz and got RC=2.42e-6 so it's better than the previous value gotten by 21kHz.
Anyway I don't have my own PC. I use public computer. Can you please suggest an online application that can help me with this? I think, however, that my calculations are not too bad as they are...
I applied this equation for the 1dB attenuation:
20xlog(Xc/sqrt(R^2+Xc^2)=-1 dB
At the end I got RC=3.79x10^-6 which is for 21kHz
Then I computed f3dB=1/(2*pi*RC)= 41993 Hz
which is quite similar to the textbook solution: 41270 Hz
Then I computed the attenuation for 820kHz and got 25.81 dB...
Hello,
I want to ask this question stated in my textbook:
Can an low-pass RC filter be used to provide 1dB attenuation at 21 kHz and at least 22 dB attenuation at 820 kHz?
It is written in the solution that the 3 dB frequency is 41,27 kHz. I don't know where this comes from?
Can you please...
And what about the second equation? Because I find it strange to use Thevenin's theorem here. I had a very similar example on the book with input impedance known but with amplifier included instead of ADC. Also AC voltage was considered to be the signal of interest so DC voltage to be as AC...
I didn't really get the third point. Can you explain more about it?
I think the offset of +5V adds to +3.3V but I don't know where the voltages are located. I am confused about this and I also don't know what the next step would be, I mean the general procedure of approaching the solution.
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.
