Hi again guys Circuit: http://i46.tinypic.com/5owp3l.jpg Calculate the cut off freq when a)R3 is not connected and b) When R3 is connected c)calculate the attenuation in dB when the signal frequency is 4Fc. ------------------------------- a)Fcut-off=1/(2∏(R1+Rs)C1) = 1015.15Hz b)Fcut-off=1/(2∏((R1+Rs)//R3)C1)=1137.57Hz c)[I choose Vin=1V]as nowhere it states an input voltage. Also i assume calculations for circuit with R3 and no R3 needs to be done: 1-[NO R3 connected] Xc=1/(2∏*4Fc*C1)=3015Ω Vout=Vin*Xc/(√((R1+Rs)^2 + Xc^2))=242.53mV Att(dB)=20log(Vout/Vin)=-12.3dB I then do the same with R3 connected to get the Attenuation. Which gets me -11.38dB. Formulas in a) should be correct but I included it for c). Is my cut off calculation correct with the R3 included? And do I understand c) correct? Thanks
I found the same answers as you except for the voltage gain (attenuation) with R3 included at 4 times fc. I got -13.3 db @ 4550 Hz.
Thanks mlog So my other calculations calculations are correct? I do realise I used the freq value with no R3 included as a mistake. However I do not get the -13.3dB . With R3 included: Xc=1/(2*pi*4*1137.57) = 2690.51 Ohm Rparallel=12060//100 000 = 10762 Ohm Vout=Vin*Xc/(√((Rparallel)^2 + Xc^2)) = 0.24253 V Atenuation = 20log(Vout/Vin) = 20log(0.24253) = -12.3dB which means the same answer as for when R3 is not included. Am I missing something?
For R3 included, you cannot use the same equation for finding the voltage gain (attenuation). The generalized equation for the voltage gain is: A = Z/(R+Z). where R=Rs+R1. With no R3, the value of Z includes only C1. However, with R3, the value of Z includes both R3 and C1. In the 1st case, Z = 1/(jωC). In the 2nd case, Z = R/(1+jωRC). In both cases, R=R3 and C=C1, and j is the imaginary operator.