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.

 

Mlog im nog getting those values. The j is screwing my calculations around.