Determine the filter's cutoff frequency

Thread Starter

Lhemming

Joined Dec 18, 2018
7
I need help to calculate the cutoff frequency for the below RC- circuit, and to know what impact R4 have.

upload_2019-10-2_12-38-56.png
Thanks in advance!
 

ericgibbs

Joined Jan 29, 2010
8,752
hi L,
Welcome to AAC.
As this is Homework, please post your attempt at solving, then we can guide you thru the problem.
E
 
Last edited:

crutschow

Joined Mar 14, 2008
23,371
Here's the LTspice simulation verification of that:
It shows a -3dB drop from the value at 100Hz at a frequency of 1753Hz (Cursor 2, limited by the graph resolution).

upload_2019-10-2_7-27-51.png
 

MrAl

Joined Jun 17, 2014
6,510
Hi,

That looks right and you can see the effect the 2nd resistor has on the frequency it acts as if it was in parallel with the first resistor.

However, in more complicated circuits we dont usually have a formula on hand that covers all cases so we have to do the entire math. It's not that difficult for this circuit though. Using the impedances for both the resistors and the capacitor we can use the voltage divider formula.

For a resistor:
zR2=R2
zR4=R4

and for the cap:
zC=1/(j*w*C2)

First we put the cap in parallel with the resistor R4 and get:
z2=R4/(j*w*C2*R4+1)

now the using the voltage divider formula we have:
Vout=V2*z2/(z1+z2)

which comes out to:
Vout=R4/(j*w*C2*R2*R4+R4+R2)

The amplitude of that is:
|Vout|=R4/sqrt((R4+R2)^2+w^2*C2^2*R2^2*R4^2)

Now first find the response to DC:
Voutdc=R4/(R2+R4)

Now we can equate the amplitude that divided by sqrt(2) to find the cutoff frequency:
R4/sqrt((R4+R2)^2+w^2*C2^2*R2^2*R4^2)=Voutdc/sqrt(2)

and solving that for w we get:
w^2=(R4+R2)^2/(C2^2*R2^2*R4^2)

or:
w=(R4+R2)/(C2*R2*R4)

Plugging in the values we get:
w=11000 rads/sec

dividing by 2*pi we get:
1750.704374010849 Hz

so rounding up we get 1751 Hz which is what you got which must be correct.

That method is a very general method which works on many types of circuits. Obtaining the DC response first is what you do for a low pass filter. For a high pass filter you have to get the infinite frequency response instead, and for a bandpass you have to get the center frequency first. So what you solve for first depends on the nature of the response of the circuit in question. Keep in mind also that sometimes you want to know the 3db down frequency referenced to some other point too which could be on a peak for example, in that case it depends on the application.
 
Last edited:

Thread Starter

Lhemming

Joined Dec 18, 2018
7
Hi,

That looks right and you can see the effect the 2nd resistor has on the frequency it acts as if it was in parallel with the first resistor.

However, in more complicated circuits we dont usually have a formula on hand that covers all cases so we have to do the entire math. It's not that difficult for this circuit though. Using the impedances for both the resistors and the capacitor we can use the voltage divider formula.

For a resistor:
zR2=R2
zR4=R4

and for the cap:
zC=1/(j*w*C2)

First we put the cap in parallel with the resistor R4 and get:
z2=R4/(j*w*C2*R4+1)

now the using the voltage divider formula we have:
Vout=V2*z2/(z1+z2)

which comes out to:
Vout=R4/(j*w*C2*R2*R4+R4+R2)

The amplitude of that is:
|Vout|=R4/sqrt((R4+R2)^2+w^2*C2^2*R2^2*R4^2)

Now first find the response to DC:
Vout=R4/(R2+R4)

Now we can equate the amplitude that divided by sqrt(2) to find the cutoff frequency:
R4/sqrt((R4+R2)^2+w^2*C2^2*R2^2*R4^2)=1/sqrt(2)

and solving that for w we get:
w^2=(R4+R2)^2/(C2^2*R2^2*R4^2)

or:
w=(R4+R2)/(C2*R2*R4)

Plugging in the values we get:
w=11000 rads/sec

dividing by 2*pi we get:
1750.704374010849 Hz

so rounding up we get 1751 Hz which is what you got which must be correct.

That method is a very general method which works on many types of circuits. Obtaining the DC response first is what you do for a low pass filter. For a high pass filter you have to get the infinite frequency response instead, and for a bandpass you have to get the center frequency first. So what you solve for first depends on the nature of the response of the circuit in question. Keep in mind also that sometimes you want to know the 3db down frequency referenced to some other point too which could be on a peak for example, in that case it depends on the application.
The teacher has not corrected it yet, but I will take the opportunity to thank you for your commitment :)
 

MrAl

Joined Jun 17, 2014
6,510
The teacher has not corrected it yet, but I will take the opportunity to thank you for your commitment :)
Hello again,

Ok great, and i just want to mention that i forgot to show Voutdc in the numerator in one of the equations. It is fixed now however. The solution was correct anyway though.
 

Thread Starter

Lhemming

Joined Dec 18, 2018
7
Hello again,

Ok great, and i just want to mention that i forgot to show Voutdc in the numerator in one of the equations. It is fixed now however. The solution was correct anyway though.
I didn't doubt it for a second, thanks again!
 
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