# Determine the filter's cutoff frequency

#### 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:

#### Lhemming

Joined Dec 18, 2018
7
hi L,
Welcome to AAC.
As this is Homework, please post your attempt at solving, then we can guide you thru he problem.
E
Can I put together R1 and R2 (R1xR4/R1+R4) and then make a cut-off frequenc calculation 1/(2*pi*R*C) ?

#### ci139

Joined Jul 11, 2016
725
• Lhemming

#### Lhemming

Joined Dec 18, 2018
7
Thanks, that was much information.
But I still wonder of the impact of R2 in the cutoff frequency.

#### LvW

Joined Jun 13, 2013
857
Why not using the DEFINITION of cut-off?
In this case, you must not guess ...but calculate!

• ci139

#### 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:

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:
• Lhemming

#### 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:

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 #### Lhemming

Joined Dec 18, 2018
7
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).

View attachment 187212
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.