The file attached is part of a lab I am writing a formal report on. Part of the report notes that -3dB frequency occurs at 1/(R2*C). Can someone explain how this can be derived?

Thanks

 

You need to show your best attempt to either solve the problem or show that you at least understand the problem. Neither needs to be perfect, but they need to be present so that we have a starting point to help you from.

In this case, describe what you know about the notion of a -3dB frequency.

Also, what is the output as a fraction of the input as a function of frequency?

 

Should be in the textbook.

 

Also, what is the output as a fraction of the input as a function of frequency?

....and what is the well-known and established name of the whole circuit?

 

Blah....to many questions. LvW...this is obviously a lowpass filter. shteii01...thanks for the thought provoking and engaging insight. WBahn, this problem is pretty straight forward. We know all voltages but Vi and Vo -->
1) (o-Vi)/Ra + (0 - Vo)/(R2||C) = 0.
2) R2 = 2*R1
3) R2||C = 1/(1/R1 + jwc)
Damnit I found my error. Factoring is the problem here!

 

shteii01...thanks for the thought provoking and engaging insight.

LOL

20*log10(|H(jw)|)=-3 dB
where w=2*pi*cutoff frequency of your filter

I really like my textbooks. They are full of answers.

 

LOL

20*log10(|H(jw)|)=-3 dB
where w=2*pi*cutoff frequency of your filter
I really like my textbooks. They are full of answers.

Are you sure that you have picked up the right answer from your textbook?
Don`t forget the DC gain Ao of the first-order lowpass.
Thus, the cut-off frequency will be at a point where |H(jw)| is 3 db below Ao (rather than |H(jw)|=-3 dB).

 

Are you sure that you have picked up the right answer from your textbook?
Don`t forget the DC gain Ao of the first-order lowpass.
Thus, the cut-off frequency will be at a point where |H(jw)| is 3 db below Ao (rather than |H(jw)|=-3 dB).

They said -3 dB.

Maybe you should discuss the difference with the OP. They might benefit.

 

They said -3 dB.
Maybe you should discuss the difference with the OP. They might benefit.

It depends, of course, on the DC gain Ao.
There is nothing to discuss - it´s basic.
If Ao=1 (0 dB) the cut-off is at a frequency which gives |H|=-3 dB, otherwise it is at a frequency for which |H|=Ao-3 dB.
For the circuit as given in post#1 we have |Ao|=10/5.1.

 

Blah....to many questions. LvW...this is obviously a lowpass filter. shteii01...thanks for the thought provoking and engaging insight. WBahn, this problem is pretty straight forward. We know all voltages but Vi and Vo -->
1) (o-Vi)/Ra + (0 - Vo)/(R2||C) = 0.
2) R2 = 2*R1
3) R2||C = 1/(1/R1 + jwc)
Damnit I found my error. Factoring is the problem here!

See.

By forcing yourself to walkthrough and explain a "pretty straight forward" problem to someone else, you will very often find the mistake in your reasoning or in your math all on your own.

Now, having said that, I don't see where you are getting some things.

In equation 1, what is Ra? Did you mean R1?

In equation 2, R2=2*R1 is approximately true -- and possibly within the tolerance of the components -- but keep in mind that there are circuits in which a few percent difference between components is what is driving the behavior of the circuit, so be a bit careful in relying on such simplifications without justifying them. However, in this case you don't need to make use of any relationship between R1 and R2. You can leave them as just generic resistances and you get the same results.

In equation 3, did you really mean R2||C = 1/(1/R1 + jwc), or did you mean to use R2 on the right hand side?

I think you need to slow down and be sure that you aren't making stupid mistakes (which we will ALL always make from time to time, but you want to go at a pace and a level of detail that results in the rate at which you make them being very, very low).

 

LOL

20*log10(|H(jw)|)=-3 dB
where w=2*pi*cutoff frequency of your filter

I really like my textbooks. They are full of answers.

The cutoff frequency of the filter and the point at which |H(jw)| = -3 dB are not always one and the same. The cutoff frequency of the filter is almost always interpretted as being where the filter response to -3dB below the nominal passband response. That is what is being sought in this case.

 

I think you need to slow down and be sure that you aren't making stupid mistakes (which we will ALL always make from time to time, but you want to go at a pace and a level of detail that results in the rate at which you make them being very, very low).

Daniel, WBahn gave you some general recommendations how to proceed.
In addition, let me tell you that the definition of the 3-dB cut-off frequency is derived from the transfer function of the whole circuit.
As you have already recognized - it is a lowpass.
Thus, your first goal must be to find the transfer function (frequency-dependent gain formula) for your circuit.
This shouldn`t be a big problem if you remember the simple inverting gain formula.
Then, as a next step, you have to find the condition under which the MAGNITUDE of the transfer function is 3 dB down. I suppose, you know what the corresponding factor is (absolute value)?
This leads immediately to the wanted cut-off frequency.