How much of that reading assignment did you do?

With a spread sheet, you should be able to graph the results (a bode plot).

 

How much of that reading assignment did you do?

With a spread sheet, you should be able to graph the results (a bode plot).

graph the result of what? of the RC filter?

 

graph the result of what? of the RC filter?

Graph Vo of the circuit you posted.

Did you read all the links?

 

Graph Vo of the circuit you posted.

Did you read all the links?

why need to grpah Vo?
or Av vs Frequency low and high R-C filters

 

The Vo graph is the bode plot ... don't bother ... it's not necessary.

After five pages, your still playing this game.

Did you read all the links in the filters section?

The question remains ... what happens when R1 is increased.

Did you "fix" your calculation of the fc for R and C? Do you know what your mistake was?

 

The Vo graph is the bode plot ... don't bother ... it's not necessary.

After five pages, your still playing this game.

The question remains ... what happens when R1 is increased.

Did you "fix" your calculation of the fc for R and C? Do you know what your mistake was?

no what was the mistake...I put the calculation how I did it...so what was the mistake thr?
1/(2pi)(10*10^3)(560*10^-9)=28.42

 

I read abt the low pass and high pass

 

no what was the mistake...I put the calculation how I did it...so what was the mistake thr?
1/(2pi)(10*10^3)(560*10^-9)=28.42

The capacitor is incorrect. The capacitor is 560 pF ... as I previously stated. You do not have the capacitor as picofararads.

I read abt the low pass and high pass

That too was obvious to me. You need to read all of them including the summary. At least you were honest to admit to reading only those two.

When your done, you should be able to answer all the questions, including the one you originally asked.

 

The capacitor is incorrect. The capacitor is 560 pF ... as I previously stated. You do not have the capacitor as picofararads.



That too was obvious to me. You need to read all of them including the summary. At least you were honest to admit to reading only those two.

When your done, you should be able to answer all the questions, including the one you originally asked.

so it will be 1/(2pi)(10*10^3)(560*10^-12)=28.42kHz
in my class we only did these tow filters...n I thought shud be from these two filters...what kind of filter is that circuit?
in A lowpass filter capacitor goes to ground.
A highpass filter resistor goes to ground.

 

so it will be 1/(2pi)(10*10^3)(560*10^-12)=28.42kHz
in my class we only did these tow filters...n I thought shud be from these two filters...what kind of filter is that circuit?
in A lowpass filter capacitor goes to ground.
A highpass filter resistor goes to ground.

Your fc answer is now correct. That is why it's important for you to "show" your work. Showing your work is not to demean you, but to help you identify where you made the mistake.

You need to read the remaining links and then think what your circuit is doing. You have the two fc's.

 

The capacitor is incorrect. The capacitor is 560 pF ... as I previously stated. You do not have the capacitor as picofararads.

Once again underlining the value and important of properly tracking units. If done correctly, that part of the expression should have been something like:

(560pF)(1F/10^12pF)

That might not have stop someone from using 10^9nF/F, but tracking down where the mistake was made becomes a lot easier and more obvious, not to mention being able to walk the correction from that point on out to the final answer.

In addition, the prefixes are also factors that can be combined and canceled out without having to explicitly express them as values.

 

Your fc answer is now correct. That is why it's important for you to "show" your work. Showing your work is not to demean you, but to help you identify where you made the mistake.

You need to read the remaining links and then think what your circuit is doing. You have the two fc's.

one fc for high pass and another one for low pass
for the example I was doing
C = 0.002 μF same as C1, R1 = 22 kΩ and R = 2.2 k Ω.. and in order to plot I have to calculate the fc...
which I got to be
Low pass filter= FC=1/2πRC=1/ (2π) (2.2kΩ) (0.002µF) =36.17 kHz
High pass filter=FC=1/2πR1C1=1/ (2π) (22kΩ) (0.002µF) =3.61 kHz

 

Funny, you never mentioned any values when you posted the circuit.

What is the filter called that comprises both a high pass and a low pass section, or did you disregard my advice to read the remaining links?

 

Funny, you never mentioned any values when you posted the circuit.

What is the filter called that comprises both a high pass and a low pass section, or did you disregard my advice to read the remaining links?

nvm..I kind of figured it out...
thanks for the help..really appreciate...

 

It's tough helping when you hold back information. It's like pulling teeth.