Is this a high pass or low pass active filter?

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In summary, the conversation discusses a problem with a solution involving a low pass filter. The individual is confused about the gain and transfer function at different frequencies and whether the filter is a high pass or low pass. Another individual helps clarify the concept of a non-inverting op amp and explains how to determine the frequency at which the filter allows signals to pass. The conversation also mentions bode plots and their relevance to understanding the filter's behavior.
  • #1
theBEAST
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Homework Statement


Here is the problem with the solution:
21Ctf.png


The Attempt at a Solution


I don't understand why the filter is low pass. How I solve this usually is I plug in ω=0 and ω=∞. When I plug in 0 I get the gain of a non inverting op amp. When I plug in ∞ I get that the transfer function (gain) is equal to 1. Since it is equal to 1 doesn't that mean it let's high frequencies pass? I remember in passive filters when you get 1 for ω=∞ it means it is a high pass filter. HOWEVER, the tricky thing about active filters is that when ω=0, the gain is not 0, instead it is 1+R1/R2 which I think is why I am confused.
 
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  • #2
theBEAST said:

Homework Statement


Here is the problem with the solution:


The Attempt at a Solution


I don't understand why the filter is low pass. How I solve this usually is I plug in ω=0 and ω=∞. When I plug in 0 I get the gain of a non inverting op amp. When I plug in ∞ I get that the transfer function (gain) is equal to 1. Since it is equal to 1 doesn't that mean it let's high frequencies pass? I remember in passive filters when you get 1 for ω=∞ it means it is a high pass filter. HOWEVER, the tricky thing about active filters is that when ω=0, the gain is not 0, instead it is 1+R1/R2 which I think is why I am confused.

That gain equation looks wrong. First of all, it's an inverting amp, so the gain has to be negative. Second, the feedback impedance goes to zero at high frequencies, so the gain should go toward -0 at high frequencies...
 
  • #3
berkeman said:
That gain equation looks wrong. First of all, it's an inverting amp, so the gain has to be negative. Second, the feedback impedance goes to zero at high frequencies, so the gain should go toward -0 at high frequencies...

Hi,

I compared it with wikipedia and it looks like a non inverting amp to me? :S

http://en.wikipedia.org/wiki/Operational_amplifier#Non-inverting_amplifier

I also solved it on my own, as you can see, when I plug in a high frequency I do not get 0. Instead I get 1.

photo_19.jpg
 
  • #4
berkeman said:
That gain equation looks wrong. First of all, it's an inverting amp, so the gain has to be negative.

Have another look.. it is not inverting :) Vi is applied to the + t-erminal and will also appear at the - terminal. This causes current to flow from the top of R2 to ground, so the op amp will be pushing current from the output toward ground through R2 and R1||C, making the output at higher voltage than the voltage at the - terminal which is Vi. So Vo > Vi.

thebeast, what you did was right. You just need to be more careful about your conclusion.

At w->0, the capacitor has infinite impedance so it is out of the circuit. The output is Vo = (1+R1/(R1+R2))*Vi.

At w->∞, the capacitor is a short so R1 is out of the circuit. Now Vo=Vi.

The output at low frequencies is higher than the output at high frequencies so this is a low pass filter. It's a little different because the gain does not approach 0 as w→∞You can also see this from the suggested solution. There is both a zero and a pole. The pole occurs at a lower frequency so a bode plot will show the gain flat, then head down at -20db/decade at the low pass corner frequency (the pole) and then become constant when the high pass corner frequency is met (the zero adds a rising slope of 20db/decade). This is what you found -- higher gain at 0 and lower but non-zero gain at infinity.
 
  • #5
aralbrec said:
Have another look.. it is not inverting :) Vi is applied to the + t-erminal and will also appear at the - terminal. This causes current to flow from the top of R2 to ground, so the op amp will be pushing current from the output toward ground through R2 and R1||C, making the output at higher voltage than the voltage at the - terminal which is Vi. So Vo > Vi.

thebeast, what you did was right. You just need to be more careful.

At w->0, the capacitor has infinite impedance so it is out of the circuit. The output is Vo = (1+R1/(R1+R2))*Vi.

At w->∞, the capacitor is a short so R1 is out of the circuit. Now Vo=Vi.

The output at low frequencies is higher than the output at high frequencies so this is a low pass filter. It's a little different because the gain does not approach 0 as w→∞


You can also see this from the suggested solution. There is both a zero and a pole. The zero occurs at a lower frequency so a bode plot will show the gain flat, then head down at -20db/decade at the low pass corner frequency (the zero) and then stay constant when the high pass corner frequency is met (the pole adds a rising slope of 20db/decade). This is what you found -- higher gain at 0 and lower but non-zero gain at infinity.



Thank you! Your first paragraph really helped me in understanding how the non inverting op amp works.

Also... So whichever one is higher is the frequency that is allowed to pass? Unfortunately our professor did not teach us bode plots so I think that is the reason to why I am having difficulty in understanding this. I will definitely check it out though!
 
  • #6
aralbrec said:
Have another look.. it is not inverting :)

Oops! You are correct, my mistake. I usually draw my opamp circuits with the inverting terminal on top, but that's still not a good excuse for missing the non-inverting configuration.
 
  • #7
Unfortunately you quoted me before I finished an edit :) I had zero and pole swapped in my last paragraph. My post is corrected but the quoted text still has it wrong.

theBEAST said:
Unfortunately our professor did not teach us bode plots so I think that is the reason to why I am having difficulty in understanding this. I will definitely check it out though!

Alright well the solution is depending on a bode plot argument so that may be a problem for you to follow.

theBEAST said:
Also... So whichever one is higher is the frequency that is allowed to pass?

All frequencies are allowed to pass since there is no w where the gain is zero; it's just that some frequencies are emphasized and some are not, making them smaller in comparison. You found the low frequencies get high gain and the high frequencies get (relatively) low gain.
 

Related to Is this a high pass or low pass active filter?

1. What is an active filter?

An active filter is a type of electronic filter that uses active components, such as transistors or op-amps, to amplify and shape the input signal. This allows for greater control and flexibility in the filtering process compared to passive filters, which only use passive components such as resistors, capacitors, and inductors.

2. What is the difference between a high pass and low pass active filter?

A high pass active filter allows high frequency signals to pass through while attenuating low frequency signals. On the other hand, a low pass active filter allows low frequency signals to pass through while attenuating high frequency signals. The choice between a high pass or low pass filter depends on the specific filtering needs of the circuit.

3. How does a high pass or low pass active filter work?

A high pass active filter works by using a combination of active components and passive components to create a frequency-dependent voltage divider. This means that at certain frequencies, the output voltage will be higher or lower than the input voltage. The specific components and their values are chosen to achieve the desired filtering characteristics. A low pass active filter works in a similar manner, but with a different arrangement of components to allow low frequencies to pass through.

4. When would I use a high pass or low pass active filter?

A high pass active filter is commonly used in audio and RF applications to remove unwanted low frequency noise from a signal. This is useful for applications such as audio amplifiers and radio receivers. A low pass active filter is often used in power supply circuits to eliminate high frequency noise and provide a clean DC output. It is also used in audio applications to remove high frequency noise from a signal.

5. How do I design a high pass or low pass active filter?

The design of a high pass or low pass active filter involves choosing the appropriate components and their values based on the desired cut-off frequency and filter characteristics. This can be done using equations and circuit analysis techniques, or through the use of specialized software. It is important to consider factors such as component tolerances, stability, and noise when designing an active filter.

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