use a frequency range from 10 Hz to 10 kHz and show that the circuit operates as a high-pass filter with a ‘knee’ point at 1 kHz
I have to be honest i an not 100% on what the question is asking me,
should i be working out Frequency resonance first using all the information provided, then i need to input this into a graph ?

 

If you use a simulator, then you don't need to do any calculations.
Why do you think you might need to?

 

not sure, but then i dont know what simulator to use ? what type of simulation program should i be looking for online?

 

Several of us here use the free analog circuit LTspice simulator from Analog Devices.
It has a somewhat steep learning curve, but there are good tutorials and sample circuits available, and we can help with any questions you may have.

 

ok well i am happy to give it a try is an online sim or do you download it ? the links i was given to plot this was all for pc programs but im on a mac

 

LTspice (from Analog Devices) has a free download for Mac.

 

use a frequency range from 10 Hz to 10 kHz and show that the circuit operates as a high-pass filter with a ‘knee’ point at 1 kHz
I have to be honest i an not 100% on what the question is asking me,
should i be working out Frequency resonance first using all the information provided, then i need to input this into a graph ?View attachment 294892

How can you determine if it is a filter when it doesn't say where the output is?

 

How can you determine if it is a filter when it doesn't say where the output is?

Perhaps this is part of the task: To specify the correct high-pass output which has a "knee point".

 

A high-pass filter with a knee or corner frequency at 1kHz will attenuate frequencies blow 1 kHz and it will pass without modification frequencies above 1 kHz. Since your source is at 50 Hz., the circuit should attenuate any energy from that circuit.

Assuming a ground reference on the negative end of the AC source and an output at the junction of the two capacitors and the resistor you should be able to find the attenuation of a 50 Hz. signal from what you know. Just turn everything into an impedance at 50Hz. and see what happens to the RMS value of your 10V source.

If you do an AC sweep you won't find a "knee" at 1 kHz, it will be mostly flat there and above but there is something going on in between 50Hz and 1kHz.

Hint: What happens to the impedance of L1 & C2 between 50hz and 1 kHz?

 

use a frequency range from 10 Hz to 10 kHz and show that the circuit operates as a high-pass filter with a ‘knee’ point at 1 kHz
I have to be honest i an not 100% on what the question is asking me,
should i be working out Frequency resonance first using all the information provided, then i need to input this into a graph ?View attachment 294892

You need to specify what the output is. You don't even indicate if the output is a voltage or a current, let alone where the output is.

What is the actual assignment? Does the assignment specify that you are to use a simulator? That would seem a bit odd in a course where they haven't given any instruction about what simulators are and how to use them.

 

use a frequency range from 10 Hz to 10 kHz and show that the circuit operates as a high-pass filter with a ‘knee’ point at 1 kHz
I have to be honest i an not 100% on what the question is asking me,
should i be working out Frequency resonance first using all the information provided, then i need to input this into a graph ?View attachment 294892

Hello there,

Where did you get this circuit from? I ask because there does not seem to be any knee at 1000Hz at either of the unknown nodes. In fact, there doesn't seem to be a knee anywhere on the response from 10Hz to 1kHz.
I am assuming the output is a voltage and is taken from one of the two unknown nodes with reference to ground, and ground is at the bottom of L1 which is also directly connected to two other components.

 

use a frequency range from 10 Hz to 10 kHz and show that the circuit operates as a high-pass filter with a ‘knee’ point at 1 kHz
I have to be honest i an not 100% on what the question is asking me,

I must admit that I am not very familiar with an expression like "knee point" (in filter theory).
However, it is not a problem to find the ac-response of the circuit - assuming that the output is taken across the inductor L.
In this case, the circuit assumes a 3rd-order high-Q highpass responce with a peak at app. wp=1000 rad/s.

When the output is taken across the C-L series connection the output looks very similar, however with a real zero at the C-L series resonant frequency which is app. wo=905 rad/s. (Perhaps this is the "knee point" according to the author of the task description ?)

 

I must admit that I am not very familiar with an expression like "knee point" (in filter theory).
However, it is not a problem to find the ac-response of the circuit - assuming that the output is taken across the inductor L.
In this case, the circuit assumes a 3rd-order high-Q highpass responce with a peak at app. wp=1000 rad/s.

When the output is taken across the C-L series connection the output looks very similar, however with a real zero at the C-L series resonant frequency which is app. wo=905 rad/s. (Perhaps this is the "knee point" according to the author of the task description ?)

Hi,

A knee point is a point where the plot goes smoothly from more horizontal to more downward vertical. It looks like the knee of a person sitting down in a chair where the plot line follows their leg from hip to knee and then from knee to foot.
I don't think there is an exact definition for this but it is still descriptive and distinct from other part of the curve.

 

I wouldn't like to have Mr. Chebyshev's knees!

 

Hi,

A knee point is a point where the plot goes smoothly from more horizontal to more downward vertical. It looks like the knee of a person sitting down in a chair where the plot line follows their leg from hip to knee and then from knee to foot.
I don't think there is an exact definition for this but it is still descriptive and distinct from other part of the curve.

Yes - that´s what I have assumed: Another (unconventional) name for the pole frequency of a frequency-dependent network.
However, this view/interpretation is not in accordance with the task desription resp. the frequency behavior of the circuit.

 

The response looks pretty flat at 1 kHz. too me. No evidence of a "knee" there.

 

Yes - that´s what I have assumed: Another (unconventional) name for the pole frequency of a frequency-dependent network.
However, this view/interpretation is not in accordance with the task desription resp. the frequency behavior of the circuit.

Hi,

I just mentioned this because there was some interest in the knee and also from the first post:
"high-pass filter with a ‘knee’ point at 1 kHz ".

 

This is what i created, i think its the same

 

This is what i created, i think its the same

What is the purpose of your last post? To show us again the same circuit which you have prsented in your 1st post?

 

This is what i created, i think its the sameView attachment 296818

Hello again,

Oh ok, assuming that is the right plot, i think the word "knee" would be a very poor choice of words for this problem if not even outright just plain wrong. There are too many places we might call a knee i think.
Maybe you can ask your instructor which 'knee' they are talking about.
I suppose the point you have marked "C1" could be the knee, assuming you got the plot right which I can't check right now maybe later.