Hi all,

I have a university assignment question that looks for the capacitor value in a low pass filter which has a resistor of 1kΩ and looks to provide a cut off frequency of 7kHz.

The output of the filter then goes into a ideal amplifier with a gain of -10 and a bandwidth of 1kHz to 7kHz.

So far I have attempted to obtain the capacitor value by using the formula: C = 1 / (2πFCR)
I can then get RC by multiplying by the resistor value.

But I am unsure what to do next with the gain mentioned and bandwidth? Any help is appreciated.

(I have attached a drawing of the circuit)

 

But I am unsure what to do next with the gain mentioned and bandwidth?

the amplitier will amplify the filter output by the gain value.
If the filter output corner frequency approaches the bandwidth of the amplifier than the amplifier output will be less than you would expect by multiplying the filter output by the low frequency amplifier gain.

 

Gain=-10 tells us two things:

1) The amplifier will invert the input signal. Meaning. When input is 1 volt, output will be -1 volt.

2) The amplifier will multiply the input signal by 10. When input is 1 volt, output will be 10 volts.

Combining these two facts, we see that whatever you input into the amplifier will be inverted and multiplied by 10. When you input 1 volt, you will get -10 volts output.

It also gives you a clue which amplifier configuration to use. There is only one amplifier configuration that inverts the input signal.

 

have a university assignment question

Can you post the original question ...

 

Hi,

I have to agree that maybe we should see the original question because this sounds a little peculiar. That's because we are to design an RC low pass filter that has cutoff of 7kHz, yet the "ideal" amplifier already has a cutoff of 7kHz, according to the text.
If the wording instead went like this:
"Design a low pass RC filter with cutoff of 7kHz to feed an ideal amplifier with gain of -10"
that would sound much better. That's because if we have an "ideal" amplifier with cutoff of 7kHz, then it's not an ideal amplifier, and if it already has a cutoff at 7kHz then when we design an RC filter with cutoff of 7kHz as a front end, we'll see more cut at 7kHz so the combined cutoff frequency would move down to some unspecified frequency. That is not usually the way questions are posed, as usually we want a specific cutoff that is obtained by sheer calculation not by chance.
Of course every problem is different, so it could be that they dont care, but if we see the original wording that may help sort this out.

 

It is possible that they need to design ACTIVE low pass filter. However, since they don't know what they are doing, they are falling back on older material where they learned about passive low pass filters. That would also explain why they don't know about the gain feature of the the op amp.

 

Hi all,

Sorry for my late reply and thanks for your responses.
I will include the question below.

Determine the value of capacitor in conjunction with a resistor value of 1 kΩ in the low pass filter to provide a cut off frequency of 7 kHz. The output of the filter is fed into an ideal amplifier operating with a gain of -10 and a bandwidth of 1 kHz to 7 kHz.

 

Hi all,

Sorry for my late reply and thanks for your responses.
I will include the question below.

Determine the value of capacitor in conjunction with a resistor value of 1 kΩ in the low pass filter to provide a cut off frequency of 7 kHz. The output of the filter is fed into an ideal amplifier operating with a gain of -10 and a bandwidth of 1 kHz to 7 kHz.

Hi,

Is that the exact question word for word?

If so, then you just have to design the RC filter with 7kHz cut off which you may have done already.

 

Here is active low pass filter:





You want the Equivalent inverting amplifier filter circuit.

In the picture the R1 is 10k. Your R1 is 1 kOhm. The Gain is 10. So you need to recalculate R2.

In the picture capacitor C is 10 nF. You need to recalculate capacitor C for your assignment. Take the fc formula at the top of the pic, plug in your 7 kHz, plug in your R2 and solve for C.

Then just copy the circuit, use your R1, R2 and C. The input signal is going into the inverting input of op amp, that inverts the input signal and provides you with the minus sign in the gain description.

 

Hi all,

Sorry for my late reply and thanks for your responses.
I will include the question below.

Determine the value of capacitor in conjunction with a resistor value of 1 kΩ in the low pass filter to provide a cut off frequency of 7 kHz. The output of the filter is fed into an ideal amplifier operating with a gain of -10 and a bandwidth of 1 kHz to 7 kHz.

Hi,

Do they provide a schematic?

 

Hi,

Do they provide a schematic?

Hi,
Only what I have attached in my original post.

 

Hi,
Only what I have attached in my original post.

Hi,

Ok then it sounds like they want you to just design an RC filter that goes on the front end of another amplifier. Why they word it like that does not make it clear because they seem to mentioned things that dont affect the RC filter itself. If you can find some solved problems in the same book or ask the instructor exactly what they want that would help a lot.