# Cascaded low-pass filter followed by a buffer amplifier

• topcat123
In summary: Yes, and as a consequence what is the rule of thumb for the potential difference between the inputs (when negative feedback is present)?The feedback is negative so the potential difference between the inputs should be small.
topcat123

## Homework Statement

3. FIGURE 3(a) shows a simple low-pass filter followed by a buffer
amplifier.
(a) Write down the transfer function for the filter.
(b) Determine the 3db frequency (fc) if R = 10 kΩ and C = 10 nF.
(c) If four such stages are cascaded as shown in FIGURE 3(b),
determine the gain and phase of the overall transfer function at
(i) 0.1fc
(ii) 10fc.

## Homework Equations

The Transfer function
$$\frac{V_{out}}{V_{in}}=\frac{-1}{1+j\frac{{\omega}}{{\omega}_c}}$$
The natural frequency at -3dB
$$f_c=\frac{1}{2{\pi}RC}$$

## The Attempt at a Solution

a) as above the transfer function
b)$$f_c=\frac{1}{2{\pi}RC}=\frac{1}{2{\pi}10000*10*10^{-9}}=1.59KHz$$

c) I am a bit stuck with this one

I know
The resulting RC filter circuit would be known as an “nth-order” filter with a roll-off slope of “n x -20dB/decade”.
So the roll off of this 4th order filter would be -80dB
The phase shift for a single low pass filter is$$-arctan(2{\pi}fRC)=-arctan(\frac{f}{f_c})$$I am asuming that the phase shifts of all 4 can be added?

All help is apreciated
Thanks

#### Attachments

• Fig 3(a).JPG
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• Fig 3(b).JPG
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Why do you expect a NEGATIVE transfer function?`The buffer provides a gain of "+1", does it not?

Check the buffer stage. What's its gain?

Yes, with the buffer amplifier in place the phase shifts can simply be added.Edit: Oops! LvW got there ahead of me!

I figured it would have -ve feedback as the output feeds back into the inverting input.

how do I workout the overall gain with the -80 dB roll off?

topcat123 said:
I figured it would have -ve feedback as the output feeds back into the inverting input.
What are the "rules" for the ideal op-amp?
how do I workout the overall gain with the -80 dB roll off?
The stage are cascaded. Thanks to the buffers they don't load each other or affect each other's individual transfer functions. So each transfer function applies, one at a time, as you pass from one to the next...

gneill said:
What are the "rules" for the ideal op-amp?

am ideal op-amp will have infinite input impedance infinite gain and zero output impedance.

gneill said:
So each transfer function applies, one at a time, as you pass from one to the next

so the output of the first becomes the input to the second and so on?

thanks

topcat123 said:
am ideal op-amp will have infinite input impedance infinite gain and zero output impedance.
Yes, and as a consequence what is the rule of thumb for the potential difference between the inputs (when negative feedback is present)?
so the output of the first becomes the input to the second and so on?
Yes.

gneill said:
Yes, and as a consequence what is the rule of thumb for the potential difference between the inputs (when negative feedback is present)?

As the feed back is negative and at unity then 0V potential difference.
But I believe there is actual a small amount. Because the gain is large this small difference is amplified so the output become almost equal to non-inverting input voltage at a point of equilibrium.

topcat123 said:
As the feed back is negative and at unity then 0V potential difference.
But I believe there is actual a small amount. Because the gain is large this small difference is amplified so the output become almost equal to non-inverting input voltage at a point of equilibrium.
Okay. So then, taking that into consideration, is the gain of the stage positive or negative?

gneill said:
Okay. So then, taking that into consideration, is the gain of the stage positive or negative?

Ah I get it the gain is clearly positive +1 and so must the TF
And the feedback is negative.

Thanks

gneill
so as far as I figure the TF will be $$\frac{V_{out}}{V_{in}}=\left(\frac{1}{1+j\frac{{\omega}}{{\omega}_C}}\right)^4$$

And the phase shift is $$(-arctan(2{\pi}fRC))*4$$

Last edited:
That looks reasonable.

topcat123

## What is a cascaded low-pass filter?

A cascaded low-pass filter is a type of electronic filter that is used to pass low-frequency signals while attenuating high-frequency signals. It consists of multiple stages of low-pass filters connected in series to achieve a sharper roll-off and better frequency response.

## Why is a buffer amplifier used after a cascaded low-pass filter?

A buffer amplifier is used after a cascaded low-pass filter to prevent the filter from loading the signal source. This helps to maintain the integrity of the filtered signal and prevent any unwanted distortion or attenuation.

## What is the purpose of a cascaded low-pass filter followed by a buffer amplifier?

The purpose of a cascaded low-pass filter followed by a buffer amplifier is to filter out unwanted high-frequency signals and amplify the desired low-frequency signals. This is commonly used in audio applications to improve the quality of the signal and remove any unwanted noise or interference.

## How do you calculate the cut-off frequency of a cascaded low-pass filter?

The cut-off frequency of a cascaded low-pass filter can be calculated by taking the reciprocal of the total resistance and capacitance values in the filter. It can also be calculated using the formula f_c = 1/(2πRC), where R is the total resistance and C is the total capacitance of the filter.

## What are the advantages of using a cascaded low-pass filter followed by a buffer amplifier?

The main advantage of using a cascaded low-pass filter followed by a buffer amplifier is that it provides a high-quality, low-noise signal with a steep roll-off. It also allows for more control over the frequency response of the filtered signal and can be easily adjusted to fit specific needs.

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