# Cascaded low-pass filter followed by a buffer amplifier

1. Nov 20, 2016

### topcat123

1. The problem statement, all variables and given/known data
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.

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

3. 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

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2. Nov 20, 2016

### LvW

Why do you expect a NEGATIVE transfer function?`The buffer provides a gain of "+1", does it not?

3. Nov 20, 2016

### Staff: Mentor

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!

4. Nov 22, 2016

### topcat123

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?

5. Nov 22, 2016

### Staff: Mentor

What are the "rules" for the ideal op-amp?
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...

6. Nov 24, 2016

### topcat123

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

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

thanks

7. Nov 24, 2016

### Staff: Mentor

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

8. Nov 24, 2016

### topcat123

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.

9. Nov 24, 2016

### Staff: Mentor

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

10. Nov 25, 2016

### topcat123

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

Thanks

11. Nov 25, 2016

### topcat123

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: Nov 25, 2016
12. Nov 25, 2016

### Staff: Mentor

That looks reasonable.