Cascaded low-pass filter followed by a buffer amplifier

[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

 

[LvW]

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

 

[gneill]

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!

 

[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?

 

[gneill]

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...

 

[topcat123]

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

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

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

 

[gneill]

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.

 

[topcat123]

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.

 

[gneill]

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?

 

[topcat123]

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

 

[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

 

[gneill]

That looks reasonable.