Implementation of a circuit from knowing only the poles and zeroes and gain

1. Oct 23, 2011

Ry122

If i know just these 3 values is it possible to implement a circuit using just OP-AMPS, capacitors, and resistors? How would I go about determining what values the capacitors and OP-amps should have?

The circuit is a low-pass filter and thus needs to amplify and filter simultaneously.

2. Oct 23, 2011

Staff: Mentor

If you cascade three single-pole LPF opamp circuits and a gain circuit, then all you are left with is the tradeoff between the R & C values for each to give you the RC pole for each. Are you further constrained? Like doing it all in one opamp stage, or two opamp stages, or in some other way?

3. Oct 24, 2011

Ry122

there's no further constraints.
Basically, what's happening is that I have some input voltage to an unknown circuit, and the output voltage from that unknown circuit needs to pass into this LPF, and the resulting output from the LPF needs to be what the input voltage into the unknown circuit was.
So the LPF is just undoing the effect the unknown circuit had on the input.

I'm only wanting to implement a first order LPF so it won't be very complicated.
Knowing this, is the method you suggested the easiest to do this?

4. Oct 24, 2011

Ry122

Also I know that after attenuation occurs at the cut-off frequency, the gain stabilizes again at some specific frequency.

Wikipedia says that poles = 1/RC for RC circuits with the zero located at origin.
Because this is an LPF the zero and pole will be located at some point after the origin,
so I'm not really sure if that equation still applies.

5. Oct 25, 2011

jim hardy

low pass filter?

write its transfer function

remember the transfer function for an opamp is Zfeedback/Zinput
Z(s) for a capacitor is 1/sc
and for a resistor it's R
and they combine in series and parallel just like they did in time domain just you write s instead of jω

see if you can arrange an opamp's feedback and input impedances to give it the transfer function you want

6. Oct 25, 2011

Ry122

So already having the poles, zeroes and gain which together give me the transfer function of the LPF, I just apply Zfeedback/Zinput?

For example if the transfer function of the LPF was (1/k)*(1+(s/300))/(1+(s/100))

for the numerator 1+(s/300)
this can be taken as the feedback impedance of the op-amp
and (1+(s/100) can be taken as the input impedance?

This is probably wrong, I'm not really sure how the op-amp transfer function relates to the LPF transfer function.

Last edited: Oct 25, 2011
7. Oct 25, 2011

jim hardy

well let me clear something first.

I do NOT understand the Laplace transform. I wish i did.
but i learned to use simple ones just as a cave-man uses a rock.
Said cave-man can crack bones to get at the marrow using a rock without understanding equations of momentum, impulse, or fractuire mechanics.

Similarly, I can use Laplace to solve simple circuits without understanding what that genius was doing.

I'd rather admit my weakness than pretend its not there.

Now, the transfer function you gave is i believe a little more than a low-pass filter. It is what we call in control systems a "Lead Lag". That's okay, we're just demonstrating a technique.

Since there's a 1 in transfer function's numerator it'll have DC gain.
A practical opamp circuit establishes DC gain by having resistors in both its feedback and input networks.

So let's see what we can cook up here.

Let's try a parallel resistor-capacitor in the feedback network

Remember from elementary circuits that paralleling two elements gives Z1*Z2/(Z1+Z2),, product over sum? ?
let subscripts 'fb' and 'in' be feedback and input repectively..
and instead of jω we write s...

okay Rfb parallel with Cfb gives Zfb = Rfb*(1/sCfb)/ (Rfb +1/sCfb) ,,, that's product over sum..
multiply numerator and denominator both by sCfb
and we get Zfb = Rfb/(Rfb*sCfb + 1)

and similarly if we place a R and C in parallel for input network,
Zin will = Rin/(Rin*sCin +1)

and we'll make an opamp circuit with that Zfb and Zin;

Zfb/Zin = Rfb/Rin * (Rin*sCin +1)/(Rfb*sCfb+1)

so to achieve your transfer function of
(s/300 + 1)/(s/100+1) * 1/k
you'd pick a capacitor-resistor pair for input whose product equals 1/300.
and feeedback product of 1/100
and Rfb/Rin = 1/k

i think....

i'm rusty - been away from this for 20 years and as i say never mastered it to begin with.

But see if you can make the technique work

and if you figure out Laplace please explain him to me!

yungman and sophie have inspired me. i am retired , and searching for my old calculus book........

maybe one of them can explain what Laplace was thinking when he invented his transform.

bottom line is - yes you can design a circuit from its transfer function. that's the beauty of op-amps.

Last edited: Oct 25, 2011
8. Oct 25, 2011

psparky

I was never too concerned how the Laplace transfer was obtained or what not. Simply pluggin and chuggin in the chart always worked well for me. We cant know and understand everything!

1/(1+s) = e^-t

Works for me!