Difficult Cascaded Op-amp frequency response problem


by Learnphysics
Tags: cascaded, difficult, frequency, opamp
Learnphysics
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#1
Nov22-13, 02:08 AM
P: 92
1. The problem statement, all variables and given/known data


http://imgur.com/tJEPgbA

2. Relevant equations



3. The attempt at a solution
I started by focusing on the left Op-amp provided.

My methodology to solve this question was essentially to assume a virtual short circuit so that the negative input terminal of the op-amp goes to Vi. Using Vi, we can find current through C1.

Current through C1 is also the current through the parallel connection of resistances and capacitances.

Once we know the current between Vi and Vo, and we know the total resistance (impedence) we can solve for the voltage difference between Vi and Vo...
and also Vo/Vi. (which is the transfer function).

From that point I believe I need to get great the break frequency, (I'm not too sure how to do this, could anyone help?).

After I get the break frequency i can simply find the magnitude of the transfer function, express that on a logarithmic scale, and play around with the component values to get a 500radiens/sec break frequency. (Another question, how do I get my design to produce the 20db/decade (6per octave) rise? I mean all of the components fall into 'fb', and fb is 500. So what determines the gradient of the transfer function past the break frequency?)

Then, once I have that, I can repeat with the other circuit to get an idea how that one operates generally...

Then play around cascading them until I can create a 6db gain, and a 12db fall acting at the specific frequencies.

So i suppose what I'm asking is, once I get the transfer function, how do I go about pulling a break frequency out of it?

Il'l post working for this shortly (let me neaten it up a bit)
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
http://imgur.com/mZRJuyS

This is my attempt at the solution
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NascentOxygen
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#2
Nov22-13, 09:21 AM
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This is an interesting exercise.

I don't like your term virtual short circuit. Is that something your lecturer uses, or is it your invention? It's not necessary to label it with a name, anyway. (It sounds too close to "virtual earth" for my liking. )

The 20db/decade slope is characteristic of any first-order transfer function filter when operating well away from the corner frequency. So long as it's a first-order system it will exhibit that slope.

You massage the gain equation into a form (........) / ( 1 + j(ω/ωo) )

and you can see when the ω/ωo term has a value of 1 the denominator has a value of √2 and you know that gain dropping to 1/√2 of its flat gain corresponds to -3db. Thus ωo is termed the break frequency.
Learnphysics
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#3
Nov22-13, 03:46 PM
P: 92
Quote Quote by NascentOxygen View Post
This is an interesting exercise.

I don't like your term virtual short circuit. Is that something your lecturer uses, or is it your invention? It's not necessary to label it with a name, anyway. (It sounds too close to "virtual earth" for my liking. )

The 20db/decade slope is characteristic of any first-order transfer function filter when operating well away from the corner frequency. So long as it's a first-order system it will exhibit that slope.

You massage the gain equation into a form (........) / ( 1 + j(ω/ωo) )

and you can see when the ω/ωo term has a value of 1 the denominator has a value of √2 and you know that gain dropping to 1/√2 of its flat gain corresponds to -3db. Thus ωo is termed the break frequency.
It's kind of coming together in my mind.
So, in my working out, just before the last 3 lines I have it in
(........) / ( 1 + j(ω/ωo) form.

with W0 being (1/(R2*C2))


Fair enough.

What about the numerator. It's got some term "C1", which doens't appear at all in the denominator. The denominator terms are defined by W0 and the 'required break frequency'. How would I find the value of C1?

eg.

1/(R2*C2) = 500

Say C2 is a 10microFarad Capacitor, R2 will be a 31.8 ohm resistor (theoretically at least).


In the numerator we have:

J*W*R2*(C1+C2) +1


I get R2, and C2, but what value must C1 take? How do we go about finding it?

I could consider only magnitudes, then put the whole thing into log10 form and then say the gain is equal to 20db/dec. Then from that point solve for C2. But I'm not sure if that's the way to go.

NascentOxygen
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#4
Nov22-13, 05:22 PM
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Difficult Cascaded Op-amp frequency response problem


Quote Quote by Learnphysics View Post
So, in my working out, just before the last 3 lines I have it in
(........) / ( 1 + j(ω/ωo) form.

with W0 being (1/(R2*C2))
It's always wise to check the dimensions (think, "units").

This big fraction has no units, because it is the expression for voltage gain, and volts/volt has no units.
with W0 being (1/(R2*C2))
That checks out, because 1/(RC) has units of ???? and these units cancel with the units of ωo which are ????

What is the ???? in this sentence?

What about the numerator. It's got some term "C1", which doens't appear at all in the denominator.
If that's what you have, it means your numerator has units (i.e., Farads). But the numerator should be dimensionless (to cancel with that "1" in the denominator which has no units), meaning that you have made a mistake somewhere.

The denominator terms are defined by W0 and the 'required break frequency'. How would I find the value of C1?
Once you have the equation for voltage gain, you examine it and summarize some important details in terms of those element values, details such as filter type (high pass, low pass, etc), low-frequency gain, corner frequency, high-frequency gain, fall-off slope (db/decade), etc. The numerator features in this.
Learnphysics
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#5
Nov22-13, 06:22 PM
P: 92
Quote Quote by NascentOxygen View Post
It's always wise to check the dimensions (think, "units").

This big fraction has no units, because it is the expression for voltage gain, and volts/volt has no units.

That checks out, because 1/(RC) has units of ???? and these units cancel with the units of ωo which are ????

What is the ???? in this sentence?


If that's what you have, it means your numerator has units (i.e., Farads). But the numerator should be dimensionless (to cancel with that "1" in the denominator which has no units), meaning that you have made a mistake somewhere.


Once you have the equation for voltage gain, you examine it and summarize some important details in terms of those element values, details such as filter type (high pass, low pass, etc), low-frequency gain, corner frequency, high-frequency gain, fall-off slope (db/decade), etc. The numerator features in this.
I don't believe I've made an error elsewhere. (The working out is up there, if you have time to double check).

Instead I think the units are correct because it factorizes into:
JWR2(C1+C2).

SO the units would be radiens*Ohms*Farads.
Same as the denominator.



So my gain equation looks like this.

(J*W*R2*(C1+C2) + 1)/((J*W*C2*R2) +1) = Vo/Vin

The units appear to cancel.

I can look at the denominator and given a ωo of 500 radiens/sec I can select values for C2 and R2.

But as you can see I have a variable C1 in the numerator that's an unknown. It cancels out, so the expression is valid. I jsut haven't selected a value for it. Could I choose any value? What significance does it have?

Sorry If we're going in circles with this, I'm not sure I understand just yet. Thanks for taking the time to help out.

And 1/RC has units of (ohms^-1 * farads*^-1). 1/RC is the ωo.
NascentOxygen
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#6
Nov22-13, 08:59 PM
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The answer to your questions lies in my final paragraph. This being your assignment, there's not a lot more I can add at this stage. You have to compare the transfer function with what you know the graph will look like, and discover the connections for yourself. That is the key to effective learning, being self-guided.

Good luck.
rude man
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#7
Nov23-13, 01:56 AM
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It should be obvious by just looking at both given circuits that neither one will give you a 20dB/decade rolloff for high frequencies.

However, hint: by adding 2 components to the 1st circuit you can get the Bode plot you want.
Learnphysics
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#8
Nov23-13, 03:19 AM
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Quote Quote by rude man View Post
It should be obvious by just looking at both given circuits that neither one will give you a 20dB/decade rolloff for high frequencies.

However, hint: by adding 2 components to the 1st circuit you can get the Bode plot you want.
Neither one will give you it?
My intuition I don't think is strong enough just yet to make that call.

I did the math, and following Nascant's advice. Got a transfer function in decibels. Put it into excel (the circuit on the right). I seem to be able to produce a bode plot that is flat until 1000radiens/sec and then suddenly drops at what I think is 20db/Dec. (Not sure how to evaluate that it is, but the graphs shape looks right.)....

I repeated the process for the circuit on the right and it produced (conveniently enough) a bode plot that tapers UP (the exact opposite) at 500radiens/sec.

My problem is now that If I cascade them, they don't exactly provide what I'm looking for as I'd need a 40db/dec drop to offset the original 20 and bring the line back down at 1000db/dec.

Again i'm not sure if I've got this conceptually wrong, So il'l spend some time visiting some textbooks.
rude man
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#9
Nov23-13, 10:50 AM
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You can't design an op amp circuit with a gain of less than +1 if you go into the + input of the op amp directly (unless you employ active feedback which is the "advanced course"). So no way will you achieve unlimited rolloff at high frequencies with either of your given circuits, or by cascading them together.

I have given you another hint as to how to proceed in the statement above.
The Electrician
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#10
Nov23-13, 11:30 AM
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Quote Quote by Learnphysics View Post
Again i'm not sure if I've got this conceptually wrong, So il'l spend some time visiting some textbooks.
These references may help:

http://en.wikipedia.org/wiki/Bode_plot

http://www.ece.utah.edu/~ee3110/bodeplot.pdf
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#11
Nov23-13, 06:57 PM
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Quote Quote by The Electrician View Post

These references, especially the second one help a lot.

How do I go from the standard form of H(s).. into the circuit diagram which produces it?
Is there any easy technique, or do I need to intuitively familiarize myself to know that a capacitor in parralell with a resistor in the feedback network of a positive op-amp will have a very specific effect on the equation.

Also I know that to get the first part of the bode plot we simply know the zero looks something like 1+(jw/w).

with w being 500.

The second one i'm not sure of. I assume that the First equation will rise at 500radiens per second, at 20db/decade. At 1000radiens, I can put in a pole as: (1+jw/w) Causing the line to flatten out. But to send it in the reverse direction Il'l need yet another pole. Is this a valid way to do it?

How would this then manifest itself in a circuit diagram?

(1+jw/w1)/((1+jw/w2)(1+jw/w2))
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Nov23-13, 07:19 PM
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Quote Quote by Learnphysics View Post

(1+jw/w1)/((1+jw/w2)(1+jw/w2))
That is the correct expression for the transfer function you need. With the requirement that w1 < w2 by the correct amount.
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#13
Nov23-13, 08:29 PM
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Quote Quote by rude man View Post
That is the correct expression for the transfer function you need. With the requirement that w1 < w2 by the correct amount.
True, but how does that manifest itself in terms of circuit topology? How can we find out what that transfer function will end up looking like in terms of resistors and capacitors and so on.?
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Nov24-13, 12:58 AM
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Quote Quote by Learnphysics View Post
True, but how does that manifest itself in terms of circuit topology? How can we find out what that transfer function will end up looking like in terms of resistors and capacitors and so on.?
I gave you two hints already. Why not think about them & propose one or two ideas? I'll help along the way but you need to work this ...
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Nov24-13, 02:15 AM
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Quote Quote by rude man View Post
I gave you two hints already. Why not think about them & propose one or two ideas? I'll help along the way but you need to work this ...
My apologies, I've left an exam to the last moment and am trying to cram these things.

Our transfer function is:

(1+jw/w1)/((1+jw/w2)(1+jw/w2))


I know that when we cascade circuits together we multiply their transfer functions out.

The transfer functions of the left op-amp provided in the question appears to be useful in that the numerator of the transfer function is in a similar form.

The numerator being: J(w/wo) +1 , and w0 being 1/(C1R1+C2R2).


the denominator however appears to cause a different pole to form at a different radiens/sec.

The denominator being Jw/w1 , w1= 1/(R2*C2).

This is all based on those initial calculations for the left op amp being correct.


So with my current understanding, and the calculations I've done before the left circuit, has a zero and a pole, in the form of (jw/w +1).

In essence this transfer looks like this:
http://imgur.com/AoXFCDy


If we could cascade an element with a single pole, occuring at W1, we could send the bode plot back down and create the desired response.


Essentially we are looking for a circuit with the transfer function: 1/(jw/w +1).

Now I have to consider what circuit makes this kind of function. I'm not sure.

Could you validate my thinking so far, tell me if I'm on the right track, and if you're feeling especially gracious; how to design a circuit with a single pole, and nothing else?
The Electrician
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#16
Nov24-13, 02:44 AM
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What do you get with a series R and shunt C?

How about a series C and shunt R?

You might look at the several tutorials starting here:

http://www.electronics-tutorials.ws/.../filter_1.html

Also have a look here:

http://www.linkwitzlab.com/Removed%2.../x-filters.htm
rude man
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Nov24-13, 11:21 AM
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Quote Quote by Learnphysics View Post
My apologies, I've left an exam to the last moment and am trying to cram these things.

Our transfer function is:

(1+jw/w1)/((1+jw/w2)(1+jw/w2))


I know that when we cascade circuits together we multiply their transfer functions out.

The transfer functions of the left op-amp provided in the question appears to be useful in that the numerator of the transfer function is in a similar form.

The numerator being: J(w/wo) +1 , and w0 being 1/(C1R1+C2R2).


the denominator however appears to cause a different pole to form at a different radiens/sec.

The denominator being Jw/w1 , w1= 1/(R2*C2).

This is all based on those initial calculations for the left op amp being correct.


So with my current understanding, and the calculations I've done before the left circuit, has a zero and a pole, in the form of (jw/w +1).

In essence this transfer looks like this:
http://imgur.com/AoXFCDy


If we could cascade an element with a single pole, occuring at W1, we could send the bode plot back down and create the desired response.


Essentially we are looking for a circuit with the transfer function: 1/(jw/w +1).

Now I have to consider what circuit makes this kind of function. I'm not sure.

Could you validate my thinking so far, tell me if I'm on the right track, and if you're feeling especially gracious; how to design a circuit with a single pole, and nothing else?
You're on the right track.

So, what does a series-R and shunt-C transfer function give you?

Could you add such a simple circuit to the left-hand circuit? Remember what I said, you can't get a gain of less than 1 if you go directly into the + op amp input. That means you can't get your infinite rolloff at high frequencies by going directly into the + input.

Your hint is the word "directly"!
NascentOxygen
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#18
Dec5-13, 12:42 AM
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Quote Quote by Learnphysics View Post
If we could cascade an element with a single pole, occuring at W1, we could send the bode plot back down and create the desired response.


Essentially we are looking for a circuit with the transfer function: 1/(jw/w +1).

Now I have to consider what circuit makes this kind of function. I'm not sure.

Could you validate my thinking so far, tell me if I'm on the right track, and if you're feeling especially gracious; how to design a circuit with a single pole, and nothing else?
Did you finally succeed with this, Learnphysics?


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