# Create analog function with op amps

Discussion in 'Analog & Mixed-Signal Design' started by Nikos Mparoutis, Oct 17, 2016.

1. ### Nikos Mparoutis Thread Starter New Member

Oct 8, 2016
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Hello ,i am building a system and I have to create the analog signal V(t)=c1*t^2+c2*t - 1/t with op amps and analog components.I have to minimise the cost so i beleive i have to use only one multiplier ,so i want to know if i should use capacitors to store the signals and reuse the multiplier and do the final operations.Or to use buffers or suggest me an other way.
Thanks

2. ### crutschow Expert

Mar 14, 2008
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It could likely be done more easily using digital techniques.
Why must it be done using analog components?
What is the timeframe for "t"?
What are c1 and c2?

3. ### Alec_t AAC Fanatic!

Sep 17, 2013
7,127
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1/t isn't practical unless non-zero time limits are defined. At time zero you would need an infinite voltage .
Is this a Homework problem?

4. ### Nikos Mparoutis Thread Starter New Member

Oct 8, 2016
19
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I stated ,no homework.I finished with homework years before but i have not experience with op amps,it is my first attempt to build somethings big!
I want to keep the cost to minimum but i realizes that it is imposible to use only one multiplier.So i will use 2 multiplier to do the x^2 and now my question is to do the C1*t^2 (c =constant).
I believe to do this C1*T^2 is better with gain manipulation of an op amp than usage of a multiplier.

5. ### crutschow Expert

Mar 14, 2008
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You can do X² with just one multiplier by connecting X to both multiplier inputs (giving X * X).
Then you run the multiplier output through an op amp with a gain of C1.

6. ### #12 Expert

Nov 30, 2010
17,897
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@crutschow
Could you dumb this down for me?
I think it's interesting, but I can't imagine how to use C1 (capacitance?) to multiply (time squared).
If my question seems stupid, I confess. I am lost!

7. ### atferrari AAC Fanatic!

Jan 6, 2004
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Is C1 a cap's value?

Apr 5, 2008
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9. ### crutschow Expert

Mar 14, 2008
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In post #4 the OP said c was a constant so I interpreted C1 as a gain value
But that assumption could very well be wrong.

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10. ### kubeek AAC Fanatic!

Sep 20, 2005
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Using C and t sure confuses things. Can you please explain what exactly is that equation supposed to calculate, and what each symbol means?

11. ### Nikos Mparoutis Thread Starter New Member

Oct 8, 2016
19
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c (=constant) ,no capacitors.
Also i had the same though...The gain of a op amp can be the constant
If i decide to use ONE multiplier is it possible to do the function?

12. ### ramancini8 Active Member

Jul 18, 2012
462
130
I believe that you need 2 multipliers; one to do t^2 and one to do 1/t. You probably can't use storage because of the time delays. I did implement an Observer set of equations, and timing is critical.

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13. ### kubeek AAC Fanatic!

Sep 20, 2005
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But is t really time, or is it x?

14. ### Nikos Mparoutis Thread Starter New Member

Oct 8, 2016
19
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Yhea Right!
I done some work and now for a weird reason the Multisim does not compile the circuit and the result is wrong
Here is the Multisim citcuit and the datasheet of AD633 multiplier.As you can see the connections are right,so iis this a problem with the mutlisim?

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15. ### atferrari AAC Fanatic!

Jan 6, 2004
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Niko,

Nice of you ignoring all but one who took the time to answer your OP!

16. ### crutschow Expert

Mar 14, 2008
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What do you mean the result is wrong? What does the program do?
Could be a problem with the model.
I had no problem with my LTspice simulation (below).
I used a different op amp since I didn't have the AD711, but that shouldn't make a significant difference in the simulation.

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17. ### Nikos Mparoutis Thread Starter New Member

Oct 8, 2016
19
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Yes Right!
I uploaded the wrong pic.It actually works with small error.
Can i trim this error?
Is there a way to trim the *10 scalling ?
Thank you all
(The t is time ,the V(t) is a voltage function with the variable of time)

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18. ### crutschow Expert

Mar 14, 2008
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We can better advise you on how to reduce the error if you tell us what the error is (what is "small").

19. ### Nikos Mparoutis Thread Starter New Member

Oct 8, 2016
19
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In the pic the right result is 5 V but the output show is 5.12 V.
I know that the low cost multiplier affects this but i like the perfectionism.
The scalling *10 : i think i can put it inside the constant C =c1/10.

20. ### crutschow Expert

Mar 14, 2008
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It isn't whether you like "perfectionism" or not, it's what you need for your application that's important.

It's hard to say where the error occurs (my simulation shows only 3mV of error).
You'll have to go through the circuit, measuring all the nodes to determine where the error is occuring.

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