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Re: Integrator transfer function and arbitrary continuous input signals

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JM <sunaecoNoChoppedPork@gmail.com> wrote:
> On Mon, 11 Aug 2025 09:07:01 -0800, Christopher Howard
> <christopher@librehacker.com> wrote:
> 
>> Hi, I'm trying to work slowly through the great Op Amp book by Roberge et al
>> that was recommended earlier. I downloaded the 2nd edition v. 1.8.1. I'm
>> finding it enlightening to slowly process the paragraphs and take notes
>> on the diagrams and equations.
>> 
>> Something I'm getting hung up on though is they dive early on into
>> transfer functions, and that hasn't been covered yet in my introductory
>> DE book. I've been trying to cram in some quick Internet research on
>> laplace transforms and such but it has been a bumpy ride.
>> 
>> In chapter one (equation 1.21) they gave the transfer function of an
>> integrator, i.e., an op amp with resistor and capacitor feedback
>> network, as -1/(RCs). I found that if I replaced s with 2 pi f, I could
>> predict the gain from a steady sinusoidal signal of matching frequency,
>> and when I tried this out with my real integrators, I got matching
>> results.
>> 
>> My questions:
>> 
>> (1) so, if I replace s with a complex number, one that has both a real
>> and an imaginary part, what does that mean? Is that the same as
>> calculating the gain for a sinusoidal input of a particular amplitude
>> and frequency?
> 
> No, you can calculate the frequency response by setting s=jw, or the
> dc responce by setting to 0.
> 
>> 
>> (2) How do I use/apply this transfer function if I've got some
>> nonsinusoidal continuous input, like say a steady voltage, or a linear
>> ramping voltage?
> 
> The transfer function is the ratio of the Laplace transform of the
> system output to the Laplace transform of the systen input.  So to
> find the system output y(t) to an arbitrary input x(t) you:
> 
> 1 - calculate L[x(t)]
> 2 - multiply above by the transfer fuction
> 3 - calculate the inverse transform of the above
> 
> For example for a heaviside input (x = 1 for t > 0 otherwise x = 0)
> L[x] = 1/s.  For an integrator with a transfer function 1/s the output
> would therefore be 1/s^2 (ie L[y]).  Taking the inverse transform of
> that would yield y = t (ie a ramp).
> 

**Two-sided** Laplace, i.e. Fourier with a change of variable. 

That’s one of the many sources of confusion when EEs talk to other
technical folk. 

WDDNS two- sided Laplace. 

Cheers 

Phil Hobbs 

-- 
Dr Philip C D Hobbs  Principal Consultant  ElectroOptical Innovations LLC /
Hobbs ElectroOptics  Optics, Electro-optics, Photonics, Analog Electronics 

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Integrator transfer function and arbitrary continuous input signals Christopher Howard <christopher@librehacker.com> - 2025-08-11 09:07 -0800
  Re: Integrator transfer function and arbitrary continuous input signals Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> - 2025-08-11 17:34 +0000
  Re: Integrator transfer function and arbitrary continuous input signals "Edward Rawde" <invalid@invalid.invalid> - 2025-08-11 14:17 -0400
    Re: Integrator transfer function and arbitrary continuous input signals "Edward Rawde" <invalid@invalid.invalid> - 2025-08-11 14:19 -0400
    Re: Integrator transfer function and arbitrary continuous input signals Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> - 2025-08-11 19:43 +0000
  Re: Integrator transfer function and arbitrary continuous input signals "Edward Rawde" <invalid@invalid.invalid> - 2025-08-11 16:56 -0400
  Re: Integrator transfer function and arbitrary continuous input signals JM <sunaecoNoChoppedPork@gmail.com> - 2025-08-15 23:14 +0100
    Re: Integrator transfer function and arbitrary continuous input signals Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> - 2025-08-16 00:34 +0000

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