# Why use Fourier to find a system response to a ramp function?

• posted

I remember seeing an example where a system response was found given a differential equation as a transfer function.

Fourier analysis was used. What are the advantages to using Fourier in such a situation over s-domain/Laplace?

I remember seeing that the derivative was found, then the Fourier transform, and integration was later performed to get back to the correct form- Can anyone familiar with this technique explain it a bit more to me?

Thanks.

Daz.

• posted

One of the aphorisms used by Richard Feynman was that if you could solve a problem one way, there will be other ways of solving it as well. If you can solve it using a Laplace transform, then you can solve it in other ways. With Laplace and Fourier transforms, being like a pair of fraternal twins, it should not be surprising that a problem could be solved using either transform. Of course, according to the aphorism, other methods exist as well.

Bill

• posted

Advantages? Few, other than simpler math. A Fourier solution is a steady-state solution whereas a Laplace solution includes a transient component.

I suppose it depends on the characteristics of the input and type of solution needed. If its a repeating ramp function (like a sawtooth) and all you want is the steady state solution, Fourier will do.

A Fourier transform changes a time varying function f(t) into a function varying with frequency, f(w). In the frequency domain, solving for the system response involves simpler math than a time domain solution. A differential in the time domain becomes the multiplication of the transformed function by he frequency, w.

A Laplace transform is similar in concept, except that the s domain is two-dimensional whereas the frequency domain has a single dimension. This makes the math somewhat more involved, but gives you more information about the transient response of a system.

• posted

Fourier transforms were used to solve transient transient problems. It is just a bit trickier compared to initial value problems but in can be done. Remember that you are dealing with COMPLEX frequencies and that s=jw.

It really behooves an EE to UNDERSTAND what is happening rather than just following a recipe.

Bill

• posted

s = a+jw in a Laplace transform, where 'a' is the transient part.

I'd like to see where the transient response of a system is described in a Fourier transform.

• posted

It is not a problem. But because my eyesight is not great, I will point you in the right direction and let you do the grind.

Consider the response of an RC integration circuit to a rectangular pulse. Such a pulse is usually described with a rect function. rect(t) would be a unit height pulse lasting from time -1/2 to 1/2.

Among other places, there is a list of Fourier transform pairs in the article

The FT of the rect(t) is a sinc function sinc(x) = sin(x)/x give or take a factor of 2 or so.

The frequency response for the integrator is 1/(1 + jwRC). Multiply the source transform by this transfer function to get the FT of the response waveform. It will have the form something like sin(w)/[w*(1+jwT)] where T is the time constant RC.

Now, you have to get the inverse transform. That may actually be in a table somewhere. If not, you have to do just what you do with Laplace transforms to find the inverse; integrate over a contour in the complex plane. In this case, the transform can be resolved by finding residues at the poles (think partial fractions) w=0 and w=1/T. You have to select a contour with stability just as you have to do with Laplace transform.

Bill

• posted

There are two issues that continue to confuse me...

One is the aforementioned link between Laplace and Fourier and s =3D a + jw. Perhaps I should take a break from EE studies and focus on math to clear this up?

The other is the link between Laplace and phasor analysis where the complex variable s is replaced by jw. Am I correct in thinking that if a s-domain circuit is solved with a sinusoidal input, then the resulting steady-state equations are the same as those represented by the phasors?

I've tried to solve circuits using s-domain and a sinusoidal input but have not been successful yet. Haven't found an example where this is done anywhere either, unfortunately.

• posted

Most EE and physics courses covering Laplace and Fourier transforms go into the mathematics. Before these transform methods were in vogue, Heavyside developed operational methods without using mathematical rigor. It drove mathematicians to distraction.

Ordinarily, when you think of resonance, you think of sine waveform. The complex frequencies can be considered to be waveforms that also see resonances. These would correspond to poles and zeros that are not real frequencies but correspond to series and parallel resonances at the complex frequency. If that confuses you, skip this paragraph.

Bode, a key author on feedback amplifers have impedance functions of p in his book. The p corresponds to the s usually used in EE books.

That is correct. Because of what is called analytic continuation, two function of a complex variable that are equal equal over a complex contour must be equal everywhere. There are some picky mathematical exceptions.

You should be able to come up with a Laplace transform for the product of a step function and a sine wave. To find the steady state solution you need to insert damping (think resistor) into the circuit. Otherwise, the initial transient will ring forever rather than dying out leaving the steady state solution behind. All real passive circuits have such damping.

Bill

• posted

It's been a loooong time, but I seem to recall that what Laplace had over Fourier was that the Laplace Integral would converge over a much wider range of functions. For example, Fourier could handle a pulse, but not a step, since the step had a value a infinity. Or somethin' like that. Like I said, it's been a long time.

dave y.

• posted

I presume that you will be happen if you could find the FT of a step function, at least until you get to the next layer of the onion.

The FT of the the step function U(t) is integral from zero to infinity of exp(-jwt) wrt t. The indefinite integral does not go to zero at infinity, To make the integral converge insert a damping factor exp(-at) before integration. Then take the limit as a->0. That is your FT. Consider that any real circuit will have an attenuation that would bring the response to zero at infinite frequency.

Bill

• posted

And after you insert the damping factor, you have the Laplace Tranform, right?

• posted

That would be the case if w->0 and a becomes the transform variable. With a->0 you get the FT. Other damping factors can be used. Nevertheless, you can see that the FT and LT are closely related.

Bill

• posted

I guess that out of habit, I missed seeing that a ramp function was in the subject line. Realistic signals will not have ramps that increase with time. An exponential damping function will still take care of a linear ramp to produce a FT. When you take the limit as the damping goes to zero, I expect the dc and low frequency terms will nlot remain finite.

Using LT's instead of FT's will not be of much help. You still will need sensible inputs to get sensible outputs.

Bill

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