# Control theory: Frequency response

1. Nov 18, 2008

### Mårten

In my control theory text book, frequency response descriptions is one of three main areas of the subject (the others being Laplace transfer functions and state space representations).

My questions are: What is the purpose with frequency response in control theory? Why are not transfer functions and state space representations good enough? In what context does frequency response appear? Are there any concrete physical examples where one needs the methods of frequency response (my book doesn't give any...)?

2. Nov 18, 2008

### Pythagorean

I'm going to shoot from a musician point of view that this might have to do with equalizing your sound for a better tone quality, but also for eliminating feedback between mic and speakers (input and output)

hrm, when I was working in infrasound, we'd often test the frequency response of our equipment (predominately the microphones). If we're measuring infrasound in the forest, the mic would pic up a signal, but it might not necissarily be the exact same signal the source in the forest sent out. If I recall correctly, this is because the mic's frequency response is not perfect. It responds with a lower (or higher?) amplitude to some frequencies, even though those frequencies might have been sent from the source with the same amplitude.

Last edited: Nov 18, 2008
3. Nov 18, 2008

The Laplace and State Space representations are both more general than the frequency response/Fourier transform. I.e., given a Laplace or State-Space representation, you can derive the frequency response, but not vice-versa.

There are times when the Laplace and State-Space models are more than you need, in which case it's simpler to use the Fourier representation. For example, if you're only interested in steady-state behavior. Regardless, the frequency response exposes intuitively useful system properties in a nice, graphical way that makes it handy in ways that the more abstract Laplace and State Space versions aren't.

Also, I'd say that "frequency response" is a misnomer in this case. What you're (presumably) talking about is the Fourier transform. The Laplace transform is also a frequency response, just the frequencies in question can also exhibit damping...

4. Nov 19, 2008

### Pythagorean

I don't know how relevant this is, but in the context of differential equations, the fourier transform only works on functions that converge, while the laplace transform forces divergence (it doesn't have the i in the exponential)

Cocerning limits, we also generally apply laplace transforms on the interval o to infinite (so called semi-infinite interval) while the fourier transform is reserved for intervals on negative infinite to infinite (infinite intervals)

5. Nov 19, 2008

### Mårten

Okey, my picture of the whole area starts to get a little more brighter...

Just so we speak of the same thing. We have an input $u(t) = sin (\omega t)$ and the output then becomes $y(t) = |G(i\omega)| sin (\omega t + \phi)$.

I can imagine this makes sense in the example from the forest's infrasound above. I there imagine that you have multiple inputs u(t), one for each frequency, and these inputs all become superposed (a sort of Fourier series). Then for each u(t), i.e., for each frequency (or for each term in the Fourier series), you receive an output y(t). So, the equipment in the forest seemed to have a Bode plot, which suppressed some of the frequencies which actually were out there, and amplified frequencies that was not out there (or more correct, was not really as loud).

This is okey. But how is this relevant in a control theory context? Then you often want a constant output, by controlling the input with a controller (as for example when controlling the temperature of a room). There are not so much frequencies involved there, as far as I can see.

6. Nov 19, 2008

### Pythagorean

Well, I can stick to the acoustics, since I don't know where frequency would arise in temperature (I guess frequency in acoustics arises from amplitude vs. time data so you could theoretically analyze the frequencies in temperature vs. time data and try to make the line more flat so that temperature isn't fluctuating. I'd assume you'd do this by damping the frequencies with large amplitudes so that the temperature stayed more constant. This is a guess though).

In acoustics, we could use control theory. Here's wikipedia's definition:

so in acoustics, let's go back to music. I'm a musician, and let's say my genre is 100% pure egomaniac shredding, but I have crappy technique so you can always hear my fingers moving around the strings in between notes.

The EE answer would be the noise gate, which could identify the signals I send through the guitar and only let the notes through, and not the scraping sounds of my fingers moving around the strings. If you look at the fourier transform of my signal, you'll see frequency vs. amplitude (instead of time vs. amplitude) and you can just cut down the spikes in the frequencies you don't want to hear, then transform back to the original signal and play it, and now the sounds pertaining to those frequencies will be dampened.

7. Nov 19, 2008

### Proton Soup

i think maybe you guys are arguing about signal processing vs. control theory. but frequency analysis is critical to control theory if you hope to build controllers that are stable. and yes, even the acoustic example can be looked at from a control perspective because you need to filter out frequencies that create squealing when fed back into your mic. for a live performance where you want a stable amplifier, look at it as a control system. for a recording where there is no feedback, it's more of a signal processing problem.

8. Nov 20, 2008

### Pythagorean

that clarifies a bit. I was kind of lumping the two together.

To be picky though, you can still generate significant feedback while recording if you're monitor is a speaker and your input is a microphone (which I did frequently with acoustic sessions. Never owned an acoustic/electric; only one or the other. Obviously you can patch your signal right in with an electric guitar, and even if you use a speaker for a monitor, it generally won't feedback into the pickups... generally! If you're trying for it, it's not difficult to pull off: just turn everything up and play near the speaker...)

Anyway, I'm getting the feelng that control theory involves simultaneous input and output while signal processing is more "in series:" input, computation (transformation), output.

9. Nov 21, 2008

### Mårten

Okey, that seems reasonable. But still, I would like to have more examples from technical applications, where I think control theory is more often used. Anybody?

10. Nov 21, 2008

### Proton Soup

11. Nov 21, 2008

### saltine

I remember a robust control problem where the plant is given in Laplace domain, but the controller design is done using Nichols chart, because the region of instability has a simple shape on the Nichols chart.

Lead, lag controllers are also designed in frequency domain. This is related to tolerating time delay and uncertainty in the gain (phase margin and gain margin).

12. Nov 22, 2008

### Mårten

Okey, thank's so far.

My main insight here has been, that a frequency response diagram (i.e. a Bode plot) is the response not for just one input u(t) but for all possible u(t) at the same time (provided that u(t) = sin(wt)). You put all these responses in the same diagram so to speak. This in contrary to when dealing with normal transfer functions G(s) - then you just look at one response at a time. And with normal transfer functions, u(t) is often not sinusoidal, more often a step or something like that. I hope all this is correct, otherwise tell me.

But then, wouldn't it be possible, if you have a u(t) = a step, to say that u(t) = cos(wt), with w=0? Then you can look to the leftmost part of the Bode plot, and find the response.

Another thing that confuses me is that when you have a normal transfer function G(s), which may for instance describe the dynamics of the temperature in a house, then you say that G(s) is in the frequency domain. I cannot see any frequencies there. G(s) is for me just a Laplace transform which I will quickly inverse transform back to get the solution y(t).

13. Nov 22, 2008

### saltine

Re:

The step function does not correspond to cos(wt) when w goes to zero; that would actually be the constant function, and the gain you get there is the DC gain.

"Another thing that confuses me is that when you have a normal transfer function G(s), which may for instance describe the dynamics of the temperature in a house, then you say that G(s) is in the frequency domain."

I don't think that it is accurate to say that G(s) was describing the dynamics of the temperature. The complex variable s corresponds to a damped sinusoidal signal. The transfer function was not written "to describe the dynamics", that would be the differential equations. The moment you start using s, you assert this:

"Suppose the input is a damped sinusoid, then the system can be described by this transfer function, which is designed to be meaning when the input is a damped sinusoid. What if the input is not a damped sinusoid, you ask? Well, then we express the input as a summation of damped sinusoids and use linearity to sum the outputs."

The moment you replace d/dt by s in a differential equations, you have made the decision to, "let's see what happens if the input is only sinusoids!"

- This is the way I understand it at the moment.

14. Nov 25, 2008

### Mårten

My text book uses an engine as an example of a dynamic system, and it says that the engine has the transfer function G(s). So in that way, I mean that G(s) indirectly describes the dynamics of the engine.

Btw, what do you mean by a damped sinusoid? Is that $e^{-at} sin(\omega t), a > 0$?

That sounds very strange for me. Is it really so? What I have learned is that you could have a step as an input, u(t). Then you Laplace transform this to get U(s) = 1/s. And as output you get Y(s) = G(s)U(s), which will give you y(t) = g(t)*u(t). Where are the sinusoids here? Okey, you can find the Fourier series for u(t), but is that really what happens here, and if so, why?

15. Nov 25, 2008

### saltine

Yes, that is what I meant. But now I am confused too. What I was trying to say was this:

As a function, the input of H(s) can only be a complex number, and its output can only be a complex number. H(s) itself is describing the input-output ratio of damped sinusoids.

Suppose x(t) = etsin(t), and you run it through a system that has the Laplace transform H(s) = s. Then in this case, since 's' is the combination of the exponent and the frequency, s = 1 + i. H(s) evaluated at s=1+i is 1+i. The magnitude is sqrt(2), the phase is 45o. This means that the magnitude of x(t) will be modified by sqrt(2) and its phase would change by 45o. So

y(t) = sqrt(2) et sin(t +45o).

In this example, there is no need to do Laplace transform of x(t), multiply it to H(s), and do inverse Laplace transform of Y(s) to get y(t), because this is done using the meaning of H(s), and you get y(t) by reading the gain and phase shift.

I don't know how to explain the case with unit step function. It works but there seems to be no direct meaning while the unit step input is respresented by 1/s in s-domain. It is like,

"Let's treat the unit step input as if it is part of the system, and that the input to this arbitrary system is some imaginary damped sinusoid described by the shorthand complex notation s; then we observe that the system is composed of linear operators that can be multiplied together, and use inverse Laplace transform to find the overall impulse response--which is really the output y(t) we wanted to find originally. We do it like this because the math becomes easier, not because there is a direct meaning."

In my way of seeing it, the Laplace transform of signal x(t) = eatsin(wt)u(t), which is X(s) = w/( (s-a)2+w2 ) is not really a representation of the signal x(t), but a representation of a system with impulse response x(t).

Last edited: Nov 25, 2008
16. Nov 26, 2008

### TheAnalogKid83

Hmm, how are you separating frequency response from laplace domain? They are basically concerning the same concept. If you're looking at frequency response on Bode plots and other graphical methods (root-locus), then you can focus on specific properties and control theory ideas about the relationship between magnitude and phase which concerns stability and response, and this is what I did in my first part of my first control theory course. But you can derive the frequency response from laplace and transfer functions, so it is a more mathematical tool to use laplace or transfer functions.

As far as your concern using s operator and laplace domain to represent dynamical systems as opposed to just sinusoids, you must realize that laplace is a cousin of fourier, and that you can be addressing complex signals with laplace. for instance, the delta function is represented as s, which is equal to the sum of all frequencies at a set magnitude (0 to infinity).

I'm sure you're well beyond this, but imagine a transfer function G(s) as a filter that can have any signal pass through it, and this function will treat each frequency component of this complex signal as the equations applies and then the filtered frequencies add up at the output as a new filtered complex signal. Standard control theory is just a special application of filters and signal processing . . they are just different approaches to the same math and concepts.

Final note to remember, s operator is a linear operator, so your transfer function assumes a LINEAR input signal. Maybe this is where your problem accepting its applications to real systems is at?

Last edited: Nov 26, 2008