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Comments on H(jω) does not exist for unstable systems, but we still use it when designing controllers - contradiction?

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H(jω) does not exist for unstable systems, but we still use it when designing controllers - contradiction?

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According to Signal processing and linear systems by Lathi, the transfer function $H(j\omega)$ does not exist for systems with poles in the RHP: -

Lathi

This makes sense to me, since $H(j\omega) = \frac{Y(j\omega)}{X(j\omega)} $. However, since the system is unstable $Y $ is unbounded (and growing) and the fourier transform doesn't exist for such functions. So $Y(j\omega)$ doesn't exist and therefore, $H(j\omega) $ must be meaningless for unstable systems.

BUT! we use $H(j\omega) $ when designing controllers for unstable systems anyway - and it works. We look at the Bode Plot, we look at the Nyquist Plot both of which you need to know $H(j\omega)$ and design a controller based on what we see - and the controller actually works!

How can this be? How can there be this contradiction between systems and signals theory and control theory? It seems that concepts like region of convergence and existence of fourier integral are only dealt with on coursework and once that's done, you don't hear from them ever again.

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Please note that the text refers to the integral, or the mathematical evaluation through integration which, indeed, cannot be obtained. But that doesn't mean you can't obtain the Laplace transfer function directly. A hypothetical RLC filter with a negative resistor is very much possible. In fact, it can be made in practice with emulated elements, and its transfer function will give a pole in the RHP:

$$H(s)=\dfrac{\dfrac{1}{LC}}{s^2-\dfrac{1}{RC}s+\dfrac{1}{LC}}$$

This can be obtaind by considering the raw Laplace equivalents of the elements. Is it unstable? Yes. Can it be done practically? Yes, here is the concept of it:

-RLC

You can clearly see the phase going up. Does that violate the textbook? No -- the textbook only picks on the mathematical aspect (Olin says it better: semantics).

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But it does violate the textbook. The Bode Plot on the left is a plot of \$H(j\omega)\$ which should ... (2 comments)
But it does violate the textbook. The Bode Plot on the left is a plot of \$H(j\omega)\$ which should ...
Carl‭ wrote almost 3 years ago

But it does violate the textbook. The Bode Plot on the left is a plot of $H(j\omega)$ which should be meaningless according to Lathi. But it is clearly not meaningless.

a concerned citizen‭ wrote almost 3 years ago

Carl‭ I think the author may be saying something else, and that is dependent on different details that are not shown (I don't have the book). But, clearly, any transfer function can be plotted by replacing s->jw, since it's basic complex arithmetic. It's probably still related to the integral part which, mathematically, cannot converge (true), but there clearly exist RHP transfer functions, they can be created randomly (even only for amusement), and they can be evaluated anywhere in the complex plane. If you try to use the Fourier integral then you can't due to the lack of convergence, but the transfer function, itself, has no such problems.