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Consider the LTI system, a series RL circuit (initially at rest). enter image description here If we define voltage as input and current as output, whatever be the forcing function, output contains a component \$ e^{-Rt/{L}}\$ corresponding to the particular system.

http://www.intmath.com/differential-equations/5-rl-circuits.php

We also know \$ e^{st}\$ is an eigen function of LTI system which means if we apply an input of \$ e^{st}\$, a scaled version H(s)* \$ e^{st}\$ will appear as output. Output contains only the frequency corresponding to input (forcing function). Hence we hear statements like 'Linear systems do not introduce new frequencies(harmonics)'.

Isn't the first case and second argument contradictory? How can we explain the absence of natural frequency(\$ e^{-Rt/{L}}\$) term in latter case? Also how can we impose initial conditions at t= \$-\infty\$ for these everlasting inputs of kind \$ e^{st}\$?

Note: A similar situation arises for LC circuit analysis, where output contains a terms corresponding to natural frequency \$ \omega_0\$.

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    \$\begingroup\$ There is no natural frequency produced by a first order system. \$\endgroup\$ – Andy aka Dec 21 '16 at 10:46
  • \$\begingroup\$ I'm stuck on the point that the output contains an exponential despite the forcing function. Can you demonstrate that for me with diff eqs? \$\endgroup\$ – Scott Seidman Dec 21 '16 at 13:58
  • \$\begingroup\$ @Scott : link added to illustrate the point. \$\endgroup\$ – Divya K.S Dec 22 '16 at 7:36
  • \$\begingroup\$ @Andy: Yes, I understand \$ e^{-Rt/L} \$ is not truly a frequency in traditional sense. But the point is response is having a term other than source frequency. \$\endgroup\$ – Divya K.S Dec 22 '16 at 7:41
  • \$\begingroup\$ That happens because there''s a step in the input that contains many frequncies. Initial conditions can do the same thing. \$\endgroup\$ – Scott Seidman Dec 22 '16 at 11:31
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The eigenvalues are the roots of the characteristic equation, so \$s\$ needs a value in your first \$e^{st}\$. Generally, if \$s=-\sigma_1\pm j\omega _1\$, is a complex conjugate pair, then the corresponding eigenvalues will be \$e^{-\sigma_1 t}(cos(\omega _1 t)\pm jsin(\omega_1 t))\$, and \$\omega _1\$ will be the natural frequency.

In your second \$e^{st}\$, \$s=j\omega\$ represent a forcing sinusoid at an arbitrary frequency, \$\omega\$.

If there are no complex or imaginary eigenvalues, there is no natural frequency, just 1st order real exponential terms.

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"If we define voltage as input and current as output, whatever be the forcing function, output contains a component e−Rt/Le−Rt/L corresponding to the particular system."

This is wrong. Consider input voltage 0. That's cheating? Consider the input \$\delta(t) + \frac{Ru(t)}{L} \$ then.

Here's what's going on: when both your transfer function and the Laplace transform of your input signal are pretty polynomial fractions, you can use partial fraction decomposition, and one of the terms will be your natural response. But it's coefficient may be 0. A similar trick is used for the Fourier transform.

Your second statement about not introducing new frequencies is correct, but you seem to be confused about how to interpret it when using the Laplace transform and initial conditions. I suggest reading about the bilateral Laplace transform to clear it up but, long story short, when you are talking about eigenfunctions, you are assuming \$e^{-st}\$ extends to both sides (and so does your solution), but when you take the (one sided) transform, you are actually talking about \$u(t)e^{-st}\$, which has different frequencies.

As an exemple, \$u(t)sin(\omega t)\$ has frequency components that \$sin(\omega t)\$ does not have, so a LTI system at rest excited by the former may produce a response with frequency components other than \$\omega\$. Solving for the latter, however, only produces the steady state response.

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