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Transfer functions always seem to take the form $$F(s) = F(j\omega)$$

However, going back to the original Laplace Transform used to obtain the transfer function, \$s\$ is said to be a complex number, \$s = \sigma + j\omega\$.

What is the justification for dropping sigma, or the real part of \$s\$?

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    \$\begingroup\$ AC coupling.... \$\endgroup\$ Jan 29 '21 at 19:11
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    \$\begingroup\$ @PeteW How can w be complex when it represents physical frequency, an inherently real quantity? Frequency is directly related to wavelength via a real constant, which immediately implies wavelength is complex, which makes no sense. And, so on. \$\endgroup\$
    – EthanT
    Jan 29 '21 at 19:43
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    \$\begingroup\$ @Ethan, it comes out of fourier transform definition, is just math. example: take FT of d/dt[f(x)] . . . \$\endgroup\$
    – Pete W
    Jan 29 '21 at 20:11
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    \$\begingroup\$ d/dt[f(t)] ... not f(x). here ... it is complex. FT and Laplace differ by just that rotation. a pure sinusoid in s plane is on imaginary axis. there is no real part to throw away, in that special case. \$\endgroup\$
    – Pete W
    Jan 29 '21 at 20:32
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    \$\begingroup\$ in w plane, pure sinusoid has no imaginary part. but it does have a peak at -w and +w. this corresponds to it being a complex conjugate (+jw and -jw) in s plane. \$\endgroup\$
    – Pete W
    Jan 29 '21 at 20:40
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Transfer functions always seem to take the form $$F(s) = F(j\omega)$$

No not really..

The transfer function, usually denoted \$ H(s) \$, is generally speaking defined for all \$ s= \sigma + j\omega \$ in the complex plane (s-domain). When analyzing things like BIBO-stability and causality of an LTI system we need to cosider the entire s-domain.

As an example if the transfer function has a pole at \$ \bar s = \bar \sigma + j \bar \omega \$, ie. \$ H(\bar s) \rightarrow \infty \$.

Then \$ \bar \sigma > 0 \$ means that the system is unstable (it has a pole in the right-half-plane).

What is the justification for dropping sigma, or the real part of \$ s \$?

There is no justification for "dropping" the real part \$ \sigma \$, which we don't generally do, this is the wrong phrasing.

The reason you usually see the transfer function evaluated only at imaginary values of \$ s \$ is that this corresponds to a real input excitation, ie. a sinusoid.

Or in plain English; real-world signals lie on the \$ \sigma = 0 \$ line of the complex plane..

A lot of theory goes into explaining the s-domain in detail, I am very much brushing over it here, it is just to give you an idea of the concept.

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    \$\begingroup\$ This is perfect, thank you. It is what I was beginning to suspect, per my comment above to Andy. This makes me feel better too. It gives context to why we sometimes see L(jw), which means it's not being as loose with the math, as it first appeared. \$\endgroup\$
    – EthanT
    Jan 29 '21 at 21:27
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If all you are interested in is the bode plot magnitude or phase then sigma is irrelevant and should be made zero so that the resulting formula for the transfer function is defined only for the bode plot plane of the pole zero 3D diagram. Example: -

enter image description here

The resulting bode plot shown below has \$\sigma = 0\$ : -

enter image description here

In other words, it's the \$\sigma = 0\$ plane

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  • \$\begingroup\$ Can similar arguments be made for Root Locus, Nyquist Plots, etc? \$\endgroup\$
    – EthanT
    Jan 29 '21 at 19:14
  • \$\begingroup\$ @EthanT Dunno. Make a case and ask a formal question like you did above. This is a Q and A site so we have to be clear about the rules and avoid lengthy speculation about this or that. \$\endgroup\$
    – Andy aka
    Jan 29 '21 at 19:22
  • \$\begingroup\$ I thought I had done that. My question wasn't about a specific type of analysis. It was about transfer functions in general, as they are treated in control theory. Still, I found your response very interesting. It just doesn't fully answer my original question. \$\endgroup\$
    – EthanT
    Jan 29 '21 at 19:44
  • \$\begingroup\$ @EthanT I gave 1 example and thought that would be enough to convince you. However, if you looked at root locus it would be meaningless without the sigma axis. And, if this doesn't tally with your thoughts it's because you have asked a rather open ended question in the comment above. \$\endgroup\$
    – Andy aka
    Jan 29 '21 at 19:47
  • \$\begingroup\$ Good point on rlocus. It makes me think that the sigma term is NOT dropped, or set to zero, when computing the transfer function during a Laplace Transform. Seems like it's always there. But, ignored when not needed, like w/ Bode Plot, in your example. Explicit when needed, as in Root Locus. Overall, that would make writing a transfer function as L(jw), rather sloppy, loose treatment mathematically speaking. Not sure if that is the correct conclusion to draw, but it's where I'm heading right now, unless I can see some other justified way of viewing this. \$\endgroup\$
    – EthanT
    Jan 29 '21 at 20:36

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