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I'm given the TF of a system, and I can figure out the magnitude for the Bode plot (just switch to jw and grind out for the requested w), but I don't understand in layman's terms how a transfer function affects the phase. Not an electrical person (mechanical engineer major), but hoping someone can put this stuff into something I can understand.

Can you explain this, or link something? I'm really just trying to understand how to plot out a phase diagram based solely on a TF, which before I was always given copious amounts of data, but none of my 2 semester notes mention how to deal with not being given that copious amount.

(Please feel free to roast me, but this is the TF: \$\frac{.1(s+1)}{(s+10)(s+1000)}\$. I have Matlab and can see the results, so I'm not asking for the answer. I'm asking for why it's the answer.)

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  • \$\begingroup\$ A super hint: \$s = \sigma + j\omega\$, the phase (\$\phi\$) = \$atan(\frac{j\omega}{\sigma})\$. Another super hint: denominator = +atan, nominator = -atan. Sum them up. \$\endgroup\$
    – Harry Svensson
    Commented Oct 23, 2017 at 4:28
  • \$\begingroup\$ Not an answer since it's short, but \$H(s)\$ is, in general, a complex number. A complex number can be written in real-imaginary format (\$C = a + jb\$) or in polar format (\$C = Ae^{j\theta}\$). In polar format, the phase is given by the value of \$\theta\$. As Harry mentioned above, \$s = \sigma + j\omega\$, where typically \$\sigma\ = 0\$ for a steady-state situation. \$\endgroup\$
    – Shamtam
    Commented Oct 23, 2017 at 4:39

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A transfer function is a complex valued function relating the output of a system to its input. It is complex-valued because \$s\$ is a complex variable.

A complex number (the value of the transfer function at some frequency or for some \$s\$) can be expressed as either a real part and an imaginary part, or as a magnitude and a phase.

When making a Bode plot we use the magnitude-phase representation, not the real and imaginary representation. The phase plot in a Bode plot presents the phase of the transfer function as a function of frequency (for \$s=j\omega\$ if you're working in the Laplace domain).

So the transfer function doesn't "affect" the phase. It has a phase, and the Bode phase plot is way of showing how that phase varies with frequency, just like the magnitude lot shows how the magnitude varies with frequency.

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