How transfer functions become Bode plots

[kostoglotov]

Homework Statement

I have a question related to taking the logs of transfer functions, getting the individual Bode plots of each subsequent factor, and adding those plots graphically.

I'm working from Fundamentals of Electric Circuits, 5th edt. Let me start with the following screen capture.

imgur link: https://i.imgur.com/NOqNK69.png

Homework Equations

Transfer function standard form:

imgur link: https://i.imgur.com/24GBArb.png

The Attempt at a Solution

I can see how taking the natural log spits out the phase, but isn't the Bode plot created from taking the decibel measure, ie, multiply by 20 the log base 10 of the transfer function.

This would produce:

$$log_{10}(H) + j\phi log_{10}(e)$$

the imaginary part of that is $\phi log_{10}(e)$, not $\phi$.

So I thought maybe we use log base 10 to get the decibel magnitude by taking the real part of that, and then get the phase by taking the natural log and taking only the imaginary part.

Which would make sense to me since it is convenient to use decades on the frequency axis for the magnitude vs frequency plot, but I also noticed that the phase vs. frequency plot also uses decades on the frequency axis...

What exactly is the mathematical procedure that bridges the conversion of the transfer function into the two Bode plots; gain vs freq and phase vs freq ?

 

[BvU]

Hi there,

It's just a matter of establishing the vertical scale. dB for the amplitude and radians or degrees for the phase is pretty common and legible. the $\log_{10}$ factor isn't practical.

Using a logarithmic scale for the frequency makes sense: in general a characteristic spans serveral decades and the limiting behaviour shows up as a straight asymptote: $\ \pm$3 dB/octave (or 20 /factor 10) for a first order and double for a second order.

 

[LvW]

BvU - if I am not wrong we have for a first order function a slope of 6dB/octave (not 3dB/octave).

 

[BvU]

BvU - if I am not wrong we have for a first order function a slope of 6dB/octave (not 3dB/octave).

Absolutely correct. Something with power $\propto$ amplitude squared. My bad. See the examples