So when you derive the Bode Plot from the transfer function you plug jw into the transfer function H(s). You then get the magnitude of H(jw) and plot it to get the Bode Plot. My question is why does this work? My book simply tells me to plug in jw into the equation without giving me any intuition on why I need to do so.

  • \$\begingroup\$ Because that is what you want to have on the x and y axis. Or are you asking why having these things in a bode plot makes sense and why you want to look at that plot at all? \$\endgroup\$ – PlasmaHH Jun 14 '16 at 15:37
  • \$\begingroup\$ "Why" question aren't the greatest in engineer... Because I would say algebra and mathematic works? \$\endgroup\$ – MathieuL Jun 14 '16 at 15:39
  • \$\begingroup\$ I think his question is what difference does H(jw) and H(S) have and why one should consider H(jw) while doing frequency domain analysis and not H(S). \$\endgroup\$ – Bhuvanesh Narayanan Jun 14 '16 at 16:22

This is how I see the bigger picture...

The transfer function and bode plot are intimately related. More specifically the pole zero diagram and bode plot are related. See this example of a 2nd order low pass filter: -

enter image description here

Top three diagrams show the amplitude part of the bode plot. Bottom left shows how this fits into the 3 dimensional picture and bottom right is the plan view of the 3D picture (pole zero diagram).

So when you replace s with jw you are just focusing on one aspect of the 3D big picture.

  • \$\begingroup\$ Excellent illustration! +1 \$\endgroup\$ – MathieuL Jun 15 '16 at 12:19
  • \$\begingroup\$ @Andy: could you tell me where did you get the picture, from which book or website? \$\endgroup\$ – anhnha Nov 7 '16 at 7:03
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    \$\begingroup\$ @anhnha I drew it i.e. it's home made. Have you spotted a problem with it maybe? \$\endgroup\$ – Andy aka Nov 7 '16 at 8:18
  • \$\begingroup\$ No, I see it interesting so I want to study more about this. Which software did you use to draw this? \$\endgroup\$ – anhnha Dec 19 '16 at 11:32
  • \$\begingroup\$ @anhnha I wish I could recommend somewhere to look. I had to draw this because I could find nowhere on the web that pictorially showed this relationship! \$\endgroup\$ – Andy aka Dec 19 '16 at 11:34

S is nothing but σ + jw . If you consider the σ =0 then you are left out with just jw. When you want to see how the system behaves for different frequencies which is what the bode plot shows ,you dont consider σ because you only need jw (w = 2*pi*f, is the angular frequency).This is one main reason why we do so.

Also another way of saying the same is that, the fourier transform H(jw) describes how the system behaves for different sinsoidal frequencies. The H(s) describes how the system works for different sinusoidal frequencies and exponential signals. When you plug H(jw) to a H(S) you have the fourier transform from the laplace transform, removing the exponential factor gives you pure frequency depended function alone upon which you can now perform anylsis such as magnitude response and phase response which nothing but the bode plot.


jw is frequency. w = 2piF. I assume s was the symbol Laplase chose to use in his functions? The impedance of capacitors and inductors change with frequency. Frequency is your X axis. As impedance changes, you would expect to see different magnitudes. Magnitude is your Y axis. The plot is logarithmic to avoid exponential curves running off the page. Draw 100 bode plots for various circuits and you will understand. Use sites like wolframalpha to verify your work.


To explain really simply,

Laplace's transform is but a specific case of Fourrier's transform where the imaginary operator (j) has been removed.

By posing s = jw in Fourrier's transform, you obtain Laplace's transform. (NB : supposing σ = 0)

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    \$\begingroup\$ Other way around; Fourier transforms are a special case of the Laplace transform; Fourier did his analysis for purely imaginary modes (i.e. oscillations only), while Laplace extended it to the full complex plain since s in general can have a real and imaginary component. Also, s=jw is not a Laplace transform, but a Fourier transform. \$\endgroup\$ – helloworld922 Jun 14 '16 at 16:37
  • \$\begingroup\$ False, Laplace isn't a special case of Fourrier. \$\endgroup\$ – MathieuL Jun 14 '16 at 16:42

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