2
\$\begingroup\$

Noob question here but my books talks about the magnitude of the transfer function |H(jw)| a lot but I don't understand the meaning of |H(jw)| itself. Because of this I can't interpret the meaning of Bode plots of filters as it is the Y-axis of these Bode plots.

\$\endgroup\$
2
\$\begingroup\$

A transfer function takes a complex number input (or in this case, pure imaginary \$j \omega\$), and produces a complex number output.

The "magnitude" of \$H\$ is simply the complex absolute value:

\begin{gather} |H(j \omega)| = \sqrt{\mathrm{real}(H(j \omega))^2 + \mathrm{imag}(H(j \omega))^2} \end{gather}

As far as a physical meaning, consider an input signal with an amplitude \$V_0\$ and frequency \$\omega\$. \$H(j \omega)\$ "transforms" this input signal into an output signal, with final amplitude \$V_1 = |H(j \omega)| V_0\$, but it also can shift the phase of the output signal with respect to the input signal.

To summarize, the two plots of the Bode plot are:

  1. \$|H(j \omega)|\$, which tells you the ratio of input and output amplitudes (the "gain", or "attenuation")
  2. \$\tan^{-1}\left(\frac{\mathrm{imag}(H(j \omega))}{\mathrm{real}(H(j \omega))}\right)\$, which tells you the phase shift (note: this formulation is only accurate in the 1st quadrant; see atan2 for the definition which extends to the entire complex plane).
| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

\$H(j\omega)\$ is a complex number, say, \$a+jb\$, where \$a\$ and \$b\$ are functions of frequency, \$\omega\$. The magnitude of the complex number, \$\sqrt {a^2+b^2}\$, is the gain of the transfer function at \$\omega \: rad\:s^{-1}\$. And the phase angle at \$\omega \:rad\:s^{-1}\$ is: \$arctan\large \left(\frac{b}{a}\right)\$

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.