0
\$\begingroup\$

Everyone knows that \$ e^{-30 t} \$ is a rapidly decaying function. In fact, a little too fast to be useful.

enter image description here

Which is amusing why this function would have the following properties as displayed by the bode plot.

enter image description here

Can someone explain how the Bode plot of this function correspond what we see in the time domain?

  • Why do we have a large negative drop off in the Bode Plot when we have a large positive rise in the time domain?

  • Is it possible to observe the effect of the pole in the time domain?

  • Finally, how do we understand the phase of this function? I can't see any phase effect in the time domain at all.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ You might have a conceptual error. The phase and magnitude plots describe the entire time domain. Magnitude is always positive. The drop off in magnitude is a drop off with frequency, not time. Intuitively, there is a drop off in frequency because the function is not zero width and not periodic, and thus higher frequencies are 'smaller' than the function. \$\endgroup\$ Apr 6 '15 at 2:39
  • \$\begingroup\$ To intuitively understand a Bode plot, start with the pole-zero plot. \$\endgroup\$
    – Matt Young
    Apr 6 '15 at 2:48
  • \$\begingroup\$ Since time is the inverse of frequency, you should understand that the second two plots are expansions in an inverted manner of the first plot. They begin at 1 and go left on the first plot towards 0. So just from that, you can see that your scale of the first plot way too large to compare well with the second two plots. \$\endgroup\$
    – horta
    Apr 6 '15 at 5:07
  • \$\begingroup\$ And then you'd need to expand the time domain plot in a logorithmic manner to closer match the magnitude plot in frequency. \$\endgroup\$
    – horta
    Apr 6 '15 at 5:09
  • 1
    \$\begingroup\$ Just wanted to say that there's no such "a little too fast function". My control theory teacher once asked a student "if the answer time of a control system is 1s, is it fast or slow?" it took some time but the answer is: it depends. \$\endgroup\$ Apr 6 '15 at 9:03
2
\$\begingroup\$

Why do we have a large negative drop off in the Bode Plot when we have a large positive rise in the time domain?

Because those plots belong to two different domains. A large peak in time domain does not guarantee a peak in frequency domain. Consider unit impulse function as an example and think.

Is it possible to observe the effect of the pole in the time domain?

Yes. Depending on the position of pole various effects can be observed in the time domain. But it doesn't mean that the effect of a pole at s=5 can be observed at specific time like t=5 or t=1/5. See examples:

  1. If there is a pole at right half plane then signal in time domain will be exponentially increasing.

  2. If there are poles on the imaginary axis, then oscillations can be seen in the time domain.

Finally, how do we understand the phase of this function? I can't see any phase effect in the time domain at all.

The phase plot of a transfer function gives the phase delay caused by that system for different frequencies. To see an effect, feed a sinusoidal signal to a system with impulse response \$e^{−30t}\$ and measure the phase difference between input and output.


To understand better, I suggest you to perform an experiment. The impulse response \$e^{−30t}\$ corresponds to a RC low pass filter with RC = 1/30. So implement a this circuit on a breadboard or simulator. Do a frequency sweep and plot the gain and phase with respect to the frequency. Compare with the bode plot.

\$\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.