I have a transfer function of the following form:

\$\ T(jw)= a0 / (1+ s/w1) (1+ s/w2) (1+ s/w3) \$

where a0 = 3600 w1= 1MHz w2= 4MHz w3= 40MHz

I drew the the bode plot and double checked with matlab and I found some discrepancy. I found out the problem happens between 1MHz and 4MHz as the poles are close to each other (Less than 1 decade difference).

So since at 0 Hz, the gain is around 71 db, I expected that at 1MHz, the plot will start declining with 20db/dec. Between 4Mhz and and 1Mhz, there is ( \$\ log(4M/1M) = 0.6 \$ decade (not a full decade), and hence at 4MHz the gain is 71 - log(4M/1M)*20 = 71 - 12 = 59 dB.

However, in MATLAB, at 4MHz the gain is 69 db. Which means the gain dropped by 8dB not 12 dB. Can you please tell me where is the flow in my understanding?

I know that this problem happens because the poles are close to each other with less than 1 decade difference, and I am not sure how such cases are handled.

So the purpose of My main point is what's the gain at 4M (2pi) rad/s? Or in other words, how much the gain will fall between 2pi*1M rad/s and 2pi*4M rad/s and why?

My plot (The x axis is multiplied by 2pi and it is in rad/s):

enter image description here

MATLAB plot:

enter image description here

  • \$\begingroup\$ So, what does MATLAB say and why not use excel and plot it in more detail - why are you asking stack exchange ? \$\endgroup\$ – Andy aka Oct 17 '15 at 10:41
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    \$\begingroup\$ Check the dB scale (your drawing is not linear) as well as the frequency scale (also not correct). \$\endgroup\$ – LvW Oct 17 '15 at 10:41
  • \$\begingroup\$ @Andyaka In MATLAB, at 4MHz the gain is 71 - 8 = 63dB.. so it says basically from 1MHz to 4MHz, it is only 0.4 decade not 0.6 decade as in my calculations \$\endgroup\$ – HaneenSu Oct 17 '15 at 12:39
  • \$\begingroup\$ Download some log-Lin graph paper from papersnake \$\endgroup\$ – Chu Oct 17 '15 at 12:42
  • \$\begingroup\$ You need to show what matlab says diagramatically. I know what is happening but you need to do some work to make this a better question that might be useful to others. \$\endgroup\$ – Andy aka Oct 17 '15 at 12:45

Have you considered that the MatLab plot has been miss interpreted? For example all the pole locations in the question are labeled in MHz, but the MatLab plot axis shows rad/sec. If you multiply frequencies by 2\$\pi\$, pole locations become 6.28Mrad/sec, 25.1Mrad/sec, and 251Mrad/sec.

Now, you are performing an asymptotic analysis, so the numbers you get will be a little off of the MatLab result, but should be correctly done after the change in frequency scaling. For example, after correcting for asymptotic error the first two poles should have magnitudes of 68.1dB at 6.28Mrad/sec, and 55.8dB at 25.1Mrad/sec.

Note, you won't have a complete Bode plot until you add a phase plot too.

  • \$\begingroup\$ @Andy , Thanks. You're not the only one who may need an eye test, I had to zoom the MatLab plot by 200+% to read the axis units. \$\endgroup\$ – gsills Oct 24 '15 at 16:51

This answer missed the point (very tiny picture in the op's question) that the frequency response was in radians per second instead of Hz. However, I'm leaving it up here because it's still quite an important point about what happens when two frequency breakpoints are close - they tend to merge to a single 2nd order breakpoint at logarithmically half way between the two original break points.

Original answer: -

OK, now you have the MATLAB picture, I can confirm you are potentially getting a 2nd order "peaking" effect. Forget about the term at 40 MHz - this is irrelevant to the range 1 MHz to 4MHz. So now your TF becomes: -

\$\dfrac{A_0}{(1 +s/\omega_1)(1 + s/\omega_2)}\$

If you multiply this out you get: -


In other words, it has the form of a classic 2nd order, low pass filter having a net resonance at \$\sqrt{\omega_1\omega_2}\$ or 2 MHz. How much that peak emerges above the falling-slope baseline is dependant on the middle term \$\omega_1+\omega_2\$: -

enter image description here

The middle term (\$\omega_1+\omega_2\$) incorporates zeta (\$\zeta\$) with the whole equation of the form: -

\$H(s) = \dfrac{\omega_0^2}{s^2 + s(2\zeta\omega_0) + \omega_0^2}\$

Where \$\omega_0\$ is \$\sqrt{\omega_1\omega_2}\$ or 2 MHz

Basically, if the two frequencies are close zeta reduces and a combinational effect will become noticable. In your case zeta is 1.25 and this will not produce noticable peaking to the uninformed eye but it will tend to push the two 3dB points together to become a 3dB point at more like 2MHz i.e. just enough to confuse you in the predicted (but too simple) bode plot.

Note that although I've shown the general equation for this type of 2nd order low passs filter, due to the nature of the original formula, zeta can never get lower than 1 therefore the two seperate -3dB points combine to give a new break point logarithmically between 1 MHz and 4 MHz i.e. 2 MHz.

  • \$\begingroup\$ Basically for frequencies at 1 and 4 the effect is small. Numerically those two frequencies give a zeta of 5/(2*2) which is 1.25 and as zeta gets bigger the effect gets less and less. Zeta can only get bigger as the frequencies get wider so I guess at 1 and 4 you could make a case for saying this is a threshold point but for other applications it could be further away. It is what it is and different dependent on your point of view. \$\endgroup\$ – Andy aka Oct 17 '15 at 13:35
  • \$\begingroup\$ Hey you calculated zeta LOL \$\endgroup\$ – Andy aka Oct 17 '15 at 13:36
  • \$\begingroup\$ Yes but I removed it I thought it is wrong. Thanks for the confirmation! \$\endgroup\$ – HaneenSu Oct 17 '15 at 13:36
  • \$\begingroup\$ Whole books can be written on this subject of course and the implications are profound in all walks of life - mechancial resonance and the Tacoma bridge disaster (youtube.com/watch?v=3mclp9QmCGs) skip to 1 minute in! \$\endgroup\$ – Andy aka Oct 17 '15 at 13:42
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    \$\begingroup\$ But this TF has three real 1st order poles, so there can't be a resonance peak. \$\endgroup\$ – Chu Oct 18 '15 at 0:31

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