bode plot troubles

I'm trying to plot the transfer function 10/(s*(s+5)*(s+10)) by hand. I know that the poles all decrease at 20 db/dec, crossing 0 dB point at 1, 5, and 10 rad/s respectively. However, I don't know what the contribution of the gain is. According to Matlab, it seems that the contribution is 20*log(1/10). However, I think that the gain contribution should be 20*log(10). Which one should be correct?

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None of the above.

Your poles are at 0, 5 and 10 radians/second. Your complex gain at 1 radian per second is 10/(j * (j + 5) * (j + 10)). You should be able to work it out from there.

If you're learning to sketch Bode plots it's probably a good idea to calculate a few points by hand and dot them onto good old graph paper, then sketch in the rest around those points.

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Tim Wescott


Thanks for your reply. Apparently, I found out that I have missed the j term in the bode plot, hence causing the difficulty that I have been experiencing (been a while since I looked at Bode plots).

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Another way to handle this situation is to first draw the Bode plot for your 10/S term and then add the lags at the |S| = 5 and |S| =10 terms. When f = 1 cps S = 2*pi*1*J |10/S| = 10/(2*pi). The magnitude in db is 20*log(10(2*pi)) = 4.04 db. Next plot the 4 db point at a frequency of 1 cps on a log log scale graph and draw a line through the point with a slope of -20 db per decade. The break frequency for a lag at |S| = 5 is 0.8 cps and the break frequency for the lag |S| =

10 is 1.6 cps. Next determine where 0.8 cps intersects the -20 db per decade plot and draw a line through the intersection with a slope of

- 40 deb per decade. Finally determine where 1.6 cps intersects your

-40 db per decade plot an draw a line through the intersection with a slope of at -60 db per decade.


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I found and really liked some course notes a while back, and have used them when teaching asymptotic Bode plotting, especially for systems with no quadratic terms like the transfer function (TF) you've presented. The method presented in the notes does not require individual plotting of each term's response and the subsequent addition that most texts teach, but rather a method of a "running sum" of the slope contributions from each term. With practice, it's a very quick and accurate method. I find students are less likely to make mistakes than when they try adding component responses, which is what typically happens with the first method I described. Here's a link:

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You've provided the Evans form of the TF. I find it easier for my students to make Bode plots by first putting the TF in Bode form, which is how the author of the notes presents the problem:

All binomial terms containing s (this excludes integrators and differentiators, i.e., poles and zeros at 0 rad/s) should be put into the form (1+s/p), where p is the pole frequency. All constants that are factored out of each term should be combined into the gain, which happens to be the DC gain of the system if no integrators or differentiators are present. (Integrators contribute infinite gain at DC as in your example; differentiators cause 0 gain at DC.)

Putting your example in Bode form, I get 0.2/(s*(s/5+1)*(s/10+1)). No zeros; poles at {0, 5, 10} rad/s. Infinite DC gain due to the integrator

I always find the nonzero poles and zeros from my binomial terms, then select boundary frequencies to plot from at least one decade below the lowest pole or zero frequency to at least one decade above the highest pole or zero frequency to show the phase plot properly. If one or more integrator or differentiator appears, I shift my lower frequency boundary one decade lower so I can show the slope contributed by those terms to the gain plot. (In your example, your lowest frequency pole is at 5 rad/s, one decade below is 0.5 rad/s, but you can round down to start the plot at 0.1 rad/s. Your highest frequency pole is at 10 rad/s, so end your plot at 100 rad/s.)

Substitute the lower boundary frequency you selected in for s to find the magnitude of the gain plot at that frequency to use as a starting point, then simplify by ignoring binomial terms whose values are close to one at that frequency as described in the notes. Break the frequency range into intervals with constant slopes as shown in the notes. (For your example, substitute 0.1 rad/s in for each s term, which produces 0.2/(0.1*(0.1/5+1)*(0.1/10+1)), which simplifies to

0.2/0.1, which is 2. Next, find 20*log(2), which is approximately 6 dB. This is the starting magnitude for the plot at 0.1 rad/s. Since there is a pole at 0 rad/s, the magnitude plot slopes at -20 dB/decade from the 6 dB point. Therefore, at 1 rad/s, the magnitude should be

-14 dB. Keep this slope until the frequency hits 5 rad/s, at which point the slope becomes -40 dB/decade. Estimate out one decade to 50 rad/s, and mark 40 dB down from your magnitude value at 5 rad/s, then start drawing a straight line between these magnitude points . Stop when you hit 10 rad/s. At 10 rad/s the slope becomes -60 dB/decade. Estimate out one decade to 100 rad/s, and mark 60 dB down from your value at 10 rad//s, then start drawing a straight line to connect these points. Continue this slope to infinite frequency).

The phase plot slopes at -45 deg for each pole for a decade either side of the pole (+ slope for each zero), then it contributes no additional shift. Break down the frequency range into intervals as shown in the notes, and follow the slope for each interval. (For your example, the plot starts at -90 deg at 0.1 rad/s, due to the single integrator. The intervals are 0.1-0.5 rad/s, 0.5-1 rad/s, 1-50 rad/s and 50-100 rad/s. The corresponding slopes for each interval are 0 deg/ decade, -45 deg/decade, -90 deg/decade, and -45 deg/decade. The phase plot then becomes asymptotic to -270 deg above the upper boundary of your plot).

Quadratic terms add some complexity, but you're example doesn't require them (and the notes don't either).

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