# Bode plot of Transfer Function

can anyone explain why the graph is shifted up vertically by 40dB? is that what we have to do by summing all lines and then shift them up by the amount of DC gain? I know that the DC gain is 20log10(100)=40dB and I can see where poles and zeros are. Also, how can I find the slopes for each line? I know there is a way to find it mathematically but I don't know how.

Thanks

To determine slopes:

Starting from the left, only the transfer function zero is active as the two poles start contributing at their respective break frequencies (100/1000 rad/s). Slope of a system with a single zero is +20 db/decade. The first pole (lower frequency, 100 rad/s) begins to contribute at its break frequency (100 rad/s) and reduces the overall gain slope to 0 dB/decade [+20 (zero) -20 (pole)]. The second pole at 1,000 rad/s then changes the gain slope to +20-20-20 = -20 dB/decade.

Each pole reduces the slope by 20 dB/decade and each zero increases the slope by 20 dB/decade.

To figure out the vertical position of the plot, one can look at 1/10th of the frequency of the lowest frequency pole/zero (but not located at the origin), calculate the gain at the point, and go from there using the slope criteria.

In this particular case, the lowest frequency pole is at 100 rad/s. So choose 10 rad/s (1/10th of 100 rad/s) and plug the value in the transfer function:

$H(10) = 100*10 = 1,000$

In [dB] : $20 * \log_{10}(1,000) = 60 \textrm{dB}$

Note that the contribution of the two poles can be neglected:

$H(10) = 100*10 / [(1+10/100)*(1+10/1000)] \approx 1,000$

This particular graph starts 2 decades below the lowest break frequency:

$H(1) = 100*1 = 100$

In [dB] : $20 * \log_{10}(100) = 40 \textrm{dB}$

Further help: Example 2 in this link shows a very similar transfer function and the way to develop its Bode plot: http://lpsa.swarthmore.edu/Bode/BodeExamples.html

• thanks for the answer, but it is not clear why "In this particular case, the lowest frequency pole is at 100 rad/s. So choose 10 rad/s and plug the value in the transfer function:" ? also why the peak is at 80dB? I don't see that in your math clculations Oct 3 '15 at 11:49
• you say "the lowest frequency pole is at 100 rad/s. So choose 10 rad/s" why do you choose 10 instead of 100 ? Oct 3 '15 at 11:50
• It is common to choose a frequency that is far away from break frequencies of any poles/zeros. The actual Bode plot is quite smooth and the curves really turns into lines at frequencies 10x lower and 10x higher than the break frequency. If you choose to calculate the gain at the break frequency, you will get a significant error. Hence, it is customary to start from the left and choose the initial frequency as 10% (or in this case 1%) of the lowest break frequency. Oct 3 '15 at 12:48
• how the peak 80dB is obtained? thanks for the effort but I don't see the order in your math calculations. Oct 3 '15 at 12:50
• Ok, we have determined the gain at 10 rad/s, which is 60 dB. At this point no pole is yet active (they start contributing at 100 rad/s and 1,000 rad/s) so the slope is +20 dB/dec due to the zero at origin. Coming from 10 rad/s to 100 rad/s, we get 60 dB + 20 dB = 80 dB. At this point, at 100 rad/s, the first pole kicks in, so the slope becomes zero (+20-20dB). Draw a horizontal line from 100 rad/s to 1,000 rad/s. At 1000 rad/s, the second pole kicks in. The slope is -20dB/dec. As a result, the gain at 10,000 rad/s is 80-20=60 dB, the gain at 100,000 rad/s is 80-20-20=40dB. Oct 3 '15 at 12:56