# Plotting bode plot of the transfer function two poles p1=1 and p2=25 two zeros are z1=100 and z2=origin and the DC gain is 20log10(10)=20dB I know that that the Dc gain Is 20dB which is a straight line with slope of 0 , The zeroes have 20dB slope, but I don’t know how to find amplitude of poles ( vertical positions) and and at what vertical position does the graph start ?

• It is very easy plotting bode plot on Matlab. In my humble opinion, it can help you, maybe.
– seda
Oct 3, 2015 at 14:57
• I cannot use MATLAB in this course. Oct 3, 2015 at 14:59

First, cancel out the pole and zero:

$H(s) = 250s/(s+25)$

Next, determine the pole location:

$H(s) = 10s/(1+s/25)$

The break frequency is 25 rad/s.

Since there is a zero at origin, the slope until 25 rad/s is +20 dB and the slope after 25 rad/s is 0 dB.

We can start the graph at 1/10th of the lowest break frequency: 2.5 rad/s but that's quite an odd number. Let's select 1 rad/s instead.

The gain at 1 rad/s is:

$H(1) = 10*1/(1)\ = 10 = 20 \textbf{dB}$

This does not directly tell us the gain at and beyond the 25 rad/s break frequency. We know that the gain difference between the gains at the break frequency and at 1/10th of the break frequency is 20 dB.

Thus, let's determine the gain at 2.5 rad/s:

$H(2.5) = 10*2.5/(1)\ = 25 = 28 \textbf{dB}$

Gain at 1/10th of the break frequency is 28 dB. One can draw vertical line from 20 dB at 1 rad/s through 28 dB at 2.5 rad/s to 48 dB at 25 rad/s. At 25 rad/s, the gain slope becomes zero and hence one can draw a horizontal line at 48 dB to ~1,000 rad/s.

Also, notice that the DC gain is 0, which is a number that cannot be represented on a logarithmic scale.

$H(0) = 10*0/(1)\ = 0$

MatLab: • can you share the MATLAB command ? Oct 3, 2015 at 18:14
• s = tf('s'); H = 250*s*(s+100)/(s+100)/(s+25); bode(H) Oct 3, 2015 at 19:35