How to make a slope of 20db/decade for bode plots?

Question

The y-axis starts from -50 to 50. The fuction is 3125*(s^2+20s)/(s^3+150s^2+7500s+125000).I have already worked it out to this....0.5s*((50)^3/(s+50)^3)*((s+20)/20). Not sure how to find the coordinates using the slope of 20db/decade. I have a sheet that has bode plots for functions such as a/(s+a);(s+a)/s etc...

Answer 1

You take 20LOG base 10 of the entire equation:

Simplify further and substitute jw for s:

0.5s * (50^3 / (s+50)^3) = 0.5 jw * 1/ (jw/50 + 1)^3

1. Magnitude in dB = 20LOG(0.5 |jw| * |jw/50 + 1|^-3)

2. Use property of LOG: LOG(AB) = LOG(A) + LOG(B) to simplify.

3. Use property of LOG: LOG(x^y) = y LOG(x) as well.

4. Mag in dB = 20LOG(0.5) + 20LOG |jw| - 60LOG (|jw/50+1|)

5. Mag in dB = -0.3 + 20LOG w - 60LOG (w/50)

First term is a constant amplitude of -0.3 dB
Second term is a line with a slope of 20 dB/dec crossing through 0 dB at w = 1
Third term approximates the response with a line that starts at w = 50 with a negative slope of 60 dB/dec for all w > 50.
Result is sum of all three lines on the graph.

When plotting this, there should be a positive rising 20dB/dec line that is -0.3 dB below the 0 dB axis at w = 1 which continues to rise until w = 50. At w = 50, the +20 dB/dec is canceled out by -60 dB/dec slope with a resulting -40 dB/dec line for all w > 50.

20dB/dec on semi-log paper should increase/decrease 20 on the vertical scale for every 10x change in frequency.... jw for instance results in a line which passes through w = 1 with a +20dB/dec slope so that it also passes through w = 10 at 20 dB and w = 0.1 at -20 dB and at w = 100 at +40 dB...etc.