# Bode diagram---magnitude calculation

I am doing a Bode diagram plot of the following transfer function:

$$G(s)=\frac{100 \, (s+2)}{s\, (s+10)}$$ that can be put in the form

$$\frac{20 (1+s/2)}{s\, (1+s/10)}$$

At $\omega =100$, can the magnitude in dB be calculated as

$$20 \log(20)+20 \log \left (\sqrt{1^2 + 50^2} \right)-20 \log(100) - 20 \log \left (\sqrt{1^2 + 10^2} \right)$$

The above gives $-0.0415$ dB as opposed to $0$ dB that I get on a Bode magnitde plot. At $\omega =100$, $G(j\omega)$ has contributions from all terms, so the magnitude calculation should include all terms. Is the above calculation ok?

Your answer is correct, it's just that you only verified visually the magnitude plot. If you measured it, you would have found out that the results coincide:

• Yes I agree! I am plotting the asymptotic bode plot. The real graph will not totally agree with the asymptotic plot. May 29 '18 at 6:54
• May I ask which software output is that? Thanks. May 29 '18 at 6:55
• @user11206 LTspice. It's free, but there are plenty other free simulators, this is just one of them. May 29 '18 at 7:21
• @a concerned citizen, you are correct (and my answer was not). I have deleted my answer.
– LvW
May 29 '18 at 11:20
• @LvW I'm sorry if it turned out this way, but maybe you shouldn't have deleted it. After all, your answer did describe the usual way of calculating. A minor modification to point this out would have been all there is needed to keep it. May 29 '18 at 11:26