Electrical – Draw the amplitude and phase diagram through the transfer function

bode plottransfer function

My function in terms of s (Laplace variable) is:


The calculations I've done so far:

$$G(s)=\frac{1000(s+200)}{(s+20)(s+2000)}=\frac{\frac{1000}{11}}{s+20}+\frac{\frac{10000}{11}}{s+2000}=\frac{\frac{1000}{11}}{20+1\omega j}+\frac{\frac{10000}{11}}{2000+1\omega j}=\frac{\frac{1000}{11\times 20}}{1+\frac{1}{20}\omega j}+\frac{\frac{10000}{11\times 2000}}{1+\frac{1}{2000}\omega j}=\frac{\frac{50}{11}}{1+\frac{1}{20}\omega j}+\frac{\frac{5}{11}}{1+\frac{1}{2000}\omega j}$$
$$G(s)\approx 4,55\frac{1}{\sqrt{1+\big(\frac{\omega}{20}\big)^2}}e^{-j\,arctan\big(\frac{\omega}{20}\big)}+0,45\frac{1}{\sqrt{1+\big(\frac{\omega}{2000}\big)^2}}e^{-j\,arctan\big(\frac{\omega}{2000}\big)}$$



I simulated the Bode Plots in Scilab:

enter image description here

I know that:


My problem is that I do not know how to calculate the points of the Bode diagram through calculations, since I have a sum of two distinct individual transfer functions (one with \$\omega_1=20\$ rad/s and other \$\omega_2=2000\$ rad/s).
How can I apply the previous expression to calculate in dB the amplitude of the sum of the two functions?

My attempt:

enter image description here

Best Answer

The trick is to not transform the transfer function the way you did into a sum. Just use the original form

$$G(s) = \frac{A(s-z_1)(s-z_2)\ldots(s-z_n)}{(s-p_1)(s-p_2)\ldots(s-p_n)}$$

From this form, you can read off directly the DC gain, the pole frequencies and the zero frequencies.

That is enough to draw the Bode plot.