Draw the amplitude and phase diagram through the transfer function

Question

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

$$G(s)=\frac{1000(s+200)}{(s+20)(s+2000)}$$

The calculations I've done so far:

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

Then,

$$\omega_1=20\,rad/s$$ $$\omega_2=2000\,rad/s$$

I simulated the Bode Plots in Scilab:

I know that:

$$1\,dB=20log\Big|\frac{u_0}{u_g}\Big|$$

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:

Answer 1

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.

Answer 2

The thing is to rewrite your transfer function in a so-called low-entropy which will give you immediate insight on the position of gains, zeros and poles. This is part of the fast analytical techniques or FACTs: the leading term (if any) carries the unit of the transfer function (if any) while the quotient \$\frac{N(s)}{D(s)}\$ is unitless. In your expression, \$G(s)=\frac{1000(s+200)}{(s+20)(s+2000)}\$, factor 200 in the numerator, then 20 and 2000 in the denominator. You should obtain, \$G(s)=\frac{1000\times200}{20\times2000}\frac{1+\frac{s}{200}}{(1+\frac{s}{20})(1+\frac{s}{2000})}\$. This follows the low-entropy format defined as \$G(s)=G_0\frac{1+\frac{s}{\omega_z}}{(1+\frac{s}{\omega_{p1}})(1+\frac{s}{\omega_{p2}})}\$ in which you have the poles and the zero positions as shown below: