# Draw the amplitude and phase diagram through the transfer function

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: • Do you know how to find the pole and zero frequencies? – The Photon Nov 29 '17 at 17:46
• @ThePhoton The zero is $\omega=200$ rad/s and the poles are $\omega_1=20$ rad/s and $\omega_2=2000$ rad/s (I think) – Jose Marin Nov 29 '17 at 17:51
• Okay, now you have enough information to draw a Bode plot. – The Photon Nov 29 '17 at 17:52
• Let $s\rightarrow j\omega$ in the first equation, then find magnitude and phase angle of each factor. Multiply/divide all the magnitudes and add/subtract the phase angles, as appropriate. Plenty of references to this on the web. – Chu Nov 29 '17 at 17:56
• @Chu I put my attempt to draw the bode plot in the post, but it is not the same as the simulation and I do not understand why. – Jose Marin Nov 29 '17 at 18:12

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.

• I put my attempt to draw the bode plot in the post, but it is not the same as the simulation and I do not understand why. – Jose Marin Nov 29 '17 at 18:11
• @JoseMarin, In your amplitude plot you didn't include the effect of the 2nd pole in your plot (though you drew the construction line). You also didn't include the DC gain factor. If you did, you'd get the right result. Hand-drawn bode plots are only an approximation --- it's hard to get the exact curves so we only try to get a piecewise linear approximation. – The Photon Nov 29 '17 at 18:55

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: • I put my attempt to draw the bode plot in the post, but it is not the same as the simulation and I do not understand why. – Jose Marin Nov 29 '17 at 18:11
• I have to be able to draw the Bode plot from my calculations, I can only use the simulation to confirm my results. – Jose Marin Nov 29 '17 at 18:14
• You have a dc gain for $s=0$ then follows a pole (-1 slope). The zero kicks in and brings the magnitude to a 0-slope then the second pole shows up and forces another -1-slope. The phase goes down because of the 1st pole, goes up because of the zero and finally lags again thanks to the second pole. The hi-frequency asymptote is -90°. – Verbal Kint Nov 29 '17 at 18:15
• Start with your low frequency (DC) gain. Each pole is a -20dB/decade slope contribution and each zero is +20dB/decade contribution. Phase is -45 degrees at the pole, asymptotic to -90 degrees. The inverse for the zero. – John D Nov 29 '17 at 18:17