Control problem. Transfer function given. Sketch Bode plot

[rowardHoark]

1. For the control system described by the transfer function H(s)=Y(s)/U(s)=10/(s^2+11s+10) Sketch the bode plot: amplitude and phase diagram. What is the bandwidth of the system?



2.
General form of a second order system H(s)=wn^2/(s^2+2*zeta*wn*s+wn^2)
Magnitude characteristic in logarithmic form is A(w)[dB]=-20log[(2*zeta*w/wn)^2+(1-w^2/wn^2)^2]^0.5
The phase of second-order system is theta(w)=-tan^(-1) [2*zeta*w/wn]/[1-(w/wn)^2]




3.
I really want to understand the algorithm (step by step process) how to do this.

1. Using the general form of a transfer function, I can find resonance frequency wn=10^0.5. I use it as my cut of frequency when plotting magnitude. Before wn magnitude is a straight line at 0 dB. After wn=10^0.5 it is a line going down.
I can also find zeta, which is = 11/(2*10^0.5). I know that it influences the shape of the function near the corner frequency.
No ideas how to do the phase plot though.

 

[xcvxcvvc]

H(s)=\frac{10}{s^2+11s+10}
As you probably know, bode plots deal with frequency response, so s = jw
H(j\omega)=\frac{10}{(j\omega)^2+11j\omega+10}
H(j\omega)=\frac{10}{-\omega^2 + 11\omega j + 10}
When w is extremely small, the equation becomes
H(j\omega) = \frac{10}{10} = 1
which has a phase of 0.
when omega is very big we have
H(j\omega)= \frac{10}{-\omega^2}
which has a phase of -pihttp://ocw.mit.edu/courses/mechanic...ntrol-i-spring-2005/exams/secondorderbode.pdf

from google, here is an example. You can see the angles start at 0 degrees and approach -180 degrees or -pi radians, as we just explored. This is how all phases for second order underdamped systems will turn out. The larger zeta is, the faster the transition from 0 to -pi.

 

[rowardHoark]

H(s)=\frac{10}{s^2+11s+10}
As you probably know, bode plots deal with frequency response, so s = jw
H(j\omega)=\frac{10}{(j\omega)^2+11j\omega+10}
H(j\omega)=\frac{10}{-\omega^2 + 11\omega j + 10}
When w is extremely small, the equation becomes
H(j\omega) = \frac{10}{10} = 1
which has a phase of 0.
when omega is very big we have
H(j\omega)= \frac{10}{-\omega^2}
which has a phase of -pi


http://ocw.mit.edu/courses/mechanic...ntrol-i-spring-2005/exams/secondorderbode.pdf

from google, here is an example. You can see the angles start at 0 degrees and approach -180 degrees or -pi radians, as we just explored. This is how all phases for second order underdamped systems will turn out. The larger zeta is, the faster the transition from 0 to -pi.

Thank you for some useful insights.
However, I can understand why H(j\omega) = \frac{10}{10} = 1 results in phase 0 and H(j\omega)= \frac{10}{-\omega^2} in phase -pi ?

 

[xcvxcvvc]

Thank you for some useful insights.
However, I can understand why H(j\omega) = \frac{10}{10} = 1 results in phase 0 and H(j\omega)= \frac{10}{-\omega^2} in phase -pi ?

omega is just a real-valued number. What is the phase of a negative, real-valued number? It is pi radians (aka on the negative real axis). However, we are dividing by this negative, real number. When you divide a complex numerator by a complex denominator, you do regular division of the numerator's and denominator's magnitudes and you subtract their angles.

In that example, you do |10|/|-w| = 10/w to find the new magnitude. 10 has an angle of 0 and -w has an angle of pi (remember that we are dealing with positive omegas). So 0 - pi = -pi