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

 

[DeeNos]

As a scientist, it is important to have a clear understanding of the control problem and the transfer function given. The control problem is a fundamental issue in engineering and involves designing a system that can regulate and maintain a desired output in response to external disturbances. The transfer function, in this case, represents the relationship between the input and output of the system, and provides important insights into the behavior and performance of the system.

To sketch the Bode plot for the given transfer function, we can follow these steps:

1. Convert the transfer function into its general form, H(s) = wn^2 / (s^2 + 2*zeta*wn*s + wn^2), where wn is the resonance frequency and zeta is the damping ratio.

2. Calculate the resonance frequency, wn, which is the frequency at which the magnitude of the transfer function is maximum. In this case, wn = 10^0.5.

3. Calculate the damping ratio, zeta, which determines the shape of the magnitude characteristic curve near the resonance frequency. In this case, zeta = 11/(2*10^0.5).

4. Plot the magnitude characteristic using the logarithmic form, A(w)[dB] = -20log[(2*zeta*w/wn)^2 + (1-w^2/wn^2)^2]^0.5. The magnitude will be a straight line at 0 dB before the resonance frequency, and will decrease after the resonance frequency with a slope of -40 dB/decade.

5. To plot the phase characteristic, use the equation theta(w) = -tan^(-1)[2*zeta*w/wn]/[1-(w/wn)^2]. The phase will be 0 degrees before the resonance frequency, and will decrease to -180 degrees after the resonance frequency.

6. Based on the Bode plot, we can determine the bandwidth of the system, which is the range of frequencies over which the system can effectively regulate the output. In this case, the bandwidth is approximately 10^0.5 - 11 = 0.9 rad/s.

I hope this step-by-step process helps in understanding the algorithm for sketching the Bode plot for the given transfer function. It is important to note that the Bode plot is a powerful tool for analyzing the behavior of control systems and can provide valuable insights for designing and optimizing these systems.