2.14 Lecture 13. Tuesday Oct 13, 2015, Noise rejection, second order systems

Today we will finish noise attenuation, then start on second order systems. The classical method of noise attenuating is based on this feedback system block diagram

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We often have systems where noise or disturbance comes in between the input and the output. Without \(N\left ( s\right ) \) we would have perfect open loop. The classical approach to noise attenuation is as shown in the above diagram, which is to add \(H_{1}\left ( s\right ) \) and \(H_{2}\left ( s\right ) \) with the idea to reduce the effect of \(N\left ( s\right ) \) while at the same time to preserve \(R\left ( s\right ) \) input signal and not affect it. For \(H_{1}\left ( s\right ) \) we use large pure gain \(\alpha \), this is for attenuation. For \(H_{2}\left ( s\right ) \), we start with what is called the inversion method, which is to use \(H_{2}\left ( s\right ) =\frac{1}{G_{1}G_{2}}\). As discussed in last lecture, this method looks good in math, but not good in practice, since \(H_{2}\left ( s\right ) \) becomes improper transfer function.  Now we will explain a more practical method, which is to introduce a low pass filter \(H_{LP}=\frac{1}{\left ( \varepsilon s+1\right ) ^{k}}\) which will reject noise frequency and also make \(H_{2}\left ( s\right ) \) become proper. We will use \(H_{2}\left ( s\right ) =\frac{1}{G_{1}G_{2}}H_{LP}\left ( s\right ) \) instead of just \(H_{2}\left ( s\right ) =\frac{1}{G_{1}G_{2}}\) as before.

We need to pick \(\varepsilon ,k\). Both are positive. To design for \(H_{LP}\left ( s\right ) \) we need to know something about \(N\left ( s\right ) \). We need to know the frequency content of \(N\left ( s\right ) \) so we can design \(H_{LP}\left ( s\right ) \) to block most of frequency content of \(N\left ( s\right ) \) while allowing all the content of \(R\left ( s\right ) \) to pass through. We assume \(R\left ( s\right ) \) frequency is all in the passband of the low pass filter. This is done in frequency domain. \[ \left \vert H_{LP}\left ( j\omega \right ) \right \vert =\left \vert \frac{1}{\left ( \varepsilon j\omega +1\right ) ^{k}}\right \vert =\frac{1}{\left \vert \varepsilon j\omega +1\right \vert ^{k}}=\frac{1}{\left ( \sqrt{\varepsilon ^{2}\omega ^{2}+1}\right ) ^{k}}=\frac{1}{\left ( \varepsilon ^{2}\omega ^{2}+1\right ) ^{\frac{k}{2}}}\] The plot of \(\left \vert H_{LP}\left ( j\omega \right ) \right \vert \) might now look like this

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We can make the filter closed to desired by boosting \(k\) and decreasing \(\varepsilon \).

Second order systems

We will now start on second order systems. We want to study transient response. So far, we said nothing about transient response. Final value theorem give the steady state response (when it exists, if the system is stable) but not what happens in between. The system could have undesired transient response before getting to the steady state. For example, we could want to send the response to zero very quickly, but this can cause bad transient response.

Why consider only second order systems?

1.
Many physical systems are second order system
2.
Many systems can be well approximated by second order system, using the method of dominant poles.
3.
Math is much simplified when using second order system than higher order

When we design, say RLC circuit we get second order system. Same for mass spring damper. When the system is higher order, we use dominant pole method to approximate the system to second order. But after approximate to second order and doing the analysis on the second order, we should go back and simulate the original higher order system numerically (say using simulink) and compare the second order approximation with the full order system to make sure the approximation used produces close enough results.

Dominant pole method

Imagine 6th order system. We can ignore poles much further away from the imaginary axis, since these indicate modes that attenuate very fast

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In many practical systems, \(\beta \gg \alpha \) and the poles further to the left can be ignored since these are modes which disappear very quickly. So we are left with the two dominant poles \(s_{1,2}=-\alpha \pm j\omega \). Generic second order system is given by \[ G\left ( s\right ) =\frac{\omega _{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\] Where \(\omega _{n}\) is the natural frequency and \(\xi \) is the damping ratio. Consider a unit step input \(R\left ( s\right ) =\frac{1}{s}\). For practical system, \(0<\xi <1\). \begin{align*} Y\left ( s\right ) & =G\left ( s\right ) R\left ( s\right ) \\ & =\frac{\omega _{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\frac{1}{s} \end{align*}

The inverse Laplace transform of the above is\begin{equation} y\left ( t\right ) =1-\frac{e^{-\xi \omega _{n}t}}{\sqrt{1-\xi ^{2}}}\sin \left ( \omega _{d}t+\phi \right ) \tag{1} \end{equation}

To plot the above in Matlab, here is small code

t=0:.1:45;  
z=0.707;  
wn=.2;  
y=@(t,z,wn) 1- exp(-z*wn*t)/sqrt(1-z^2).*sin(wn*sqrt(1-z^2)*t+acos(z));  
plot(t,y(t,z,wn));  

To use Matlab step() command, here is small code

z=0.707;  
wn=.2;  
s=tf('s');  
sys= wn^2/(s^2+2*z*wn*s+wn^2);  
step(sys);

In (1), \(\omega _{d}\) is the damped natural frequency given by \(\omega _{d}=\omega _{n}\sqrt{1-\xi ^{2}}\) and \(\phi =\cos ^{-1}\xi \). (In our textbook, \(\phi \) was defined as \(\phi =\tan ^{-1}\frac{\sqrt{1-\xi ^{2}}}{-\xi }\), but this seems strange to me. I will use the more common definition of \(\sin \phi =\sqrt{1-\xi ^{2}}\) and \(\cos \phi =\xi \) as used by Nise text and other, hence \(\phi =\tan ^{-1}\frac{\sqrt{1-\xi ^{2}}}{\xi }\) from now on).

What if we are not given a standard second order system transfer function such as \(G\left ( s\right ) =\frac{25}{s^{2}+5s+10}\), we can convert this to standard by doing \(G\left ( s\right ) =\frac{\left ( \frac{25}{10}\right ) 10}{s^{2}+5s+10}=2.5\frac{10}{s^{2}+5s+10}\) and now apply the result to \(\frac{10}{s^{2}+5s+10}\) and then scale the output by \(2.5\).

For undamped case, \(\xi =0\), the response is pure harmonics with no damping. The harmonics have \(\omega _{n}\) frequencies. We will now look at poles and zeros of \(G\left ( s\right ) \) in complex domain. Poles of \(G\left ( s\right ) =\frac{\omega _{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\) are\begin{align*} s_{1,2} & =-\xi \omega _{n}\pm \sqrt{\omega _{n}^{2}\left ( \xi ^{2}-1\right ) }\\ & =-\xi \omega _{n}\pm \omega _{n}\sqrt{\xi ^{2}-1} \end{align*}

For underdamped, \(\xi <1\) the above can be written as\begin{align*} s_{1,2} & =-\xi \omega _{n}\pm j\omega _{n}\sqrt{1-\xi ^{2}}\\ & =-\xi \omega _{n}\pm j\omega _{d} \end{align*}

The following diagram shows the main components on the \(s\) space plot

pict

\[ \phi =\tan ^{-1}\frac{\omega _{d}}{\xi \omega _{n}}\]

\(\phi \) is called the damped phase.