Question 1:What is the reason for a BIBO stable LTI system that all the roots of the characteristic equation have negative real parts?
Y(s)=T(s)*R(s)= N(s)/D(s), where T(s) is closed loop transfer function and R(s) is input to the system. For y(t) to be bounded, all the roots of D(s) should lie in left half of the s plane. It can have roots on imaginary axis (but not multiple root on same point). According to BIBO output it should be bounded for every bounded input. So poles of system transfer function should lie in the left half of the s plane(excluding imaginary axis as bounded input can have poles on imaginary axis). Is the laplace transform table only proof for this?
Question 2:
It is the same thing extended for Polar and Bode also.They both are not complete plots. For an unstable open loop system, they cannot predict the stability of a closed loop system. In books they mention polar and bode can be applied to minimum phase systems, but stable open loop systems may have zeros in the right half of the plane. So in polar and bode open loop transfer function restricted to minimum phase systems or systems only with poles only in the left half of the plane?
In Nyquist it is clear that it is an extension as it comes from Cauchy's argument principle.
For closed loop systems having repeated roots on the imaginary axis is the Nyquist theorem valid?
Please elaborate and correct me if I'm wrong somewhere.
1. Ignore the input for a moment. If the poles of the closed-loop LTI system are in left half plane (LHP), the plant is asymptotically stale. In other words, the plant will go to zero from any nonzero initial conditions.
It is because if it has the factor 1/(s+a) with a>0, its time response, its impulse response (Laplace transform of impulse is 1) is exp(-at). For a couple complex conjugate poles, you get something like exp(-at)sin(bt+fi). Both are converging time functions. If the transfer function (TF) contains a product of many LHP poles, the decomposition gives you a sum of converging time functions.
Then, if the plant without inputs is asymptotically stable, we can show that it is also BIBO.
If you have a bounded input, such as step (for simplicity) which gives you 1/s, it only gives you another factor to the product and, after decomposition, this factor gives you a term which results in a bounded time-function, while all "stable" poles give you vanishing terms.
2. If you know the closed-loop poles, you don't need any plots.
Bode, Nyquist (polar) and Nichols plots allow you to plot the open loop G(s) and "think" about closed loop T(s)=G(s)/(1+G(s)).
Bode is the best and simplest plot exactly then, when the plot is simple, almost straight lines. When gain and phase start jumping up-and down, however, it is almost impossible to get any conclusion about any stability.
Nyquist is always right. However, in real systems, the gain that may go from 0 to 1000 makes it difficult to get all results from one plot.
Nichols plot is like Nyquist, only the dB scaling allows you to get all results from one plot (-40 to 40 dB makes an easier reading). Nichols is less intuitive in the beginning and maybe less taught then the others, yet these days, in MATLAB you just write nichols(sys) and, after playing with it and digesting it for a while, you may reach the conclusion that nichols is the best.
I hope it helps. If not, please feel free to come back with any question, comment or objection.
1st question:
basically a system is modeled as set of differential equations. that means the solutions of these equations give the states of the system. So in LTI systems, when we solve the system we end up to the exponential solutions in which the system's eigenvalues (let's say poles of the system) are as the exponents of the solutions. so now you can see that by having a pole in the left-hand side of the plane (negative eigenvalue) the exponential term of the solution damps and vice versa for positive eigenvalues (poles in right hand side of the plane) the system goes unstable as the exponential term tends to infinity by the time. a system with non-positive eigenvalues (it can have eigenvalue=0) are called stable while systems with negative eigenvalues are called asymptotically stable (or more precisely about LTI systems exponentially stable). I suggest you to read more about internal stability . that gives you more insight about the number of poles on jw axis.
2nd question:
firstly i should say that all of these diagrams are for BIBO stability and each are helpful in different cases. for example nyquist is pretty much helpful in stability problems when we may face some difficulties in bode diagram in cases that we have high overshoots in magnitude diagram. Then it is upto the type of control problems i believe.
Anyway your question is somehow covering a vast area of the control systems and it is very difficult to explain deeply :)
Please elaborate about Bode and Polar. How they ensure BIBO criteria?
Can we Bode and Polar to any LTI system?
i could answer about the basic idea of stable system. we all know that transfer function is just a tool that can explain any LTI system easier than we use an exponential form.
suppose we have a first order LTI system, with a given bounded input, or let can just use : a step input. in exponential form, we could say step response of this system mathematically : Y(t) = (1 - exp(jAt)) U(t).
we have A parameter there. so, if we want a stable system, we should have negative value of A. why? because if A has a positive value, the response of the system at t = infinite, will have an infinite response, which is not suitable with BIBO criteria.
A parameter is related with root characteristic, while practically we always say that A is (-1/tau) in first order system without knowing the stability of the system. and in first order system (1/(tau.s+1)), its root = (-1/tau). so basically, A is the system root characteristic.
Bode diagram is something that developed to explain the system in spectral form (in frequency domain). you should read about a linear system theory.
Answer to first question: Consider a system with pole located at -2 (LHP), the time response (using inverse Laplace Transform is) e^(-2t), this decays to zero at t approaches infinity. However, a pole located at RHP say +2 leads to a time response e^(2t) which blows up with time. In terms of system, it means the output increases as time increases, or in other words is unstable. Therefore, poles on RHP means system is unstable.
BIBO stability criterion implies that a bounded signal (one that has finite energy) should result in bounded output. One can verify that, a system having RHP pole does not result in bounded output and is therefore unstable in BIBO sense as well.
Answer to question 2: As for, Bode and polar plots, they can be easily applied to non-minimum phase or even delay systems. For non-minimum systems, the magnitude plot is unchanged. There is a change only in the phase plot, where a phase shift of -180 (approximate) gets added, thereby bringing the system to the verge of instability. Gain margin is the amount of gain that can be tacked to the system before it gets unstable, and phase margin is the delay that can be added to the system to be at verge of instability. So, non-minimum phase system usually brings the system to the verge of instability. However, it depends on the lead components in the numerator. Hope, this answers your question. You can refer to sections of my book as well regarding stability.
Answer to question 2: Some systems cannot be easily expressed as linear transfer functions, such as the systems which include some delay fenomena. So then, the roots of D(s) give us not enough information about estability. In such cases, Bode and Nyquist plots give us all the information related to analyse the stability. Often, such "real" systems are difficult to model with differential equations and (in many cases) can be modelled only after experimental data, that is to say, Bode and Nyquist plots.
I am not sure if the following was already mentioned in the foregoing contributions:
To verify the relation between time and frequency domain (diff. equation and transfer function) one must realize that the DENOMINATOR of the transfer function is identical to the CHARACTERISTIC POLY NOM (left side of the charact. equation) which belongs to the solution in the time domain.
Please elaborate on for what type of system bode and polar plot are applied. what is reason behind that they cannot be extended to any system.
@Seshadhri Srinivasan you have mentioned bounded signal (one that has finite energy) .
sin(wt) or unit step are bounded signal but energy is infinite.
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