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Q1: The term b times x dot is a force as you say. The coefficient b is the viscous friction damping factor. It represents the real loss in the system and is analogous to resistance in an electrical circuit.
Q2: The m is in the denominator because the matrix elements in the state space system equations have the m in the denominator. Hence, this is a natural representation. The x2 dot equation is the second equation in the state space system. Simply apply that equation to the input of the x2 integrator and that's what you get.
Q3: The open loop response does not use the R(s) or the summer, so there is no feedback. Actually, if the output signal is disconnected from the summer R(s)=E(s), so you can also say that B(s)/R(s) is the open loop response if the output is not fed back to the summer. You consider the case where the output is not actually fed back to close the loop. So, G(s)/E(s) is also the open loop response. If you close the loop, then the closed loop response is B(s)/R(s). So, we are dealing with two different block diagrams. The open loop response disconnects the output from the summer and their is no feedback. The closed loop response connects the output to the summer and there is feedback.
Q2b: "How do we get x2(dot) after the summing junction?"
That's not the right way to look at it. x2(dot) should have been there before the summing junction that's where it came from. It's not your fault however, it's the way they seem to present the signal flow technique. The better way to present this information is to state that we would START with two integrators and label both inputs and outputs. That's before we do anything else. The integrators are usually connected together too because we want the following integrator to get it's input from the previous integrator output which makes the equations work well. So we more or less look at the integrators backwards where we know what the output is first, so that means we know the input must be the derivative.
So before we got that complete block diagram we would have started with this:
You can see here that because the OUTPUT of the first integrator is x2, the input must necessarily be x2(dot) which is the derivative. That follows from the fact that the derivative of x2 integrates to the variable itself x2. Ditto for the second integrator, and note that x2 and x1(dot) are now the same thing, and since they are connected in tandem this means that x2(dot) is also x1(dot dot) because if we integrate x1(dot dot) twice we get x1. That's really the heart of the block diagram or signal flow graph.
So the best way is to start with that and then add whatever else is needed.
If we only had one integrator in the system regardless of how complex the rest of the system was, and 'x' was the variable, we would start with:
x(dot) o----[INTEGRATOR]----o x
and then add everything else as needed. We might then end up with a summing junction in front of that [+]:
Code:
In o----[+]---x(dot)---[INTEGRATOR]----x----+----o Out
| |
+-------[-G]-----------------------+
Also for one of the other questions, if you think about how friction acts on a body...it doesnt do anything unless the body is moving. It has no effect unless the body is in motion. Hence the friction coefficient is applied to the velocity which is the first time derivative of the displacement and might be represented as y(dot) for example.
A key point about block diagrams is that (just like state space representations) there is no ONE correct way to write it. You could have used the direct equation, but usually when we go through the trouble of defining the system in terms of the state space method, we prefer to write the block diagram that matches the state space equations straightforwardly.
For the Q3 clarification, I was not able to follow your comment on why you are confused. Let's make it as simple as possible. The open loop system does not connect the output to anything. It just hangs free. The closed loop response is defined to be the case where you take the output and feed it back to the input. If you do this with a negative sign (as we typically do) then it is negative feedback.
Notice that if you don't feed the output back, then R(s) directly connects to E(s), and there is no difference between these signals. If you do feed the output back, then E(s) is the error signal made from the command (or reference) R(s) minus the output B(s).
The open loop system does not connect the output to anything. It just hangs free. The closed loop response is defined to be the case where you take the output and feed it back to the input.
Yes, I also think so. The author is referring to Figure 3-5 for the topic "Open Loop Transfer Function and Feedforward Transfer Function". The figure's title says "Closed-loop system". Perhaps, the author has different view of "open loop transfer function" and "open loop system". Otherwise, it's not an open loop system because they have connected the output to the summer and the signal B(s) is direct result of this feedback. **broken link removed**, I think they are also discussing the same issue. Please have a look. Thank you.
which is a second order system and you wanted to start from that. Well, in a way, that's what you do, just not as direct as you might prefer.
The simplest way to get from the second order system to the block diagram or signal flow graph is to first convert that second order system to a coupled first order system. This is not hard to do really as you follow a general procedure that works with a system of any order 2 and above. For a second order system you go from one equation to two equations, for third order from one equation to three equations, so for an Nth order system you end up with a set of N first order equations from which you can immediately see the required matrixes.
Do a quick search for "reduction to a first order system" or something like that. If you cant find anything we can show an example here later (see below).
For Q3 i am not entirely sure i understand either. If you break the feedback you obviously have a single line flow which no longer has any feedback. But when you do break the feedback you would also set that summer minus input (-) input to zero (the input gets set to zero). That means you get the entire open loop response. But since E then equals R they took the short cut of not bothering with the actual input this time. Bad example if you ask me, as the better way to do it is to set the input removed to zero. If there had been other things ahead of that we would have been forced to set the summer minus input to zero and could not get away with the short cut.
So in the future when you open a loop like that, set the removed input to zero and then analyze the WHOLE circuit not just part of it.
Also, when you said that the op amp somehow had feedback even when we call it open loop, i dont follow this logic because the op amp does have an open loop response which is much different than the closed loop circuit response. For example, many op amps have an open loop DC gain of 100,000 or more. But once we close the loop, we always leave some of that for the proper operation of the op amp and so it cuts down to maybe 10. That's a big difference between open and closed loop response wouldnt you say?
Here's a quick view of converting from second order to a set of first order ODE's:
Perhaps, the author has different view of "open loop transfer function" and "open loop system". Otherwise, it's not an open loop system because they have connected the output to the summer and the signal B(s) is direct result of this feedback.
OK, I think I now understand what is confusing you.
The author is claiming that the transfer function B(s)/E(s) is the open loop transfer function, and he says this in the context of a closed loop system diagram. Don't let the fact that the loop is closed confuse you. His logic is based on what MrAl and I have already explained. Once you open the loop, R(s) is the same as E(s), hence either B(s)/R(s) or B(s)/E(s) is the open loop transfer function if the loop is open.
However, when the loop is closed, B(s)/E(s) is still the open loop response, but B(s)/R(s) is the closed loop response. The two are no longer equal. The open loop response is always embedded in the closed loop system, and it is possible to identify it. Do you see it now?
I would say that those diagrams are different. Certainly, the potential at the red dot and the blue dot is not the same, even in the case where the signals are actual voltages.
The diagram (Fig.1) with B(s) would often be said to have open loop response B(s)/E(s), as opposed to C(s)/E(s). This is useful for talking about feedback theory because the intent is to close the loop through H(s). Hence, H(s) is important and people can say something about what will happen when the loop is closed, once they know the open loop response B(s)/E(s). In particular, the ideas of phase and gain margines are used.
Another person might consider the C(s) to be an output, and would be tempted to call C(s)/E(s) the open loop response. This is OK as long as H(s)=1, but if H(s) is not equal to one, that definition is in conflict with feedback theory terminology. Note that there is nothing wrong with defining things differently. We are always free to do that if we are clear about it. I expect one can find books and people that use differing terminology, so always be careful to look and see how things are defined.
Anyway, when talking about feedback and closing a loop, the open loop response is the response you get by breaking the loop and looking at the round trip response at the point that you break it. Often this is just a conceptual thing, and other times one might actually break the loop in an experiment to characterize the system. Once you know the open loop response, you can predict what will happen if you close the loop. The gain and phase margin that will result tells you a lot, and if you don't like what will happen, you can add gain/attenuation to the loop, or even add additional compensators (such as H(s)) to improve the feedback system.
I would say that those diagrams are different. Certainly, the potential at the red dot and the blue dot is not the same, even in the case where the signals are actual voltages.
I believe according to you Figure 1 is the correct one. If H(s) is a resistor and the loop is broken then I would still claim that the potential is same at both dots. But perhaps H(s) is not a resistor; it might be something similar to op-amp (or, some other blackbox component). Somewhere in my mind from the very beginning I had this concept of same potentials going on and this was what confusing me. Now if H(s) can't be taken to be a resistor then my confusion is solved. So, what do you say? Please let me know. Thank you.
Ahhh, he he, now your question is much more clear. I am surprised i did not pick up on this possibility sooner
The short answer is that they are BOTH open loop responses, but they are obviously not the SAME open loop responses. Thus this means we must make a clear distinction in the signal path in order to define a given open loop response.
But i must point out that this is what the text actually did. They defined the open loop response path for you. But your curiosity again got the better of you and you decided to try to define it yourself. This is both good and bad: good because it brought out other points of interest, but bad because you could not follow the text without question. But heck being new at this the question is certainly a reasonable one i must also say.
So now we see where your comparison to the op amp came into play. You already knew the op amp open loop output is just the upper path (usually the gain is called G(s)) and the lower path (usually the gain is called H(s)) is just not there yet (we havent created a circuit yet so it cant be there yet). When we 'close' the path with the op amp we are actually adding parts to the circuit and thus we then create the G(s) part so we can have two different "open loop" responses. However, in control theory the open loop response is often measured for the feedforward and feedback paths as this problem illustrates. That's so that you can estimate the closed loop response. So it really boils down to what you are doing that matters. Note we could also have a gain block AFTER the pickoff point where the output is then taken from the output of a gain block that is NOT included in the feedback path. What shall we call the open loop then. So keep an eye out for what the application (or the text) requires.
As mentioned, the gain in the upper part is usually identified as G(s) or just G while the feedback path is usually identified as H(s) or just H. The entire upper part (no matter how complex) is called the feedforward path, and the feedback path is called the feedback path (chuckle) although there could be several feedback paths we'd have to call feedback path 1, feedback path 2, etc. We could however reduce this to a single feedback path and even a single feedforward path. If we had multiple feedback paths though we'd have to consider which ones to break and which ones not to break. So you can see this gets a little application specific, although for your text they defined this already so you have to go with that until they move on to other circuits.
Using the above nomenclature, the transfer function of the single loop negative feedback system is then:
R/C=G/(1+G*H)
In a flow graph however we dont use summers that are actually subtractors, we just use summers, although we might still write:
R/C=G/(1+G*H)
because then we assume that H is negative. If we dont assume that H is negative, then we write:
R/C=G/(1-G*H)
where G and H are both positive. The only difference really is that the gain block H which is positive with a subtractor in the block diagram becomes simply a negative gain with a pure summer. In any case, either G*H or -G*H must come out positive in a negative feedback system.
This is much simpler looking at the attached diagram.
I believe according to you Figure 1 is the correct one. If H(s) is a resistor and the loop is broken then I would still claim that the potential is same at both dots. But perhaps H(s) is not a resistor; it might be something similar to op-amp (or, some other blackbox component). Somewhere in my mind from the very beginning I had this concept of same potentials going on and this was what confusing me. Now if H(s) can't be taken to be a resistor then my confusion is solved. So, what do you say? Please let me know. Thank you.
OK, the idea of potentials and resistors or other components is completely the wrong view. These block diagrams are system level block diagrams and the transfer functions (like H(s) , G(s) etc.) are Laplace domain transfer functions. Unless H(s)=1, B(s) can not be the same as C(s). Clearly, B(s)=C(s)H(s) because Laplace domain trasfer functions multiply when the are concatenated in series as shown. Remember, mulitplication in the s-domain is convolution in the time domain.
You just need to get used to all of this new theory. As I mentioned to you before, there is a lot of details that must be digested. In the end, this will all seem trivially simple to you, once you master it and use it in practice.
Could you please help me with this query? Thanks a lot.
Regards
PG
PS: I'm sorry if someone is in process of replying. I have solved it. It was not difficult; I'm really time pressed that's why it wasn't clear to me. Anyway, thank you for giving it a look.
Q1: Of course, those matrix elements are taken into account, but it's hard to see them. Elements of zero are not shown because including it would be silly. You would have a branch that comes from a state or input and it would be multiplied by zero. Since it does not do anything to add something that is multiplied by zero, why put it in the diagram? The matrix elements with 1 are definitely in the diagram, but we don't need to put a box with a 1 in it because multiplying by 1 does not change the value. So, there is a branch, but no box when you have a 1, and there is no branch and no box when you have a zero.
Q2: Yes you can do that and often that is the preferred way. I generally like to have all summing nodes add branches only (no subtracting), and I keep the negative signs in the boxes.
Q1: Of course, those matrix elements are taken into account, but it's hard to see them. Elements of zero are not shown because including it would be silly. You would have a branch that comes from a state or input and it would be multiplied by zero. Since it does not do anything to add something that is multiplied by zero, why put it in the diagram? The matrix elements with 1 are definitely in the diagram, but we don't need to put a box with a 1 in it because multiplying by 1 does not change the value. So, there is a branch, but no box when you have a 1, and there is no branch and no box when you have a zero.
No. There are two problems here. A big problem makes the diagram incorrect, and the small problem is not representing the matrix elements directly as your question relates.
The big problem is that the diagram is not mathematically equivalent because you added an extra branch. It would be helpful for you to add the variable x1_dot to the diagram. Then the meaning of the first state equation will be clearer. The matrix element A12=1, is already shown as the branch that goes from x2 to x1_dot.
The small problem is that when you added the 1-block at the output, the feedback branch that goes back should come from the input side of the 1-block. The ouptut should be labeled y and the input should be labeled x1.
Q1: While the order of matrices is important when multiplying, the actual order that you choose to multiply those matrices does not matter. If it did matter, we would be required to use parentheses to make it clear. But, we dont' need to do that.
To be more clear, ABC is different than BCA, CAB, ACB, CBA and BAC. However, (AB)C=A(BC).
Q2: You are using a different C matrix than the book. The book shows C=[1 0], while you are using C=[0 1].
No. There are two problems here. A big problem makes the diagram incorrect, and the small problem is not representing the matrix elements directly as your question relates.
The big problem is that the diagram is not mathematically equivalent because you added an extra branch. It would be helpful for you to add the variable x1_dot to the diagram. Then the meaning of the first state equation will be clearer. The matrix element A12=1, is already shown as the branch that goes from x2 to x1_dot.
The small problem is that when you added the 1-block at the output, the feedback branch that goes back should come from the input side of the 1-block. The ouptut should be labeled y and the input should be labeled x1.
Shouldn't I add that branch? I added it because all the element of A should be fed to the summing point just before the integral box. Where do I have it wrong? Could you please tell me? Thanks.
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