block diagrams etc.

PG1995 said:

Thank you very much.



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.

Regards
PG

Click to expand...

I have to assume that you don't yet understand how to make block diagrams from the state-space equations. I think MrAl did a good job of explaining this above.

Basically, you should start with an integrator for each state. Then, label the inputs to the integrators as x1_dot, x2_dot etc. and the outputs as x1, x2 etc. Now apply each state equation to the x1_dot, x2_dot etc. In this way you will fully define all states in the block diagram. Then you can generate all outputs using the output equations. The inputs will automatically be generated in the block diagram as you need them.

So, in this specific place, you have it wrong because you are trying to apply the x1 state equation element A12 to the x2_dot input in the block diagram.

Thank you for the help, and especially for the patience.

I have tried to state how I see it. I know that I don't yet understand it conceptually. If I'm still quite wrong then just leave it. Perhaps, it will make sense in some days. Thanks.

Regards
PG

Hello there,

I agree with Steve in that you are making a tiny mistake here. I think what is happening is you are not correlating the proper derivatives to the feedback but are insisting that all the feedbacks must get fed back to the same inputs. This is not how it works but rather each input has it's own feedback. Let me illustrate.

There are two state vars and two derivatives, not just one derivative:

x1'= upper A x
x2'= lower A x

The elements of A that get fed back to x2' are shown in your diagram, but you have the elements of A that should get fed back to x1' currently shown as fed back to x2'. All you need to do is recognize that the elements of A that get fed back to x1' are on the top row and the ones that get fed back to x2' are on the bottom row.

This means that the only thing that gets 'fed' back to x1' is x2 because the coeff of x1 is zero. And since x2 already connects to x1' we're done so there is nothing to add for that part.

So you are right about the lower row of A feeding back to x2', but the *upper* row of A feeds back to x1' not to x2'. If there were more rows, the top row feeds back to x1', the second row feeds back to x2', the third row to x3', and so on.

In all of the above x1' refers to x1(dot) while x1 refers to the variable itself.


Lastly, the output equation for y is separate from the state equation with Ax and Bu. This means we can think of the equation for y being totally independent of the equation for the x'. The most important part is the x'=Ax+Bu part, and that solves for all the state variables. The secondary part is the y=Cx+Du part, and that solves for the outputs. So the output part with y will always tap off of places within the block diagram of x'=Ax+Bu.
So we can illustrate this as:
In o----[x'=Ax+Bu]---->----[y=Cx+Du]---->---o Out

with the line going from the first equation to the second equation representing any lines that need to be connected. So really we are solving for the x' part first and then the y= part. But there is never any feedback from the y= part. It's a one way street from state equation to output equation, and the block diagram will always reflect this by only having lines leading from state variables or other points within that part to the output part.

Just to note, sometimes for brevity we dont even consider the y= part but just the x'=Ax+Bu part, assuming that any outputs are simply the state variables. When studying this stuff however they almost always include the output equation for completeness.

If you have time you can look up Mason's Signal Flow Graph Gain Formula which is where they would have gotten the formula for the block diagram.

PG1995 said:

I know that I don't yet understand it conceptually. If I'm still quite wrong then just leave it. Perhaps, it will make sense in some days. Thanks.

Click to expand...

Yes, you are somehow missing a key concept. However, this is not something you can leave and make sense of someday. You need to get it right now. So, let's just take it step by step until you get it. If MrAl's explanation doesnt make it clear, keep asking questions. I'm sure you will get it soon.


Unfortunately, this is one of those questions that would be much easier to answer at a blackboard with one-on-one discussion. This format makes it more difficult to explain, but, no problem, you'll get it soon.

Thank you very much, MrAl, Steve.

All you need to do is recognize that the elements of A that get fed back to x1' are on the top row and the ones that get fed back to x2' are on the bottom row.

Click to expand...

Okay. I get it now. But I don't say I understand it completely. I think the diagram is misleading for a beginner. It shows that all the elements of matrix A are fed back to the same point. Thanks.

Regards
PG

Hello,

Well we have to keep track of the x(dot)'s too as this diagram shows...

Also note that a gain of 1 can be represented with just a line with no gain block drawn with it.

Also notice that "1 x1" (which is 1*x1) is in the line with x1(dot) not in the line with x2(dot), so it gets connected to x1(dot) not x2(dot).

Yes that block diagram in your last post is a rough outline of a general block diagram We could draw a better one
But keep in mind that when only one x and one x(dot) is shown it sometimes means they are showing the vector x and vector x(dot) which represents ALL of the state variables and ALL of their derivatives, not just one of each.

PG1995 said:

I think the diagram is misleading for a beginner.

Click to expand...

That diagram is misleading because it is a vector based diagram. The problem is that a general diagram for the general case is very messy and confusing (unless you only have one input and one output) if you show the individual signals rather than the vectors only. However, you can look at special cases of state-space representations. For example, the controlable-cononical form and observable-cononical form are two examples that lead to neat and understandable block diagrams.

Think of the vector form of block diagram as nice compact representations that get the general concept across. When you get into more advanced control designs, you will find the need to think in terms of vectors and matrices, rather than signals and constants only. Often, you will form matrices and vectors from other matrices and vectors simply because the details, at the signal level, are too cumbersome. However, this takes much experience and is not critical when you are first learning.

Thank you, MrAl.

But do you agree that the diagram is misleading? Just curious to know your or Steve's opinion. Thanks.

Regards
PG

PS: I see Steve has already given his opinion. Thanks

Hi again,


Well it is a little, but then again the variables are vectors so it does make a little sense. I think it should be taken as a rough diagram and maybe we can look at a full diagram next. Even a 3rd order system takes quite a bit of work to draw in full however, and how many inputs do we show.

Here's a lame attempt at drawing such a monster...

Thanks a lot for the opinions, Steve, MrAl.

@MrAl: Thank you for the drawing and it's not that lame, in fact quite pretty!

Best wishes
PG

Hi,

Did the drawings in post #26 help?

MrAl said:

Hi,

Did the drawings in post #26 help?

Click to expand...

Sorry for the late reply, MrAl.

It did help but I believe it would take me some more time and practice to understand this stuff. Thank you.

Best regards
PG

Hi

Could you please help me with this query? Thank you.

Regards
PG

Hi,


You could use Mason's Flow Graph Gain Formula as i mentioned somewhere else here.

Since the block diagram is algebraic you can also use algebra which i'll show here...

Since the diagram is not readable, i will assign gains to each block from the top block down to the lower block. The top block we'll call A, the middle block B, and the lower block we'll leave as H. I'll also call the input R and output C. We want to solve for C.

First we have the upper feedforward section:
Cupper=R*A (and that should be obvious following the upper path from input to output)

The output of the first summer sums R and H, but H is multiplied by C, so we have for the output of the first summer:
S=R-H*C (and this should be somewhat obvious too)

Notice by doing this last step we've eliminated the lower feedback path implicitly, so now all we have left is the upper forward path where C=R*A and the middle path with modified feedback which because it's just a gain now with input S and output C we have:
Cmidd=S*B=(R-H*C)*B

Now the output sums both C's so we have:
C=Cupper+Cmidd
or
C=R*A+(R-H*C)*B

Solving this explicitly for C we get:
C=((B+A)*R)/(B*H+1)

which is the expression for this diagram.

BTW, if we factor that and expand it we get:
C=(B*R)/(B*H+1)+(A*R)/(B*H+1)

where we can see the output is just the sum of the two forward loops working independently with the feedback and mid gain block.

Take note of the algebraic structure of the block diagram...it's basically just multiplication and addition and subtraction and sometimes division.

If there is still any question, just yell

PG1995 said:

Hi

Could you please help me with this query? Thank you.

Regards
PG

Click to expand...

I'll reiterate what MrAl said about signal flow graphs and Mason's gain rule. This is the best way I know of to deal with these calculations/derivations quickly. This also allows a nice pictorial view of linear system feedback loops and gives good insight into a linear system. Personally, I like to use block diagrams for nonlinear systems and signal flowgraphs for linear systems. However, it seems that more people are familiar with block diagrams, and very few engineers master the flowgraph approach nowadays.

I've attached another way to approach this problem that might be more in line with what you originally asked. I don't actually recommend this approach because, in general, it is tedious. It's better to have a general method that gives answers quickly. Still, seeing this way is instructive and perhaps illustrates why you would not want to go through this process with a more complicated system.

Hello again,


Steve, PG:

Here is my take on Steve's drawing. The tiny newly created feedback loop is easily eliminated and thus renders the graph into one with only one feedback path. It should be easily evaluated after that.

Thank you, MrAl, Steve.

Q1: In this figure the key step was taken in 2nd block diagram from the top where the pickoff point was moved toward the right of summing junction and as a result of it a new feedback block "HGfG1Gp" was added. Could you please tell me exact rule for moving a pickoff point toward the right or left of a summing junction? Like I know that what to do when moving a block toward right or left of a summing junction etc.

Q2: Could you please also help me with this query? Thanks a lot.

Q3: It would be really kind of you if you can help me with this query too?

Regards
PG

MrAl said:

Hello again,


Steve, PG:

Here is my take on Steve's drawing. The tiny newly created feedback loop is easily eliminated and thus renders the graph into one with only one feedback path. It should be easily evaluated after that.

Click to expand...

That's definitely a nice way to go, and in a more general case would be preferable too. Reducing block diagrams or signal flowgraphs in steps usually allows many different options. In my case I was trying to cater to the specific comment from PG that he wanted to eliminate the summing node S2, so I thought that progression of steps did that in a clear way.

The problem with both of these approaches is that it requires a number of steps and then requires additional algebra to get it in the proper form. But, as anyone who has mastered signal flowgraphs and Mason's gain rule knows, this can all be done very quickly pretty much in one step, and the result is in the preferred form immediately.

In this case, the flowgraph does not need to be drawn because the block diagram is simple. One sees two transmission paths (GC G1 Gp and Gf G1 GP) and the feedback loop is clearly -H Gc G1 Gp, which gives determinant 1+H Gc G1 Gp. Hence, in 3 seconds one sees the answer as follows, with almost no work at all.

T(s)=((Gc +Gf) G1 Gp)/(1+H Gc G1 Gp)

Surprising, many experienced engineers I know would take quite a bit of time working this out, but that's because they did not commit themselves to mastery of a very useful tool.

PG1995 said:

Thank you, MrAl, Steve.

Q1: In this figure the key step was taken in 2nd block diagram from the top where the pickoff point was moved toward the right of summing junction and as a result of it a new feedback block "HGfG1Gp" was added. Could you please tell me exact rule for moving a pickoff point toward the right or left of a summing junction? Like I know that what to do when moving a block toward right or left of a summing junction etc.

Q2: Could you please also help me with this query? Thanks a lot.

Q3: It would be really kind of you if you can help me with this query too?

Regards
PG

Click to expand...

Q1: For block diagrams, I don't use formal rules. I'm sure they are tabulated somewhere, but I don't have them handy. I've attached some PDFs of my signal flowgraph notes from decades ago. Flowgraphs provide a better viewpoint for applying rules, and the formal rules and methods are contained here.

I generally don't need rules, and eventually you wont either. You can just use common sense and visualization. In this case, when I moved the "pickoff" to the right, an extra unwanted term was added due to the feedback path. So, I simply subtracted this contribution off with another feedback at the output.

Q2: This is just algebra. Rearrange the equation using normal algebraic rules.

Q3: What you wrote is the same as what the book wrote. They just changed the order of terms, but it is the same thing.

Thank you.

Yes, you are right that it's simple algebra for Q2 but for some reason my simple algebra isn't working! If possible, please help me with it. Thanks.

Regards
PG