signal flow graphs etc.

[PG1995]

Hi

This shows how to transform a state space model of a system is transformed into a state space block diagram. If we are given a state space model, how do we transform it into signal flow graph? My guess is that we simply translate the state space block diagram into equivalent signal flow graph because there is one-to-one correspondence between a block diagram and signal flow graph. Please let me know. Thanks.

Regards
PG

 

[steveB]

Hi

This shows how to transform a state space model of a system is transformed into a state space block diagram. If we are given a state space model, how do we transform it into signal flow graph? My guess is that we simply translate the state space block diagram into equivalent signal flow graph because there is one-to-one correspondence between a block diagram and signal flow graph. Please let me know. Thanks.

Regards
PG

This is an interesting question, and the answer might take some discussion to figure out for sure. I'll first give some information that I feel confident about, and then I'll give an opinion that will either require someone else to verify, or if no one else knows the answer, I may need to derive a proof (or disproof) myself.

First, I would agree that there is a one to one correspondence between signal flow graphs SFGs and block diagrams BDs, provided the system is linear. In that case, it is relatively trivial to go from one to the other. Note that integrators in a BD become transfer functions of "1/s" in a SFG, or in a digital system sample delays in a BD , become transfer functions of "1/z" in a SFG.

However, BDs are more general than SFGs because they can represent either the time domain or the frequency domain mathematics. They can also represent nonlinear systems as well, because signals can be multiplied and divided and blocks can have nonlinear operations on the signals. For nonlinear systems, the BD would only be a time domain representation.

Now the diagram you provided has an added complication in that the signals are vectors and the blocks are time domain matrix operators. Now, one can take this full system and generate a SFG using the individual signals and individual frequency domain transfer functions, but this will get very very messy, and applying Mason's gain formula MGF might become impractical if the vectors and matrices are of large dimensions, and if the matrices are not diagonal or with many zero elements in them.

However, one can write a vector based SFG without a problem and this will look just as neat as the vector based BD. But, now comes the tricky part. And, I would ask anyone who knows if the following can be done; can we now derive and use a matrix form of Mason's Gain Formula? I would guess the form might be as follows.

[latex]\frac{y_j}{u_i}=(\Delta)^{-1}\cdot \sum_k \Delta_k \cdot T_k [/latex]

or maybe it's

[latex]\frac{y_j}{u_i}=(\Delta)^{-1}\cdot \sum_k T_k \cdot \Delta_k[/latex]

or maybe something more complicated is needed.

Notice that instead of dividing by the determinant Δ as in MGF, I wrote it as multiplication of the inverse of the determinant in front, because we are dealing with matrices, not scalars now. Also, the number 1 in the determinant becomes the identity matrix.

If I apply either of the two formulas to the vector-based SFG converted from the state-space block diagram you showed, I get the correct answer for the transfer function. Now this is not a proof that one of these formulas is correct because more complicated systems might fail to obey this formula, so I would ask anyone who knows to chime in here. I'm going to do my own search, and failing that might try to derive the formula. I have a personal interest in answering this question, and it might be interesting to others too.

EDIT: Also note that state space systems can be put into a number of different canonical forms, and most of these lend themselves to simple and standard scalar SFG representations. Many books on control systems actually diagram these out explicitly.

 

[MrAl]

Hi,

The system can get pretty complicated because not all the gains are where we want them to be for every system we might run across. For example, sometimes we have gains between integrators and that prevents a one to one translation as you were expecting. You got that idea because that block diagram is already simplified to a general form. But if it is not like that, then you need a more general approach.

There's a general approach where we start with the transfer function and work it out through algebra. The transfer function would come from Mason's for example or else move the gain blocks, pickoff points, etc., until you get a simpler form. Having the transfer function, we then proceed to turn this into a bunch of interconnected integrators.

Vout=Vin*(A*s+B)/(C*s^2+D*s+E)

Multiply out the denom:
Vout*(C*s^2+D*s+E)=Vin*(A*s+B)

Divide by the highest power of s:
Vout*(C*s^2+D*s+E)/s^2=Vin*(A*s+B)/s^2

Expand:
(Vout*E)/s^2+(Vout*D)/s+Vout*C=(Vin*B)/s^2+(Vin*A)/s

Subtract all but the constant term with Vout:
Vout*C=-(Vout*E)/s^2-(Vout*D)/s+(Vin*B)/s^2+(Vin*A)/s

Divide by the constant:
Vout=-(Vout*E/C)/s^2-(Vout*D/C)/s+(Vin*B/C)/s^2+(Vin*A/C)/s


[LATEX]Vout=-\frac{Vout\,E/C}{{s}^{2}}-\frac{Vout\,D/C}{s}+\frac{Vin\,B/C}{{s}^{2}}+\frac{Vin\,A/C}{s}[/LATEX]

[LATEX]Vout=-\frac{Vout\,\frac{E}{C}}{{s}^{2}}-\frac{Vout\,\frac{D}{C}}{s}+\frac{Vin\,\frac{B}{C}}{{s}^{2}}+\frac{Vin\,\frac{A}{C}}{s}[/LATEX]

And now it is in a form which is convenient to turn into a signal flow graph. The forward terms have Vin in them, and the feedback terms have Vout in them.

We can start to draw this now by starting with the number of integrators equal to the highest power of s, which is 2 here:

Vin o-----Int---->---o------Int--->----o Vout

and looking at the transformed equation above it should be easy to draw the rest of the lines and gains. But one thing to note is that we always start with the integrators connected directly together with one output connected to the next input. And since every Int is equal to 1/s, that means a signal that goes through both Int's gets multiplied by 1/s^2.

I posted a good example here several years back.