# Signal Flow Graph and Integration

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I'm trying to convert the basic Mass-spring-damper ODE into a signal flow graph (in it's frequency domain representation) as a function of the Mass position. The problem I'm having is that I need to apply gain to the signal in the time domain (i.e. scale it by the coefficients of the terms in the differential equation). The equation is:

So I need to scale by M, b and k. The integration terms are obviously just 1/s but how do I represent multiplication by a coefficient in the time-domain as multiplication (or addition) in the frequency domain (just using Signal Flow Graphs)?

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Multiplication by a scalar in one domain is equivalent to multiplication by that same scalar in the other. The Laplace transform is a linear operator, which means that, if f is a function and k is a scalar, then L(k f) = k L(f).

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Have we gotten rid of the "EDIT" feature? I would have just added more to my last post, but, if that's still possible, I'm not seeing the button to do it.

Anyway...

I'll define

a = y''
v = y'
x = y.

Knowing only that much, we can draw this signal flow graph:

a --[1/s]--> v --[1/s]--> x

Now, we also know one more thing: The equation that you wrote tells us,

a = (1/m) [ -B v - k x]

which we can represent by adding two more edges feeding back from v and x, weighted by -B/m and -k/m, respectively.

Finally, I should add that there are many equivalent signal flow graphs, all of which represent the same ODE. E.g., you could move that common (1/m) factor somewhere else in the graph -- or even define your nodes to be some other set of variables that contain the same information as a,v,and x. Say, you could use

z1 = a - v
z2 = a + v
z3 = a + v + x

and figure out the relationships between these three.

Each of these different representations is called a "realization," and the different variables that you're using for your nodes are called "state variables." The particular representation that I showed you with a, v, and x is the simplest one to construct from an ODE like the one you wrote, and is called the controllable canonical form.

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I was thinking about that right after I posted, the Laplace transform is a Linear transform and thus should satisfy the Homogeneity property like you stated. I think I'm getting confused because I've heard that "Multiplication" in the frequency domain is equivalent to Convolution in the time domain but now that I think about it it refers to multiplication with another signal rather than just a single scalar. And if you perform the inverse Laplace transform of a single scalar you get an Impulse function which when Convolved with the time signal should produce the scaled time signal?

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Yes, when people say "multiplication in the frequency domain is equivalent to convolution in the time domain," they're talking about multiplying and convolving signals. It goes the other way as well; convolving in the frequency domain is equivalent to multiplication in the time domain.

When you talk about the Laplace (or inverse Laplace) transform of a "single scalar..." I'd think of it this way: The Laplace transform doesn't take scalars as its input. That's not the data type it works on. It works on functions, which it maps to other functions.* So when you say "single scalar," what you really mean is "constant-valued function."** And yes, the constant-valued function is mapped to the Dirac delta by the Laplace transform.

* Well, not quite technically. It works on tempered distributions. Think "functions, but also with stuff like Dirac deltas being well defined."

** If it's not clear, I mean the difference between (to speak C/C++ for a moment), "double x = 7.0;" and "double f(double t){return 7.0;}."

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