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I'm studying the space-state representation and I'm not understanding a simple situation that I created. I created the following system, on the frequency domain:

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By writing the time domain representation of the bottom block, we get:

$$3y_1'+5y_1=u_1'+u_1 \,\,\,\,\,\,\,\,\,(1)$$

where y1 and x1 are, respectively, the output and the input of this block. Let's call the input and output of the integrator by x2 and y2.

We can see that:

$$u_1'=-y_2'$$ since u1=constant-y2

Replacing that in (1), we get:

$$3y_1'+5y_1=-y_2'+u_1 $$

But y2'=u2=2y1, so: $$3y_1'+5y_1=-2y_1+u_1 \\y_1'=-\frac{7}{3}y_1+\frac{u_1}{3}$$

But if i replace the bottom block by the frequency domain of this equation:

$$\frac{y_1(s)}{u_1(s)}=\frac{1/3}{s+7/3}$$

we don't get the same result as the first situation (at least it is what my simulation shows). Can someone explain me why? Is that something about initial conditions? I created this simple system because i was trying to do the same thing to a bigger system (where there are some relations between inputs and outputs of subsystems), to manipulate state equations, but I was getting different results, like on this example.

(EDIT) I simulated that last equation using the state representation and using the output as the state variable with initial conditions equal to 1/3 (just by guessing) and i got the same system as the first one. The step function value i used was 1. Does anyone know how i can derive the initial conditions i need for the output of this block so the system would be the same as the first one? By simulations i can see that it is the step value divided by 3, but i cant understand why.

Thanks.

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  • \$\begingroup\$ With an integrator feedback (1/s) , the initial condition ought to be 0 for output and 0 for input. Since a step function contains infinite frequencies (s), the lower forward G(s) block has a ratio of 1s/3s=1/3 and the integrator will ramp till output saturation \$\endgroup\$ – Sunnyskyguy EE75 Apr 2 '18 at 2:05
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Your notation is confusing.

Let \$ \small y(t)\rightarrow Y(s)\$ be the system output signal from the integrator, and let the system input signal be \$\small u(t) \rightarrow U(s)\$. The CLTF is then: $$\small\frac{Y(s)}{U(s)}=\frac{2(s+1)}{3s^2+7s+2}$$

To proceed via controller canonical form (other forms are available - see literature), introduce a dummy variable, \$\small x(t) \rightarrow X(s)\$: $$\small\frac{Y(s)}{U(s)}=\frac{Y(s)}{X(s)}.\frac{X(s)}{U(s)}=\frac{2(s+1)}{3s^2+7s+2}=\frac{\frac{2}{3}s+\frac{2}{3}}{s^2+\frac{7}{3}s+\frac{2}{3}}$$

Now, partion the TF as below:

$$\small\frac{X(s)}{U(s)}=\frac{1}{s^2+\frac{7}{3}s+\frac{2}{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \small \frac{Y(s)}{X(s)}=\frac{2}{3}s+\frac{2}{3}$$

Derive the differential equation from each expression (\$\small sX(s)\rightarrow\dot{x}\:\:etc.)\$:

$$\ddot{x}= u-\small\frac{7}{3}\dot{x}-\small\frac{2}{3}x \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:y=\frac{2}{3}\dot{x}+\small\frac{2}{3}x$$ Finally, express in state space form: $$ \begin{bmatrix} \ddot{x} \\ \dot{x} \end{bmatrix} =\begin{bmatrix} -\frac{7}{3} & -\frac{2}{3} \\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} \dot{x} \\ x \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \end{bmatrix}u $$

$$y =\begin{bmatrix} \frac{2}{3} & \frac{2}{3} \end{bmatrix} \begin{bmatrix} \dot{x} \\ x \end{bmatrix} $$

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