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I'm learning control theory. Block diagram for transfer function:

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is:

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What would be block diagram for these two transfer functions:

$$H(s)=\frac{1}{s^{2}+1}$$ and $$H(s)=s\frac{1}{s+1}$$?

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    \$\begingroup\$ \$\large{H(s)=s\frac{1}{s+1}=\frac{s}{s+1}=\frac{1}{1+1/s}}\$ that is feedforward gain is 1 and feedback gain is 1/s. For \$\large{H(s)=\frac{1}{s^2+1}=\frac{1}{1+s^2}}\$ feedforward gain is 1 and feedback gain is \$s^2\$ \$\endgroup\$
    – K. Rmth
    Commented Dec 24, 2015 at 6:06
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    \$\begingroup\$ Normally, in a primitive block diagram each block would represent an individual physical component or characteristic. As an academic exercise, a number of structures are possible, e.g., in addition to those already given, forward=s; feedback=-1 will do s/(s+1) \$\endgroup\$
    – Chu
    Commented Dec 24, 2015 at 7:37
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    \$\begingroup\$ First example (1/(s²+1)=(1/s²)/(1+1/s²): Two intergrators in series with unity feedback (100%). \$\endgroup\$
    – LvW
    Commented Dec 24, 2015 at 11:11

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first thing to do is divide both numerator and denominator by \$s\$ or \$s^2\$, whatever power of \$s\$ so that no positive powers exist in the denominator. (we want integrators, \$s^{-1}\$, in the blocks, not differentiators.) then your transfer function will appear like

$$ H(s) = \frac{Y(s)}{X(s)} = \frac{b_0 + b_1 s^{-1} + b_2 s^{-2}... + b_N s^{-N}}{a_0 + a_1 s^{-1} + a_2 s^{-2}... + a_N s^{-N}} $$

then divide both numerator and denominator by the constant term in the denominator, \$a_0\$, so that now your transfer function will look like

$$ H(s) = \frac{Y(s)}{X(s)} = \frac{b_0 + b_1 s^{-1} + b_2 s^{-2}... + b_N s^{-N}}{1 + a_1 s^{-1} + a_2 s^{-2}... + a_N s^{-N}} $$

(the other \$a_n\$ and \$b_n\$ are different now than they were before because they all were divided by \$a_0\$.)

now rearrange things:

$$ Y(s) \left(1 + a_1 s^{-1} + a_2 s^{-2}... + a_N s^{-N} \right) = X(s) \left(b_0 + b_1 s^{-1} + b_2 s^{-2}... + b_N s^{-N} \right) $$

...

$$ Y(s) + a_1 s^{-1}Y(s) + a_2 s^{-2}Y(s)... + a_N s^{-N}Y(s) = \\ \quad b_0 X(s) + b_1 s^{-1}X(s) + b_2 s^{-2}X(s)... + b_N s^{-N}X(s) $$

...

$$ Y(s) = b_0 X(s) + (b_1 (s^{-1}X(s)) + b_2 (s^{-2}X(s))... + b_N (s^{-N}X(s)) \\ - a_1 (s^{-1}Y(s)) - a_2 (s^{-2}Y(s))... - a_N (s^{-N}Y(s)) $$

that defines \$Y(s)\$ in terms of \$X(s)\$, integrated versions of \$X(s)\$ and integrated versions of \$Y(s)\$, the latter is feedback from the output.

try drawing a block diagram out of that.

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  • \$\begingroup\$ Any particular reason why we would only want integrators and not differentiators? \$\endgroup\$
    – K. Rmth
    Commented Dec 24, 2015 at 6:29
  • \$\begingroup\$ the reason they always told us when i was in skool was that differentiators are noisy (they're high-pass filters). from a DSP point-of-view, an integrator depends on the input at the present time and in past times. a differentiator, without delay, would have to depend on the input at a time that is a milli-smidgen in the past and another input value that is a milli-smidgen in the future. future values of an input are not realizable in a real-time system. \$\endgroup\$ Commented Dec 24, 2015 at 6:40
  • \$\begingroup\$ but you can build op-amp circuits that are differentiators. i guess they're noisy. \$\endgroup\$ Commented Dec 24, 2015 at 6:44

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