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According to the Commutativity Property of block diagrams, we can place b0 at the entrance or at the exit of the block diagram, it wouldn't matter.

My question is, if we placed it at the entrance, we should have the following difference equation:

$$y[n]=b_0x[n]-a_1y[n-1]$$

But if we placed it at the exit, we should have this difference equation:

$$y[n]=b_0x[n]-a_1b_0y[n-1]$$

Which don't correspond to the same transfer function, can anybody explain to me how it won't matter where to put b0 and that both schematics would lead to the same transfer function.

Block Diagram

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If you apply the multiplication with $b_0$ at the output, then that multiplier is after the feedback loop. Consequently, if the output (after the multiplier) is $y[n]$ then the input to the delay element is $y[n]/b_0$. The corresponding difference equation is

$$y[n]=b_0\left(x[n]-\frac{a_1}{b_0}y[n-1]\right)=b_0x[n]-a_1y[n-1]\tag{1}$$

which is identical to the one with the multiplier at the input, as expected.

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