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Could anyone help me with the start, I don't how to begin?

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Hints: -

State what Y equals by naming node values: -

Y = aC - bE where a is the node_value on the input of C and b is the node_value on the input of E. Then you'd likely evaluate a from b and Y via D

So, a = c - DY then, go back to the first equation and get rid of the a term.

Just keep doing this and it should fall-out.

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My approach for solving the problem is as follows:

1.) Redraw the circuit with the goal not having any cross-couplings. That means: We have only simple feedback loops.

2.) Therefore: Shift the input node for E to the right (after the summing junction). As a consequence the function of this block now must be (E+D)

3.) Shift the input of B to the left (before the summing junction). Now, this block must have the function (B-D) instead of B only.

4.) Now we have one simple forward path (E+D) in parallel to C resulting in (C+D+E) and two simple negative feedback loops which can be solved separately (Blacks formula).

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There is more than one way in which you could use these rules to solve the problem. Here is one:

  • Move the input of B to the left, before the summing point.

schematic

simulate this circuit – Schematic created using CircuitLab

  • Move C to the left, before the summing point.

schematic

simulate this circuit

  • Interchange the summing points.

schematic

simulate this circuit

  • Simplifications: blocks in parallel(C - E) and feedback loop in CD.

schematic

simulate this circuit

  • Move (C - E)/(1 + CD) to the left, before the take-off point.

schematic

simulate this circuit

  • Simplification.

schematic

simulate this circuit

  • Simplify the feedback loop and there you have it: $$\frac{AC - AE}{1 + CD + AB + ABDE}$$
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