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

## 3 Answers

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.

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).

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. simulate this circuit – Schematic created using CircuitLab

• Move C to the left, before the summing point. simulate this circuit

• Interchange the summing points. simulate this circuit

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

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

• Simplification. simulate this circuit

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