Transfer Function from input to node.

Question

From the IIR filter flow graph below i don't understand how the transfer function is calculated in every node:

• The circles contains 'X' inside are multiplications.
• The circles contains 'Σ' inside are additions.
• The triangles are delayers.

For example for node1, my textbook states that the transfer function is:

For node2:

For node3:

For node4:

I really don't understand how we have calculate this.

Answer 1

OK, I've had a try. You need to write down equations for what happens in the circuit. There is some feedback loop in the circuit which will help you solve the circuit in the end.

Let's call the state at the nodes $n_i$. Then, the state at $n_4$ is clearly the sum of $bx$ and what comes in from node 3 times some delay $z^{-1}$, so

$$n_4=z^{-1} n_3 + bx$$

Similarly, at node 2 we have

$$n_2 = bc_1 x -a_1 n_4$$

and at node 1 we have

$$n_1 = bc_2 x - a_2 n_4 \; .$$

Finally node 3 gives

$$n_3 = n_2 + z^{-1} n_1$$

This gives 4 linear equations in 4 unknowns. Solving them, I got the following for $n_1$

$$n_1=bx\frac{(c_2-a_2)+(c_2a_1-c_1a_2)z^{-1}}{1+a_1z^{-1}+a_2z^{-2}} \; .$$