# How do I interpret this block diagram correctly?

I am reading notes ahead of the class and have encountered this particular slide:

While I completely agree with the first block diagram, I am at a loss trying to understand how the second block diagram is equivalent to the first.

Here's how I interpret the first diagram: x[n] enters H1 and H1 outputs b0 x[n] + b1 x[n-1]. The output of H1 enters H2, and H2 outputs -a1 y[n-1] + b0 x[n] + b1 x[n-1]

Here's how I interpret the second diagram: I am actually lost! I have no idea how to make sense of it.

I understand how the "simpler" diagrams work, for example the multiplication and addition components. But this is Greek to me, and it seems like I cannot find any list of rules to interpret this.

It is better to "parse" these networks from the output back towards the input, calling their input some general $$x$$ and performing substitutions and/or compositions.

So, let's call these networks $$U$$pper and $$L$$ower.

From the upper diagram:

$$UH_2[n] = x[n] + x[n-1] \cdot -a_1$$

and

$$UH_1[n] = x[n] \cdot b_0+x[n-1] \cdot b_1$$

Now, the output of $$U$$pper is given by the composition of the two:

$$UH_y[n] = UH_2[UH_1[n]]$$

This is because, the output of one, becomes the input to the other.

Or...

$$UH_y[n] = UH_1[n] + UH_1[n-1] \cdot -a_1$$

...and if you substitute...

$$UH_y[n] = x[n] \cdot b_0 + x[n-1] \cdot b_1 + (x[n-1] \cdot b_0 + x[n-2] \cdot b_1) \cdot -a_1 \Rightarrow \\ x[n] \cdot b_0 + x[n-1] \cdot b_1 + x[n-1] \cdot b_0 \cdot -a_1 + x[n-2] \cdot b_1 \cdot -a_1$$

Notice here that if you are trying to get to the $$-1$$ of the $$x[n-1]$$, then that would be the $$x[n-2]$$.

And this concludes the $$U$$pper part.

For the $$L$$ower part, we are not going to go through the whole thing, because of two reasons:

1. If you notice, the $$H2, H1$$ are reversed (The $$L$$ower network calls $$H2$$ what the $$U$$pper network calls $$H1$$).

2. We have already mapped $$H2, H1$$.

So, the lower network's response is:

$$LH_y[n] = LH_1[LH_2[n]]$$

Or...

$$LH_y[n] = LH_2[n] \cdot b_0+LH_2[n-1] \cdot b_1$$

And if you substitute:

$$LH_y[n] = (x[n]+ -a_1 \cdot x[n-1]) \cdot b_0 + (x[n-1]+ -a_1 \cdot x[n-2]) \cdot b_1 \Rightarrow \\ x[n] \cdot b_0 + -a_1 \cdot b_0 \cdot x[n-1] + x[n-1] \cdot b_1 + b_1 \cdot -a_1 \cdot x[n-2]$$

Which, after you re-arrange, looks exactly the same.

Hope this helps.