3
$\begingroup$

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

enter image description here

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.

$\endgroup$
3
$\begingroup$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.