Recursive implementation of an FIR Filter

[Master1022]

Homework Statement

An FIR filter can also be implemented using a recursive structure. Can you find such an example?

Relevant Equations

FIR Filter
Recursive Structure

Hi,

I have a question related to filter structures. Question: An FIR filter can also be implemented using a recursive structure. Can you find such an example?

I don't really know how to proceed here.

Method:
FIR filter is a 'finite impulse response', which means that the response doesn't last forever and is usually non-recursive. A recursive structure, however, has dependence on the output as follows:
$$ y[k] = \sum_{j=0}^{N} a_j x[k - j] + \sum_{j = 1}^{M} b_j y[k - j] $$

For this to be a representation of an FIR filter, do we just want $b_j < 1$ for all $b_j$? Is that correct? This makes sense intuitively to me as this will cause the output signal to decay over time.

Thanks in advance.

 

[collinsmark]

Homework Statement:: An FIR filter can also be implemented using a recursive structure. Can you find such an example?
Relevant Equations:: FIR Filter
Recursive Structure

Hi,

I have a question related to filter structures. Question: An FIR filter can also be implemented using a recursive structure. Can you find such an example?

I don't really know how to proceed here.

Method:
FIR filter is a 'finite impulse response', which means that the response doesn't last forever and is usually non-recursive. A recursive structure, however, has dependence on the output as follows:
$$ y[k] = \sum_{j=0}^{N} a_j x[k - j] + \sum_{j = 1}^{M} b_j y[k - j] $$

For this to be a representation of an FIR filter, do we just want $b_j < 1$ for all $b_j$? Is that correct? This makes sense intuitively to me as this will cause the output signal to decay over time.

Thanks in advance.

Simply keeping $b_j < 1$ isn't enough.

As $k$ becomes larger than the larger of $M$ and $N$, $y[k]$ needs to equal zero, not just approach it. The finite part of "FIR" means that when the input is an impulse, the output must be zero after some finite $k$, not requiring that $k \rightarrow \infty$.

Fortunately, you can contrive such a filter.

Of course, you can always pick your $b_j = 0$, for all $j$. But I'm guessing you won't get full points for that. That's kind of the trivial case (sort of cheating). So let's contrive a filter such that not all of the $b_j$s are zero.

You know that $x[k]$ is an impulse. You can and should use this to your advantage.

$x[k] = 1, 0, 0, 0, \cdots$

Now pick something simple for your $a_j$s. This is going to be your choice of $a_j$s, since this filter is your own personal creation. But keep it pretty simple. I suggest making $a_0$ and $a_1$ nonzero, and that's it. (That should make a simple example, but not too simple).

Now step through $k$, one at a time (starting from $k = 0$), and choose your $b_j$ such that $y[k] = 0$ after some finite $k$. If you want to keep things really simple, you can start forcing $y[k] = 0$ as soon as $k = 1$. It's up to you.

Do this by choosing your $b_j$s one $j$ at a time; one $b_j$ for each iteration.
For example, suppose you want to start forcing $y[k] = 0$ as soon as $k = 1$. In that case, fist assume that the filter's initial output is zero. So you can start with

$$y[0] = \sum_{j = 0}^N a_j x[0 - j]$$

(Don't forget that $x[k] = 1, 0, 0, 0, \cdots$, it will make things a lot simpler.)
Now, in this example (where we're forcing $y[k] = 0$ as soon as $k = 1$), you can start forcing your $y[k] = 0$. Choose your $b_1$ such that

$$y[1] = \sum_{j = 0}^N a_j x[1 - j] + b_1 y[0] = 0$$

Now that you've chosen your $b_1$ you can move on to $b_2$.

$$y[2] = \sum_{j = 0}^N a_j x[2 - j] + b_2 y[0] + b_1 y[1] = 0$$

Then keep going with $b_3, b_4, \cdots$ and so on.

You can simplify the above equations significantly by realizing that $x[k] = 1, 0, 0, 0, \cdots$. I'll let you do that.

Whatever the case, since you've forced $y[k] = 0$ for some finite number of $k$s, then eventually, all the relevant inputs and outputs will be equal to zero, and the output will remain equal to zero forever after. At that point, you can stop.

[Edit: Corrected some indexing mistakes with the $b_j$ coefficients.]

 

[Master1022]

Simply keeping $b_j < 1$ isn't enough.

As $k$ becomes larger than the larger of $M$ and $N$, $y[k]$ needs to equal zero, not just approach it. The finite part of "FIR" means that when the input is an impulse, the output must be zero after some finite $k$, not requiring that $k \rightarrow \infty$.

Fortunately, you can contrive such a filter.

Of course, you can always pick your $b_j = 0$, for all $j$. But I'm guessing you won't get full points for that. That's kind of the trivial case (sort of cheating). So let's contrive a filter such that not all of the $b_j$s are zero.

You know that $x[k]$ is an impulse. You can and should use this to your advantage.

$x[k] = 1, 0, 0, 0, \cdots$

Now pick something simple for your $a_j$s. This is going to be your choice of $a_j$s, since this filter is your own personal creation. But keep it pretty simple. I suggest making $a_0$ and $a_1$ nonzero, and that's it. (That should make a simple example, but not too simple).

Now step through $k$, one at a time (starting from $k = 0$), and choose your $b_j$ such that $y[k] = 0$ after some finite $k$. If you want to keep things really simple, you can start forcing $y[k] = 0$ as soon as $k = 1$. It's up to you.

Do this by choosing your $b_j$s one $j$ at a time; one $b_j$ for each iteration.
For example, suppose you want to start forcing $y[k] = 0$ as soon as $k = 1$. In that case, fist assume that the filter's initial output is zero. So you can start with

$$y[0] = \sum_{j = 0}^N a_j x[0 - j]$$

(Don't forget that $x[k] = 1, 0, 0, 0, \cdots$, it will make things a lot simpler.)
Now, in this example (where we're forcing $y[k] = 0$ as soon as $k = 1$), you can start forcing your $y[k] = 0$. Choose your $b_1$ such that

$$y[1] = \sum_{j = 0}^N a_j x[1 - j] + b_1 y[0] = 0$$

Now that you've chosen your $b_1$ you can move on to $b_2$.

$$y[2] = \sum_{j = 0}^N a_j x[2 - j] + b_2 y[0] + b_1 y[1] = 0$$

Then keep going with $b_3, b_4, \cdots$ and so on.

You can simplify the above equations significantly by realizing that $x[k] = 1, 0, 0, 0, \cdots$. I'll let you do that.

Whatever the case, since you've forced $y[k] = 0$ for some finite number of $k$s, then eventually, all the relevant inputs and outputs will be equal to zero, and the output will remain equal to zero forever after. At that point, you can stop.

[Edit: Corrected some indexing mistakes with the $b_j$ coefficients.]

Thank you very much for the reply - this makes much more sense!