Fir filtering operation? Also convolution?

I came across a concept which indicated that fir filtering operation is also convolution As shown in attached photo

In figure caption line,convolution is written in bracket after fir filtering

If yes?then what about IIR filtering operation? Is that also convolution? • FIR and IIR systems are Linear and Time-invariant. For such systems convolution is the operation by which you apply them to your input.
– jojek
Apr 14 '20 at 12:01

Yes in general both IIR and FIR filters can be derived from the convolution summation, but since and IIR by definition has an impulse response that goes to infinity, to implement this directly would require an infinitely long convolution. The FIR can be implemented from the convolution expression directly, but a recursive subset from difference equations is used to implement IIR filters. So convolution still takes place in both cases (the output is the convolution of the input with the impulse response), except in the case of the IIR filter the generalized convolution expression cannot be used directly for implementation.

Given the generalized convolution summation as:

$$y[n] = x[n] \star h[n] = \sum_{k=-\infty}^\infty h[k]x[n-k]$$

$$h[n]$$ is the impulse response of the filter and $$x[n]$$ is our input signal.

Here is a convolution with a Finite Impulse Response (FIR), which is causal (starts at $$k = 0$$, and has a finite impulse response (summation is over $$M$$ samples) As written this is directly realizable as an FIR filter:

$$y[n] = \sum_{k=0}^{M-1} h[k]x[n-k]$$ In comparison here is a convolution with an Inifinite Impulse Response (IIR), which is also causal but the impulse response extends to infinity:

$$y[n] = \sum_{k=0}^{\infty} h[k]x[n-k]$$

This is impossible to create a direct realization, but a subclass of IIR systems can be implemented using recursive difference equations:

$$y[n] = x[n] - \sum_{k=0}^{N} h[k]x[n-k]$$ • Please kindly elaborate your sentence "the key is only the recursive subset of the IIR filters are realizeable" Apr 14 '20 at 12:29
• @Man do you see how the first IIR expresssion that goes to $\infty$ is the convolution with an infinitely long impulse response? Clearly we can't implement it. By feeding the output back recursively as is subsequently done, we can achieve a realization of the filter that will have an impulse response that last infinitely long--- consider a simple accumulator which is feeding back the output and adding it to the next input-- we put in an impulse and the output (response) will stay at one forever. Apr 14 '20 at 12:33

Input - Output (I/O) relationships of systems can be given in a number of ways. The most general form is :

$$y[n] = T\{x[n]\}$$ where the output $$y[n]$$ is described by some formula for the given input $$x[n]$$. This representation includes all sorts of systems; linear, nonlinear, time-invariant, time-varying, causal, noncausal, stable, non stable etc...

A second form of I/O relationship happens specifically for LTI (linear time invariant) systems, known as a convolution sum : $$y[n] = T\{x[n]\} = \sum_{k=-\infty}^{\infty} h[k] x[n-k] = h[n] \star x[n]$$ Where $$h[n]=T\{\delta[n]\}$$ is the impulse response of the LTI system, and when it's finite length, that's called an FIR (finite impulse response) system, or IIR when it's infinite length.

A third way of expressing an I/O relation for a system is given by the difference equation LCCDE form : $$\sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{M} b_k x[n-k]$$ and with suitable initial rest conditions, this representation is also equivalent to an LTI system with convolution output.

Finally, all I/O relationships for systems describe an operation of processing the input and producing an output, which is called as the filtering operation in the most general sense. As it can be seen, for LTI systems, filtering operation is equivalent to convolution operation. This is true for both FIR and IIR systems, however it's computabe only for FIR systems due to practical reasons.