Fir filtering operation? Also convolution?
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:

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