I am trying to understand a step taken in the derivation of convolution as a linear system applied to a signal.

The derivation goes as follows:

  1. A signal $f$ can be decomposed into a sum of impulse functions (aka the sifting property): $f(t) = \int f(\tau)\delta(t - \tau)d\tau$
  2. By way of linearity, the linear operator $L$ applied to signal $f$ is defined as $L\{f(t)\} = L\{\int f(\tau)\delta(t - \tau)d\tau\} = \int f(\tau)L\{\delta(t - \tau)\}d\tau$
  3. Define the impulse response $h$ as $h(t) = L\{\delta(t)\}$
  4. Substitute h into step 2, and out comes the definition of convolution: $L\{f(t)\} = \int f(\tau)h(t - \tau)d\tau = (f \circledast h)(t)$

From past experience, I know there exist impulse responses (aka kernels) whose range contains an arbitrary number of elements (e.g. a gaussian).

I can't reconcile this fact with the definition of $h$ via the dirac function. How is it possible that $h(n) \neq h(m)$ s.t. $m \neq 0$, $n \neq 0$, $m \neq n$? Since $h(t) = L\{\delta(t)\}$, wouldn't $h(m) = L\{\delta(m)\} = L\{0\}$ and $h(n) = L\{\delta(n)\} = L\{0\}$. Thus, both $h(m)$ and $h(n)$ are equal to $L\{0\}$? Based on how impulse functions have been defined, it seems to me like an impulse function should only ever be able to map to at most two unique values: $h(t=0)$ and $h(t\neq0)$

  • $\begingroup$ Leaving aside the usual caveats about the Dirac function not being a function, there is no reason to expect a linear transformation of a function to have the same support as the function. That’s like saying a linear transformation of a vector in a finite vector space has to have exactly the same components nonzero. $\endgroup$ – spaceisdarkgreen Aug 15 '18 at 23:36
  • $\begingroup$ Yes, I guess I've slightly mis-phrased this question. What I mean to ask is how can $L\{\delta(t)\}$ map to more than two unique values since $\delta(t)$ will only be 1 or infinity? $\endgroup$ – bronxbomber92 Aug 16 '18 at 0:51
  • $\begingroup$ How can the unit vector $(1,0,0)$ which has only two unique values $1$ and $0$ map under a $30$ degree rotation in the x-y plane (which is a linear transfomation) to $(1/2, \sqrt 3/2,0)$ which has three unique values? $\endgroup$ – spaceisdarkgreen Aug 16 '18 at 1:00
  • $\begingroup$ I've tried updating the original question to be more clear about what I misunderstand. Sorry for the initial confusion. $\endgroup$ – bronxbomber92 Aug 16 '18 at 1:15
  • $\begingroup$ You can't think of it point-wise. The linear transformation maps the entire function to another function. If you write the components of vectors $v_i,$ then the $t$ is roughly analogous to the $i$. It is an index. The example I just gave is analogous to what you're saying. It's plainly wrong for finite dimensional vectors, so it's not true for infinite-dimensional vectors either. $\endgroup$ – spaceisdarkgreen Aug 16 '18 at 1:20

Summarizing the answer from the comments..

My mistake was thinking that $L\{\delta(t)\}$ meant:

  1. $temp := \delta(t)$
  2. $L\{temp\}$

in other words, perform function application of the direct to obtain a scalar, and then use that scalar to perform functional application of the linear system.

In reality, what is actually meant is that $L$ is linear map from one infinite-dimensional vector to another infinite-dimensional vector, and $t$ picks out a coordinate from the resulting vector that $L$ mapped to.


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