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Say we have a function of time ($S(t)$) of the length $T$, and then a customized impulse response (say $I(t)$) of the length $T+N$. The question is, when $S(t)$ is convolved with $I(t$), what are the possible frequency domain (or alternately time domain) representation of all of the possible resulting functions?

For the case when the convolution is performed circularly, the solution appears immediately clear. If $S(t)$ has zero magnitude components in the frequency domain, they will stay zero no matter how the impulse response is configured. Thus, it's impossible to for example, convert a sinusoidal waveform into any non-sinusoidal waveform using circular convolution.

But the result is a lot less clear for the case when linear convolution is performed. If we limit our horizon to the period of $T$, and completely disregard the tail that the impulse response will generate (for cases where $T>1$, and length of the impulse response $ > 1$ ), what will the answer be then? Are there any limits, or can you transform any function into any other function (provided the tail is thrown away), using linear convolution?

(EDIT: for non-casual impulse responses the "pre-echo" is also thrown away, time shifting is allowed also for such cases).

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    $\begingroup$ Welcome to DSP.SE! I've edited your question to make the equations a little clearer, but I was unsure what you meant by "IR > 1". Can you please review and let me know? $\endgroup$ – Peter K. Oct 12 '15 at 11:59
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    $\begingroup$ @PeterK Apologies for confusion. I meant that the length of the impulse response and the signal itself has to be larger than one, for the convolution product to spread in time domain. $\endgroup$ – Dole Oct 12 '15 at 12:08
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If the function is zero-valued throughout then so will be the output. But if you exclude this trivial homogeneous case, then your question can be reformulated as:

Given any function $f$ of compact support and any list $l$ of frequency domain zeros, does a function $g$ always exist that equals $f$ within the support of $f$ and has a Fourier transform that has the zeros given in $l$, with other zeros allowed?

Starting at $g_0 = f$, it would be enough to demonstrate a procedure $p$ such that $g_{i+1} = p(g_i, l_i)$ that adds a single given frequency domain zero $l_i$ to function $g_i$ without modifying it within or to the left of the support of $f$ and without altering the existing frequency domain zeros. We choose a causal filter as that procedure. The procedure being a filter guarantees that existing frequency domain zeros are preserved. To not change the function within or to the left of the support of $f$, the filter must have an impulse response that has a Dirac delta function followed by a zero-valued segment of the same length as the support of $f$. A comb filter consisting of two Dirac delta functions, the second one inverted and located after the zero-valued segment, matches this description. The comb filter can be made to have a zero at any wanted frequency by placing the inverted Dirac delta to an integer multiple of the period of that frequency. Choosing a large enough multiplier moves the inverted Dirac delta to outside the zero-valued segment.

If the functions are discretely sampled sequences rather than functions of a real argument, a Dirac delta function that lands between the sample points gives a sampled sinc function that has no compact support. To circumvent this problem, instead of a sinc impulse, a fractional, unity gain delay at any given frequency can be created by setting to certain values the two samples surrounding a multiple of the period of that frequency. The delay and gain at other frequencies will be off but this matters not.

So (unless I missed something), the answer to your question is: No, there are no limitations if the (allowed to be long enough) tail is thrown away.

Note that successive application of the above procedure does not generate the impulse response of your original question but an output (one of many possible) that enables a realizable impulse response.

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  • $\begingroup$ Thank you. I don't completely understand the terminology used, but will come back to the answer after studying more. Unless no one finds a mistake, it's an interesting result to be sure, meaning that any manipulation can be done via convolution (provided there is some signal) in the digital domain, not mere multiplication of frequencies. $\endgroup$ – Dole Oct 13 '15 at 16:23

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