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For example, how can I determine if the convolution of $x(t)$ with $y(t)$ is equal to $0$?

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    $\begingroup$ Circular or linear convolution? And if linear, do you care about the initial/ending transients being zero or not? $\endgroup$ – hotpaw2 Apr 27 '16 at 4:38
  • $\begingroup$ linear convolution. I just need an example to understand under what circumstances I can end with 0 after convolving to signals together. The output should be zero. thanks! $\endgroup$ – Carlos M. Navarro Apr 27 '16 at 4:41
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    $\begingroup$ Consider a rectangular pulse rect$\left(\frac tT\right)$ of duration $T$ and a DC-free periodic signal of period $T$. For any delay, the convolution integral evaluates to $T$ times the average value of the periodic signal, that is, the convolution integral has value $0$. $\endgroup$ – Dilip Sarwate Apr 30 '16 at 14:07
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Time-domain convolution is frequency-domain multiplication. If at all frequencies at least one of the signals is zero-valued in frequency domain, then the convolution of the two signals will be zero-valued at all frequencies, and at all times. Except for a zero signal, no finite-length signal has a continuous run of frequency domain zeros, so your choices are limited to infinite-length signals such as periodic signals with no coinciding non-zero harmonics and ideal lowpass–highpass filter impulse response pairs.

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    $\begingroup$ For periodic signals with no coinciding non-zero harmonics, e.g, $\cos t$ and $\cos \pi t$, there is also the question of whether the convolution integral converges. $\endgroup$ – Dilip Sarwate Apr 30 '16 at 14:11
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For two non-zero sequences $x_n$ and $h_n$ given that their convolution is zero, At least one of them must be infinite length.

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  • $\begingroup$ Very true but some explanation, or a reference to where the result might be found with more details, would be helpful. $\endgroup$ – Dilip Sarwate May 6 at 18:36
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tl; dnr version
The convolution of $x(t)$ and $y(t)$ is zero for all time instants if their Fourier transforms $X(f)$ and $Y(f)$ have the property that for each real number $f, -\infty < f < \infty$, at most one of $X(f)$ and $Y(f)$ is nonzero.

As pointed out in the answer by Olli Niemitalo, convolution in the time domain corresponds to multiplication in the frequency domain. Thus, if the convolution $x \star y$ equals $0$ for all $t, -\infty < t <\infty$, it must be the case that $X(f)Y(f)$ equals $0$ for all $f, -\infty < f < \infty$. So, at least in the classical Fourier transform theory (no Dirac deltas a.k.a. impulses allowed in the time domain or the frequency domain), $x \star y = 0$ corresponds to $X(f)$ and $Y(f)$ having disjoint support, that is, for all choices of $f$, at most one of $X(f)$ and $Y(f)$ can be be nonzero: both being nonzero is prohibited except possibly at isolated points (which can ignored when computing the inverse Fourier transform). So, a simple example of two signals whose convolution has value $0$ everywhere is \begin{array}{ll} X(f) = \operatorname{rect}(f) &\implies x(t) = \operatorname{sinc}(t)\\ Y(f)= \operatorname{rect}(f-1)+\operatorname{rect}(f+1)&\implies y(t) = \operatorname{sinc}(t)\cdot2\cos(2\pi t).\end{array} Note that the support of the signals is the entire real line. And of course, there are infinitely many signal pairs that we can come up with that satisfy the "disjoint support of $X(f)$ and $Y(f)$" property

Is it possible to have two finite-duration signals whose convolution is zero everywhere? Alas, No. The support of the Fourier transform of a finite-duration signal is the entire frequency axis and so the signals fail the "disjoint support of $X(f)$ and $Y(f)$" requirement. OK, how about one finite-duration signal and one infinite-duration signal? Well, that is possible if we allow periodic signals which have Fourier series (and hence impulses in their Fourier transforms). As suggested in my comment on the main question, if $x(t) = \operatorname{rect}\left(\frac tT\right)$ is a finite-duration signal and $y(t)$ is a DC-free periodic signal of period $T$, then \begin{align} x \star y\big |_{t} &= \int_{-\infty}^\infty x(\tau)y(t-\tau) \,\mathrm d\tau\\ &= \int_{-\frac T2}^{\frac T2} y(t-\tau) \,\mathrm d\tau &{\scriptstyle{x(\tau) = 1~\text{iff }\tau~ \in \left(-\frac T2, \frac T2\right)}}\\ &= T\cdot C_0 \end{align} where $C_0$ is the zeroe-th term in the Fourier series for $y(t)$ (a.k.a. the DC value of $y(t)$) which we have assumed to be $0$. Hence, $x\star y$ equals $0$ for all time instants $t$. "But, but, but", you sputter, what about all that crap about nonoverlapping Fourier transforms that you were spouting before?". Well, here \begin{align}X(f)Y(f) &= T\cdot \operatorname{sinc}(Tf)\sum_{n=-\infty}^\infty C_n \delta\left(f-\frac nT\right)\\ &= T \sum_{n=-\infty}^\infty C_n \cdot\operatorname{sinc}(Tf) \delta\left(f-\frac nT\right)\tag{1}\end{align} which sure looks nonzero to everybody, but observe that $\operatorname{sinc}(Tf) = 0$ exactly when $f = \frac nT$ where $n$ is a nonzero integer and $C_0 = 0$ by assumption. Hence, every impulse in that sum in $(1)$ has a coefficient of $0$, and so although $X(f)Y(f)$ seems to be nonzero, it is in fact equivalent to the Fourier Transform of the zero signal. One could make this more explicit if one chooses to use the "fact" that $\delta\left(f-\frac nT\right) = 0$ when $f \neq \frac nT$ to say that for each nonzero real number $f$, either $\operatorname{sinc}(Tf)=0$ or $\delta\left(f-\frac nT\right) = 0$ and so each term in $(1)$ (except the $n=0$ term) has value $0$. And, $C_0 = 0$ by assumption.....

Note that the recent claim by Melih Bağçeli that at least one of the two signals must have infinite duration gives a necessary condition for the convolution to equal $0$ for all $t$ but it is hardly sufficient.

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  • $\begingroup$ How does classical Fourier theory not allow Dirac deltas in the frequency domain if it is based on decomposing signals into sinusoids, i.e., Dirac deltas in the frequency domain? Do you mean the Fourier series, rather than the Fourier transform? $\endgroup$ – Rodrigo de Azevedo May 6 at 7:49
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    $\begingroup$ @RodrigodeAzevedo Sorry, that should have read "classical Fourier transform theory" and not classical Fourier theory. Classical Fourier transform theory deals with signals for which $x(t)$ and $X(f)$ are ordinary integrable functions (so $\int_{-\infty}^\infty |x(t)| dt$ is finite and no Dirac deltas are permitted), and so periodic functions don't qualify. Fourier series are a representation of an arbitrary $x(t)$ over a finite interval and result in sinusoids of period $\frac nT$. If $x(t)$ is indeed periodic with period $T$, this representation holds for all time. $\endgroup$ – Dilip Sarwate May 6 at 13:27
  • $\begingroup$ Thank you for the clarification. $\endgroup$ – Rodrigo de Azevedo May 6 at 16:45

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