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$$x(t)=\int_{-\infty}^{\infty}x(\tau)\delta(\tau-t)d\tau=\int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau$$ Due to linearity, $$y(t)=T\{x(t)\}=\int_{-\infty}^{\infty}x(\tau)T\{\delta(\tau-t)\}d\tau=\int_{-\infty}^{\infty}x(\tau)T\{\delta(t-\tau)\}d\tau$$ considering the impulse response of system to be h(t),due to time invariance of unit impulse signal , $$y(t)=\int_{-\infty}^{\infty}x(\tau)h(\tau-t)d\tau=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$$ here, i know i have done something wrong, because
$y(t)=\int_{-\infty}^{\infty}x(\tau)h(\tau-t)d\tau$ is the formula for cross-correlation between x(t) and h(t) and
$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$ is the formula for convolution between x(t) and h(t)
but i cannot find, what exactly i have done wrong. could you help?

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closed as unclear what you're asking by Marcus Müller, lennon310, MBaz, jojek Mar 11 at 14:21

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ correct me, but aren't the last equations of each of your three large formulas totally redundant? $\endgroup$ – Marcus Müller Mar 5 at 17:58
  • $\begingroup$ Could you explain what $T$ is? $\endgroup$ – Marcus Müller Mar 5 at 17:58
  • $\begingroup$ since,cross-correlation and convolution are two separate concepts, they have different formulas. but due to some logical inconsistency, i have arrived at the conclusion that both cross-correlation and convolution have the same formulas $\endgroup$ – abhishek Mar 5 at 18:26
  • $\begingroup$ the last parts are not redundant. look closely, the order of t and tau is interchanged $\endgroup$ – abhishek Mar 5 at 18:27
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    $\begingroup$ ah, so that equality is wrong! $\endgroup$ – Marcus Müller Mar 5 at 18:27
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Your confusion stems from abusing the time invariance property.

Let's restate it for a simple 1D system:

$$ \text{if} ~~ T\{ x(t) \} = y(t) \implies T\{x(t-d) \} = y(t-d) $$ is correct but the following

$$ T\{ x(-t) \} = y(-t) ~~~\text{is wrong in general} $$

Now in your derivation you replace the input $\delta(t-\tau)$ with time reversed input $\delta(\tau-t)$ and claim the same on the output as $h(t-\tau) = h(\tau-t)$ which is not true as shown above.

The solution comes simply by recognizing the fact that $\delta(t-\tau) = \delta(-(t-\tau)) = \delta(\tau-t)$ and then proceed only with the $\delta(t-\tau)$ case as Marcus have stated.

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