$$x(t)=\int_{-\infty}^{\infty}x(\tau)\delta(\tau-t)d\tau=\int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau$$ 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?

  • $\begingroup$ correct me, but aren't the last equations of each of your three large formulas totally redundant? $\endgroup$ – Marcus Müller Mar 5 '19 at 17:58
  • $\begingroup$ Could you explain what $T$ is? $\endgroup$ – Marcus Müller Mar 5 '19 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 '19 at 18:26
  • $\begingroup$ the last parts are not redundant. look closely, the order of t and tau is interchanged $\endgroup$ – abhishek Mar 5 '19 at 18:27
  • 1
    $\begingroup$ ah, so that equality is wrong! $\endgroup$ – Marcus Müller Mar 5 '19 at 18:27

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.

| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.