# Confusion in the derivation of convolution formula [closed]

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

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

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

$$\text{if} ~~ T\{ x(t) \} = y(t) \implies T\{x(t-d) \} = y(t-d)$$ 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.