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I have phrased the question as such, to confirm that convolution of the two functions raises the dimensionality of the convolution product. So, if I do convolution of velocity and time, then the resultant should have units of metre x time. Am I correct?

Or an alternative example can be charge Q stored in a capacitor is capacitance (a constant) times the voltage, V, applied across the plates. So, Q(t) = CV(t)

But if the capacitance becomes time varying, then is it correct to say that,

Q(t) = C(t) * V(t),

where * means convolution operation.

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So, if I do convolution of velocity and time, then the resultant should have units of metre x time. Am I correct?

Yes. If you integrate over time, dimensionally you get the unit [length][time].

$$ \left[\int \text{velocity}\times\text{time}\times dt\right] = [v][t][dt] = [\text{length}][\text{time}] $$

But if the capacitance becomes time varying, then is it correct to say that, Q(t) = C(t) * V(t), where * means convolution operation.

No. Simply $Q(t) = C(t)V(t)$, you don't need convolution. And the dimensions would not match anyway if you use convolution.

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  • $\begingroup$ @ Zeick Thanks. I agree with your comment. Please have a look at the question once again. $\endgroup$ – Vikash May 18 at 11:46
  • $\begingroup$ I have edited the answer to include the capacitor part of your question. $\endgroup$ – Zeick May 18 at 11:49
  • $\begingroup$ Thanks, Zeick. I had the same opinion. $\endgroup$ – Vikash May 18 at 11:57
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No, since $x(t)\neq\int\operatorname{\text{d}\tau} v(t-\tau)\tau$? Maybe I didn't understand your question correctly?

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  • $\begingroup$ I think his question is "why is your expression an inequality rather than an equality?". $\endgroup$ – thermomagnetic condensed boson May 18 at 11:43
  • $\begingroup$ @ Ivan, You understood it correctly. I have added some extra information to include what I am trying to ask actually. $\endgroup$ – Vikash May 18 at 11:48

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