# Can 'distance' be mathematically described as the convolution of velocity and time, in time domain?

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.

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.

• @ Zeick Thanks. I agree with your comment. Please have a look at the question once again. – Vikash May 18 at 11:46
• I have edited the answer to include the capacitor part of your question. – Zeick May 18 at 11:49
• Thanks, Zeick. I had the same opinion. – Vikash May 18 at 11:57

No, since $$x(t)\neq\int\operatorname{\text{d}\tau} v(t-\tau)\tau$$? Maybe I didn't understand your question correctly?

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