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How come are the units of convolution for LTI system not consistent?

$$y(t)= \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau$$

$y(t)$ unit is volt, whereas the RHS has volt.time unit!!!!

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impulse responses, $h(t)$, of continuous-time systems (those with output that is the same species of animal of the input) have dimensions 1/time.

consider a wire which has impulse response

$$ h(t) = \delta(t) $$

the area of the $\delta(t)$ is equal to the dimensionless $1$. the width of the nascent delta is in dimension of time, so the height must be in dimension of 1/time.

the impulse response for a simple RC low-pass filter is

$$ h(t) = \tfrac{1}{RC} e^{-^{\, t}/_{RC}} \ u(t) $$

the dimension of $h(t)$ is in 1/time because $RC$ is in units of time.

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    $\begingroup$ the height of $x(t)=K \cdot \delta(t)$ would be volts. so $K$ would be volt-seconds. what comes out is $K \cdot h(t)$. $\endgroup$ – robert bristow-johnson Aug 2 '17 at 20:22
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    $\begingroup$ yes now that makes a lot more sense (I've checked the library for a few circuit books and reached the same result, which I was almost typing! :-) ) $\endgroup$ – Fat32 Aug 2 '17 at 20:27
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    $\begingroup$ that value of $K$ would depend on how wide your nascent delta is. so if the input was a thin pulse that is 2 volts tall and 3 $\mu$s wide, the value of $K$ is 6 $\mu$V-s. $\endgroup$ – robert bristow-johnson Aug 2 '17 at 20:29
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    $\begingroup$ I've still doubts however about the book, this time :-) impulse response can also be measured as voltage-in current-out form? or vice versa? I'll check their units as well... $\endgroup$ – Fat32 Aug 2 '17 at 20:30
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    $\begingroup$ yes, i was assuming that the species of animal coming out is the same species going it. that makes the transfer function dimensionless (a simple ratio of like-dimensioned physical quantities). if it's voltage in and current out, then the transfer function is in units of conductance and the impulse response would be conductance/time. $\endgroup$ – robert bristow-johnson Aug 2 '17 at 20:32

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