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From what I understand the convolution of the impulse response of a system with the input to that system gives the output.

Now if the impulse response is a rectangular function and the input is also a rectangular function we get a triangular function as the output.

What I dont understand is: How could this ever be the case? As the rectangular function of the impulse response has only two levels as does the input. How can the system ever produce a value that is between these levels let alone a ramp of values between these levels.

It's obvious I am missing something but I cant quite put my finger on what that is.

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    $\begingroup$ just because the output due to an impulse applied to the input takes only two levels, that does not mean that the output due to a rectangular pulse applied to the input takes on only two levels. $\endgroup$ – robert bristow-johnson Apr 5 '20 at 17:54
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    $\begingroup$ Above is exactly where you're getting confused. The impulse response is the output given the input is a single impulse, but a rectangular function input is "like an infinite number of impulses" separated by dt time so if you think about it in this way (probably too loosely but hoping it gets a point across), then of course the output grows like ramp until the response due to the first impulses starts to die off (overlap becomes zero) like shown in @EdV's answer $\endgroup$ – Engineer Apr 5 '20 at 19:13
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    $\begingroup$ You may find this trick amusing. $\endgroup$ – Rodrigo de Azevedo Apr 5 '20 at 19:25
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A convolution integral is an overlap integral, i.e., for any given shift of the two aperiodic functions being convolved, the convolution integral is simply the overlap area. McGillem and Cooper [1, p. 58] defined the convolution integral of $x_1$ and $x_2$ as

$$\mathrm {x_3 =x_1*x_2 =\int_{-\infty}^{\infty}x_1(\lambda)x_2(t-\lambda)\,\mathrm d\lambda \tag{1}}$$

As a simple graphical illustration of the defining integral, they considered the following two rectangular pulses:

Rectangular pulses

With $x_1$ and $x_2$ as shown in the above figure, their convolution is shown in the figure below:

Convolution of rectangular pulses

This figure is redrawn from [1, p. 59]. The shaded areas are the overlap areas as a function of the shift, $t$, and the resulting convolution has a trapezoidal shape. If the rectangular pulses had had equal width, then the convolution would havec simplified to an isosceles triangular shape.

1 C.D. McGillem, G.R. Cooper, "Continuous and Discrete Signal and System Analysis", 2nd Ed., Holt, Rinehart and Winston, NY, ©1984, pp. 58-59.

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If you understand that the input and output of an LTI (linear time-invariant) system are related by convolution, then you should also be able to understand that a rectangular input and a rectangular impulse response result in a triangular signal, if you know what convolution means, namely:

$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau\tag{1}$$

where $y(t)$ is the output signal, $x(t)$ is the input signal, and $h(t)$ is the impulse response.

Assuming that $x(t)$ has a constant value $A$ in the interval $t\in[0,T]$ (and is zero otherwise), and $h(t)$ has a constant value $B$ in the same interval (and is zero otherwise), then $(1)$ becomes

$$y(t)=AB\int_{\max\{0,t-T\}}^{\min\{t,T\}}d\tau=\begin{cases}AB\int_0^td\tau=ABt,&0<t<T\\AB\int_{t-T}^Td\tau=AB(2T-t),&T<t<2T\\0,&\textrm{otherwise}\end{cases}\tag{2}$$

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  • $\begingroup$ (+1) Elegant and concise, as usual! $\endgroup$ – Ed V Apr 5 '20 at 19:40
  • $\begingroup$ Thx @EdV :) ${ }$ $\endgroup$ – Matt L. Apr 5 '20 at 19:42
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A discrete time signal can have any finite amplitude. It doesn't have to be either 0 or 1. Ex: you could have a signal of value 2 or 3 or 100 at different points of discrete time

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