Electrical – Solving RC circuits with impulse or unit step input by using convolution integral

I do not believe there is any problem with the notation used by the author. He just wanted to express the following (\$x\$ input, \$y\$ output - the response to a eigenfunction \$e^{st}\$):

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

Thus, the response to \$e^{st}\$ is of the form

$$y(t) = H(s)e^{st}$$

where \$H(s)\$ is a complex constant whose value depends on \$s\$ and wich is related to the impulse response by

$$ H(s) = \int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau $$

There is no conflict time x frequency here.

Also, note:

If \$\lambda\$ is a pole of \$H(s)\$, the mode \$e^{{\lambda}t}\$ can be generated at the output by an initial state without the application of any input. If \$\lambda\$ is not a pole of \$H(s)\$, then this is not possible. In this case, the only way to generate \$e^{{\lambda}t}\$ at the output is to apply \$e^{{\lambda}t}\$ at the input. In other hand, if the input \$x(t)\$ is of the form \$e^{{\lambda}t}\$, where \$\lambda\$ is not a pole of \$H(s)\$, then the output due to input \$x(t) = e^{{\lambda}t}\$ is equal to \$y(t) = H(\lambda)e^{{\lambda}t}\$.

When they say:

$$y(n)=x(n)\ast h(n)=\sum_{m=-\infty}^{\infty}x(m)\cdot h(n-m)$$

they are correct, but in the next passage they oversimplify a bit.

Let's do it step by step:

\begin{align*} &\sum_{m=-\infty}^{\infty}x(m)\cdot h(n-m) = \sum_{m=-\infty}^{\infty}u(m)\cdot u(n-m) = \\ &= \sum_{m=0}^{\infty} 1 \cdot u(n-m) = \sum_{m=0}^{\infty} u(n-m) \\ \end{align*}

The function to be summed is \$u(n-m)\$, which is equal to 1 if and only if \$n-m>0 \Leftrightarrow m < n\$.

Since the sum is carried out only for \$m>=0\$, if \$n>=0\$ the result is actually \$n+1\$, but if \$n<0\$ the result is 0, therefore they should have written:

\[ \sum_{m=-\infty}^{\infty}x(m)\cdot h(n-m) = (n+1)u(n) \]

rightaway.

The following image may help:

the red dots are the function values, the green shaded area highlights the value range that will be summed. The image was drawn with \$n = 3\$. You may infer that if \$n<0\$ the green area will disappear, indicating there is nothing to sum.