Adding two signals and finding the RMS and Peak to peak value of the output signal

I think you can do all of this with just two op-amps (although someone will probably correct me).

The first stage is the peak detector and decay. The op-amp is wired to emulate a perfect diode. We take the feedback signal from after the diode so that the op-amp drives hard enough to overcome the voltage drop of the diode. Being a peak detector, it doesn't really eed a rectified signal. The capacitor holds the peak value, while the resistor decays it. You can make the decay slower by choosing a larger capacitor and resistor. And vice versa.

The second stage is simply 15x gain. It has low output impedance.

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.