What is the output of a cascaded amplifier?

Question

I wanted to make an comparator with cascade of amplifiers of equal gain, output impedance etc. We know that we can approximate it as single pole system , thus a resistor and a capacitor in the output will do the approximation. I was able to calculate the gain for first stage i.e $$-g_m * R_o * \frac{t }{R_o C_o} $$

This is when t<< time constant. Now as stages increases , I want to derive the speed and delay. Can someone help me or share materials that contain this derivation?

Answer 1

Assume that \$V_i = 1\$. It doesn't change the essence of the problem since a linear time invariant system is assumed. Just multiply the final answer with \$V_i\$.

The step response is approximated as \$k\cdot t\$ in equation (6.2), where \$k\$ is some constant (\$\frac{g_m}{C_0}\$). So the corresponding (approximate) impulse response is \$k\ u(t)\$, the unit step function. So, you need to convolve the output of previous stage with this impulse response to get output of next stage.

Input of first stage is unit step. Output of first stage is \$k\ t\$ equation (6.2).

So, the input of second stage is \$k\ t\$. When convolved with \$k u(t)\$ we get the output as $$ \int_0^t k \cdot u(\tau) \times k\cdot (t-\tau)\ d\tau = k^2 t^2 - k^2 t^2/2 = k^2 t^2/2 $$

This is the input of the third stage. When convolved again with \$k u(t)\$ we get the output as (checked via Wolfram Alpha) $$ \int_0^t k \cdot u(\tau) \times k^2 \cdot (t-\tau)^2/2\ d\tau = k^3 t^3/(2\cdot 3) $$

Continuing, the n^th stage output is \$k^n t^n / n!\$.

This is what is given in (6.4).

note

If you are familiar with Laplace transform, the difficult to perform convolutions simplify to multiplication. Try that route if convolutions are difficult for you.