Confusion regarding stability & step response?

Question

Let's say we have a system (for example an integrator) whose step response has a negative slope as shown by the MATLAB code below:

clc;clear ;close all
R=47*10^3
C=1*10^-7
num1=[-1]
den=[R*C 0]
sys1=tf(num1,den)
num2=[1]
sys2=tf(num2,den)
subplot 211
pzmap(sys1)
title('pole zero map for pure integrator, with -1 in numerator')
subplot 212
pzmap(sys2)%plotting poles and zeros
title('pole zero map for integrator, with +1 in numerator')
figure
subplot 211
step(sys1)
title('step response for pure integrator, with -1 in numerator')
subplot 212
step(sys2)
title('step response for integrator, with +1 in numerator')

sys1 is an ideal integrator and we know that ideal integrators are stable systems.

But what if we replace -1 in the numerator with 1 (sys2): then the step response slope will be positive: is sys2 unstable because of the positive slope?

Answer 1

An ideal integrator is not BIBO-stable, i.e., there are bounded input signals which result in an unbounded output. A step at the input is such a signal.

Clearly, changing the sign of the transfer function doesn't make any difference. In both cases, the system is not BIBO-stable.

As mentioned in a comment, the ideal integrator is only marginally stable because it has a pole on the imaginary axis. This makes it BIBO-unstable. Any DC at the input will result in an output signal with linearly increasing magnitude.