# Confusion regarding stability & step response?

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=
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?

• What do you mean by "pure" integrator, an ideal integrator? Anyway, changing the sign of a transfer function will definitely not change its stability properties. Sep 7, 2022 at 7:47
• An ideal integrator with a transfer function of $1/s$ is only marginally stable since it has pole at $s=0$ and infinite gain at DC. That's why the step response of the integrator grow to inifinity Sep 7, 2022 at 11:12
• @MattL. please check updated question Sep 7, 2022 at 11:20