Housekeeping

POLES IN THE LHP:

Consider the transfer function with a single pole in the left-half-plane. By taking the Inverse Laplace of we see that the system behavior will be governed by indicating an exponential decay. This can also be observed by rearranging the equation as and taking the Inverse Laplace to find the differential equation in the absence of an input we can observe that the solution is indeed the solution. Based on these observations we can interpret the change in the output state will be proportional to negative value of its current state scaled by the pole magnitude 𝑝. Regardless of the current value of this will ensure it's next value will "chase" zero. Indeed the equation is the general solution to the equation (where A is TBD) and is guaranteed to decay towards zero exponentially.

time = 0:0.01:10;

 

f = figure();

for p = -1*(1:2:10)

plant = 1/(s-p);

y = step(plant, time);

legendlabel=string(p);

plot(time,y, 'DisplayName','pole = '+legendlabel);

hold on

end

f.Position = [0 0 1300 500];

title('Step Response Based on Pole');

xlabel('time (s)'); xlim([0,10]);

ylabel('magnitude'); ylim([0,1.5]);

legend('FontSize',15, 'Location', 'northwest')

hold off

POLES IN THE RHP:

Consider now the transfer function with a single zero in the right-half-plane. By taking the Inverse Laplace of we see that the system behavior will be governed by indicating an exponential decay. This can also be observed by rearranging the equation as and taking the Inverse Laplace to find the differential equation in the absence of an input we can observe that the solution is indeed the solution. Based on these observations we can interpret the change in the output state will be proportional to negative value of its current state scaled by the pole magnitude 𝑝. Regardless of the current value of this will ensure it's next value will deviate further from 0. Indeed the equation is the general solution to the equation (where A is TBD) and is guaranteed to grow exponentially.

time = 0:0.01:2; %note the difference in time!!!

 

f = figure();

for p = 1*(1:2:10)

plant = 1/(s-p);

y = step(plant, time);

legendlabel=string(p);

plot(time,y, 'DisplayName','pole = '+legendlabel);

hold on

end

f.Position = [0 0 1300 500];

title('Step Response Based on Pole');

xlabel('time (s)'); xlim([0,2]);

ylabel('magnitude'); ylim([0,100]);

legend('FontSize',15, 'Location', 'northwest')

hold off

WHY POLES DETERMINE STABILITY:

Any engineer that has taken a course on control theory knows that poles determine the stability of a system, and most will also remember that a system is only stable if all of its poles are located in the left-half-plane of a system. In the absence of an input the idea in the first paragraph will suffice, but here we examine two situations that include inputs.

Consider again a transfer function with a single pole at . We can rearrange this equation and use the Inverse Laplace Transform to restore it to its differential equation . We can observe that the solution in the first paragraph is indeed the solution to the non-homogeneous form of this equation. Suppose now that we have an input . The non-homogeneous form is then . Using the method of undetermined coefficients we can attempt a solution of the form , and solve for A and B to find the non-homogeneous solution . The solution to our entire problemis then the linear combination of both solutions yielding .

From this we can see that a sinusoidal input signal driving the non-homogeneous form of our system can induce oscillations, but the exponential decay of the pole will always be included because it is "baked into" the dynamics of the system itself. We can see then that any bounded input will produce a bounded output yielding a BIBO stable system. Indeed if the pole were in the right-half plane, an exponential growth term would be included guaranteeing exponential growth and instability regardless of the input used. The only option then is to introduce a controller of such a design to move that pole into the left-half-plane.