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July 2, 2015 compiled on — Sunday June 25, 2017 at 11:39 AM

Let us consider a closed loop control system with proportional controller and with plant transfer function given by standard second order mass-spring-damper system where is the mass and is the damping coefficient and is the stiffness coefficient of the spring. In block diagram, the system is

The error is where is the reference input that we want the output to track. We are now interested in finding how adding the controller and closing the loop changes the dynamics of the plant by viewing the changes in the differential equation of the new system. Without the controller and the closed loop, the system was

In time domain this is represented by the differential equation where is the Laplace transform of . We now ask, how does this differential equation changes by adding the controller and closing the loop? From the first diagram, or , hence the plant now appears as

Hence

And in time domain, the differential equation now becomes

Considering the case where the reference signal is a unit step (with amplitude that has units of distance or length), then and the above reduces to

| (1) |

We see now that the effect of adding a proportional controller is to increase the stiffness of the plant by an amount and also forcing the plant with constant force (actuating force). Increasing the effective stiffness also means the natural frequency of the plant will increase, since and the mass remained the same. A note on the units: The controller has units of Newton per unit length in this case (stiffness coefficient units) and the force will have units of Newton only (since we multiplied it by the unit step reference single which had units of length). Even though the term appears as stiffness on the left side and as force in the right side, it represents only the magnitude, and the units will depend on the context.

It is easy to verify the above. We can connect the system, view the output of the closed loop for a unit step input , and compare it the solution of the differential equation as given by (1). The differential equation is solved using zero initial conditions. We see below that the same result shows up.

m = 1; c = 1; k = 20; kp = 400;

plant = TransferFunctionModel[1/(m s^2 + c s + k), s];

controller = TransferFunctionModel[kp, s];

sys = SystemsModelSeriesConnect[plant, controller];

sys = SystemsModelFeedbackConnect[sys];

o = OutputResponse[sys, UnitStep[t], {t, 0, 6}];

p1 = Plot[o, {t, 0, 6}, Frame -> True, PlotRange -> All,

FrameLabel -> {{"y(t)", None}, {"t (sec)",

"response of closed loop system"}}, BaseStyle -> 14, ImageSize -> 350,

Epilog -> {Red, Line[{{0, 1}, {10, 1}}]}];

sol = First@DSolve[{m y’’[t] + c y’[t] + (k + kp) y[t] == kp,

y[0] == 0, y’[0] == 0}, y, t];

p2 = Plot[Evaluate[y[t] /. sol], {t, 0, 6}, Frame -> True,

PlotRange -> All, FrameLabel -> {{"y(t)", None},

{"t (sec)", "solution of closed loop differential equation"}},

BaseStyle -> 14, ImageSize -> 350];

Grid[{{p1, p2}}]

plant = TransferFunctionModel[1/(m s^2 + c s + k), s];

controller = TransferFunctionModel[kp, s];

sys = SystemsModelSeriesConnect[plant, controller];

sys = SystemsModelFeedbackConnect[sys];

o = OutputResponse[sys, UnitStep[t], {t, 0, 6}];

p1 = Plot[o, {t, 0, 6}, Frame -> True, PlotRange -> All,

FrameLabel -> {{"y(t)", None}, {"t (sec)",

"response of closed loop system"}}, BaseStyle -> 14, ImageSize -> 350,

Epilog -> {Red, Line[{{0, 1}, {10, 1}}]}];

sol = First@DSolve[{m y’’[t] + c y’[t] + (k + kp) y[t] == kp,

y[0] == 0, y’[0] == 0}, y, t];

p2 = Plot[Evaluate[y[t] /. sol], {t, 0, 6}, Frame -> True,

PlotRange -> All, FrameLabel -> {{"y(t)", None},

{"t (sec)", "solution of closed loop differential equation"}},

BaseStyle -> 14, ImageSize -> 350];

Grid[{{p1, p2}}]

The differential equation given by (1) has an analytical solution when the force is constant and assuming the damping ratio is (under-damped), where . The analytical solution is given by

Where is the damped natural frequency and is the undamped natural frequency . We see now that as and hence the steady state solution will never reach unity (which is the reference signal in this example) but will get closer to it as is made larger and larger. This is the reason why response to a unit step using a proportional controller will always have a steady state error given by where is the controller gain and is the plant original stiffness coefficient. The term is called the static deflection. The final steady state can also be found in Laplace domain using final value theorem as follows.

Assuming is unit step, then

And using final value theorem

Which is the static deflection from the analytical solution of the differential equation found above.

Let us now look at the dynamics of the plant when the controller is a PID given by . The block diagram of the system is

The error is where is the reference input that we want the output to track. As was done for the case of the proportional controller, the plant now appears as

Hence

And in time domain, the differential equation now becomes

For the special case when the reference signal is a unit step

The term is the integral of the error . When is unit step the above becomes

Therefore, using a PID controller has the effect of making the system more damped, since the effective damping coefficient has now become compared to just before, and making the system more stiff by adding to the stiffness coefficient. The actuating force has two components (for the special case of unit step reference), which is constant force of and a force which is proportional to the error: .

PID controller allows the steady state to become zero. This can be seen by using final value theorem in Laplace domain as follows. The transfer function is

be found in Laplace domain using final value theorem as follows.

Assuming is unit step, then

And using final value theorem

Hence is a unit step. So using PID controller it was possible to achieve the same value as the desired tracking signal. Hence the steady state error is zero.

Plot of the step response of the above is given below. was made large to force the steady state error to go to zero in about 6 seconds.

m = 1; c = 1; k = 20; kp = 400; kd = 1; ki = 200;

plant = TransferFunctionModel[1/(m s^2 + c s + k), s];

controller = TransferFunctionModel[kp + kd s + ki 1/s, s];

sys = SystemsModelSeriesConnect[plant, controller];

sys = SystemsModelFeedbackConnect[sys];

o = OutputResponse[sys, UnitStep[t], {t, 0, 6}];

p1 = Plot[o, {t, 0, 6}, Frame -> True, PlotRange -> All,

FrameLabel -> {{"y(t)", None}, {"t (sec)",

"response of closed loop system, PID controller"}},

BaseStyle -> 14, ImageSize -> 400,

Epilog -> {Red, Line[{{0, 1}, {10, 1}}]}]

plant = TransferFunctionModel[1/(m s^2 + c s + k), s];

controller = TransferFunctionModel[kp + kd s + ki 1/s, s];

sys = SystemsModelSeriesConnect[plant, controller];

sys = SystemsModelFeedbackConnect[sys];

o = OutputResponse[sys, UnitStep[t], {t, 0, 6}];

p1 = Plot[o, {t, 0, 6}, Frame -> True, PlotRange -> All,

FrameLabel -> {{"y(t)", None}, {"t (sec)",

"response of closed loop system, PID controller"}},

BaseStyle -> 14, ImageSize -> 400,

Epilog -> {Red, Line[{{0, 1}, {10, 1}}]}]