Cruise Control

Pretty interesting, the teacher in MATH213 gave a formal introduction to this.

We have a car to moves at a given velocity. We know Newton’s Laws of Motion $f=ma$, so $f(t)=m{v}^{\prime }(t)$. Taking into account drag, we’ll get this equation $m{v}^{\prime }(t)+bv(t)=f(t)$

• $m$ is the mass
• $b$ is a drag coefficient
• $v(t)$ is the velocity of the car
• $f(t)$ is the velocity forcing term

You should be very comfortable with the Transfer Function.

So if we want the car to go at a certain velocity, we can control $f(t)$.

Breaking down $f(t)$ into 2 parts: $f(t)=u(t)+d(t)$

• $u(t)$ is the control input
• $d(t)$ is the disturbance input

The controller controls $u(t)$.

Now, how does this work exactly? Like if $u(0)=0$, and then $u(0.001)=1000$, does the velocity instantly spike up? like the function has to be differentiable right. It’s a differential equation.

The teacher gets this problem to be solved in frequency space.

In time-domain $m{v}^{\prime }(t)+bv(t)=u(t)+d(t)$ In frequency-domain $V(t)=\frac{U(t)+D(t)}{ms+b}$

• how the conversion is done is simply through the rules of the Transfer Function

We end up with our first Open Loop Controller.

It’s open-loop because there is no feedback into the system.

In standard form, the transfer function $\frac{1}{ms+b}=\frac{1/b}{(m/b)s+1}$

Proportional Controller

Now, we design our first closed loop controller.

• $R(s)$ is the Laplace transform of desired velocity function $v(t)$

So instead of directly feeding in a control input $u(t)$ blindly, we know first look at the error $e(t)=r(t)-v(t)$, and that is used to dynamically adjust the control input $u(t)$.

Now, we obviously don’t know $C(s)$, but we can define it $C(s)=k_{p}E(s)$. The control input is some proportion to the error term.

We can write

$$\begin{array}{rl}V(s)& =\frac{1}{ms+b}F(s)\\ & =\frac{1}{ms+b}U(s)\text{since D(s) = 0}\\ & =\frac{1}{ms+b}{k}_{p}E(s)\\ & =\frac{1}{ms+b}{k}_p(R(s)-V(s))\end{array}$$

Therefore, $V(s)=\frac{\frac{k_p}{ms+b}}{1+\frac{k_p}{ms+b}}R(s)$ We write it in Standard First Order System: $V(s)=\frac{\frac{k_p}{b+k_p}}{\frac{m}{b+k_p}s+1}R(s)$

We know have our velocity in terms of some reference. But then how does it actually get working in real life??

• I’m confused

Alright, so let’s go back into time domain by taking the $v(t)$.

Standard First Order System.

Integral Controller

For the integral controller, we get that $V(s)=P(s)C(s)E(s)$ where

• $P(s)$ is the car transfer function, the same as before $P(s)=\frac{1}{ms+b}$
• $C(s)$ is the controller transfer function
• $E(s)=(R(s)-V(s))$

Notice how this is where we compose a transfer function.

So we get $V(s)=\frac{P(s)C(s)}{1+P(s)C(s)}R(s)$

• Actually, this is a more generic form. For proportional controller, the transfer function $C(s)=k_p$ (because remember we defined $C(s)=k_{p}E(s)$, for integral controller its a $\frac{k_i}{s}$.

Next

• Adaptive Cruise Control

Related

• PID