Introduction, Task 1,2

${\displaystyle W(s)=\frac{Y(s)}{U(s)}=\frac{km}{Tms+1}}$

${\displaystyle Y}$:output, which actually present the speed the motor rotates. ${\displaystyle U}$:input, for this lab, it actually is a constant.

To build the model up in MATLAB, there are two ways we can use:

• one in the form of input-output differential equation
• one in the way of the transfer function

Input-Output Differential Equation

So we firstly need to change the transfer function into the differential equation, using the method mentioned in Lecture 2

We have one operator ${\displaystyle s}$ in the denominator, indicates that the differential equation is in the first order at the side of output. And we can note:

${\displaystyle TmsY(s)+Y(s)=kmU(s)}$${\displaystyle Tm{y}^{\prime }+y=kmU}$

So, replace y with ${\displaystyle \omega }$, we get:

${\displaystyle {\omega }^{\prime }=\frac{km}{tm}U-\frac{1}{Tm}\omega }$

When beginning, the speed is 0 and we can take ${\displaystyle \omega }$ out of our consider, we just give the system an input ${\displaystyle U}$, and as the motor starts to rotate, the speed ${\displaystyle \omega }$ starts to increase and then be put into the circuit.

So, at first we only have one input ${\displaystyle U}$, and the feedback ${\displaystyle -\frac{1}{Tm}\omega }$ is generated by the input ${\displaystyle U}$.

Finally the flow chart is just like below.

![[Pasted image 20250923163306.png]]

Transfer Function

It is simpler than the input-output differential equation, since it comes from the Laplace-transformed input-output differential equation, with the input and out put variables out of consideration, it just depict the inherent properties of the system(depends on the type of input, like step, impulse etc.).

P-control, Task 3