Find the differential equation to the closed-loop system

Question

I know that the closed-loop transfer function is equal to Y/Ysp = (P controller * DC motor) /(1+(P controller * DC motor)) then I am struck. Can anyone give me some hints on how to proceed?

Answer 1

You don't need to use the transfer function for this problem, just substitute the known equations in the time domain to get a differential equation for the closed loop. You have

$$ V(t) = K_p e(t) \qquad e = Y_{sp} - Y(t) \qquad \tau_L = 0$$ $$ T \frac{d^2Y(t)}{dt^2} + \frac{dY(t)}{dt} = K V(t) - K_B \tau_L(t)$$

Just insert the first three equations into the fourth and you have your closed loop differential equation.

$$ T \frac{d^2Y(t)}{dt^2} + \frac{dY(t)}{dt} = K K_p (Y_{sp} - Y(t))$$

Rearrange to get into the standard differential equation form with your states on the left-hand-side and your inputs on the right-hand-side:

$$ T \frac{d^2Y(t)}{dt^2} + \frac{dY(t)}{dt} + K K_p Y(t) = K K_p Y_{sp}$$

Now you can analyze the control system as a simple second-order DE with a constant input.

Edit: Steady state value

The steady state value can easily be derived from the differential equation. We know that when the system reaches steady state, by definition, $\frac{d^2Y(t)}{dt^2} = 0$ and $\frac{dY(t)}{dt} = 0$. Therefore, substituting those values in the DE and setting $Y(t)=Y_{ss}$:

$$ T \cdot 0 + 0 + K K_p Y_{ss} = K K_p Y_{sp}$$ $$Y_{ss} = Y_{sp}$$