# Find the differential equation to the closed-loop system   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?

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.

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 $$Y(t)=Y_{ss}$$:
$$T \cdot 0 + 0 + K K_p Y_{ss} = K K_p Y_{sp}$$ $$Y_{ss} = Y_{sp}$$
• @LearningLaSo if $Y_{sp}$ changes then the system will respond according to the differential equation. I don't know what exactly it is you want to know Feb 27, 2017 at 22:12