# From transfer function to differential equation

I have the below detailed solution (boxed in blue) that I don't understand completely:

I can reconstitute the differential equation from: $$(1+Ts) X(s) = K_v U(s)$$ $$x(t) + T\dot x(t) = K_v u(t)$$ but I don't get how do they get $$x(t)$$ from the initial conditions?

I realize that my question is more about differential equations but it is something I obviously missed before...

Given a transfer function

$$G_v(s) = \frac{k_v}{1 + sT} \tag{1}$$

the corresponding LCCDE, with $$y(t)$$ being the solution, and $$x(t)$$ being the input, will be

$$T ~\dot{y}(t) + y(t) = k_v ~x(t) \tag{2}$$

Your formulation replaces $$x(t)$$ with a unit-step $$u(t)$$, and $$y(t)$$ with $$x(t)$$, yielding

$$T ~\dot{x}(t) + x(t) = k_v ~u(t) \tag{3}$$

or equivalently

$$\dot{x}(t) + \frac{1}{T} x(t) = \frac{k_v}{T} ~u(t) \tag{4}$$

Eq.4 represents a typical first order, constant coefficient, linear, ordinary differential equation (abbr LCCDE) whose solution procedure is as follows:

First, find the homogeneous solution to the Eq.4 with RHS being zero, as

$$x_h(t) = A e^{-t/T} \tag{5}$$

where $$A$$ is an unknown coefficient to be determined by the initial condition $$x(0)$$.

Second, find the particular solution of Eq.4 with the RHS being $$u(t)$$. For this you can use the method of undetermined coefficients, by assuming a solution of the type

$$x_p(t) = B u(t) \tag{6}$$

an letting $$t \to \infty$$, will yield in Eq.4 as

$$0 + \frac{B}{T} = \frac{k_v}{T} \tag{7}$$

where you find $$B$$ as

$$B = k_v \tag{8}$$

Then you find the complete solution by adding the homogeneous part to the particular part, and solving for the unknown constant $$A$$ from the initial condition $$x(0)$$

$$x(t) = x_h(t) + x_p(t) = A e^{-t/T} + k_v ~~~~,~~~~ t > 0\tag{9}$$

The initial condition $$x(0)$$ is given as $$0$$ in the problem, hence

$$x(0) = 0 = A + k_v \implies A = -k_v \tag{10}$$

And the overall solution of the initial value problem in Eq.4 with $$x(0)=0$$ becomes:

$$x(t) = -k_v e^{-t/T} + k_v = k_v (1 - e^{-t/T}) u(t) \tag{11}$$

Note that we look for a causal solution, unless otherwise stated, hence assume $$t>0$$.

• I am sorry I missed something the first time, why does u(t) disappears in equation (9)? Feb 17, 2021 at 11:16
• $u(t)$ is equivalently expressed in $t >0$ in Eq.9. The unit-step function $u(t)$ has the following standart definition :$$u(t) = \begin{cases}{ ~~~1~~~,~~~t >0\\~~~0~~~,~~~t <0}\end{cases}$$ and as you can see, it's equivalent to saying $t > 0$. Feb 17, 2021 at 11:28

The solution to the differential equation is given by the sum of a particular solution and the solution of the homogeneous differential equation. The particular solution is a solution to the non-homogeneous equation

$$T\dot{x}(t)+x(t)=k_vu(t)\tag{1}$$

which is easily found as

$$x_p(t)=k_v,\qquad t>0\tag{2}$$

The homogeneous solution is the solution of the homogeneous equation

$$T\dot{x}(t)+x(t)=0\tag{3}$$

which is given by

$$x_h(t)=Ce^{-t/T}\tag{4}$$

The complete solution is the sum of $$x_p(t)$$ and $$x_h(t)$$:

$$x(t)=k_v+Ce^{-t/T},\qquad t>0\tag{5}$$

With the initial condition $$x(0)=0$$, the constant $$C$$ is easily determined as $$C=-k_v$$, which results in the given solution.