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

enter image description here

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...


2 Answers 2


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$.

  • $\begingroup$ I am sorry I missed something the first time, why does u(t) disappears in equation (9)? $\endgroup$
    – Dovendyr
    Feb 17, 2021 at 11:16
  • $\begingroup$ $u(t)$ 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 $t > 0$. $\endgroup$
    – Fat32
    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


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


which is given by


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.


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