# How to calculate step response for $y''(t) - y(t) = x'(t) - x(t)$ in time domain?

How to calculate step response for $y''(t) - y(t) = x'(t) - x(t)$ in time domain? So, without Laplace or Fourier Transforms.

This is what I tried:

The Homogeneous solution of the differential equation would be: $y_\text{homogeneous} = A*e^t + B*e^{-t}$.

$u(t)$ is the unit step function.

$$u(t) \triangleq \begin{cases} 1, & \text{if }t>0 \\ 0, & \text{if }t<0 \end{cases}$$ When I try to fill in the stepfunction for $x: x(t)=u(t)$. So $x'(t) = u'(t) = \delta(t)$

And I try a step response $y(t) = y_\text{homogeneous}(T)*u(t) = (A*e^t + B*e^{-t})*u(t)$

$$\frac{\partial ^2 ( \left(A e^t+B e^{-t} \right)u(t) )}{\partial t^2}-\left(A e^t+B e^{-t}\right) u(t)=\delta(t)-u(t)$$

So, I will get:

$2 (A-B) \delta(t)-A \delta(t)+A \left( \delta ' (t) \right )+B \delta(t)+B \left( \delta ' (t) \right)=\delta(t)-u(t)$

But there's no $u(t)$ at the left hand side.

• just a matter of notation, here in the Electrical Engineering world we use the notation "$u(t)$" to denote the unit step function, otherwise known as the Heaviside step function. we use capital letters to denote the frequency domain Laplace or Fourier transform of a time function and we often use "$h(t)$" to denote the impulse response of a system. so "$H(s)$" would be the Laplace transform of the impulse response and is normally called the "transfer function" of the system. want me to edit your question and fix the notation? Nov 18 '15 at 20:39
• I edited the post and changed it to the unit step function u(t) Nov 18 '15 at 20:55
• You can bring $x'(t)$ on the left side and substitute $y'_2(t)=y''(t)-x'(t)$ and $y_1(t)=y(t)$ to convert the dfe to a nonhomogeneous system of 1st order differential equations that can be solved using various methods in time domain, even numerically. Nov 18 '15 at 23:13

It is indeed possible to convert the system to an equivalent first order system, but let's first directly solve the given problem. The system's step response satisfies

$$y''(t)-y(t)=\delta(t)-u(t)\tag{1}$$

This implies that we can compute the step response of the given system by considering another system

$$y''(t)-y(t)=x(t)\tag{2}$$

and compute its impulse response ($x(t)=\delta(t)$) and its step response ($x(t)=u(t)$), and obtain the step response $(1)$ as the difference of the impulse response and the step response of the new system $(2)$.

For the step response we need to solve

$$y''(t)-y(t)=u(t)\tag{3}$$

Assuming the system is initially at rest, we get initial conditions $y(0)=y'(0)=0$. The homogeneous solution of $(3)$ is $y_h(t)=Ae^{t}+Be^{-t}$, $t>0$, whereas a particular solution is $y_p(t)=-1$, $t>0$. The complete solution is then

$$y(t)=y_h(t)+y_p(t)=Ae^{t}+Be^{-t}-1,\quad t>0\tag{4}$$

The constants $A$ and $B$ are determined by the initial conditions:

\left.\begin{align}y(0)&=A+B-1=0\\y'(0)&=A-B=0\end{align}\right\}\Rightarrow A=B=\frac12\tag{5}

Consequently, the step response of system $(2)$ is

$$y(t)=\frac12\left(e^t+e^{-t}\right)-1,\quad t>0\tag{6}$$

Its impulse response satisfies

$$y''(t)-y(t)=\delta(t)\tag{7}$$

and it can be obtained by differentiating the step response $(6)$:

$$y(t)=\frac12\left(e^t-e^{-t}\right),\quad t>0\tag{8}$$

Finally, according to $(1)$ and $(2)$, the step response of the original system is the difference between the impulse response and the step response given in $(8)$ and $(6)$:

$$y(t)=\frac12\left(e^t-e^{-t}\right)-\frac12\left(e^t+e^{-t}\right)+1=1-e^{-t},\quad t>0\tag{9}$$

Another approach would be to see that the original system can also be represented by a first-order system. Note that

$$y''(t)-y(t)=(y(t)+y'(t))'-(y(t)+y'(t))\tag{10}$$

So if $y''(t)-y(t)=x'(t)-x(t)$, it follows from $(10)$ that

$$y(t)+y'(t)=x(t)\tag{11}$$

With $x(t)=u(t)$ you obtain from $(11)$ the same step response as in $(9)$:

\begin{align}y_h(t)&=Ce^{-t}, &t>0\\ y_p(t)&=1, &t>0 \end{align}\\y(t)=y_h(t)+y_p(t)=Ce^{-t}+1,\quad t>0

From $y(0)=0$ we get $C=-1$, and the step response becomes

$$y(t)=1-e^{-t},\quad t>0$$

just like in $(9)$.

As a final check it is useful to see what happens when you solve the problem using the Laplace transform. The original difference equation transforms to

$$Y(s)(s^2-1)=X(s)(s-1)$$

which gives for the transfer function

$$H(s)=\frac{s-1}{s^2-1}=\frac{1}{s+1}$$

This shows that you have a pole-zero cancellation, and the system is actually a first-order system, just as described by Eq. $(11)$. For $x(t)=u(t)$ you get $X(s)=1/s$, and, consequently, for the Laplace transform of the step response

$$Y(s)=\frac{1}{s}\frac{1}{s+1}=\frac{1}{s}-\frac{1}{s+1}$$

which corresponds to the time domain function

$$y(t)=u(t)-e^{-t}u(t)=(1-e^{-t})u(t)$$

which is the same as $(9)$.