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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.

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  • $\begingroup$ 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? $\endgroup$ – robert bristow-johnson Nov 18 '15 at 20:39
  • $\begingroup$ I edited the post and changed it to the unit step function u(t) $\endgroup$ – Codart Nov 18 '15 at 20:55
  • $\begingroup$ 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. $\endgroup$ – Harris Nov 18 '15 at 23:13
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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 $(9)$.

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