Consider a system for which the input $x(t)$ and output $y(t)$ are related by the differential equation $$\frac{d^2y(t)}{dt^2} + \frac{3}{2}\frac{dy(t)}{dt} - y(t) = x(t) \tag{1}$$Determine $h(t)$ for each of the following cases:

  1. The system is stable.
  2. The system is causal.

My attempt: I know that this question can be solved easily using the Laplace transform but I'm trying to solve it in time-domain. Since it's assumed that system is linear, causality is equivalent to: For any time $t_0$ and any input $x(t)$ such that $x(t) = 0$ for $t \lt t_0$, the corresponding output $y(t)$ must also be zero for $t < t_0$. This is the initial rest condition and we can easily solve $(1)$ by methods like this.

Stability for an LTI system is equivalent to $$\int_{-\infty}^{+\infty}|h(\tau)|d\tau<\infty$$

How this condition can be used to solve $(1)$ in time-domain? Can we get initial conditions using this condition?


1 Answer 1


Solving the characteristic equation

$$s^2+\frac32 s-1=0\tag{1}$$

gives the following homogeneous solution:


The requirements of causality and stability are taken into account by choosing at least one of the constants $c_1$ and $c_2$ to be zero for either $t>0$ or $t<0$. If we're looking for a causal solution we know that $y(t)=0$ for $t<0$, i.e.,


which can be interpreted as choosing $c_1=c_2=0$ in $(2)$ for $t<0$.

For a stable solution, we require $c_2=0$ for $t>0$ because stability means that there can't be exponential growth. For the same reason we require $c_1=0$ for $t<0$:


Eqs $(3)$ and $(4)$ are the general forms of the solutions given the requirements of causality or stability, respectively. In both cases, the constants $c_1$ and $c_2$ are determined by requiring that $y''+\frac32 y'-y$ equals a Dirac delta impulse with weight $1$.

Note that when calculating the derivatives of $(3)$ and $(4)$ you need to use the product rule:


Solving such problems using the Laplace transform is indeed much less tedious.

As an example I'll show how to arrive at the values of $c_1$ and $c_2$ for the causal solution $(3)$. Define $f(t)=c_1e^{-2t}+c_2e^{t/2}$. With $y_c(t)=f(t)u(t)$, the derivatives of $y_c(t)$ are


We need to satisfy the differential equation $y_c''(t)+\frac32 y_c'(t)-y_c(t)=\delta(t)$. Consequently, the term $f(0)\delta'(t)$ must vanish:


Furthermore, the coefficients associated with the Dirac delta impulses must add up to $1$:

$$f'(0)+\frac32 f(0)=1$$

This results in the requirement $$-2c_1+\frac12 c_2+\frac32 (c_1+c_2)=1$$

which leads to $c_1=-\frac25$ and $c_2=\frac25$.

In exactly the same way you can derive the coefficients of the stable solution $(4)$, which results in $c_1=c_2=-\frac25$.

  • $\begingroup$ Would you elaborate more, please? I know that we should solve $\frac{d^2h(t)}{dt^2} + \frac{3}{2}\frac{dh(t)}{dt} - h(t) = \delta(t)$ in homogeneous and particular parts. We can easily find $y_h(t)=c_1e^{-2t}+c_2e^{t/2}$ as you did. $c_1$ and $c_2$ depends on initial conditions. So why "we may need to choose different constants for $t\gt 0$ and for $t\lt 0 $."? $\endgroup$
    – S.H.W
    Commented Nov 2, 2020 at 11:36
  • $\begingroup$ @S.H.W: Because we want a causal solution and a stable solution. As I've mentioned, for causal we need $y(t)=0$ for $t<0$, but for stable we need the exponential terms to decay, so each of the constants is zero for either $t>0$ or $t<0$. With "different" I meant either zero or non-zero. This gives us the general forms of the solutions $(3)$ and $(4)$. The values of $c_1$ and $c_2$ are now determined by satisfying the differential equation with $x(t)=\delta(t)$. $\endgroup$
    – Matt L.
    Commented Nov 2, 2020 at 11:59
  • $\begingroup$ Thank you so much. As an engineering point of view, your solution is perfect but I'm not sure whether it's a rigorous mathematical solution since $\delta(t)$ is not a function and it should be understood as a distribution. $\endgroup$
    – S.H.W
    Commented Nov 2, 2020 at 15:19
  • 1
    $\begingroup$ @S.H.W: It is treated as a distribution. The product law is also valid for distributions. $\endgroup$
    – Matt L.
    Commented Nov 2, 2020 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.