# Stability of a system in time-domain

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?

Solving the characteristic equation

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

gives the following homogeneous solution:

$$y(t)=c_1e^{-2t}+c_2e^{t/2}\tag{2}$$

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.,

$$y_c(t)=\big[c_1e^{-2t}+c_2e^{t/2}\big]u(t)\tag{3}$$

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 $$t<0$$:

$$y_s(t)=c_1e^{-2t}u(t)+c_2e^{t/2}u(-t)\tag{4}$$

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:

$$\big[f(t)u(t)\big]'=f'(t)u(t)+f(t)u'(t)=f'(t)u(t)+f(t)\delta(t)\tag{5}$$

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

\begin{align}y_c'(t)&=f'(t)u(t)+f(t)\delta(t)=f'(t)u(t)+f(0)\delta(t)\\y_c''(t)&=f''(t)u(t)+f'(t)\delta(t)+f(0)\delta'(t)=f''(t)u(t)+f'(0)\delta(t)+f(0)\delta'(t)\end{align}

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:

$$f(0)=c_1+c_2=0$$

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 $$c_1=c_2=-\frac25$$.

• 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 $t\lt 0$."? – S.H.W Nov 2 '20 at 11:36
• @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 $x(t)=\delta(t)$. – Matt L. Nov 2 '20 at 11:59
• 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. – S.H.W Nov 2 '20 at 15:19
• @S.H.W: It is treated as a distribution. The product law is also valid for distributions. – Matt L. Nov 2 '20 at 15:36