1
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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?

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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 $c_1=c_2=-\frac25$.

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  • $\begingroup$ 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 Nov 2 '20 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. Nov 2 '20 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 Nov 2 '20 at 15:19
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    $\begingroup$ @S.H.W: It is treated as a distribution. The product law is also valid for distributions. $\endgroup$ – Matt L. Nov 2 '20 at 15:36

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