How to conclude LTI, causality and BIBO stability of a system represented by a differential equation?

Question

I have started to learn about systems represented by differential equations in Oppenheim's Signals & Systems, and I got really confused about it. I am trying to understand how I can show that a system is LTI (or not), causal (or not) and BIBO stable (or not) for any input.

I know the following (please correct me if I'm wrong):

1. If the differential equation has zero auxiliary conditions then the system is linear. Therefore, showing linearity is easy.

2. A system is defined by the response to the Dirac input.

3. Causal, LTI $\iff$ initial rest (mathematically meaning if $x(t)=0$ for $t<0$ then $y(t)=0$ for $t<0$).

I am not sure if I got it right but if I'm willing to use #3 in order to show that the system is causal and LTI, then it has to have initial rest for any input. How can something like that be shown if as I say it has to be for any $x(t)$? I thought that maybe using #2 will be helpful (solving for the differential equation for delta function and then somehow to show), can it be?

In conclusion, I would thankful if you explain to me how show those features when I got a system that represented by differential equation. For example, in case that we got that simple ODE:

$$\frac{d^2y}{dt^2}-3\frac{dy}{dt}+2y(t)=x(t) \ \ \ ,y(0)=\frac{dy}{dy}=0$$

Answer 1

I take your equation: $\frac{d^2y}{dt^2}-3\frac{dy}{dt}+2y(t)=x(t)$

Laplace transform will be: $ s^2Y(S) - 3sY(S) +2Y(s) = X(s) $

Now I can find transfer function: $H(s) \triangleq \frac{Y(s)}{X(s)}$

$H(s)=\frac{1}{s^2-3s+2}$

Poles are the solution of denominator = 0 and zeros are the solution of numerator = 0. In this case there are not zeros because numerator is always equal to 1, denominator = 0 is a second order equation that can be easily solved.

$ Poles[H(s)] = \{s=1 ; s=2\} \\ Zeros[H(s)] = \{\} $

There are two real poles.

To determine if the system is BIBO stable you can see if all poles have real part $ < 0 $. In this is example there is at least one pole that has real part $ \geq 0 $ (both in reality) so this system is UNSTALBE.

To be causal the number of zeros must be $\leq$ of the number of poles in this case we have $ 0 \leq 2 $ that it's true and means that your system is CAUSAL.

Time invariant system is a system that doesn't depends on absolute time but only on relative time from time origin. In a transfer function generally there are not information to determine if it's a time invariant system or not beacuse it should be an external specification of your problem. Btw in DSP you consider systems time invariant almost ever.

Finally this is only a quick overview of zero-pole analysis you should study the argument as well and the prove of the concepts I used.