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

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:

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

• Not all text books agree that zero initial conditions are necessary for a system to be LYI – Stanley Pawlukiewicz Apr 16 '18 at 21:38
• You have to study Laplace transform and zero-pole analysis. Btw if you have poles with positive real part your system is not bibo stable, otherwise is stable. To be causal you must have more poles than zeros. – Andrea Apr 16 '18 at 22:09
• @StanleyPawlukiewicz, first thank you for the comment. I think that in my course it's necessary because it's somehow taught using this book. Assuming it is, does I understand 3) (the right to left side) as it should be? that I should show that it's true for all x(t). if indeed, then how something like can be done. Moreover, I thought about casualty and BIBO stability, and I think that finding the response to Dirac will give me someway to show it (for casualty it is immediate). which left me with how to show time invariant in case like this. thank you. – Mr.OY Apr 16 '18 at 22:16
• @Andrea, true i didn't get there yet. does studying it will reveal also a way to show time invariant, or that it's not mention in your comment in porrupse? – Mr.OY Apr 16 '18 at 22:19
• Time invariant means only that your system doesn't change if you use it in a different moment. For example if you put a CD in your CD player you make a time invariant system, if you tap "play" now or tomorrow the output will be the same. In DSP you consider your system time invariant almost ever. – Andrea Apr 16 '18 at 22:26

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.