# Causality Condition(Continuous time LTI systems)

1. Jun 27, 2008

### alextsipkis

Can anybody prove or give a bit detail about the causality condition i.e

h(t) = 0, t<0 for continuous-time LTI system.

And based on this how a continuous time LTI system is causal if,

x(t) = 0, t<0.

2. Jun 28, 2008

### rbj

okay, Alex, i'm gonna try to deal with both questions together. besides getting a good book on what we used to call Linear System Theory (what is currently most often labeled "Signals and Systems"), maybe take a look at this thread before moving on. also, just because it's easier, i'm gonna move your question from continuous-time LTI systems to discrete-time LTI systems.

now, solely because of the fact that an LTI system is linear:

$$\mathbf{L} \left\{ x_1[n] + x_2[n] \right\} = \mathbf{L} \left\{ x_1[n] \right\} + \mathbf{L} \left\{ x_2[n] \right\}$$

(which is synonymous with "superposition applies") and time-invariant:

if $$y[n] = \mathbf{TI} \left\{ x[n] \right\}$$

then $$y[n-m] = \mathbf{TI} \left\{ x[n-m] \right\}$$,

solely because of those axioms, we can show that, for any general input x[n], the output y[n] is

$$y[n] = \sum_m x[m] \ h[n-m] \right\}$$

where h[n] is the impulse respone

$$h[n] = \mathbf{LTI} \left\{ \delta[n] \right\}$$.

this could be causal or acausal. if it is causal, the output cannot depend on any input values that are in the future, that means all of h[n] for any n<0 must be zero for those terms not to show up in the summation. if any one of h[n]<>0 for any n<0, that means the output will be different if some future value of x[n] changes and the output is able to anticipate future input values (which can't happen in a real-time physical system, but we can define processing of one file of data into another where the algorithm "looks ahead" to "future" samples in the input file, but those samples were really recorded sometime in the past).

this could be memoryless or not memoryless, if it is not memoryless, then only h[0] is non-zero. if, in a causal system, any other h[n] (for n>0), then memory is involved because somehow the present output value has to know about past input values (which is remembering them), thus memory is needed.

to convert this to continuous time, look at that thread. what happens is the summation becomes a Riemann integral an these discrete sample values get infinitesimally close to each other.

3. Jun 30, 2008