Causality Condition(Continuous time LTI systems)

  • #1

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
rbj
2,226
7
The condition for LTI system to be memoryless is of the form y(t) = Kx(t), and the impulse response will be the same with x(t) replaced by delta function.

Further, it says that if h(tx) is not equal to zero for tx is not equal to zero, the continuous-time LTI system has memory. Can somebody put more insight into this?

I wanted to have a more detail about LTI systems with and without memory
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.

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:

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

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

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

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

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

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

where h[n] is the impulse respone

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

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
Thanks rbj for your detailed reply.

Ever since you posted, i was reading and trying to get my concepts clear and finally almost everything is clear then before.
 

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