Quick question about Causal LTI system

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In summary, the conversation discusses the concept of causality in LTI systems and how it relates to the impulse function h(t). Being causal means that the system cannot depend on the future, so h(t) must be zero for all t<0. This is because an impulse function represents the response of the system to a unit impulse at t=0, and a causal system cannot have a non-zero response before t=0. Various examples and explanations are given to illustrate this concept.
  • #1
ElijahRockers
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Homework Statement



I am reading a little bit on Causal LTI systems. I know causality means that a system cannot depend on the future, but why does that mean an impulse function, say

h(t) = 0, for all t<0

I mean, if t is negative, then it's not really depending on the future, but the past right? Or am I visualizing this wrong?
 
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  • #2
ElijahRockers said:

Homework Statement



I am reading a little bit on Causal LTI systems. I know causality means that a system cannot depend on the future, but why does that mean an impulse function, say

h(t) = 0, for all t<0

I mean, if t is negative, then it's not really depending on the future, but the past right? Or am I visualizing this wrong?

h(t)=0 for all t<0 isn't really depending on the past either. It doesn't depend on anything. You just said what it is. There's no evolution equation to specify any time evolution.
 
  • #3
Dick said:
h(t)=0 for all t<0 isn't really depending on the past either. It doesn't depend on anything. You just said what it is. There's no evolution equation to specify any time evolution.

It doesn't depend on anything? But I thought being being a function of t by definition means it depends on t. Sorry if that sounds obtuse, I just can't relate anything you said to anything I've learned..., apparently I am struggling with this.

I just don't see why the system, being causal, means that h(t) has to be 0 for all t<0...

assuming I have some impulse h[n] = h1[n]+h1[n-1] and I know h[n] = 2 for n>0

then at n=0, 2 = h1[0] + h1[-1]. since the system is causal, i could say h1[0] = 2, because h1[-1] = 0. what i don't get is WHY i can say h1[-1]=0...

I hope that makes sense.. it's hard for me to give an example of something I don't really understand in the first place
 
  • #4
ElijahRockers said:
It doesn't depend on anything? But I thought being being a function of t by definition means it depends on t. Sorry if that sounds obtuse, I just can't relate anything you said to anything I've learned..., apparently I am struggling with this.

I just don't see why the system, being causal, means that h(t) has to be 0 for all t<0...

assuming I have some impulse h[n] = h1[n]+h1[n-1] and I know h[n] = 2 for n>0

then at n=0, 2 = h1[0] + h1[-1]. since the system is causal, i could say h1[0] = 2, because h1[-1] = 0. what i don't get is WHY i can say h1[-1]=0...

I hope that makes sense.. it's hard for me to give an example of something I don't really understand in the first place

I'm just saying you haven't presented the system you are evolving. You need to look at that. Say,
[tex]y(t)=\int_{-\infty}^{\infty} x(t-\tau)h(\tau) d\tau[/tex]
where x is the input, y in the output and h is the response function. If you want to calculate ##y(0)## then that doesn't depend, for example, on ##x(1)## because the contribution of ##x(1)## to the integral happens when ##\tau=(-1)## and being causal says ##h(-1)=0##.
 
  • #5
It was explained to me like this, earlier today:

[itex] y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n-k][/itex]

[itex] y[n] = \sum_{k=-\infty}^{n}x[k]h[n-k] + \sum_{k=n+1}^{\infty}x[k]h[n-k][/itex]
then let h[n] = 0 for n<0

=> h[n-k] = 0 for k>n

=> [itex]\sum_{k=n+1}^{\infty}x[k]h[n-k] = 0[/itex]

So now x[k] only depends on values from -infinity to n, which is the past and present.
 
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  • #6
ElijahRockers said:
It was explained to me like this, earlier today:

[itex] y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n-k][/itex]

[itex] y[n] = \sum_{k=-\infty}^{n}x[k]h[n-k] + \sum_{k=n+1}^{\infty}x[k]h[n-k][/itex]



then let h[n] = 0 for n<0

=> h[n-k] = 0 for k>n

=> [itex]\sum_{k=n+1}^{\infty}x[k]h[n-k] = 0[/itex]

So now x[k] only depends on values from -infinity to n, which is the past and present.

Sure, same thing basically. That's the discrete case.
 
  • #7
Ya, we are doing convolutions with discrete time signals so this is the only kind I've really had exposure to, other than a tiny bit in diff eq. Makes sense now, thanks for your help!
 
  • #8
Another way to think of it is that h(t) is the response of the system to a unit impulse at t=0. A causal system can't have a non-zero response before t=0, otherwise it wouldn't be causal.
 
  • #9
See that's what is tripping me up...

When it's explained simply like that, I just can't picture it. Seeing the sums broken up helps, but I feel there is something crucial about impulses that I am not understanding.
 
  • #10
You want h(t) to be the response only to the impulse, so the system, by assumption, needs to be at rest before the impulse occurs. At t=0, an impulse is imparted to the system. If h(t) isn't zero when t<0, that means either the system wasn't at rest, which by assumption isn't true, or the system began responding to the impulse before it was imparted, which means the system is not causal.
 

1. What is a Causal LTI system?

A Causal LTI (Linear Time-Invariant) system is a type of system in which the output response depends only on the current and past input values, and is not affected by any future input values. It is a fundamental concept in signal processing and is used to analyze and design systems in various fields such as engineering, physics, and economics.

2. What are the characteristics of a Causal LTI system?

The main characteristics of a Causal LTI system are linearity, time-invariance, causality, and stability. Linearity means that the output response is a linear combination of the input signals. Time-invariance means that the system's behavior does not change over time. Causality means that the output does not depend on future inputs. Stability means that the output remains bounded for all bounded inputs.

3. How is a Causal LTI system different from a Non-Causal LTI system?

A Causal LTI system only depends on past and current inputs, while a Non-Causal LTI system can also depend on future inputs. This means that a Causal LTI system has a physical realization, while a Non-Causal LTI system is purely theoretical. Additionally, a Causal LTI system is stable and has a causality property, while a Non-Causal LTI system may not have these properties.

4. What are some examples of Causal LTI systems?

Some examples of Causal LTI systems include an RC circuit, a first-order control system, a linear filter, and a linear time-invariant electronic amplifier. These systems are commonly used in electronic circuits, communication systems, and control systems.

5. How is a Causal LTI system represented mathematically?

A Causal LTI system can be represented mathematically using differential equations, difference equations, or transfer functions. In differential equations, the system's input and output signals are described using derivatives. In difference equations, the signals are described using discrete values. Transfer functions represent the relationship between the input and output signals in the frequency domain.

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