Causal Systems: Understanding the Basics

In summary, the system defined by y(t) = x(t-1) is causal although x(t-1) is something else than x(t).
  • #1
Tom McCurdy
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We have been going over causal systems and I am still having trouble determining what defines a system to be causal.

I was told that if the input is anything besides x(a*t) where a=1 then the system is non causal. I can kind of see this, but it is still a bit blurry for me. I also was wondering if that would still apply if you removed t directly from the input equation...

say like if you had [tex] y(t) = \int_{-\infty}^{t}x(5{\tau}) d\tau [/tex]

then is this automatically not causal because of the the 5 coefficient on the inside of x()
 
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  • #2
Tom McCurdy said:
I was told that if the input is anything besides x(a*t) where a=1 then the system is non causal.

That's not true.

The system defined by [tex]y(t) = x(t-1)[/tex] is causal although [tex]x(t-1)[/tex] is something else than [tex]x(t)[/tex].

The general definition for a causal system (linear or non-linear, time-invariant or time-variant) is:

Given 2 input signals [tex]x_1(t)[/tex] and [tex]x_2(t)[/tex] such that [tex]x_1(t) = x_2(t)[/tex] for any [tex]t < t_0[/tex], the system is causal if the output signals [tex]y_1(t) = y_2(t)[/tex] for any [tex]t < t_0[/tex]

If the system is linear then if we apply a signal [tex]x(t) = x_1(t) - x_2(t)[/tex] the output should be [tex]y(t) = y_1(t) - y_2(t)[/tex], so the condition for the system to be causal (in the case of linear systems) reduces to:

if [tex]x(t) = 0[/tex] for [tex]t<t_0[/tex] then [tex]y(t) = 0[/tex] for [tex]t<t_0[/tex]

If the system is linear and time invariant, the condition for causality reduces to:

[tex]h(t)=0[/tex] for [tex]t<0[/tex]

So depending on the kind of system and your known data you should check one of these conditions.

In the case of [tex] y(t) = \int_{-\infty}^{t}x(5{\tau}) d\tau [/tex] I know that it's linear because it's defined by an integral which is a linear operation so I will check the second condition.
I pick an instant [tex]t_0[/tex] at which the output will be [tex] y(t_0) = \int_{-\infty}^{t_0}x(5{\tau}) d\tau [/tex]

So we see that the output depends on values of [tex]x(t)[/tex] till [tex]5t_0[/tex] but
we know that [tex]x(t) = 0[/tex] only for [tex]t<t_0[/tex] and thus the output will not be 0 for any [tex]t<t_0[/tex] which means that the system is not causal.
 
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  • #3
i sort of like the Wikipedia definition of causal system (i had a hand in it before they kicked me out of Wikipedia):

A causal system (also known as a ... nonanticipative system) is a system where the output [itex]y(t)[/itex] at some specific instant [itex]t_{0}[/itex] only depends on the input [itex]x(t)[/itex] for values of [itex]t[/itex] less than or equal to [itex]t_{0}[/itex] . Therefore these kinds of systems have outputs and internal states that depends only on the current and previous input values.

The idea that the output of a function at any time depends only on past and present values of input is defined by the property commonly referred to as causality.
antonantal said:
If the system is linear and time invariant, the condition for causality reduces to:

[tex]h(t)=0[/tex] for [tex]t<t_0[/tex]

i think you can conclude that for an LTI system, causality is equivalent to the impulse response h(t) being zero for all t < 0. t0 is not a parameter of the impulse response. the impulse response is the LTI to a unit impulse applied at t=0.
 
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  • #4
rbj said:
i think you can conclude that for an LTI system, causality is equivalent to the impulse response h(t) being zero for all t < 0. t0 is not a parameter of the impulse response. the impulse response is the LTI to a unit impulse applied at t=0.

You're right of course. I just copied the latex expression above it and forgot to replace t0 with 0 as well. I'll edit it now. Thanks!
 

1. What is a causal system?

A causal system is a system in which the output depends only on the current and past inputs, and not on any future inputs. This means that the output of the system is determined by the inputs that have already been given to it, and not by any inputs that will be given in the future.

2. How do causal systems differ from non-causal systems?

Causal systems differ from non-causal systems in that the output of a non-causal system can depend on future inputs, while a causal system only depends on current and past inputs. This means that a causal system can be analyzed and understood using only historical data, while a non-causal system requires knowledge of future inputs to fully understand its behavior.

3. What are some real-world examples of causal systems?

Some real-world examples of causal systems include weather forecasting models, economic models, and medical diagnosis systems. In each of these examples, the output is determined by current and past inputs, such as previous weather patterns, economic data, or medical symptoms.

4. Why is understanding causal systems important?

Understanding causal systems is important because it allows scientists and researchers to accurately predict and model the behavior of complex systems. By understanding the cause-and-effect relationships within a system, we can make informed decisions and potentially intervene to improve outcomes.

5. How do we mathematically represent a causal system?

A causal system can be represented mathematically using a mathematical model. This model typically consists of input variables, output variables, and a set of equations that describe how the inputs affect the outputs. This allows us to analyze and understand the behavior of the system and make predictions about future outcomes.

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