Quick question for you electrical engineers

[RichieZ]

If a system is time-invarient, is it also causal?

 

[josphII]

uh oh flashbacks of EE120 at berkeley

i used to know all that stuff but i cant even remember what casual is anymore

define casual for me

 

[RichieZ]

a system in which the response to an input signal is not dependent upon future input values.

 

[MrDingleDangle]

no all causal means is that it is there is no output before 0 time...

time-invarient, i believe, means nothing in the future affects the past (could be slightly off on that def.)

 

[SuperTool]

Yeah, that class made for good sleeping at Cornell. So I don't remember jack from it.

 

[tikwanleap]

(um... man it's been a long time since i've done this stuff.... so i might be way off)

time-invariant means it doesn't matter at what point in time you put in an input signal. the output will still be the same.

See if there is an exception to the rule: look for a time-invariant system that is non-causal.

(i forgot the notation is it like input and output arrays?)

y[n] = x[n] + x[n+1] <--- this is time invariant but not causal...

hope that helps...

 

[RichieZ]

Causal has been defined to me as:
Output does not start before the input starts
and
Output at time n does not depend on inputs" in the future"

 

[ArmenK]

<< If a system is time-invarient, is it also causal? >>



causal means that you don't consider output after the present time t (at least in my class it does), its not the same thing as time invarient

 

[MrDingleDangle]

<< (um... man it's been a long time since i've done this stuff.... so i might be way off)

time-invariant means it doesn't matter at what point in time you put in an input signal. the output will still be the same.

>>



that i believe is the right for time invarient

and it is right that causel means no output till input

 

[tikwanleap]

Ok time to break out the dusty old EE32 books...

Taken from Signals and Systems - Oppenheim and Willsky

Causality


<<
A system is causal if the output at any time depends only on values of the input at the present time and in the past. Such a system is often referred to as being nonanticipative, as the system output does not anticipate future values of the input. Consequently, if two inputs to a causal system are identical up to some point in time t0 or n0, the corresponding outputs must also be equal up to this same time.
>>



Time Invariance


<<
Conceptually, a system is time invariant if the behavior and characteristics of the system are fixed over time. For example, the RC circuit of Figure 1.1 is time invariant if the resistance and capacitance values R and C are constant over time: We would expect to get the same results from an experiment with this circuit today as we would if we ran the identical experiment tomorrow.........

......The property of time invariance can be described very simply in terms of the signals and systems language that we have introduced. Specifically, a system is time invariant if a time shift in the input signal results in an identical time shift in the output signal. That is, if y[n] is the output of a discrete-time, time-invariant system when x[n] is the input, then y[n-n0] is the output when x[n-n0] is applied.
>>



whew, that's alot of stuff there.

if you understand the discrete time notation they are using, here it is in a nutshell:

causal: y[n] = x[n] + x[n-1] + x[n-2]
(all the x input's are in the past)
not causal: y[n] = x[n+1]
(this x is one step in the future)

time invariant: y[n] = x[n+1]
(if x[1] = 10 then y[0] = 10.... and if x[1000] = 10 then y[999] = 10, etc... they are the same input, same output with a shift in time)
not time invariant: y[n] = n * x[n+1]
(if x[1] = 10 then y[0] = 0.... but x[1000] = 10 then y[999] = 999 * 10 = 9990... so different at different points in time)

This one is time-invariant and non-causal which contradicts your conjecture: y[n] = x[n+1]

[edit: fixed typos]

 

[tikwanleap]

So the quick answer to your question is: no