Time Variant System: y(t) = x(-t)

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The discussion centers on determining whether the system defined by y(t) = x(-t) is time invariant or time variant. The method for checking time invariance involves applying a delay to the input and comparing the output with the output of the system after applying a delay. The original poster believes the system is time invariant based on their calculations, but the textbook states it is time variant. Further analysis reveals that when comparing delayed outputs, the results differ, confirming the system is indeed time variant. The confusion arises from the interpretation of the transformation involved in y(t) = x(-t).
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Homework Statement


Is the system: y(t) = x(-t) time variant or time invariant?

Homework Equations


Condition for time invariant:
1) Apply delay to input and check output as Y1
2) Apply input to system without delay and apply delay to output and name it as Y2

If Y1 = Y2, system is Time invariant.

The Attempt at a Solution



We have y(t) = x(-t)
So I tried with example:
T -3 -2 -1 0 1 2 3
x(t) 1 2 3 4 5 6 7
y(t) 7 6 5 4 3 2 1

At time t = 2
First apply delay to input.
Delay of 1
So time becomes 1
x(1) = 5
y(1) = x(-1) = 3-----equation 1

Now at time t = 2
Apply this to system
y(2) = x(-2) = 2
Apply delay of 1
y(1) = 3-----equation 2

Equation 1 and equation 2 are same.
So this is time invariant.
Book says it's time variant equation.
How?

upload_2018-2-6_11-45-51.png

If I go by red line (first delay and then system)
or if I go by blue line (first system and then delay) I get same answer,
so how is this time variant?
 

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jaus tail said:

Homework Statement


Is the system: y(t) = x(-t) time variant or time invariant?

Homework Equations


Condition for time invariant:
1) Apply delay to input and check output as Y1
2) Apply input to system without delay and apply delay to output and name it as Y2

If Y1 = Y2, system is Time invariant.

The Attempt at a Solution



We have y(t) = x(-t)
So I tried with example:
T -3 -2 -1 0 1 2 3
x(t) 1 2 3 4 5 6 7
y(t) 7 6 5 4 3 2 1

At time t = 2
First apply delay to input.
Delay of 1
So time becomes 1
x(1) = 5
y(1) = x(-1) = 3-----equation 1

Now at time t = 2
Apply this to system
y(2) = x(-2) = 2
Apply delay of 1
y(1) = 3-----equation 2

Equation 1 and equation 2 are same.
So this is time invariant.
Book says it's time variant equation.
How?

View attachment 219783
If I go by red line (first delay and then system)
or if I go by blue line (first system and then delay) I get same answer,
so how is this time variant?
I don't quite follow your logic with the red and blue arrows.

Instead, take a step back and write out the sequences, one by one. It's fine to start with your existing example sequences,

Code:
T       -3    -2     -1     0      1      2      3
x(t)     1     2      3     4      5      6      7
y(t)     7     6      5     4      3      2      1

Delay the "y(t)" above to obtain your Y2

Now write out x(t-1).

Then write out y(t-1) based on what you just did with x(t-1). From that you have your Y1.

At this point, compare your Y1 to Y2.
 
Okay. Let's take time = 3.
Apply x(3) to system.
We get output:
y(3) = 1
Delay it by 1 to get y(2) and we get 2.

Now delay time. So t = 3 - 1 = 2.
Apply x(2) to system and we get output:
y(2) = 2.
They still turn out to be same.
 
jaus tail said:
Okay. Let's take time = 3.
Apply x(3) to system.
We get output:
y(3) = 1
Delay it by 1 to get y(2) and we get 2.
I follow that so far.

Now delay time. So t = 3 - 1 = 2.
Apply x(2) to system and we get output:
y(2) = 2.
They still turn out to be same.

Now I'm a little lost. Perhaps start with this:
Code:
T      -3  -2  -1   0   1   2   3
x(t)    1   2   3   4   5   6   7
x(t-1)  0   1   2   3   4   5   6

Now what's y(t-1)?
 
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upload_2018-2-7_10-7-27.png

Is this right? I'm still not sure.
Row 4 and 6 are different so it's time variant.
 

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jaus tail said:
View attachment 219860
Is this right? I'm still not sure.
Row 4 and 6 are different so it's time variant.
Yeah, I think that's the right idea. :smile:

(For what it's worth, I appreciate the confusion though. Personally, I think this problem is a little ambiguous in what is actually meant by "y(t) = x(-t)" and exactly how that is accomplished in the system. That said, demonstrating that rows 4 and 6 are different is probably what your instructor/coursework is expecting.)
 
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The coursework actually requires using x(t)--->y(t)---->y(t-to)
and then x(t)---> replace all t by t-t0 and then find out y(t-to)
Even google suggested replacing t by t-t0 in which I never understood how the minus sign spreads.
That's why I prefer the above method of making rows and columns.
 

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