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

[jaus tail]

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?

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?

 

[collinsmark]

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,

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.

 

[jaus tail]

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.

 

[collinsmark]

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:

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)?

 

[jaus tail]

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

 

[collinsmark]

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.

(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.)

 

[jaus tail]

The coursework actually requires using x(t)--->y(t)---->y(t-t_o)
and then x(t)---> replace all t by t-t₀ and then find out y(t-t_o)
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.