# Time invariant or time varying?

#### Zazoo

Joined Jul 27, 2011
114
The problem given in my textbook is:

y(t) = x(t-5) - x(3-t)
Show that this system is time-invariant.

Um, isn't this system time varying?
The second term involved a time inversion, doesn't this make it time varying?
e.g. g(3-t) = g(-(t-3))

That is:
Let x1(t) = g(t), Then y1(t) = g(t-5) - g(3-t)
Let x2(t) = g(t-t0), Then y2(t) = g(t-t0-5) - g(3-t-t0)

y1(t-t0) = g((t-t0)-5) - g(3-(t-t0)) = g(t-t0-5) - g(3-t+t0) ≠ y2(t)

Thus the system is time-varying.

Am I wrong or is the book?

Thanks

#### steveb

Joined Jul 3, 2008
2,436
The problem given in my textbook is:

y(t) = x(t-5) - x(3-t)
Show that this system is time-invariant.

Um, isn't this system time varying?
The second term involved a time inversion, doesn't this make it time varying?
e.g. g(3-t) = g(-(t-3))

That is:
Let x1(t) = g(t), Then y1(t) = g(t-5) - g(3-t)
Let x2(t) = g(t-t0), Then y2(t) = g(t-t0-5) - g(3-t-t0)

y1(t-t0) = g((t-t0)-5) - g(3-(t-t0)) = g(t-t0-5) - g(3-t+t0) ≠ y2(t)

Thus the system is time-varying.

Am I wrong or is the book?

Thanks
You just have a slight algebra error in calculating y2. It should be as follows.

e.g. g(3-t) = g(-(t-3))

That is:
Let x1(t) = g(t), Then y1(t) = g(t-5) - g(3-t)
Let x2(t) = g(t-t0), Then y2(t) = g(t-t0-5) - g(3-t+t0)

y1(t-t0) = g((t-t0)-5) - g(3-(t-t0)) = g(t-t0-5) - g(3-t+t0) = y2(t)

#### Zazoo

Joined Jul 27, 2011
114
You just have a slight algebra error in calculating y2. It should be as follows.

e.g. g(3-t) = g(-(t-3))

That is:
Let x1(t) = g(t), Then y1(t) = g(t-5) - g(3-t)
Let x2(t) = g(t-t0), Then y2(t) = g(t-t0-5) - g(3-t+t0)

y1(t-t0) = g((t-t0)-5) - g(3-(t-t0)) = g(t-t0-5) - g(3-t+t0) = y2(t)
I did this on purpose because my understanding is that when placing the delay before the transfer function, any time-scaling/time-inversion should be applied to t only (and not to the shift t0). For example, in another example the function is given:

y(t) = x(t/2)

With the delay before the transfer function, e.g. x(t-t0), they give y(t) = x(t/2 - t0)
If the delay is after the transfer function, then y(t-t0) = x((t-t0)/2)
Thus the system is time-variant.

Am I wrong in this interpretation?

#### steveb

Joined Jul 3, 2008
2,436
I did this on purpose because my understanding is that when placing the delay before the transfer function, any time-scaling/time-inversion should be applied to t only (and not to the shift t0). For example, in another example the function is given:

y(t) = x(t/2)

With the delay before the transfer function, e.g. x(t-t0), they give y(t) = x(t/2 - t0)
If the delay is after the transfer function, then y(t-t0) = x((t-t0)/2)
Thus the system is time-variant.

Am I wrong in this interpretation?
Yes, if i'm understanding you correctly, you are wrong in this interpretation. You are interested in redefining the time variable by adding a shift. There are two ways to do this. You can shift this input function and run it through the system. Or, you can run the unshifted function through the system and shift the output. In a time invariant system these two methods will agree. The substitution is t=t'-t0, where t0 is the time shift.

#### Zazoo

Joined Jul 27, 2011
114
Yes, if i'm understanding you correctly, you are wrong in this interpretation. You are interested in redefining the time variable by adding a shift. There are two ways to do this. You can shift this input function and run it through the system. Or, you can run the unshifted function through the system and shift the output. In a time invariant system these two methods will agree. The substitution is t=t'-t0, where t0 is the time shift.
Ok, can you help me spot the flaw in my reasoning here:

I picked a fairly simple, arbitrary piecewise function for x(t), just to have some real numbers to play with:

$$x(t)=\begin{cases}2 & : -1 < t < 2\\1 & : 2 < t < 4\end{cases}$$

Applying a delay/shift of 2 before the transfer function:

$$g(t) = x(t-2) =\begin{cases}2 & : 1 < t < 4\\1 & : 4 < t < 6\end{cases}$$

Then running it through a function like: y(t) = g(3-t) gives:

$$y(t) =\begin{cases}2 & : -1 < t < 2\\1 & : -3 < t < -1\end{cases}$$

Now, this is clearly a different function than that obtained by a time delay/shift after the transfer function:

$$y(t-2) = x(3-(t-2)) =\begin{cases}2 & : 3 < t < 6\\1 & : 1 < t < 3\end{cases}$$

Suggesting to me that this system, y(t) = x(3-t), is time-variant.

I'm not seeing where I am making my mistake and I'm really confused. Thank you,

#### steveb

Joined Jul 3, 2008
2,436
I'm not seeing where I am making my mistake and I'm really confused. Thank you,
It is confusing. You know, I get in more trouble trying to answer confusing questions at 2:00 AM in the morning.

I see your point. Forgetting the math, just think about what time reversal does. A signal delay on the input, drives the output further into the past. So, this is anti-time-invariant, if I can make up my own terminology. This type of system is also non-causal, so we don't see them too often.

I'm going to do a Google search later just to verify with an authority, but right now I agree with you.

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#### steveb

Joined Jul 3, 2008
2,436
I'm going to do a Google search later just to verify with an authority, but right now I agree with you.
So, consultation of some text books and a Google search did not reveal a definitive statement that time reversal implies a time varying system, but I still think you are correct.

To make it more concrete, let's try to look at a simple time reversal system y(t)=x(-t), and put it in a form that makes the time dependence of the system itself more obvious. This is actually quite simple as follows.

y(t)=x(t-t0) where t0=2t

Now compare this to a transmision line whose length is varying over time sinusoidally (maybe it's stretching and compressing due to oscillatory movement).

y(t)=x(t-t0) where t0=a+b*sin(wt), where "a" is a constant and "b" is a much smaller constant

When expressed in this way, it's clear that both systems take the input and delay it by a time delay which is itself time dependent. Hence, both systems are time dependent systems.

The case of the transmission line is a causal system that can be built, but the pure time reversal case is noncausal because at negative values of time, the system depends on future values of the input signal. However, you could modify the time reversal to create a causal system as follow.

y(t)=x(t-t0) where t0=2t for t>0 and t0=0 for t<=0

This would be like a transmission line that continually stretches in length, hence causing greater and greater delay as time moved forward.

• Zazoo

#### Zazoo

Joined Jul 27, 2011
114
Thank you for your responses steveb, I'm starting to see it clearer now, this has really helped me out.

#### waterineyes

Joined Sep 6, 2014
9
This system must be Time-Variant, as x(3-t) is making it Time-variant..

You must know that y(t) = x(-t) is a Time-variant system, so y(t) = x(3-t) would certainly be a Time varying system..

y_1(t) = x_1(t) = x(t - k - 5) - x(3 -t - k) (Here, Input is shifted or delayed to (t - k))

Shifting y(t) to y(t - k),

y_2(t) = y(t - k) = x(t - k - 5) - x(3 - t + k)

So, in y_1(t) and y_2(t), there is just a difference of sign in front of "k"..

So, y_1(t) is not equal to y_2(t).

So, this system must be Time-Variant..

@Zazoo, which book are you following where this system is given as Time-Invariant?? May I know that??

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#### appsblue

Joined Sep 8, 2014
6
In my textbook it is given that y[n]=x[2n] is a time invariant system.
but when i tried solving it i got it time varient.
I am really confused. can any one help me please

#### waterineyes

Joined Sep 6, 2014
9
In my textbook it is given that y[n]=x[2n] is a time invariant system.
but when i tried solving it i got it time varient.
I am really confused. can any one help me please
Let me try it :

y_1[n] = x_1[2n] = x[2n - k]

y_2[n] = y[n - k] = x[2n - 2k], that means : y_1[n] is not equal to y_2[n]

Absolutely, it is Time-Variant system..

Which book you are studying? May I know its name??

#### appsblue

Joined Sep 8, 2014
6
It
Let me try it :

y_1[n] = x_1[2n] = x[2n - k]

y_2[n] = y[n - k] = x[2n - 2k], that means : y_1[n] is not equal to y_2[n]

Absolutely, it is Time-Variant system..

Which book you are studying? May I know its name??
Is just a local auther's book
I refered it for the sum for practice

#### appsblue

Joined Sep 8, 2014
6
It

Is just a local auther's book
I refered it for the sum for practice

#### appsblue

Joined Sep 8, 2014
6
I