My book with answers to proakis and manolakis book says that $F(x,n)=x(n)u(n)$ is time invariant while my calculation shows that it is time variant. I have repeated them many times and I am mostly convinced that book is wrong. Can somebody please check my calculations?

let $u(n)$ be a Heavside step function. First we shift output signal:

$$F(x, n-k)=x(n-k)u(n-k)$$

Let $x'(n)=x(n-k)$ be a shifted input signal. Lets calculate response to shifted input signal:

$$F(x', n)=x'(n)u(n)=x(n-k)u(n)$$

since $F(x, n - k) \neq F(x', n)$ this system is time variant. Is my method correct?

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    $\begingroup$ Why the downvote? This is a very valid question. $\endgroup$ – Jazzmaniac May 15 '15 at 20:37

The answer to this (and your other similar question) is most likely found in the somewhat unusual notation used for the system.

In ordinary terms, if you just take your system as a map between input and output signal, then you're absolutely right, it would not be time invariant.

However, Proakis defines the system as a map from an input signal and a time coordinate to the output signal. If you want, the time coordinate itself can be understood as a second signal that just happens to contain the time index at each time index. Consequently, the time shift operator would act on this system definition by both shifting the proper signal and the time coordinate, so that the the system turns out to be time invariant.

So let me summarise: The system map $F(x,n)=x(n)u(n)$ is time invariant whereas the map $F(x)=x(n)u(n)$ is not time invariant.

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  • $\begingroup$ The $F(x,n)$ notation is my own creation which makes dependence on $x$ explicit. Proakis uses more standard notation e.g. $y(n) = x(n)u(n)$. I do not understand why you claim that time invariance depends on notation. Both things $F(x,n)$ and $y(n)$ mean same thing but former one shows explicitly that formula depends also on $x$ not only on $n$. Can you explain a little bit more taking in account that fact and optionally correcting the notation? $\endgroup$ – Trismegistos May 15 '15 at 20:47
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    $\begingroup$ In mathematics, notation is everything. The difference between writing $F(x)$ and $F(x,n)$ is that the first is a map $S \to S$ and the second a map $S \times S \to S$, where $S$ is the signal space. A time shift operator would act differently on both. The key difference is that in the first definition, n as it appears in u(n) is treated as a fixed external parameter, and in the first definition as part of the preimage of the map. $\endgroup$ – Jazzmaniac May 15 '15 at 21:02
  • $\begingroup$ A similar thing can be seen for a system $y[n]=x[n]\cdot v[n]$. If $x[n]$ is the input signal, and $v[n]$ is considered part of the system (like in a modulator) then the system is linear (though time varying). However, if $v[n]$ is considered as another input signal, then the system is non-linear. $\endgroup$ – Matt L. May 16 '15 at 8:07
  • $\begingroup$ @MattL. but acording to notation without $x$ as variable i.e. $y(n)=...$ it does not make any sense to speak about time invariance because you can not change input signal $x$ to check how system react to shifted $x$ because as MattL. said $x$ is part of the system. So notation without $x$ as variable is really notational abuse though widely used. $\endgroup$ – Trismegistos May 16 '15 at 9:20
  • $\begingroup$ @MattL. I have a document which explains quirks of notation web.eecs.umich.edu/~fessler/course/451/l/pdf/c2.pdf page 7. $\endgroup$ – Trismegistos May 16 '15 at 9:23

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