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My book with answers to proakis and manolakis book says that $F(x,n)=x(-n + 2)$ 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?

First we shift output signal:

$$F(x, n - k) = x(-(n-k) + 2) = x(-n+k+2)$$

now lets put $x'(n) = x(n-k)$ and check what happen when we input shifted signal:

$$F(x', n) = x'(-n+2)=x(-n+2-k)=x(-n-k+2)$$

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

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  • $\begingroup$ Here too, why the down vote? The question is well posed, shows prior research and has a non-trivial answer. $\endgroup$
    – Jazzmaniac
    May 15 '15 at 20:38
  • $\begingroup$ Please see my answer to your other question regarding time invariance. $\endgroup$
    – Jazzmaniac
    May 15 '15 at 20:39
  • $\begingroup$ You're right, a system with input-output relation $y[n]=x[-n+2]$ is time varying. Any system $y[n]=x[an+k]$ (with integer $k$) and $a\neq 1$ is time varying. $\endgroup$
    – Matt L.
    May 16 '15 at 7:58
  • $\begingroup$ Since I don't understand why the above comments seem to favor: not time invariant; please state what mathematical definition you are using. In my experience if that formula refers to a system response it is certainly time-invariant. If it refers to measuring something like standard deviation of a signal it would be: not time invariant. Normally the term "time invariant" applied to a signal means the statistics (mean, s.d, etc) don't change. $\endgroup$
    – rrogers
    May 20 '15 at 13:37
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Your reasoning is correct and we arrive at the same conclusion.

The system's output is given by

$$y[n]=x[-n+2]\tag{1}$$

Let

$$x_2[n]=x[n-k]\tag{2}$$

The corresponding output is

$$y_2[n]=x_2[-n+2]=x[-n-k+2]\tag{3}$$

From $(1)$ the shifted output is

$$y[n-k]=x[-(n-k)+2]=x[-n+k+2]\tag{4}$$

Since $(3)$ and $(4)$ are not identical, the system is time-varying.

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  • $\begingroup$ @msm: Time reversal is replacing $n$ by $-n$, as simple as that. $\endgroup$
    – Matt L.
    Sep 20 '16 at 19:33

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