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$y[n]=T{x[n]}=\displaystyle\sum_{k=n-n_{0}}^{n+n_{0}} x[k]$

it is some sort of moving summer which computes $n^{\text{th}}$ output sample by adding all samples lying within length $n_{0}$ around some point $n$ on $k$ -axis (where k is dummy variable) so even if we shift input by amount $n_{0}$ the length of interval of summation remain unchanged which is same as if we're shifting output by $n_{0}$ from which it follows system is Time invariant but how do we prove it mathematically .

my work so far

shifting input by $n_{0}$ , i.e, $T{x[n-n_{0}]}=\displaystyle\sum_{k=n-n_{0}}^{n+n_{0}} x[k-n_{0}]$

now, $k\to k+n_{0}$ then,

LHS$=\displaystyle\sum_{k=n-2n_{0}}^{n} x[k]\neq y[n-n_{0}]$

here i'm getting the answer Time variant. can anyone help ?

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You are overloading the symbol $n$ which doesn't mean the same everywhere in your calculations. Instead, write down what the sequence of input values is and the corresponding sequence of output values when the input is $x$. DO NOT use the symbol $n$ anywhere ($n_0$ is OK to use). So, your answer should look like

\begin{align} y[-2] &= \cdots\\ y[-1] &= \cdots\\ y[0] &= x[-n_0] + x[-n_0+1] + \cdots + x[0] + x[1] + \cdots x[n_0]\\ y[1] &= x[-n_0+1] + \cdots + x[0] + x[1] + \cdots x[n_0+1] \end{align} etc.

Now, shift $x$ by $k$ places (say $k=1$ for starters) and call the result $\hat x$. Repeat the above computation for the input $\hat x$ and call the result $\hat y$. Is $\hat y$ the same as $y$ except shifted by $k$ places? Repeat for $k+1$ etc until you have managed to convince yourself that regardless of what $k$ you choose, shifting $x$ by $k$ places results in $y$ also shifting by $k$ places.

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  • $\begingroup$ +1 you are amazing $\endgroup$ – Faraday Pathak Aug 4 '19 at 7:42
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In addition to Dilip's hands on answer, let me correct the theoretical path that you wanted to follow but failed.

The input output relation of the system is given by :

$$ y[n] = T\{ x[n] \} = \sum_{k=n-n_0}^{n+n_0} x[k] \tag{1}$$

First, shift the input $x[n]$, by integer $d$, denoted as $$x_d[n] = x[n-d] \tag{2}$$ and see its effect on the output $y[n]$; denoted as $y_d[n]$ :

$$ y_d[n] = T\{x_d[n]\} = \sum_{k=n-n_0}^{n+n_0} x_d[k] \tag{3} $$

Replace $x_d[n]$ with $x[n-d]$ and $x_d[k]$ with $x[k-d]$ according to Eq.2. $$ y_d[n] = T\{x[n-d]\} = \sum_{k=n-n_0}^{n+n_0} x[k-d] \tag{4} $$

Second, shift the output $y[n]$ by $d$ and see its effect on the sum:

$$y[n-d] = \sum_{k=n-d-n_0}^{n-d+n_0} x[k] \tag{5} $$

Now, make a manipulation of dummy summation index $k$ in either of Eqs. 4 or 5 to see that they are the same; i.e.,

For example, let $m = k+d$ and place it into Eq.5: $$y[n-d] = \sum_{ m = n-n_0}^{n+n_0} x[m-d] \tag{6} $$

and simply reset the dummy variable of summation from $m$ back to $k$ in Eq.6, to convince yourself that Eq.6 becomes Eq.4:

$$y[n-d] = \sum_{ k = n-n_0}^{n+n_0} x[k-d] = y_d[n] \tag{7} $$

Finally, Eq.7 declares that $y[n-d] = y_d[n]$; the system is Time-Invariant.

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    $\begingroup$ +1 nice finish .i mistook dummy variable for $n_{0}$. Your style matches with 'Oppenheim's DSP book' $\endgroup$ – Faraday Pathak Aug 4 '19 at 7:41

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