# Why the following discrete type system,variable accumulator, is time invariant?

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

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 ?

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 &= x[-n_0] + x[-n_0+1] + \cdots + x + x + \cdots x[n_0]\\ y &= x[-n_0+1] + \cdots + x + x + \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.

• +1 you are amazing – Faraday Pathak Aug 4 '19 at 7:42

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.

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