The following sequence is given which is supposed to be time-variant:

$$y[n] = \sum_{k=n_0}^n x[k]$$

I'm having difficulties proving the time-variance or finding a counterexample for it being time-invariant. My idea (which proves it being time-invariant?) is:

\begin{align}y_1[n] &= T\{x_1[n]\}\\x_2[n] &= x_1[n-n_0]\end{align}

check, whether $T\{x_1[n]\} = y_1[n-n_0]$

$$y_2[k] = \sum_{k=n_0}^n x_2[k] = \sum_{k=n_0}^n x_1[k - n_0] = \sum_{k=n_0-n_0}^{n-n_0} x_1[k] = y_1[n-n_0]$$

Where is my mistake? Working around the summation is giving me trouble.

EDIT: Following the advice in the comments for $n = n_0$ the left equation equals to 1 while the right equation equals to 0, making both equations unequal and therefore the sequence time-variant?

Second sequence

This sequence is supposed to be time-invariant.

\begin{align}y[n] &= \sum_{k=n-n_0}^{n+n_0} x[k]\\ y_2[k] &= \sum_{k=n-n_0}^{n+n_0} x_2[k] = \sum_{k=n-n_0}^{n+n_0} x_1[k - n_0] = \sum_{k=n_0-2n_0}^{n} x_1[k] = y_1[n-n_0]\end{align}

I followed the same steps but I'm still not sure about for the case $n = n_0$ since when added in left and right equation

$$\sum_{k=0}^{2n} = \sum_{k=-n}^{n}$$

They seem unequal to me?

  • $\begingroup$ What is the relationship connecting arrays $$ x_2[.] $$ and $$ x_1[.] $$ of your expressions? $\endgroup$
    – V.V.T
    Jul 15, 2020 at 12:14
  • $\begingroup$ Added to original post. $x_2$ is a shifted $x_1$ $\endgroup$
    – Granolaboy
    Jul 15, 2020 at 12:20
  • $\begingroup$ Does the last equation hold if n < n_0? (summation from zero to negative index) $\endgroup$
    – V.V.T
    Jul 15, 2020 at 12:25
  • $\begingroup$ I also thought about that, then the summation equals to 0. However if $n < n_0$ then the left summation also equals to 0 and both the equations are equal? I could possibly find a pair of $n$ and $n_0$ for which the equations are not equal. $\endgroup$
    – Granolaboy
    Jul 15, 2020 at 12:29
  • $\begingroup$ Let n = n_0. Then y[n_0] = x[n_0]. The x series left-shifted by n_0 indices is x[n - n_0]; and the summation is y_shifted[n_0] = x[0]. Right? $\endgroup$
    – V.V.T
    Jul 15, 2020 at 12:54

1 Answer 1


Sequences that are "time-invariant" must be constants (not varying with time at all).

What the OP is actually asking about is a (discrete-time) system whose response $y$ to the input $x$ is defined to be $$y[n] = \begin{cases}\sum_{k=n_0}^n x[k], & n \geq n_0,\\??? & n < n_0\end{cases}$$ where the $???$ is there because according to some traditionalists, if $n < n_0$, then that sum is an empty sum which has value $0$ but more liberal-minded might treat that sum as representing $$x[n] + x[n+1] + x[n+2] + \cdots + x[n_0-1] + x[n_0]$$ (with more serious-minded folks even negating that sum in analogy with the standard result that for $a<b$, $\int_b^a = -\int_a^b$). So, for the traditional meaning of things and $n_0$ denoting a (finite) integer, the system is not a time-invariant system. It is time-invariant (with unit pulse response $h[n]$ being the unit step function $u[n]$) if $n_0 = -\infty$ as noted in a comment by Robert Bristow-Johnson on the main question.

  • $\begingroup$ Would you please explain how the system becomes time-invariant when $n_0 = -\infty$? The left equation is $\sum_{k=\infty}^n x_2[k]$ and the right equation $\sum_{k=0}^{n-(-\infty)} x_1[k] = \sum_{k=0}^{\infty} x_1[k]$ ? $\endgroup$
    – Granolaboy
    Jul 16, 2020 at 6:07
  • $\begingroup$ $- \infty$ on the left equation* $\endgroup$
    – Granolaboy
    Jul 16, 2020 at 6:24
  • $\begingroup$ If $n_0=-\infty$, then \begin{align}y[n]&=\sum_{k=-\infty}^nx[k]\\&=\sum_{k=-\infty}^nu[n-k]x[k]\\&=\sum_{k=-\infty}^\infty u[n-k]x[k]\\&=u\star x\big\vert_n\end{align} where we have used the fact that $u[n-k]=1$ whenever $k$ is such that $n-k\geq 0$, i.e. $k \leq n$ and $u[n-k]=0$ whenever $k$ is such that $n-k<0$, i.e. $k>n$. So, we have expressed the output as a convolution of $x$ with a fixed impulse response (the unit step function) and so we have a time-invariant system, in fact, a linear time-invariant system. $\endgroup$ Jul 16, 2020 at 17:38
  • $\begingroup$ Thank you very much! $\endgroup$
    – Granolaboy
    Jul 19, 2020 at 11:34

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