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Check if the following system is linear time invariant and BIBO stable.. $$ y[n] = ny[n-1] + x[n] $$

for $n\ge 0$. We are also given that the system is at rest (i.e. $y[−1] = 0$).

I know that to show linearity, we show that $ H[ax_1 + bx_2] = aH[x_1] + bH[x_2] $ except that the question has a $y[n-1]$ on the RHS which is throwing me off.

Edit: My solution: $ LHS = $ $$H[ax_1[n] + bx_2[n]] $$$$ = n(ay_1[n−1] + by_2[n−1]) + (ax_1[n]+bx_2[n]) $$$ RHS =$$$ aH[x_1] + bH[x_2] $$$$ = any_1[n−1]+ax_1[n]+bny_2[n−1]+bx_2[n] $$ So $ RHS = LHS $

Edit 2: New issue when I try and show time invariance.

If $ x_2[n] = x1[n-k] $ then prove $y_2[n] = y_1[n-k]$.

BUT $RHS= $ $$ (n-k)y_1[n-k-1] + x_1[n-k] $$ And $LHS = $ $$ny_2[n-1] + x_2[n] = ny_2[n-1] + x_1[n-k]$$

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  • $\begingroup$ In your attempt to show linearity, isn't there a factor $n$ missing in your RHS? $\endgroup$
    – Matt L.
    Jun 17, 2018 at 9:48
  • $\begingroup$ Apologies, yes it was missing. Fixed now. $\endgroup$
    – Nobody Special
    Jun 17, 2018 at 10:03
  • $\begingroup$ So the answer should be clear now, isn't it? $\endgroup$
    – Matt L.
    Jun 17, 2018 at 10:05
  • $\begingroup$ Thanks, its clear now for linearity, (I was unsure of my method before).. However, I still fail to see the proof for time invariance :\ $\endgroup$
    – Nobody Special
    Jun 17, 2018 at 10:13
  • $\begingroup$ Well, you did prove something by showing that RHS$\neq$LHS, i.e., the response to $x[n-k]$ does not equal $y[n-k]$, where $y[n]$ is the response to $x[n]$. What does that tell us? $\endgroup$
    – Matt L.
    Jun 17, 2018 at 10:16

1 Answer 1

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I'll give you some hints that hopefully will allow you to do your homework yourself.

In order to check linearity, write down $3$ difference equations: one for the response $y_1[n]$ to an input signal $x_1[n]$, one for the response $y_2[n]$ to an input signal $x_2[n]$, and one for the response $y_3[n]$ to an input signal $ax_1[n]+bx_2[n]$. Multiply the first difference equation by $a$, the second one by $b$, and add them. If the resulting difference equation for the sequence $ay_1[n]+by_2[n]$ is the same as for $y_3[n]$, then you can conclude $y_3[n]=ay_1[n]+by_2[n]$, and, consequently, the system must be linear.

For checking time-invariance, write down the difference equation for the response $y_2[n]$ to the input $x[n-k]$. Then replace in the original difference equation the index $n$ by $n-k$ and check if the two difference equations are the same. If they are, the system is time-invariant.

To answer the question about BIBO stability, ask yourself what the output signal will be for $x[n]=\delta[n]$ and draw your conclusion.

Now that you've figured out the solution I would like to point out that you can also find an explicit (non-recursive) expression for $y[n]$:

$$\begin{align} y[0]&=x[0]\\ y[1]&=1\cdot x[0]+x[1]\\ y[2]&=2\cdot 1\cdot x[0]+2\cdot x[1]+x[2]\\ y[3]&=3\cdot 2\cdot 1\cdot x[0]+3\cdot 2\cdot x[1]+3\cdot x[2]+x[3]\\ &\vdots\\ y[n]&=\sum_{k=0}^n\frac{n!}{k!}x[k],\qquad n\ge 0\tag{1} \end{align}$$

from which it is very straightforward to show that the system is linear and time-varying. Furthermore, from $(1)$ it should be clear that the system is not BIBO-stable.

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  • $\begingroup$ Ok so my solution for linearity is: RHS: $ay_1[n]+by_2[n] = any_1[n-1] + ax_1[n] + by_2[n-1] + bx_2[n]$ except that doesnt look like $ y_3[n] = n(ay_1[n-1]) + by_2[n-1]) + ax_1[n] + bx_2$ which is the LHS $\endgroup$
    – Nobody Special
    Jun 17, 2018 at 4:25

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