Linearity, Causality and Stability of a System

Question

Consider a system:

$$ y[n] = y[n-1] + u[n], $$

where $y[n]$ is the output and $u[n]$ is the unit step function.

Is this system causal, linear, time-invariant and stable ?

My attempt at the question is that for the systems without any input, we can never talk about all the properties above as they are related to input/output relationships of the system, whereas there are is no input in the system.

However, I'm inclined to think about the system as follows:

$$ y[n] = y[n-1] + \sum_{i=-\infty}^{\infty} 0 . x[i] + u[n]. $$

So trivially, the system is non-linear as it produces non-zero output for zero input, non-causal as it depends on future inputs, unstable as no matter the input, the output is unbounded and time-variant as if we delay the input, the output stays the same.

Answer 1

According to the interpratation (that belongs to you) that output $y[n]$ is given by the recursion relation $$y[n] = y[n-1] + u[n]$$ no matter what the input $x[n]$ is, then the following conclusions can be made about the linearity, time-invariance, stability and causality of the system.

First let's solve the output for $y[n]$

i.For $n<0$, since $u[n]=0$, we have $$y[n] = y[n-1]$$ which implies that $y[-1]=y[-2]=...y[-N]...= K$ for all $n < 0$

ii.For $n\geq0$, since $u[n]=1$, then we have $$y[n] = y[n-1] + 1$$ which can be recursively solved to get

\begin{align} y[0] &= y[-1] + 1 = K +1 \\ y[1] &= y[0] + 1 = (K+1) +1 = K+2 \\ y[2] &= y[1] + 1 = K +3 \\ &...\\ y[n] &= y[n-1] + 1 = K +(n+1) \\ \end{align}

Combining them yields: $$\boxed{y[n] = K + (n+1) u[n] }$$ for all $n$ as th fixed-unique output of the system for all inputs $x[n]$

Then we analyse the system for its properties:

1. The system is non-linear: It's so evident since the system has a fixed, unique output for every input it's a non-linear system as it cannot satisfy $$\mathcal{T}\{a x_1[n]+ b x_2[n] \} = a \mathcal{T}\{x_1[n] \} + b\mathcal{T}\{x_2[n]\} $$

2. The system is time-varying: Again shifting the input have no effect on the output (which is fixed) so it cannot satisfy $$ y[n-d] = \mathcal{T}\{x[n-d] \}$$ NOTE: Does this make any sense? The system is already a fixed one. It has a fixed output for every input. Hence effectively nothing varies in its output. Nevertheless I called it as time-varying, because evidently it cannot saitsfy the time-invariance constraint.

3. The system is causal: it only depends on the past value of the output and it does not depend on any value of the input so it does not violate causality.

4. The system is unstable: as for a bounded input $x[n] = \delta[n]$ the output is $y[n] = K + (n+1) u[n]$ which grows unbounded as $n$ goes to infinity.

5. The system is not invertible.

6. The system can be implemented with and without any memory requirement.

Answer 2

Although this question is over 2 years old now, I think it's interesting to consider the solution, assuming that the original interpretation was incorrect, and that $u[n]$ represents the system input, not the unit step sequence. If so, then the first thing to recognize is that the system is just an accumulator. For a time-domain demonstration, simply make repeated substitutions into the difference equation.

$$\begin{array}{rcl} y[n] & = & y[n-1] + x[n] \\ & = & y[n-2] + x[n-1] + x[n] \\ & = & y[n-3] + x[n-2] + x[n-1] + x[n] \\ & = & y[n-4] + x[n-3] + x[n-2] + x[n-1] + x[n] \\ & \vdots & \\ y[n] & = & \displaystyle \sum_{m=0}^\infty x[n-m] \end{array}$$

You can also show this using the $z$-transform,

$$\begin{array}{rcl} Y(z) & = & z^{-1} Y(z) + X(z) \\ H(z) = \dfrac{Y(z)}{X(z)} & = & \dfrac{1}{1 - x^{-1}} \\ h[n] & = & \mbox{unit step sequence} \\ \Rightarrow y[n] & = & \displaystyle \sum_{m=0}^\infty x[n-m]. \end{array}$$

At this point, one can use the standard techniques to test if the system is linear, time-invariant, causal and stable.

Linearity: Define two input signals $x_1[n]$ and $x_2[n]$, with corresponding outputs $y_1[n]$ and $y_2[n]$,

$$\begin{array}{rcl} y_1[n] & = & \displaystyle \sum_{m=0}^\infty x_1[n-m] \\ y_2[n] & = & \displaystyle \sum_{m=0}^\infty x_2[n-m] \end{array}$$

Now define a third input signal that is a linear combination of the others, $$ x_3[n] = a x_1[n] + b x_2[n]. $$

The output in response to $x_3[n]$ is

$$\begin{array}{rcl} y_3[n] & = & \displaystyle \sum_{m=0}^\infty x_3[n-m] \\ & = & \displaystyle \sum_{m=0}^\infty \left( a x_1[n-m] + b x_2[n-m] \right) \\ & = & a \displaystyle \sum_{m=0}^\infty x_1[n-m] + b \displaystyle \sum_{m=0}^\infty x_2[n-m] \\ & = & a y_1[n] + b y_2[n]. \end{array}$$

Therefore the system is linear.

Time-Invariance: Using $x_1[n]$ and $y_1[n]$ as an input-output pair, define the input $$x_2[n] = x_1[n-N]$$.

The corresponding output is $$ y_2[n] = \displaystyle \sum_{m=0}^\infty x_2[n-m] = \displaystyle \sum_{m=0}^\infty x_1[n-N-m] = y_1[n-N]. $$

Therefore the system is time-invariant.

Causality: The system is causal by inspection, as $y[n]$ depends on $x[n-m]$ only for $m \geq 0$.

Stability: Let $x[n]$ be the bounded input signal that is 1 for all $n$. the output $y[n]$ is the sum of all present and previous inputs, which will diverge to infinity for all $n$. Therefore the system is unstable.