Showing Linearity and Time Invariance, or Not

The filter $y(n) = 2 x^2(n)$ is nonlinear and time invariant. The scaling property of linearity clearly fails since, scaling $x(n)$ by $g$ gives $2[gx(n)]^2 = 2g^2x^2(n)$, while $gy(n) = 2gx^2(n)$. The filter is time invariant, however, since delaying $x$ by $m$ samples gives $2x^2(n-m)$ which is the same as $y(n-m)$.

The filter $y(n) = n x(n) + x(n-1)$ is linear and time varying. We can show linearity by setting the input to a linear combination of two signals $x(n) = \alpha x_1(n) + \beta x_2(n)$, where $\alpha$ and $\beta$ are constants:

$$\begin{eqnarray*} n [\alpha x_1(n) + \beta x_2(n)] &+& [\alpha x_1(n-1) + \beta ... ...[n x_2(n) + x_2(n-1)]\\ &\isdef & \alpha y_1(n) + \beta y_2(n). \end{eqnarray*}$$

Thus, scaling and superposition are verified. The filter is time-varying, however, since the time-shifted output is $y(n-m) = (n-m) x(n-m) + x(n-m-1)$ which is not the same as the filter applied to a time-shifted input ( $n x(n-m) + x(n-m-1)$). Note that in applying the time-invariance test, we time-shift the input signal only, not the coefficients.

The filter $y(n) = c$, where $c$ is any constant, is nonlinear and time-invariant, in general. The condition for time invariance is satisfied (in a degenerate way) because a constant signal equals all shifts of itself. The constant filter is technically linear, however, for $c=0$, since $0\cdot(\alpha x_1 + \beta x_2) = \alpha(0\cdot x_1) + \beta(0\cdot x_2) = 0$, even though the input signal has no effect on the output signal at all.

Any filter of the form $y(n) = b_0x(n) + b_1 x(n - 1)$ is linear and time-invariant. This is a special case of a sliding linear combination (also called a running weighted sum or moving average). All sliding linear combinations are linear, and they are time-invariant as well when the coefficients ( $b_0, b_1,\ldots$) are constant with respect to time.

Sliding linear combinations may also include past output samples as well (feedback terms). A simple example is any filter of the form

$$y(n) = b_0 x(n) + b_1 x(n-1) - a_1 y(n-1). \protect$$ (5.9)

Since linear combinations of linear combinations are linear combinations, we can use induction to show linearity and time invariance of a constant sliding linear combination including feedback terms. In the case of this example, we have, for an input signal $x(n)$ starting at time zero:

$$\begin{eqnarray*} y(0) &=& b_0 x(0)\\ y(1) &=& b_0 x(1) + b_1 x(0) - a_1 y(0) =... ...b_1 -a_1 b_0) x(1) - (a_1 b_1 - a_1^2 b_0) x(0)\\ &=& \cdots . \end{eqnarray*}$$

If the input signal is now replaced by $x_2(n)\isdeftext x(n-m)$, which is $x(n)$ delayed by $m$ samples, then the output $y_2(n)$ is $y_2(n)=0$ for $n<m$, followed by

$$\begin{eqnarray*} y_2(m) &=& b_0 x(0)\\ y_2(m+1) &=& b_0 x(1) + b_1 x(0) - a_1 ... ...(b_1 -a_1 b_0) x(1) - (a_1 b_1 - a_1^2 b_0) x(0)\\ &=& \cdots, \end{eqnarray*}$$

or $y_2(n) = y(n-m)$ for all $n\geq m$ and $m\geq 0$. This establishes that each output sample from the filter of Eq. (4.9) can be expressed as a time-invariant linear combination of present and past samples.