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Introduction to Digital Filters
    Linear Time-Invariant Digital Filters
       Showing Linearity and Time Invariance, or Not

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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 the output signal $ 2[gx(n)]^2 = 2g^2x^2(n)$, while $ gy(n) = 2gx^2(n)$. The filter is time invariant, however, because 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 when $ b_0=b_1=1/2$). 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

$\displaystyle y(n) = b_0 x(n) + b_1 x(n-1) - a_1 y(n-1). \protect$ (5.7)

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$. This establishes that each output sample from the filter of Eq.$ \,$(4.7) can be expressed as a time-invariant linear combination of present and past samples.

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Next: Nonlinear Filter Example: Dynamic Range Compression
written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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