Linear Unary Operator

A linear unary operator maps a single input signal to a single output signal linearly. Let $H$ denote a linear unary operator applied to signal $X$, written $HX$.

Operator Notation

All Linear Unary Operators can be expressed and manipulated algebraically.

Operator Expressions are Operators

A polynomial expression built from operators is itself a new operator. For example, $X-HX$ can be factored as $(1-H)X$, which is equivalent to $GX$ where $G=(1-H)$. The expression $G$ is a new operator that can be applied to any signal, composed with other operators, and manipulated algebraically like any other linear unary operator.

Composition

Cascading two linear unary operators multiplies their expressions: $Y=H_2(H_{1}X)=H_2{H}_{1}X$

Distributive Property

Multiplication distributes over addition: $H(X_1+X_2)=H{X}_1+H{X}_2$

Polynomial Expansion

Composed operators expand like polynomials. $(1-H{)}^{2}X=(1-2H+H^2)X$

Operator Reciprocals

Two operators are reciprocals if their product is the identity operator. The reciprocal of $(1-H)$ is the geometric series:

$$\frac{1}{1-H}=\sum _{k=0}^{\infty }{H}^k$$

Verified by synthetic division:

$$(1-H)(1+H+H^2+\cdots )=1$$

More generally, the reciprocal of $(1-pH)$ for scalar $p$ is:

$$\frac{1}{1-pH}=\sum _{k=0}^{\infty }{p}^k{H}^k$$

This shows that an implicit system $(1-pH)Y=X$ is equivalent to an infinite explicit sum $Y={\sum }_{k=0}^{\infty }{p}^k{H}^{k}X$.

Explicit and Implicit Forms

A system can be represented in two forms depending on whether the output is directly computed from the input or only constrained by it.

Explicit Form

An explicit operator expression is a recipe; it directly specifies how to compute the output from the input. It is imperative: it tells you what to do.

$$Y=(1-H)X$$

This corresponds to a Feed-Forward structure. The output is a function purely of current and past inputs.

Implicit Form

An implicit operator expression is a constraint; it states a rule the input and output must satisfy, without directly giving a formula for the output. It is declarative: it tells you what must be true.

$$(1-H)Y=X$$

This is the same algebraic content rearranged so that $Y$ appears on both sides. Solving for $Y$ requires inverting the operator, which produces a Feedback structure.

The accumulator $y[n]=x[n]+y[n-1]$ is the canonical implicit system:

$$Y=HY+X⟹(1-H)Y=X$$

The output depends on its own past value, so there is no closed-form recipe that avoids self-reference.