Homogeneity (system property)

A system is homogeneous if multiplying the input by any constant $K$ multiplies the output by the same constant $K$. In symbols: if $g(t)\to y_1(t)$, then $Kg(t)\to K{y}_1(t)$, for any input $g$ and any constant $K$ (including complex constants).

Pictorially: if you run $x(t)$ through the system and get $y(t)$, then running $5x(t)$ through must give exactly $5y(t)$. Not $5y(t)$ plus a little extra, not $25y(t)$.

This has to hold for every $g$ and every $K$, not just some specific choice.

How to test

Apply the system’s defining rule to two inputs and compare:

1. Pick arbitrary $x_1(t)=g(t)$. Find $y_1(t)$.
2. Pick $x_2(t)=Kg(t)$. Find $y_2(t)$.
3. Check whether $y_2(t)=K{y}_1(t)$. If yes for every $g$ and $K$, the system is homogeneous.

Worked examples

$y(t)=\exp (x(t))$: Let $x_1=g$, $y_1=e^g$. Let $x_2=Kg$, $y_2=e^{Kg}=(e^g{)}^K=y_1^K$. Is $y_1^K=K{y}_1$? Only in trivial cases, so not homogeneous. Doubling the input of an exponential doesn’t double the output, it squares it. The exponential isn’t a linear map at all.

$y(t)=x(t)+2$: $x_1=g$ gives $y_1=g+2$; $x_2=Kg$ gives $y_2=Kg+2$. But $K{y}_1=Kg+2K$, which equals $y_2$ only if $K=1$. So not homogeneous.

The second example is the surprising one. “Add 2 to the input” looks innocent, an affine offset that feels basically linear. But it fails homogeneity, because the constant 2 doesn’t scale with the input.

The lesson

A system that includes a “DC offset” or any input-independent contribution is not homogeneous. Homogeneity requires that “no input” produces “no output”: the rule transforms the input, with no extras on top.

Homogeneity is one of the two ingredients of linearity; the other is additivity. Together they give the LTI property when combined with Time-invariance.