Spatially invariant systems

Question

I was watching this video on spatially invariant systems:

And I found the definition a bit confusing, or seems vacuous, because isn't $T[x(n_1 - k_1, n_2 - k_2)] = y(n_1 - k_1, n_2 - k_2)$ always true by definition?

Answer 1

Spatial invariance simply means that shifting the input signal results in an equally shifted output signal. So if the response to an arbitrary signal $x(n_1,n_2)$ is $y(n_1,n_2)$, and the response to a shifted version of the same input signal $x(n_1-k_1,n_2-k_2)$ is $\tilde{y}(n_1,n_2)$, then the system is shift invariant if and only if

$$\tilde{y}(n_1,n_2)=y(n_1-k_1,n_2-k_2)\tag{1}$$

Any system that can be described by the following convolution sum is shift-invariant:

$$y(n_1,n_2)=\sum_{l_1}\sum_{l_2}h(l_1,l_2)x(n_1-l_1,n_2-l_2)\tag{2}$$

where $h(n_1,n_2)$ is the system's impulse response.

EDIT: As requested in the comments, here are some simple examples of shift-variant systems:

1. $y(n_1,n_2)=x(2n_1,n_2)$
2. $y(n_1,n_2)=n_1\cdot x(n_1,n_2)$
3. $y(n_1,n_2)=x(-n_1,-n_2)$

Let's look at the examples above in more detail:

1. \begin{align*} y(n_1 - d_1, n_2 - d_2) &= x(2(n_1 - d_1), n_2 - d_2) \\ & \ne x(2n_1 - d_1, n_2 - d_2) \end{align*}

2. \begin{align*} y(n_1 - d_1, n_2 - d_2) &= (n_1 - d_1) & \cdot x(n_1 - d_1, n_2 - d_2) \\ &\ne n_1 & \cdot x(n_1 - d_1, n_2 - d_2) \end{align*}

3. \begin{align*} y(n_1 - d_1, n_2 - d_2) &= x(-(n_1 - d_1), -(n_2 - d_2)) \\ & \ne x(-n_1 - d_1, -n_2 - d_2) \end{align*}