# How could I approach determining if this 2D system represented as a 2D summation formula is linear?

I have a given 2D system: $$y(m,n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} x(k_1,k_2)$$

My usual approach to determining if a system is linear is to test if it is homogeneous and additive. First I would input an arbitrary point: (0,0), (1,1), (0,1), (1,0), calculate its output. Then I would scale the input and scale the output and see if the scaling amount matches. Next I would test the additive property in a similar manner. This is fine for a simple 2D system formula, for example: $$y(m,n) = x(m,n) + 3$$

With a 2D system represented as summation, I am not sure how to apply this test.

Searching the web, it seems this system looks like a 2D convolution:

$$y[i,j]= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} h[m,n]⋅x[i−m,j−n]$$

However, I am still stuck thinking of how to prove this system is linear. I would assume the process is fundamentally the same (prove it is homogeneous and additive or not). I am just not sure of the steps.

• If you do it for a fixed set of points you've only proven linearity at those points, not globally. You want to do it algebraically. – TimWescott Feb 5 at 16:19
• @TimWescott I understand that. I think I only use a few set of points (especially with 0) to see the trend. However, that is the issue I am facing here. I am not sure how to prove this algebraically when the y(m,n) is not using x(m,n) anymore but instead it is a summation of some other point up to the points m and n. – Cit5 Feb 5 at 18:05
• @Cit5: It's a generic equation, not a value, otherwise you can't show it for the general case. Have you ever done anything like that in 1D? – Matt L. Feb 5 at 21:42
• @MattL. perhaps but I don't recall changing the point variables to using the original as the bounds to a summation. I know it can be treated as an individual term. The bounds of the summation throw me off though. This is especially confusing when attempting to show in the next step that it is time space invariant. – Cit5 Feb 5 at 22:27
• Even if it were changed to 1D, I don't think that makes it easier – Cit5 Feb 5 at 22:28

I'm actually starting this out not knowing the answer, although I have my suspicions. I think it'll be a linear system, but I'm not sure. Here's your starting point: you define the system $$y = h(x)$$ as: $$y = h(x) \mid y(m,n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} x(k_1,k_2)$$

So let \begin{aligned} y_1(m,n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} A\,x_1(k_1,k_2) = h\left(A\, x_1 \right) = A h(x_1) \\ y_2(m,n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} B\,x_2(k_1,k_2) = h\left(A\, x_2 \right) = B h(x_2) \end{aligned}

Then $$y_1(m,n) + y_2(m, n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} A\,x_1(k_1,k_2) + \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} B\,x_2(k_1,k_2)$$

Because the summations have identical limits, we can bring everything under one pair of summations: $$y_1(m,n) + y_2(m, n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} \left ( A\,x_1(k_1,k_2) + B\,x_2(k_1,k_2) \right)$$

but this is just $$y_1 + y_2 = h(A\,x_1 + B\,x_2)$$. You already know that $$y_1 = A h(x_1)$$ and $$y_2 = B h(x_2)$$, so linearity is proved.

(Which is what I thought -- I was going to just dismiss your original problem as "summations are linear, go away!" -- then I realized I wasn't sure).

• what is that h(x) in the first equation? Is that the delta function? I thought proving homogeneous/additive had nothing to do with that. – Cit5 Feb 5 at 21:26
• That first equation is just defining the system $y = h(x)$. "y equals h of x such that ..." – TimWescott Feb 5 at 21:28

Do the following:

1. Define $$y_1(m,n)$$ as the response to an input $$x_1(m,n)$$
2. Define $$y_2(m,n)$$ as the response to an input $$x_2(m,n)$$
3. Determine the response to the input $$a_1x_1(m,n)+a_2x_2(m,n)$$
4. Check if that response is equal to $$a_1y_1(m,n)+a_2y_2(m,n)$$
• @Cit5: You need to use the sum indices $k_1$ and $k_2$, just like in the definition of the input-output relation. – Matt L. Feb 5 at 18:11