# Why this system is linear?

Hi guys i'm studying signals and systems, and my professor told us that

$$y(t) = \int\limits_{ t+T }^{t-T/2} {x(a+T/2)}\mathrm{d} a$$

is a linear system.

But a primitive of $$x$$ isn't $$x^2$$ ? How it's possible that's linear ?

That's not an integral of variable $$x$$. The notation $$x(a+T/2)$$ stands for a function $$x(\cdot)$$ of variable $$a$$.

So applying the fundamental theorem of calculus, and assuming there exists a function $$G(a)$$ such that $$G'(a) = x(a)$$, then you will have :

$$\int x(a+T/2) da = \int G'(a+T/2)da = G(a+T/2) + C$$

where the constant of integration, $$C$$, will be omitted in the definite integral :

$$\int_{t+T}^{t-T/2} x(a+T/2) da = \int_{t+T}^{t-T/2} G'(a+T/2)da = G(a+T/2)|_{t+T}^{t-T/2}$$

So the system has nothing with a square function.

Coming to its linearity, you can show this in line with the linearity of the integral operator...

• Even if was the integral of $x$ times $(a + T/2)$, the integration is in $a$, so it's still linear in $x$. Nov 7, 2020 at 0:15
• @TimWescott Good catch! Nov 7, 2020 at 0:29

The apparently complicated integral bounds $$t+T$$ and $$t-T/2$$, or the shift $$(a+T/2)$$ with independent variable $$a$$ inside the integral, obfuscate the simplicity of the system. It computes a signed area, on a constantly moving window $$[T,-T/2]$$ that moves around $$t$$, for an input that has a constant shift. All those ingredients suggest that the system could be linear.

To see that in a clearer fashion, it could be useful to simplify it a bit. By a variable change $$u=a+T/2$$, the system $$S$$ becomes:

$$y(t) = S(x(t))=-\int_t^{t+3T/2}x(u)\mathrm{d}u\,.$$

Then one can verify whether $$S(\lambda x_1(t)+\mu x_2(t))$$ is equal to $$\lambda S( x_1(t))+\mu S(x_2(t))$$. It was possible to check already on the original formula, maybe it is simpler with the simplified form.