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 ?


2 Answers 2


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...

  • $\begingroup$ Even if was the integral of $x$ times $(a + T/2)$, the integration is in $a$, so it's still linear in $x$. $\endgroup$
    – TimWescott
    Nov 7, 2020 at 0:15
  • $\begingroup$ @TimWescott Good catch! $\endgroup$
    – Fat32
    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.


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