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...

  • $\begingroup$ 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 '20 at 0:15
  • $\begingroup$ @TimWescott Good catch! $\endgroup$ – Fat32 Nov 7 '20 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.