Is system $h(t) = u(t)-u(t-1)$ LTI or not?

$$y(t) = \int_{0}^{\infty} x(\tau)h(t-\tau) \ \mathrm{d}\tau$$ where $h(t) = u(t)-u(t-1)$.

I have some problem with understanding whether system LTI or LTV just looking the impulse response of the system. Is there any short-cut to identify the system time-variance looking the impulse function?

• LTV systems do not have one impulse response, they have infinitely many. Jan 12 '18 at 10:38

$h(t)$ can be rewritten as $\mathrm{rect}(t-0.5)$. So basically your output signal can be calculated as:

$$y(t) = \int_{0}^{\infty} x(\tau)\mathrm{rect}(t-0.5-\tau) \ \mathrm{d}\tau$$

Note that if $t<0$, then $t-0.5-\tau<-0.5$, as $\tau\in(0,\infty$). Ergo,

$$|t-0.5-\tau|>|0.5|$$

This means that the $\mathrm{rect}()$ function will be zero if $t$ takes negative values. Therefore, we get that $y(t)=0 \ \forall t<0$.

For $t>0$, all we have to do is convolve the rectangular window with $x(t)u(t)$, as the integral begins at $0$ and not $-\infty$.

If we call the window $w(t)$, we can express $y(t)$ as:

$$y(t)= \left\{ \begin{array}{ll} [x(t)u(t)]*w(t) & \mbox{if } t > 0 \\ 0 & \mbox{if } t < 0 \end{array} \right.$$

Now, could you determine whether the system is time-invariant or not?

Hint: try appling a delay to the input signal, and see if its output is equal to a delayed version of the output itself.

• My comment was only valid for a discrete-time system; either I misread or the notation was changed, so now it's continuous-time and the claim $u(t)-u(t-1)=\delta(t)$ is not correct anymore (if $t$ is a continuous variable). Jan 11 '18 at 15:24
• @MattL. Thanks for that, I messed up big time. I've already corrected it, I believe now the answer makes sense. Jan 11 '18 at 15:51

If a system is described by an impulse response $h(t)$, then by definition it is an LTI system. Because impulse response of the form $h(t) = \mathcal{T} \{ \delta(t) \}$ only exists for LTI systems.

• text book says it is time varying i am also thinking like you Jan 11 '18 at 12:04
• @NailTosun What textbook? Jan 11 '18 at 12:23
• I edited the question Jan 11 '18 at 13:46