I just finished reading the fundamentals chapter about signals (linearity,causality,memory and time invariance). I wanted to solve some exercises and I found this one.
We have a signal with output described by $$y(t) = \int_t^{t+1}x(\tau-a)\;d\tau ,a\in R$$
- find if the system has memory, is stable, time invariant and linear
- which values of $a$ make the system causal
I am trying to understand things here. The system has memory if $a<0$ and memoryless $a>0$? I don't know from where to start to find the others.
EDIT
To prove linearity I did this
$$y(t)=F\{x(t)\}$$ $$F\{c_1x_1(t)+c_2x_2(t)\}$$ $$\int_t^{t+1}c_1x_1(\tau -a)+c_2x_2(\tau -a)d\tau $$ $$=\int_t^{t+1}c_1x_1(\tau -a)d\tau +\int_t^{t+1}c_2x_2(\tau -a)d\tau$$ $$=c_1y_1(t) + c_2y_2(t)$$
time invariance
$$F\{x(t-t_0)\}$$ $$=\int_t^{t+1}x(\tau -a-t_0)d\tau $$ $$=\int_{t-t_0}^{t+1-t_0}x(u -a)du $$ so $$=\int_{t-t_0}^{t+1-t_0}x(\tau -a)d\tau $$
which means it is time invariant
Am I right?
First, note that the output signal at time $t$ is formed by integrating the values of the input over a unit length interval: $[t-a,t+1-a]$ (a continuous average).
Then, it should be evident that the system is linear (integration is, like the sum, a linear operator), and time-invariant (the relation of input-output does not change over time). And it's certainly not memoryless (the output at time $t$ does depend on other values besides the input at time $t$). To be causal, the output at time $t$ should depend only on "past" values of the input. For example, if $a=0$ or $a=0.5$, it 's NOT casual.
All this is rather informal, if you need to proof these facts formally, this is a start, bu t you should work this out in detail.
If you have learned about the input response $h(t)$, you can also find out that the system is indeed described by a $h(t)$, which is non-zero (actually, 1) over the interval $[a-1,a]$. Once you see this, (and you are supposed to see this quickly when you have some training in signals) you can answer immediately all the original questions.