# Determining Causality and Time-Invariance of a system

Consider the following system: $$y(t-1)=\int_{-\infty}^\infty x(𝜏)u(𝜏-t) d𝜏$$ where $$u(t)$$ is the unit step function, which is zero for $$t<0$$ and equals $$1$$ for $$t>0$$.

$$(1)$$ Is the system causal? Why or why not?

I think if $$u(t)=0$$ for all $$t<0$$. This means that $$u(τ−t)=0$$ for all $$τ or, equivalently, for all $$t>τ$$ and the integrand is zero in range $$({-\infty}, t)$$.
Therefore, we can show that: $$\int_{-\infty}^\infty x(𝜏)u(𝜏-t) d𝜏 = \int_{t}^\infty x(𝜏)u(𝜏-t) d𝜏 = \int_{t}^\infty x(𝜏) d𝜏$$ So, the system is not causal! Am I right?!

$$(2)$$ Is the system time-invariant? Why or why not?

Let $$s=t-1$$
$$y(s)=\int_{-\infty}^\infty x(𝜏)u(𝜏-s+1) d𝜏$$
For invariance, we need to show output resulted from $$x(s−s_o)$$ = $$y(s−s_o)$$.
$$y_1(s)=\int_{-\infty}^\infty x(𝜏-s_0)u(𝜏-s+1) d𝜏$$ Now when I change of variable to $$z=τ−s_o$$, $$τ=z+s_0$$ , $$dz=dτ$$ it leads to: $$y_1(s)=\int_{-\infty}^\infty x(z)u(z+s_0-s+1) dz$$
For $$y(s-s_o)$$ integral becomes: $$y_2(s-s_0)=\int_{-\infty}^\infty x(𝜏)u(𝜏-s+1-s_0) d𝜏$$
So it seems to time-variant system.
Update:I would appreciate if anyone can answer my question because it's not a homework anymore and yet I didn't find the answer.

Hint

Substitute $$s = t-1$$. That gets you the equations in a more standard form $$y(s) = ...$$ Then go through the same excercise.

Concerning causality, your conclusion is correct. When you tried to investigate time-invariance you failed because you're not sufficiently careful when substituting variables. The first error occurred already when formulating the input-output equation for $$y(t)$$. Using $$s=t-1$$ means that $$\tau-t$$ becomes $$\tau-(s+1)=\tau-s-1$$, and hence you should obtain

$$y(s)=\int_{-\infty}^{\infty}x(\tau)u(\tau-s-1)d\tau\tag{1}$$

Now you're making the same mistake when you try to evaluate $$y_2(s-s_0)$$. Replacing $$s$$ by $$s-s_0$$ doesn't result in what you wrote down. If you do things right, you should see that the system is in fact time-invariant.

Time-invariance is also directly obvious from the original input-output equation, because it can be seen that the output signal is given by a convolution integral, and only LTI systems can be represented by convolution.

• Perhaps you could include a little bit of explanation as to why you are calling $(1)$ a convolution integral when, at first glance, it looks very much like a correlation integral since the variable of integration $\tau$ has the same sign in both arguments in $(1)$ whereas in a convolution integral, it should have opposite signsl. – Dilip Sarwate Dec 31 '20 at 13:27
• @DilipSarwate: I thought I'd initially leave that leap of imagination to the OP. – Matt L. Dec 31 '20 at 13:45