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 $u(Ο„βˆ’t)=0$ for all $Ο„<t$ or, equivalently, for all $t>Ο„$ 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 $x(sβˆ’s_o)$ = $y(sβˆ’s_o)$.
$$y_1(s)=\int_{-\infty}^\infty x(𝜏-s_0)u(𝜏-s+1) d𝜏 $$ Now when I change of variable to $z=Ο„βˆ’s_o$, $Ο„=z+s_0$ , $dz=dΟ„$ 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.


2 Answers 2



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


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.

  • $\begingroup$ 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. $\endgroup$ Dec 31, 2020 at 13:27
  • $\begingroup$ @DilipSarwate: I thought I'd initially leave that leap of imagination to the OP. $\endgroup$
    – Matt L.
    Dec 31, 2020 at 13:45

Your Answer

By clicking β€œPost 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.