The impulse response functions of four linear systems S1, S2, S3, S4 are given respectively by $${h_1}\left( t \right) = 1,$$ $${h_2}\left( t \right) = u\left( t \right),$$ $${h_3}\left( t \right) = \frac{{u\left( t \right)}}{{t + 1}},$$ $${h_4}\left( t \right) = {e^{ – 3t}}u\left( t \right)$$ Where u(t) is the unit step function. Which of these systems is time invariant, causal, and stable?

The correct answer is A. S1.

A system is time-invariant if its output is the same for any two inputs that are equal except for a time shift. A system is causal if its output does not depend on future inputs. A system is stable if its output does not grow without bound for any bounded input.

The impulse response function of a system is the output of the system when the input is a unit impulse. The impulse response function of a time-invariant system is the same for all time shifts. The impulse response function of a causal system is zero for negative time. The impulse response function of a stable system is absolutely integrable.

The impulse response function of system S1 is $h_1(t) = 1$. This is a constant function, which means that it is the same for all time shifts. It is also zero for negative time, which means that it is causal. Finally, it is absolutely integrable, which means that it is stable. Therefore, system S1 is time-invariant, causal, and stable.

The impulse response function of system S2 is $h_2(t) = u(t)$. This is the unit step function, which is not the same for all time shifts. Therefore, system S2 is not time-invariant.

The impulse response function of system S3 is $h_3(t) = \frac{u(t)}{t+1}$. This is a rational function, which is not absolutely integrable. Therefore, system S3 is not stable.

The impulse response function of system S4 is $h_4(t) = e^{-3t}u(t)$. This is an exponential function, which is not the same for all time shifts. Therefore, system S4 is not time-invariant.

In conclusion, the only system that is time-invariant, causal, and stable is system S1.