Due to its convolution property, the Laplace transform is a powerful tool for analyzing
LTI systems:

Also, if an LTI system can be described by a linear constant coefficient differential equation (LCCDE), the Laplace transform can convert the differential equation to an algebraic equation due to the time derivative property:

We first consider how an LTI system can be represented in the Laplace domain.

**Causality of LTI systems**An LTI system is

*causal*if its output depends only on the current and past input (but not the future). Assuming the system is initially at rest with zero output , then its response to an impulse at is at rest for , i.e., . Its response to a general input is:

Due to the properties of the ROC, we have:**If an LTI system is causal, then the ROC of is a right-sided half plane. In particular, If is rational , then the system is causal if and only if its ROC is the right-sided half plane to the right of the rightmost pole, and the order of numerator is no greater than that of the denominator , so that the ROC is a right-sided plane without any poles (even at ).****Example 0:**Given of an LTI, find :

Consider each of the two cases:- When
, can be considered as a special polynomial
(Taylor series expansion):

As this numerator polynomial has infinite order, greater than that of the denominator (zero), there is a pole at , ROC is not a right-sided plane, is not causal. - When
, we have:

As the order of the denominator polynomial is infinite, greater than that of the numerator (zero), there is no pole at , ROC is a right-sided plane, is causal.

- When
, can be considered as a special polynomial
(Taylor series expansion):
**Stability of LTI systems**An LTI system is

*stable*if its response to any bounded input is also bounded for all :

As the output and input of an LTI is related by convolution, we have:

and

which obviously requires:

In other words, if the impulse response function of an LTI system is absolutely integrable, then the system is stable. We can show that this condition is also necessary, i.e., all stable LTI systems' impulse response functions are absolutely integrable. Now we have:**An LTI system is stable if and only if its impulse response is absolutely integrable, i.e., the frequency response function exists, i.e., the ROC of its transfer function contains -axis:**

**Causal and stable LTI systems**Combining the two properties above, we have:

**A causal LTI system with a rational transfer function is stable if and only if all poles of are in the left half of the s-plane, i.e., the real parts of all poles are negative:**

**Example 1: ** The transfer function of an LTI is

As shown before, without specifying the ROC, this could be the Laplace transform of one of the two possible time signals .

-axis inside ROC | -axis outside ROC, | |

causal, stable | causal, unstable | |

-axis outside ROC | -axis inside ROC, | |

anti-causal, unstable | anti-causal, stable |

**Example 2: ** The transfer function of an LTI is

This is a time-shifted version of , and the corresponding impulse response is:

If , then during the interval , the system is not causal, although its ROC is a right half plane. This example serves as a counter example showing it is not true that any right half plane ROC corresponds to a causal system, while all causal systems' ROCs are right half planes. However, if is rational, then the system is causal if and only if its ROC is a right half plane.

Alternatively, as shown in Example 0, we have:

Now can still be consider as a rational function of with a numerator polynomial of order , which is greater than that of the denominator , i.e., has a pole at , i.e., its ROC cannot be a right-sided half plane, therefore the system is not causal. On the other hand, if , then this polynomial appears in denominator, there is no pole at , the ROC is a right-sided half plane, the system is causal.