Contents
This section derives some useful properties of the Laplace Transform. These properties, along with the functions described on the previous page will enable us to us the Laplace Transform to solve differential equations and even to do higher level analysis of systems. In particular, the next page shows how the Laplace Transform can be used to solve differential equations. A table with all of the properties derived below is here.
The linearity property of the Laplace Transform states:
This is easily proven from the definition of the Laplace Transform
The time delay property is not much harder to prove, but there are some subtleties involved in understanding how to apply it. We'll start with the statement of the property, followed by the proof, and then followed by some examples. The time shift property states
We again prove by going back to the original definition of the Laplace Transform
Because
we can change the lower limit of the integral from 0^{} to a^{} and drop the step function (because it is always equal to one)
We can make a change of variable
The last integral is just the definition of the Laplace Transform, so we have the time delay property
To properly apply the time delay property it is important that both the function and the step that multiplies it are both shifted by the same amount. As an example, consider the function f(t)=t·γ(t). If we delay by 2 seconds it we get (t2)·γ(t2), not (t2)t·γ(t) or t·γ(t2). All four of these function are shown below.
The correct one is exactly like the original function but shifted.
Important: To apply the time delay property you must multiply a delayed version of your function by a delayed step. If the original function is g(t)·γ(t), then the shifted function is g(tt_{d})·γ(tt_{d}) where t_{d} is the time delay.
The first derivative property of the Laplace Transform states
To prove this we start with the definition of the Laplace Transform and integrate by parts
The first term in the brackets goes to zero (as long as f(t) doesn't grow faster than an exponential which was a condition for existence of the transform). In the next term, the exponential goes to one. The last term is simply the definition of the Laplace Transform multiplied by s. So the theorem is proved.
There are two significant things to note about this property:
Similarly for the second derivative we can show:
where
For the n^{th }derivative:
or
where
We will use the differentiation property widely. It is repeated below (for first, second and n^{th} order derivatives)
The integration theorem states that
We prove it by starting by integration by parts
The first term in the brackets goes to zero if f(t) grows more slowly than an exponential (one of our requirements for existence of the Laplace Transform), and the second term goes to zero because the limits on the integral are equal. So the theorem is proven
Given that the Laplace Transform of the impulse δ(t) is Δ(s)=1, find the Laplace Transform of the step and ramp.
Solution:
We know that
so that
Likewise:
The convolution theorem states (if you haven't studied convolution, you can skip this theorem)
note: we assume both
f(t) and g(t) are
causal.
We start our proof with the definition of the Laplace Transform
From there we continue:
We can change the order of integration.  
Now, we pull f(λ) out because it is constant with respect to the variable of integration, t 

Now we make a change of variables  
Since g(u) is zero for u<0, we can change the lower limit on the inner integral to 0^{}. 

We can pull e^{sλ} out (it is constant with respect to integration). 

We can separate the integrals since the inner integral doesn't depend on λ. 

We can change the lower limit on the first integral since f(λ) is causal. 

Finally we recognize that the two integrals are simply Laplace Transforms. 

The Theorem is proven 
The initial value theorem states
To show this, we first start with the Derivative Rule:
We then invoke the definition of the Laplace Transform, and split the integral into two parts:
We take the limit as s→∞:
Several simplifications are in order. In the left hand expression, we can take the second term out of the limit, since it doesn't depend on 's.' In the right hand expression, we can take the first term out of the limit for the same reason, and if we substitute infinity for 's' in the second term, the exponential term goes to zero:
The two f(0^{}) terms cancel each other, and we are left with the Initial Value Theorem
This theorem only works if F(s) is a strictly proper fraction in which the numerator polynomial is of lower order then the denominator polynomial. In other words is will work for F(s)=1/(s+1) but not F(s)=s/(s+1).
The final value theorem states that if a final value of a function exists that
However, we can only use the final value if the value exists (function like sine, cosine and the ramp function don't have final values). To prove the final value theorem, we start as we did for the initial value theorem, with the Laplace Transform of the derivative,
We let s→0,
As s→0 the exponential term disappears from the integral. Also, we can take f(0) out of the limit (since it doesn't depend on s)
We can evaluate the integral
Neither term on the left depends on s, so we can remove the limit and simplify, resulting in the final value theorem
Examples of functions for which this theorem can't be used are increasing exponentials (like e^{at} where a is a positive number) that go to infinity as t increases, and oscillating functions like sine and cosine that don't have a final value..
Some other properties that are important but not derived here are listed below.
A table of properties is available here.
© Copyright 2005 to 2019 Erik Cheever This page may be freely used for educational purposes.
Erik Cheever Department of Engineering Swarthmore College