Given that x1(t)↔X1(s)x1(t)↔X1(s) and x2(t)↔X2(s)x2(t)↔X2(s) then for any constants αα and ββ, we have

The linearity property follows easily using the definition of the Laplace transform.

The reason we call this the time delay property rather than the time shift property is that the time shift must be positive, i.e. if τ>0τ>0, then x(t-τ)x(t-τ) corresponds to a delay. If τ<0τ<0 then we would not be able to use the single-sided Laplace transform because we would have a lower integration limit of ττ, which is less than zero. To derive the property, lets evaluate the Laplace transform of the time-delayed signal

Letting γ=t-τγ=t-τ leads to t=γ+τt=γ+τ and dt=dγdt=dγ. Substituting these quantities into Equation 3 gives

where we note that the first integral in the last line is zero since x(t)=0,t<0x(t)=0,t<0. Therefore the time delay property is given by

This property is the Laplace transform corresponds to the frequency shift property of the Fourier transform. In fact, the derivation of the ss-shift property is virtually identical to that of the frequency shift property.

The ss-shift property also alters the region of convergence of the Laplace transform. If the region of convergence for X(s)X(s) is σ>σminσ>σmin, then the region of convergence for Le-atx(t)Le-atx(t) is σ>σmin-Re(a)σ>σmin-Re(a).

Let's begin by taking the derivative of the Laplace transform:

So we can write

This idea can be extended to multiplication by tntn. Letting y(t)=tx(t)y(t)=tx(t), if follows that

Proceeding in this manner, we find that

The time scaling property for the Laplace transform is similar to that of the Fourier transform:

where in the second equality, we made the substitution t=γαt=γα and dt=dγαdt=dγα.

The derivation of the convolution property for the Laplace transform is virtually identical to that of the Fourier transform. We begin with

Applying the time-delay property of the Laplace transform gives

If h(t)h(t) is the the impulse response of a linear time-invariant system, then we call H(s)H(s) the system function of the system. The frequency response results by setting s=jΩs=jΩ in H(s)H(s). The system function provides us with a very powerful means of determining the output of a linear time-invariant filter given the input signal. It will also enable us to determine a means of establishing the stability1 of a linear-time invariant filter, something which was not possible with the frequency response.

The Laplace transform of the derivative of a signal will be used widely. Consider

this can be integrated by parts:

which gives

therefore we have,

The previous derivation can be extended to higher order derivatives. Consider

it follows that

which leads to

This process can be iterated to get the Laplace transform of arbitrary higher order derivatives, giving

where it should be understood that

Let

it follows that

and g0-=0g0-=0. Moreover, we have

therefore

but since

we have

Now suppose x(t)x(t) has a non-zero integral over negative values of tt. We have

The quantity ∫-∞0-x(τ)dτ∫-∞0-x(τ)dτ is a constant for positive values of tt, and can be expressed as

it follows that

where we have used the fact that u(t)↔1s.u(t)↔1s.

The initial value theorem makes it possible to determine x(t)x(t) at t=0+t=0+ from X(s)X(s). From the derivative property of the Laplace transform, we can write

Taking the limit s→∞s→∞

There are two cases, the first is when x(t)x(t) is continuous at t=0t=0. In this case it is clear that dx(t)dte-st→0dx(t)dte-st→0 as s→∞s→∞, so Equation 33 can be written as

Since x(t)x(t) is continuous at t=0t=0, x0-=x0+x0-=x0+, the Initial Value Theorem follows,

The second case is when x(t)x(t) is discontinuous at t=0t=0. In this case, we use the fact that

For example, if we integrate the right-hand side of Equation 36 with x0-=0x0-=0 and x0+=1x0+=1, we get the unit step function, u(t)u(t). Proceeding as before, we have

The left-hand side of Equation 37 can be written as

From the sifting property of the unit impulse, the first term in Equation 38 is

while the second term is zero since in the limit, the real part of ss goes to infinity. Substituting these results into the left-hand side of Equation 37 again leads to the initial value theorem, in Equation 35.

The Final Value Theorem allows us to determine

from X(s)X(s). Taking the limit as ss approaches zero in the derivative property gives

The left-hand-side of Equation 41 can be written as

Substituting this result back into Equation 41 leads to the Final Value Theorem

which is only valid as long as the limit x(∞)x(∞) exists.