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Laplace Transforms of Common Signals

Module by: Carlos E. Davila. E-mail the author

Summary: This module derives the Laplace transform of well-known signals.

We'll next build up a collection of Laplace transform pairs which we will include in a table. It's important to keep in mind that once the transform pair has been derived, the focus should be on utilizing the transform pair found in the table rather than in recalculating the transform.

Exponential Signal

Consider the Laplace transform of x(t)=eαtu(t)x(t)=eαtu(t):

L e - α t u ( t ) = ∫ 0 ∞ e - α t e - s t d t = ∫ 0 ∞ e - ( α + s ) t d t = - 1 α + s e - ( α + σ + j Ω ) t 0 ∞ = 1 α + s , σ > - α L e - α t u ( t ) = ∫ 0 ∞ e - α t e - s t d t = ∫ 0 ∞ e - ( α + s ) t d t = - 1 α + s e - ( α + σ + j Ω ) t 0 ∞ = 1 α + s , σ > - α (1)

where σ>-ασ>-α defines the region of convergence. Notice also that if α<0α<0, X(s)X(s) still exists provided σ>-ασ>-α. Therefore,

e - α t u ( t ) ↔ 1 α + s e - α t u ( t ) ↔ 1 α + s (2)

Unit Step Function

Recall we did not attempt to compute the Fourier transform of u(t)u(t) since the Fourier transform does not converge for this signal. Fortunately, the Laplace transform easily converges. In fact, we find that since u(t)u(t) is a special case of the exponential function with α=0α=0, the simply have

u ( t ) ↔ 1 s u ( t ) ↔ 1 s (3)

The region of convergence is σ>0σ>0.

Ramp Signal

This signal is given by x(t)=tu(t)x(t)=tu(t) and also does not have a Fourier transform. The Laplace transform is given by

L t u ( t ) = ∫ 0 ∞ t e - s t d t L t u ( t ) = ∫ 0 ∞ t e - s t d t (4)

Setting u=t,u'=1,v'=e-st,v=-1se-stu=t,u'=1,v'=e-st,v=-1se-st and integrating by parts gives

L t u ( t ) = - t s e - s t 0 ∞ + 1 s ∫ 0 ∞ e - s t d t = 0 - 1 s 2 e - s t 0 ∞ = - 1 s 2 e - ( σ + j Ω ) ∞ - 1 = 1 s 2 , σ > 0 L t u ( t ) = - t s e - s t 0 ∞ + 1 s ∫ 0 ∞ e - s t d t = 0 - 1 s 2 e - s t 0 ∞ = - 1 s 2 e - ( σ + j Ω ) ∞ - 1 = 1 s 2 , σ > 0 (5)

Here, the region of convergence is σ>0σ>0, and is referred to as the right-half plane.

Cosine Signal

Even though we computed the Fourier transform of the cosine signal, x(t)=cos(Ω0t)x(t)=cos(Ω0t), the Fourier transform technically does not converge for this signal. That is why X(jΩ)X(jΩ) involves impulse functions. The Laplace transform produces quite a different result. First we use the fact that

cos ( Ω 0 t ) u ( t ) = e j Ω 0 t u ( t ) + e - j Ω 0 t u ( t ) 2 cos ( Ω 0 t ) u ( t ) = e j Ω 0 t u ( t ) + e - j Ω 0 t u ( t ) 2 (6)

Since each of the two terms is an exponential function we have

L cos ( Ω 0 t ) u ( t ) = 1 2 s - j Ω 0 + 1 2 s + j Ω 0 = s s 2 + Ω 0 2 L cos ( Ω 0 t ) u ( t ) = 1 2 s - j Ω 0 + 1 2 s + j Ω 0 = s s 2 + Ω 0 2 (7)

Here, the region of convergence corresponds to σ>0σ>0 or right-half plane.

More Transform Pairs

We can use the existing transform pairs along with the properties of the Laplace transform to derive many new transform pairs. Consider the exponentially weighted cosine signal. This signal is given by

x ( t ) = e - α t cos ( Ω 0 t ) u ( t ) x ( t ) = e - α t cos ( Ω 0 t ) u ( t ) (8)

We can use the s-s-shift property of the Laplace transform (Reference) along with the Laplace transform of the cosine signal Equation 7 to get

e - α t cos ( Ω 0 t ) u ( t ) ↔ s + α s + α 2 + Ω 0 2 e - α t cos ( Ω 0 t ) u ( t ) ↔ s + α s + α 2 + Ω 0 2 (9)

Another common signal is

x ( t ) = t e - α t u ( t ) x ( t ) = t e - α t u ( t ) (10)

Here, we use the Laplace transform of the exponential signal "Exponential Signal" and the multiplication by tt property (Reference) to get

t e - α t u ( t ) ↔ 1 s + α 2 t e - α t u ( t ) ↔ 1 s + α 2 (11)

Extending this idea one step further, we have

x ( t ) = t 2 e - α t u ( t ) x ( t ) = t 2 e - α t u ( t ) (12)

So (Reference) applies, giving

t 2 e - α t u ( t ) ↔ 2 s + α 3 t 2 e - α t u ( t ) ↔ 2 s + α 3 (13)

Example 3.1 Consider the signal x(t)=te-2tu(t)x(t)=te-2tu(t). Therefore, we get

t e - 2 t u ( t ) ↔ 1 s + 2 2 t e - 2 t u ( t ) ↔ 1 s + 2 2 (14)

Example 3.2 Consider the signal x(t)=e-2tcos(5t)u(t)x(t)=e-2tcos(5t)u(t). As seen in Table 1,

e - 2 t cos ( 5 t ) u ( t ) ↔ s + 2 ( s + 2 ) 2 + 25 e - 2 t cos ( 5 t ) u ( t ) ↔ s + 2 ( s + 2 ) 2 + 25 (15)
Table 1: Some common Laplace Transform pairs.
x ( t ) x ( t ) X ( s ) X ( s )
e - α t u ( t ) e - α t u ( t ) 1 s + α 1 s + α
u ( t ) u ( t ) 1 s 1 s
δ ( t ) δ ( t ) 1 1
t u ( t ) t u ( t ) 1 s 2 1 s 2
cos ( Ω 0 t ) u ( t ) cos ( Ω 0 t ) u ( t ) s s 2 + Ω 0 2 s s 2 + Ω 0 2
sin ( Ω 0 t ) u ( t ) sin ( Ω 0 t ) u ( t ) Ω 0 s 2 + Ω 0 2 Ω 0 s 2 + Ω 0 2
e - α t cos ( Ω 0 t ) u ( t ) e - α t cos ( Ω 0 t ) u ( t ) s + α s + α 2 + Ω 0 2 s + α s + α 2 + Ω 0 2
e - α t sin ( Ω 0 t ) u ( t ) e - α t sin ( Ω 0 t ) u ( t ) Ω 0 s + α 2 + Ω 0 2 Ω 0 s + α 2 + Ω 0 2
t e - α t u ( t ) t e - α t u ( t ) 1 s + α 2 1 s + α 2
t 2 e - α t u ( t ) t 2 e - α t u ( t ) 2 s + α 3 2 s + α 3
t n e - α t u ( t ) t n e - α t u ( t ) n ! s + α n + 1 n ! s + α n + 1

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