Chapter 10
Laplace transforms
We have seen that frequency domain representations of signals, notably the Fourier transform, are very useful for analysing and designing LTI systems. A significant drawback of the Fourier transform, is that it only applies to signals that are absolutely summable (in discrete-time) or absolutely integrable (in continuous-time).
In many applications, most notably in feedback control, we need analysis tools that work for both bounded and unbounded signals. In this chapter we study the Laplace transform, a generalisation of the continuous-time Fourier transform, that applies to signals that are not absolutely summable, but enjoys many of the useful properties of the Fourier transform.
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In this chapter we focus on continuous-time signals and systems. There is an analogue of the Laplace transform for discrete-time signals and systems. This discrete-time analogue is called the Z-transform. Just as the Laplace transform generalises the continuous-time Fourier transform, the Z-transform generalises the discrete-time Fourier transform. We will not discuss the Z-transform further in this unit.
10.1 The Laplace transform
One way to motivate the Laplace transform, is by trying to extend the definition of the Fourier transform to signals that are not absolutely integrable. If x is a CT signal that is not absolutely integrable, we might try to make it absolutely integrable by scaling it by an exponential, i.e., by considering
xσ(t) = x(t)e−σt.
If x grows as t → ∞ we might hope that by choosing a large enough positive σ, the scaled signal xσ is absolutely integrable and so has a Fourier transform. Similarly, if x grows as t → −∞, we might hope that by choosing a large enough negative σ, the scaled signal xσ is absolutely integrable, and so has a Fourier transform.
Fourier transform of a rescaled signal
Let xσ(t) = x(t)e−σt be a rescaled CT signal. Assume that σ is chosen so that xσ(t) is absolutely integrable. Its Fourier transform is then
∞ Xσ(ω) =
−σt −jωt e
∞ −(σ+jω)t dt = x(t)e
This is a function of both ω and σ. It is only valid for some values of σ.
Instead of thinking of the Fourier transform Xσ(ω) of the rescaled signal xσ(t) as a function of both σ and ω, we can think of it as a function of the complex number s = σ + jω. This leads
to the Laplace transform.
Laplace transform
If x is a CT signal, the Laplace transform of x is the function Xˆ defined by ˆ ∞ −st
is the region of convergence.
X(s) = x(t)e dt for all s ∈ RoC(x) −∞
∞ −st
RoC(x) = s ∈ C : |x(t)e | dt < ∞ −∞
In words, the region of convergence consists of the set of complex numbers s for which x(t)e−st is absolutely integrable. The following gives a slightly different alternative statement.
This holds because |x(t)e−st| = |x(t)e−Re(s)te−jIm(s)t| = |x(t)e−Re(s)t|.
It follows directly that we can decide whether or not x is absolutely integrable by examining
the region of convergence RoC(x).
This allows us to show that the Fourier transform is just the Laplace transform evaluated along the imaginary axis.
We now present the most important example of a Laplace transform.
Region of convergence
The region of convergence RoC(x) is
the set of complex numbers s such that e−Re(s)tx(t) is absolutely integrable.
Region of convergence and absolute integrability
A CT signal x is absolutely integrable if and only if RoC(x) contains the imaginary axis.
Laplace and Fourier transform
If x is an absolutely integrable CT signal then RoC(x) contains the imaginary axis and the CT Fourier transform X and the Laplace transform Xˆ of x are related by
X(ω) = Xˆ(jω) for all ω.
Example 10.1: Laplace transform of right-sided general complex exponential
Let α be a complex number and let x(t) = e−αtu(t) be a right-sided general complex exponential. The Laplace transform Xˆ of x is
ˆ ∞ −αt X(s) = e u(t)e
∞ −(s+α)t
dt = e dt.
Writing s = σ + ωj and α = a + bj we have
ˆ ∞ −(σ+a)t −j(ω+b)t
X(σ + ωj) = e e dt. 0
Let y(t) = e−(a+σ)tu(t). If y is absolutely integrable and Y is the CTFT of y then, from the Fourier transform tables,
Xˆ(σ+ωj)=Y(ω+b)= 1 = 1 . (σ + a) + j(ω + b) s + α
This is only valid if y is absolutely integrable, which occurs if and only if a + σ = Re(α) + Re(s) > 0. Expressed differently, we have found that the region of convergence is
RoC(x) = {s ∈ C : Re(s) > −Re(α)} which is shown below in the case where Re(α) > 0
The following example appears quite similar to the previous one, but there is a subtle and important difference!
Example 10.2: Laplace transform of a left-sided general complex exponential
For comparison with Example 10.1, consider x(t) = −e−αtu(−t) where α is a complex number. Note that this is now a left-sided signal. The Laplace transform then turns out
ˆ 0−αt−st 1 X(s)=− e e dt= s+α.
This time, however, the signal x(t)e−st = −u(−t)e−(s+α)t is absolutely integrable when
Re(s) + Re(α) < 0. So the region of convergence is
RoC(x) = {s ∈ C : Re(s) < −Re(α)}
which is shown below in the case where Re(α) > 0
The difference between Examples 10.1 and 10.2 illustrate an important point about Laplace transforms.
10.1.1 Rational Laplace transforms
In Examples 10.1 and 10.2, the Laplace transforms were both Xˆ(s)= 1 .
This function is a ratio of two polynomials—the constant polynomial B(s) = 1 and the degree one polynomial A(s) = s + α. Signals with rational Laplace transforms arise very often, so we introduce some terminology to describe them.
Different signals may have the same Laplace transforms. In such cases, the regions of convergence for the two Laplace transforms will be different.
Rational functions
A rational function is a ratio of polynomials, and so has the form B(s) b0 +b1s+···+bmsm
F (s) = A(s) = a0 + a1s + · · · + ansn . If A(s) and B(s) have no common factors then
• the zeros of F are the complex numbers s such that B(s) = 0;
• the poles of F are the complex numbers s such that A(s) = 0. If bm and an are both not equal to zero then
• the degree or order of the numerator is m and
• the degree or order of the denominator is n.
We also consider poles and zeros ‘at infinity’. These keep track of the behaviour of the
function as |s| → ∞.
• If n > m then there is a zero with multiplicity n − m at infinity.
• If m > n then there is a pole with multiplicity m − n at infinity.
Finally, we say that a rational function is proper if n ≥ m and strictly proper if n > m.
Example 10.3
Let x(t) = e−2tu(t) + e−t cos(3t)u(t). Using Euler’s formula we can write this as x(t) = e−2tu(t) + (1/2)e−(1+3j)tu(t) + (1/2)e−(1−3j)tu(t).
Then, using Example 10.1 we see that
ˆ ∞ −2t −st ∞ −(1+3j)t −st ∞ −(1−3j)t −st
X(s) = e e dt + (1/2) e e dt + (1/2) e e dt −∞ −∞ −∞
= 1 + 1/2 + 1/2 . s+2 s+(1+3j) s+(1−3j)
For the Laplace transform to converge, it must converge for all of the three terms. This means that the region of convergence is the intersection of {s ∈ C : Re(s) > −2} and {s∈C:Re(s)>−1},whichis{s∈C:Re(s)>−1}. InthiscasetheLaplace transform is a rational function. To see this we can make a common denominator of (s + 2)(s + (1 + 3j))(s + (1 − 3j)) = (s + 2)((s + 1)2 + 9) and simplify to obtain
Xˆ (s) = 2s2 + 5s + 12 . (s+2)(s2 +2s+10)
The zeros of Xˆ (s) are the (complex conjugate pair of ) solutions of 2s2 + 5s + 12 = 0 which
s = − 5 ± j √23 . 44
The poles of Xˆ(s) are at the solutions of (s+2)(s+(1+3j))(s+(1−3j)) = 0 which are s=−2 and s=−1±3j.
Pole-zero plots
It is customary to mark poles in the complex plane with a cross ×, and zeros in the complex plane with a circle ◦. If a pole or zero has multiplicity greater than one, we often indicate the multiplicity beside the pole or zero with a number. As an example, the pole-zero plot for
Xˆ (s) = 2s2 + 5s + 12 (s+2)(s2 +2s+10)
is shown below. In this example we have labeled the poles and zeros with their complex values. We could, alternatively, have labeled the axes.
Im(s) −1+3j
−5 + √23j 44
−2 −5 − √23j
Summary of Section 10.1
The Laplace transform is a generalisation of the continuous-time Fourier transform that is valid for signals that are not absolutely summable. Two difference continuous-time signals may have the same algebraic expression for their Laplace transform, but will have different regions of convergences. Very often we are interested in signals that have Laplace transforms that are rational functions (ratios of polynomials). In this case the roots of the numerator are called zeros, and the roots of the denominator are called poles.
10.2 The region of convergence for Laplace transforms
We have seen that the region of convergence of the Laplace transform of a signal carries im- portant information that the algebraic expression for the Laplace transform ignores. In this section, we summarise some properties of the region of convergence of the Laplace transform. In particular, we are interested in
• how the region of convergence relates to properties of the signal and
• how the region of convergence relates to the poles of the Laplace transform (if the Laplace
transform is rational).
We begin with some basic properties of the region of convergence.
Basic structure of the region of convergence
For any CT signal x, the region of convergence consists of either:
• the entire complex plane
• a half-plane to the right of a vertical line in the complex plane • a half-plane to the left of a vertical line in the complex plane • a vertical strip between two vertical lines in the complex plane
Left-, right-, and two-sided signals
A CT signal x is right-sided if
x(t) = 0 for all t < T for some time T.
A CT signal x is left-sided if
x(t) = 0 for all t > T for some time T.
A CT signal that is neither right-sided nor left-sided is called two-sided. Any two-sided signal can be written as the sum of a left-sided and a right-sided signal as
x(t) = x(t)u(−t) + x(t)u(t).
Example 10.4: A two-sided signal
The signal x(t) = e−2tu(t) − e−tu(−t) is two-sided and absolutely summable. x(t)
It can be decomposed into the right-sided signal e−2tu(t) and the left-sided signal −e−tu(t).
Region of convergence for finite duration, left-, right-, and two-sided signals
• If a CT signal x has finite duration and is absolutely integrable then RoC(x) = C.
• If a CT signal x is right-sided then the region of convergence is a half-plane to the right of a vertical line, i.e.,
RoC(x) = {s ∈ C : Re(s) > a}
• If a CT signal x is left-sided then the region of convergence is a half-plane to the left
for some real number a. of a vertical line, i.e.,
for some real number a.
RoC(x) = {s ∈ C : Re(s) < a}
• If a CT signal x is two-sided then the region of convergence is a strip between two vertical lines, i.e.,
RoC(x) = {s ∈ C : b < Re(s) < a} for some real numbers b < a.
We now summarize how the poles of a rational Laplace transform are related to the region of convergence.
Region of convergence for rational Laplace transforms
Suppose that x is a CT signal and the Laplace transform Xˆ of x is a rational function. Then
• RoC(x) does not contain any poles of Xˆ
• if x is right-sided then the region of convergence is the half-plane to the right of the
rightmost pole of Xˆ, i.e.,
RoC(x) = {s ∈ C : Re(s) > Re(p)}
where p is the rightmost pole of Xˆ.
• if x is left-sided then the region of convergence is the half-plane to the left of the
leftmost pole of Xˆ, i.e.,
where p is the leftmost pole of Xˆ.
RoC(x) = {s ∈ C : Re(s) < Re(p)}
• if x is two-sided then the region of convergence is a vertical strip between two poles
(and containing no poles) of Xˆ, i.e.,
RoC(x) = {s ∈ C : Re(pi) < Re(s) < Re(pj)}
for some pair of poles pi and pj of Xˆ.
Example 10.5
Consider the signal x(t) = e−2tu(t)+e−t cos(3t)u(t) from Example 10.3. This has a rational Laplace transform, and it is a right-sided signal. The right-most poles are −1 + 3j and −1 − 3j, both of which have real part −1. The region of convergence of the Laplace transform is then
RoC(x) = {s ∈ C : Re(s) > −1}.
Example 10.6
Consider the signal x(t) = e−2tu(t) − etu(−t). Note that this is a two-sided signal. By using the general results in Examples 10.1 and 10.2 we know that
Xˆ(s)= 1 + 1 = 2s+1 . s+2 s−1 (s+2)(s−1)
This Laplace transform is rational and has poles at s = −2 and s = 1. We expect the region of convergence to be a vertical strip containing no poles. Indeed the region of convergence must be
RoC(x) = {s ∈ C : −2 < Re(s) < 1}.
A pole-zero plot, together with the region of convergence, is shown below:
Note that the region of convergence contains the imaginary axis. This is consistent with the fact that x(t) is absolutely integrable.
Summary of Section 10.2
There are relationships between properties of a signal and the region of convergence of its Laplace transform. Whether a signal is left-sided or right-sided or two-sided is reflected in the region of convergence. Similarly, a signal is absolutely integrable if its region of convergence contains the imaginary axis. For signals with rational Laplace transforms,
the region of convergence is closely related to the locations of the poles of the Laplace transform.
10.3 Properties of the Laplace transform
If we make transformations to a signal, then the Laplace transform of the signal often transforms in a fairly simple way. In this section, we describe some of these properties of the Laplace transform.
The most important of these is the convolution property. This states that convolution of two signals corresponds to multiplication of their Laplace transforms. This greatly simplifies the task of finding the response of LTI systems to a given input.
In what follows we have left out transformations of signals that correspond to passing the signal through an LTI system. This is because those properties are subsumed by the discussion of LTI systems in the Laplace domain in Topic 11.
Time-scaling
If x is a CT signal with Laplace transform Xˆ and y(t) = x(at) for all t (where a is a positive real number) then
Furthermore
Yˆ(s)=1Xˆs foralls. aa
RoC(y) = {s ∈ C : s/a ∈ RoC(x)}.
This is true because
′ 1ˆs dt =aX a
This is true because
∞ −(s−s0)t ˆ
x(t)e dt = X(s − s0).
where the second equality holds by making the change of variable t′ = at in the integral. The statement about the region of convergence holds because if Yˆ(s) = a1Xˆ(s/a) then Yˆ(s) will converge whenever Xˆ(s/a) converges, which occurs if s/a ∈ RoC(x).
Modulation
If x is a CT signal with Laplace transform Xˆ and y(t) = es0tx(t) for all t (where s0 is a
complex number) then Furthermore
Yˆ ( s ) = Xˆ ( s − s 0 ) . RoC(y)={s∈C:s−s0 ∈RoC(x)}.
Suppose x1 and s2 are CT signals with Laplace transforms Xˆ1 and Xˆ2 and α, β ∈ C are
complex numbers. If y(t) = αx1(t) + βx2(t) for all t then Yˆ (s) = αXˆ1(s) + βXˆ2(s).
The regions of convergence are related by
RoC(y) ⊇ RoC(x1) ∩ RoC(x2).
This follows directly from the linearity of integration. The main thing to note is the region of convergence. If s ∈ RoC(x1) and s ∈ RoC(x2) then certainly s ∈ RoC(y). But it is possible that the region of convergence of the sum is larger.
Example 10.7: Region of convergence of sum
Suppose that y(t) = x1(t) + x2(t) where x1(t) = e−tu(t) and x2(t) = −e−tu(t).
Clearly y(t) = 0 for all t and so Yˆ (s) = 0 for all s. Furthermore RoC(y) = C, the entire
complex plane.
On the other hand, RoC(x1) = RoC(x2) = {s ∈ C : Re(s) > −1}. This is a case where
the region of convergence of the sum can be larger than the intersection of the regions of convergence of the parts of the sum. Admittedly the example is contrived, but it shows that some care is required!
Convolution
Suppose x1 and s2 are CT signals with Laplace transforms Xˆ1 and Xˆ2. If y(t) = (x1 ∗x2)(t) is the convolution of x1 and x2 then
Yˆ (s) = Xˆ1(s)Xˆ2(s). The regions of convergence are related by
RoC(y) ⊇ RoC(x1) ∩ RoC(x2).
To see why this is true, we use the definition of the Laplace transform and of convolution: ˆ ∞ ∞ −st
We then exchange the order of integration to obtain
−∞ −∞ ˆ ∞∞
x1(τ)x2(t−τ)dτ e dt.
Since x1(τ) is constant with respect to the inner integral, and e−st = e−s(t−τ)e−sτ we have
x1(τ)x2(t − τ)e
−sτ ∞ −s(t−τ)
ˆ ∞ Y (s) =
x2(t − τ)e
Making the change of variables t′ = t − τ in the inner integral gives
ˆ∞ −sτ∞′−st′′
Y (s) = x1(τ)e −∞
x2(t )e dt dτ.
Finally we can see that the inner integral is just Xˆ2(s) so we have ˆ ˆ ∞ −sτ ˆ ˆ
Y (s) = X2(s)
as before. If s ∈ RoC(x1) and s ∈ RoC(x2) then both integrals will converge. It is possible for
x1(τ)e dτ = X2(s)X1(s)
the region of convergence to be larger, because poles in Xˆ1 could be canceled by zeros in Xˆ2.
Conjugation
If x is a CT signal with Laplace transform Xˆ and y(t) = x(t)∗ is the complex conjugate of
and RoC(y) = RoC(x).
Yˆ ( s ) = Xˆ ( s ∗ ) ∗
To see why this is true, we begin with the definition of the Laplace transform
ˆ ∞ Y (s) =
∗ −st x(t) e
∞ −s∗t∗ x(t)e dt
[e−st]∗ = [e−(σ+ωj)t]∗ = e−σt[e−jωt]∗ = e−σtejωt = e−(σ−jω)t = e−s∗t
where we have used the fact that
for all complex numbers s and all real numbers t. Then
ˆ∞−s∗t∗ ∞ −s∗t∗ˆ∗∗ Y(s)= x(t)e dt= x(t)e dt = X(s ) .
d −st d ∞
Multiplication by t
If x is a CT signal with Laplace transform Xˆ and y(t) = tx(t) for all t then
and RoC(y) = RoC(x).
Yˆ ( s ) = − d Xˆ ( s ) ds
To see why this is true we observe that
te−st = − d e−st.
∞ −st tx(t)e
x(t) −dse dt = −ds
dt = −dsX(s).
Example 10.8
Consider the signal x(t) = tn−1 e−αtu(t). We know that the Laplace transform of x1(t) =
e−αtu(t) is
Xˆ1(s) = 1 for Re(s) > −Re(α). s+α
By applying the multiplication by t property n − 1 times, we see that ˆ 1 n−1 dn−1 1 1
X(s) = (n−1)!(−1) dsn−1 s+α = (s+α)n with region of convergence Re(s) > −Re(α).
Summary of Section 10.3
Various operations on signals (scaling, addition, convolution, multiplication) give rise to corresponding operations on the Laplace transforms of those signals. Because the Laplace transform generalizes the continuous-time Fourier transform, these properties are very closely related to the corresponding properties of the Fourier transform. By thinking of a signal as being built out of ‘simple’ signals, we can use these properties to compute the Laplace transform of a complicated signal from knowledge of the Laplace transforms of a few ‘simple’ signals.
10.4 Summary
The Laplace transform of a continuous-time signal x is the complex function defined by ˆ ∞ −st
X(s) = x(t)e dt for all s ∈ RoC(x). −∞
It generalises the Fourier transform to signals that are not absolutely integrable. Associated with the Laplace transform of a signal is the region of convergence. The region of convergence of the Laplace transform of a signal tells us qualitative information about the signal, such as whether it is left-, right-, or two-sided, and whether it is absolutely integrable.
Many signals have Laplace transforms that are rational functions, i.e., ratios of polynomials. In this case, the values of s that make the numerator vanish are called the zeros of the Laplace transform. The values of s that make the denominator vanish are called the poles of the Laplace transform. For rational Laplace transforms, the pole locations are closely related to the region of convergence.
The Laplace transform has many properties that generalise properties of the Fourier trans- form. These allows us to compute the Laplace transform of complicated signals by combining the Laplace transforms of transformations of simpler signals.
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