The Laplace Transform

MUS420/EE367A Lecture 6A: The one-sided (unilateral) Laplace transform of a signal

The Laplace Transform x(t) is defined as
Z ∞
∆ ∆
X(s) = Ls{x} = x(t)e−stdt
Julius O. Smith III (jos@ccrma.stanford.edu) 0
Center for Computer Research in Music and Acoustics (CCRMA)
Department of Music, Stanford University
• t = time in seconds
Stanford, California 94305 • s = σ + jω is a complex variable
March 15, 2010 • Appropriate for causal signals
When evaluated along the jω axis (i.e., σ = 0), the
Laplace Transform reduces to the unilateral Fourier
Outline transform: Z ∞
• Definition X(jω) = x(t)e−jωtdt
0
• Linearity and Differentiation Theorem Thus, the Laplace transform generalizes the Fourier
transform from the real line (the frequency axis) to the
• Examples of Mass-Spring system analysis entire complex plane.
The Fourier transform equals the Laplace
transform evaluated along the jω axis in the
complex s plane
The Laplace Transform can also be seen as the Fourier
transform of an exponentially windowed causal signal x(t)
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Relation to the z Transform Note that the z plane and s plane are related by

The Laplace transform is used to analyze continuous-time z = esT

systems. Its discrete-time counterpart is the z transform: In particular, the discrete-time frequency axis
∞
∆
X ωd ∈ (−π/T, π/T ) and continuous-time frequency axis
Xd(z) = xd(nT )z −n
ωa ∈ (−∞, ∞) are related by
n=0
If we define z = esT , the z transform becomes ejωdT = ejωaT
proportional to the Laplace transform of a sampled
For the mapping z = esT from the s plane to the z plane
continuous-time signal:
∞ to be invertible, it is necessary that X(jωa) be zero for
all |ωa| ≥ π/T . If this is true, we say x(t) is bandlimited
X
sT
Xd(e ) = xd(nT )e−snT
n=0 below half the sampling rate. As is well known, this
As the sampling interval T goes to zero, we have condition is necessary to prevent aliasing when sampling
X∞  
xd(tn) −stn the continuous-time signal x(t) at the rate fs = 1/T to
sT
lim Xd(e )T = lim e ∆t produce x(nT ), n = 0, 1, 2, . . .
T →0 ∆t→0
n=0
∆t
Z ∞
∆
= xd(t)e−stdt = X(s)
0
∆ ∆
where tn = nT and ∆t = tn+1 − tn = T .
In summary,
the z transform (times the sampling interval T )
of a discrete time signal xd(nT ) approaches, as
T → 0, the Laplace Transform of the underly-
ing continuous-time signal xd(t).
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 Two Laplace Transform Theorems Differentiation

The differentiation theorem for Laplace transforms:

Linearity
ẋ(t) ↔ sX(s) − x(0)
∆
The Laplace transform is a linear operator : where ẋ(t) = dtd x(t), and x(t) is any differentiable
αx(t) + βy(t) ←→ αX(s) + βY (s) function that approaches zero as t goes to infinity.
Operator notation:
Proof: Let
Ls{ẋ} = sX(s) − x(0).
w(t) = αx(t) + βy(t),
where α and β are real or complex constants. Then Proof: Immediate from integration by parts:
∆ ∆ Z ∞
W (s) = Ls{w} = Ls{αx(t) + βy(t)} ∆
Ls{ẋ} = ẋ(t)e−stdt
Z ∞ 0
∆
= [αx(t) + βy(t)] e−stdt ∞
Z ∞
0 = x(t)e−st0 − x(t)(−s)e−stdt
Z ∞ Z ∞ 0
= α x(t)e−stdt + β y(t)e−stdt = sX(s) − x(0)
0 0
∆ since x(∞) = 0 by assumption
= αX(s) + βY (s).
Thus, linearity of the Laplace transform follows Corollary: Integration Theorem:
immediately from linearity of integration Z t 
X(s)
Ls x(τ )dτ =
0 s

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Laplace Analysis of Linear Systems Force-Driven Mass Analysis

The differentiation theorem converts differential equations Note that in the electrical equivalent circuit
into algebraic equations, which are easier to solve.
• Driving force = voltage source emitting f (t) volts
Example: Force-Driven Mass • Mass = inductor of L = m Henrys.

From Newton’s second law of motion “f = ma”, we have

Consider a free mass driven by an external force along an
∆ ∆
ideal frictionless surface in one dimension: f (t) = m a(t) = m v̇(t) = m ẍ(t).
Physical diagram: Taking the unilateral Laplace transform and applying the
differentiation theorem twice yields
v(t)
F (s) = m Ls{ẍ}
f (t) m
= m [ sLs{ẋ} − ẋ(0)]
= m { s [s X(s) − x(0)] − ẋ(0)}
m s2 X(s) − s x(0) − ẋ(0) .
 
Electrical equivalent circuit: =
Thus, given
v(t)
+ • F (s) = Laplace transform of the driving force f (t),
f (t) m • x(0) = initial mass position, and
∆
- • ẋ(0) = v(0) = initial mass velocity,

we can solve algebraically for X(s), the Laplace

transform of the mass position for all t ≥ 0
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 Force-Driven Mass Analysis, Continued Mass-Spring Oscillator Time-Domain Solution

If the applied external force f (t) is zero, we obtain Consider now the mass-spring oscillator:
x(0) ẋ(0) x(0) v(0)
X(s) = + 2 = + 2 . ẋ(t) →
s s s s
k
Since 1/s is the Laplace transform of the Heaviside m
unit-step function
x=0 x(t) →
(
∆ 0, t < 0
u(t) = ,
1, t ≥ 0 Electrical equivalent-circuit:
we find that the position of the mass x(t) is given for all
time by 1
x(t) = x(0) u(t) + v(0) t u(t). k m

• A nonzero initial position x(0) = x0 and zero initial

velocity v(0) = 0 results in x(t) = x0 for all t ≥ 0 Newton’s second law of motion:
(mass “just sits there”) fm(t) = mẍ(t).
• Similarly, any initial velocity v(0) is integrated with Hooke’s law for ideal springs:
respect to time (mass moves forever at initial velocity)
fk (t) = kx(t)
In summary, we used the Laplace transform to solve for Newton’s third law of motion:
the motion of a simple physical system (an ideal mass) in
response to initial conditions (no external driving forces). fm(t) + fk (t) = 0
⇒ mẍ(t) + kx(t) = 0
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We have thus derived a second-order differential equation Mass-Spring Oscillator Analysis, Continued
governing the motion of the mass and spring. (Note that
x(t) is both the position of the mass and compression of We can quickly verify that
the spring at time t.)
1
Taking the Laplace transform of both sides of this e−atu(t) ←→
s+a
differential equation gives
where u(t) is the Heaviside unit step function which steps
0 = Ls{mẍ + kx} from 0 to 1 at time 0.
= mLs{ẍ} + kLs{x} (linearity)
By linearity, the solution for the motion of the mass is
= m [sLs{ẋ} − ẋ(0)] + kX(s) (differentiation theorem)
x(t) = re−jω0t + rejω0t = 2re re−jω0t = 2Rr cos(ω0t − θr )

= m {s [sX(s) − x(0)] − ẋ(0)} + kX(s) (diff. thm again) p
ms2X(s) − msx(0) − mẋ(0) + kX(s) v02 + ω02x20
  
= v0
= cos ω0t − tan−1
Let x(0) = x0 and ẋ(0) = ẋ0 = v0 for simplicity. ω0 ω0x0
Solving for X(s) gives If the initial velocity is zero (v0 = 0), the above formula
sx0 + v0 ∆ r r reduces to x(t) = x0 cos(ω0t) and the mass
∆
p simply
p
X(s) = 2 k = + , ω0 = k/m,
s +m s + jω0 s − jω0 oscillates sinusoidally at frequency ω0 = k/m, starting
x0 v0 ∆ from its initial position x0. If instead the initial position is
r = +j = Rr ejθr , with x0 = 0, we obtain
2 2ω0
v0
x(t) = sin(ω0t)
p
v02 + ω02x20
 
∆ ∆ −1 v0 ω0
Rr = , θr = tan
2ω0 ω0x0 ⇒ v(t) = v0 cos(ω0t).
denoting the modulus and angle of the pole residue r,
respectively.

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