laplace transform basics-1.pdf jsdaskdadda

The Laplace transform
we’ll be interested in signals defined for t ≥ 0
the Laplace transform of a signal (function) f is the function F = L(f)
defined by
F(s) =
Z ∞
0
f(t)e−st
dt
for those s ∈ C for which the integral makes sense
• F is a complex-valued function of complex numbers
• s is called the (complex) frequency variable, with units sec−1
; t is called
the time variable (in sec); st is unitless
• for now, we assume f contains no impulses at t = 0
common notation convention: lower case letter denotes signal; capital
letter denotes its Laplace transform, e.g., U denotes L(u), Vin denotes
L(vin), etc.
The Laplace transform 3–4

Example
let’s find Laplace transform of f(t) = et
:
F(s) =
Z ∞
0
et
e−st
dt =
Z ∞
0
e(1−s)t
dt =
1
1 − s
e(1−s)t
¯
¯
¯
¯
∞
0
=
1
s − 1
provided we can say e(1−s)t
→ 0 as t → ∞, which is true for <s > 1:
¯
¯
¯e(1−s)t
¯
¯
¯ =
¯
¯
¯e−j(=s)t
¯
¯
¯
| {z }
=1
¯
¯
¯e(1−<s)t
¯
¯
¯ = e(1−<s)t
• the integral defining F makes sense for all s ∈ C with <s > 1 (the
‘region of convergence’ of F)
• but the resulting formula for F makes sense for all s ∈ C except s = 1
we’ll ignore these (sometimes important) details and just say that
L(et
) =
1
s − 1

More examples
constant: (or unit step) f(t) = 1 (for t ≥ 0)
F(s) =
Z ∞
0
e−st
dt = −
1
s
e−st
¯
¯
¯
¯
∞
0
=
1
s
provided we can say e−st
→ 0 as t → ∞, which is true for <s > 0 since
¯
¯e−st
¯
¯ =
¯
¯
¯e−j(=s)t
¯
¯
¯
| {z }
=1
¯
¯
¯e−(<s)t
¯
¯
¯ = e−(<s)t
• the integral defining F makes sense for all s with <s > 0
• but the resulting formula for F makes sense for all s except s = 0

sinusoid: first express f(t) = cos ωt as
f(t) = (1/2)ejωt
+ (1/2)e−jωt
now we can find F as
F(s) =
Z ∞
0
e−st
¡
(1/2)ejωt
+ (1/2)e−jωt
¢
dt
= (1/2)
Z ∞
0
e(−s+jω)t
dt + (1/2)
Z ∞
0
e(−s−jω)t
dt
= (1/2)
1
s − jω
+ (1/2)
1
s + jω
=
s
s2 + ω2
(valid for <s > 0; final formula OK for s 6= ±jω)

powers of t: f(t) = tn
(n ≥ 1)
we’ll integrate by parts, i.e., use
Z b
a
u(t)v0
(t) dt = u(t)v(t)
¯
¯
¯
¯
b
a
−
Z b
a
v(t)u0
(t) dt
with u(t) = tn
, v0
(t) = e−st
, a = 0, b = ∞
F(s) =
Z ∞
0
tn
e−st
dt = tn
µ
−e−st
s
¶¯
¯
¯
¯
∞
0
+
n
s
Z ∞
0
tn−1
e−st
dt
=
n
s
L(tn−1
)
provided tn
e−st
→ 0 if t → ∞, which is true for <s > 0
applying the formula recusively, we obtain
F(s) =
n!
sn+1
valid for <s > 0; final formula OK for all s 6= 0

Impulses at t = 0
if f contains impulses at t = 0 we choose to include them in the integral
defining F:
F(s) =
Z ∞
0−
f(t)e−st
dt
(you can also choose to not include them, but this changes some formulas
we’ll see & use)
example: impulse function, f = δ
F(s) =
Z ∞
0−
δ(t)e−st
dt = e−st
¯
¯
t=0
= 1
similarly for f = δ(k)
we have
F(s) =
Z ∞
0−
δ(k)
(t)e−st
dt = (−1)k dk
dtk
e−st
¯
¯
¯
¯
t=0
= sk
e−st
¯
¯
t=0
= sk

Linearity
the Laplace transform is linear: if f and g are any signals, and a is any
scalar, we have
L(af) = aF, L(f + g) = F + G
i.e., homogeneity & superposition hold
example:
L
¡
3δ(t) − 2et
¢
= 3L(δ(t)) − 2L(et
)
= 3 −
2
s − 1
=
3s − 5
s − 1

One-to-one property
the Laplace transform is one-to-one: if L(f) = L(g) then f = g
(well, almost; see below)
• F determines f
• inverse Laplace transform L−1
is well defined
(not easy to show)
example (previous page):
L−1
µ
3s − 5
s − 1
¶
= 3δ(t) − 2et
in other words, the only function f such that
F(s) =
3s − 5
s − 1
is f(t) = 3δ(t) − 2et

what ‘almost’ means: if f and g differ only at a finite number of points
(where there aren’t impulses) then F = G
examples:
• f defined as
f(t) =
½
1 t = 2
0 t 6= 2
has F = 0
• f defined as
f(t) =
½
1/2 t = 0
1 t > 0
has F = 1/s (same as unit step)

Inverse Laplace transform
in principle we can recover f from F via
f(t) =
1
2πj
Z σ+j∞
σ−j∞
F(s)est
ds
where σ is large enough that F(s) is defined for <s ≥ σ
surprisingly, this formula isn’t really useful!

Time scaling
define signal g by g(t) = f(at), where a > 0; then
G(s) = (1/a)F(s/a)
makes sense: times are scaled by a, frequencies by 1/a
let’s check:
G(s) =
Z ∞
0
f(at)e−st
dt = (1/a)
Z ∞
0
f(τ)e−(s/a)τ
dτ = (1/a)F(s/a)
where τ = at
example: L(et
) = 1/(s − 1) so
L(eat
) = (1/a)
1
(s/a) − 1
=
1
s − a

Exponential scaling
let f be a signal and a a scalar, and define g(t) = eat
f(t); then
G(s) = F(s − a)
let’s check:
G(s) =
Z ∞
0
e−st
eat
f(t) dt =
Z ∞
0
e−(s−a)t
f(t) dt = F(s − a)
example: L(cos t) = s/(s2
+ 1), and hence
L(e−t
cos t) =
s + 1
(s + 1)2 + 1
=
s + 1
s2 + 2s + 2

Time delay
let f be a signal and T > 0; define the signal g as
g(t) =
½
0 0 ≤ t < T
f(t − T) t ≥ T
(g is f, delayed by T seconds & ‘zero-padded’ up to T)
rag replacements
t
t
t = T
f(t) g(t)

then we have G(s) = e−sT
F(s)
derivation:
G(s) =
Z ∞
0
e−st
g(t) dt =
Z ∞
T
e−st
f(t − T) dt
=
Z ∞
0
e−s(τ+T )
f(τ) dτ
= e−sT
F(s)

example: let’s find the Laplace transform of a rectangular pulse signal
f(t) =
½
1 if a ≤ t ≤ b
0 otherwise
where 0 < a < b
we can write f as f = f1 − f2 where
f1(t) =
½
1 t ≥ a
0 t < a
f2(t) =
½
1 t ≥ b
0 t < b
i.e., f is a unit step delayed a seconds, minus a unit step delayed b seconds
hence
F(s) = L(f1) − L(f2)
=
e−as
− e−bs
s
(can check by direct integration)

Derivative
if signal f is continuous at t = 0, then
L(f0
) = sF(s) − f(0)
• time-domain differentiation becomes multiplication by frequency
variable s (as with phasors)
• plus a term that includes initial condition (i.e., −f(0))
higher-order derivatives: applying derivative formula twice yields
L(f00
) = sL(f0
) − f0
(0)
= s(sF(s) − f(0)) − f0
(0)
= s2
F(s) − sf(0) − f0
(0)
similar formulas hold for L(f(k)
)

examples
• f(t) = et
, so f0
(t) = et
and
L(f) = L(f0
) =
1
s − 1
using the formula, L(f0
) = s(
1
s − 1
) − 1, which is the same
• sin ωt = −1
ω
d
dt cos ωt, so
L(sin ωt) = −
1
ω
µ
s
s
s2 + ω2
− 1
¶
=
ω
s2 + ω2
• f is unit ramp, so f0
is unit step
L(f0
) = s
µ
1
s2
¶
− 0 = 1/s

derivation of derivative formula: start from the defining integral
G(s) =
Z ∞
0
f0
(t)e−st
dt
integration by parts yields
G(s) = e−st
f(t)
¯
¯∞
0
−
Z ∞
0
f(t)(−se−st
) dt
= lim
t→∞
f(t)e−st
− f(0) + sF(s)
for <s large enough the limit is zero, and we recover the formula
G(s) = sF(s) − f(0)

derivative formula for discontinuous functions
if signal f is discontinuous at t = 0, then
L(f0
) = sF(s) − f(0−)
example: f is unit step, so f0
(t) = δ(t)
L(f0
) = s
µ
1
s
¶
− 0 = 1

Example: RC circuit
PSfrag replacements
u y
1Ω
1F
• capacitor is uncharged at t = 0, i.e., y(0) = 0
• u(t) is a unit step
from last lecture,
y0
(t) + y(t) = u(t)
take Laplace transform, term by term:
sY (s) + Y (s) = 1/s
(using y(0) = 0 and U(s) = 1/s)

solve for Y (s) (just algebra!) to get
Y (s) =
1/s
s + 1
=
1
s(s + 1)
to find y, we first express Y as
Y (s) =
1
s
−
1
s + 1
(check!)
therefore we have
y(t) = L−1
(1/s) − L−1
(1/(s + 1)) = 1 − e−t
Laplace transform turned a differential equation into an algebraic equation
(more on this later)

Integral
let g be the running integral of a signal f, i.e.,
g(t) =
Z t
0
f(τ) dτ
then
G(s) =
1
s
F(s)
i.e., time-domain integral becomes division by frequency variable s
example: f = δ, so F(s) = 1; g is the unit step function
G(s) = 1/s
example: f is unit step function, so F(s) = 1/s; g is the unit ramp
function (g(t) = t for t ≥ 0),
G(s) = 1/s2

derivation of integral formula:
G(s) =
Z ∞
t=0
µZ t
τ=0
f(τ) dτ
¶
e−st
dt =
Z ∞
t=0
Z t
τ=0
f(τ)e−st
dτ dt
here we integrate horizontally first over the triangle 0 ≤ τ ≤ t
PSfrag replacements
t
τ
let’s switch the order, i.e., integrate vertically first:
G(s) =
Z ∞
τ=0
Z ∞
t=τ
f(τ)e−st
dt dτ =
Z ∞
τ=0
f(τ)
µZ ∞
t=τ
e−st
dt
¶
dτ
=
Z ∞
τ=0
f(τ)(1/s)e−sτ
dτ
= F(s)/s

Multiplication by t
let f be a signal and define
g(t) = tf(t)
then we have
G(s) = −F0
(s)
to verify formula, just differentiate both sides of
F(s) =
Z ∞
0
e−st
f(t) dt
with respect to s to get
F0
(s) =
Z ∞
0
(−t)e−st
f(t) dt

examples
• f(t) = e−t
, g(t) = te−t
L(te−t
) = −
d
ds
1
s + 1
=
1
(s + 1)2
• f(t) = te−t
, g(t) = t2
e−t
L(t2
e−t
) = −
d
ds
1
(s + 1)2
=
2
(s + 1)3
• in general,
L(tk
e−t
) =
(k − 1)!
(s + 1)k+1

Convolution
the convolution of signals f and g, denoted h = f ∗ g, is the signal
h(t) =
Z t
0
f(τ)g(t − τ) dτ
• same as h(t) =
Z t
0
f(t − τ)g(τ) dτ; in other words,
f ∗ g = g ∗ f
• (very great) importance will soon become clear
in terms of Laplace transforms:
H(s) = F(s)G(s)
Laplace transform turns convolution into multiplication

let’s show that L(f ∗ g) = F(s)G(s):
H(s) =
Z ∞
t=0
e−st
µZ t
τ=0
f(τ)g(t − τ) dτ
¶
dt
=
Z ∞
t=0
Z t
τ=0
e−st
f(τ)g(t − τ) dτ dt
where we integrate over the triangle 0 ≤ τ ≤ t
• change order of integration: H(s) =
Z ∞
τ=0
Z ∞
t=τ
e−st
f(τ)g(t − τ) dt dτ
• change variable t to t = t − τ; dt = dt; region of integration becomes
τ ≥ 0, t ≥ 0
H(s) =
Z ∞
τ=0
Z ∞
t=0
e−s(t+τ)
f(τ)g(t) dt dτ
=
µZ ∞
τ=0
e−sτ
f(τ) dτ
¶ µZ ∞
t=0
e−st
g(t) dt
¶
= F(s)G(s)

examples
• f = δ, F(s) = 1, gives
H(s) = G(s),
which is consistent with
Z t
0
δ(τ)g(t − τ)dτ = g(t)
• f(t) = 1, F(s) = e−sT
/s, gives
H(s) = G(s)/s
which is consistent with
h(t) =
Z t
0
g(τ) dτ
• more interesting examples later in the course . . .

Finding the Laplace transform
you should know the Laplace transforms of some basic signals, e.g.,
• unit step (F(s) = 1/s), impulse function (F(s) = 1)
• exponential: L(eat
) = 1/(s − a)
• sinusoids L(cos ωt) = s/(s2
+ ω2
), L(sin ωt) = ω/(s2
+ ω2
)
these, combined with a table of Laplace transforms and the properties
given above (linearity, scaling, . . . ) will get you pretty far
and of course you can always integrate, using the defining formula
F(s) =
Z ∞
0
f(t)e−st
dt . . .

Patterns
while the details differ, you can see some interesting symmetric patterns
between
• the time domain (i.e., signals), and
• the frequency domain (i.e., their Laplace transforms)
• differentiation in one domain corresponds to multiplication by the
variable in the other
• multiplication by an exponential in one domain corresponds to a shift
(or delay) in the other
we’ll see these patterns (and others) throughout the course

laplace transform basics-1.pdf jsdaskdadda

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