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Lecture 4 : Transform Properties and Interpretations
Continued to the Next (Higher) Level
1. Example 1. Demo of the mult-by-t property.
(i) Consider above graphs of f (t) and g(t) = tf (t) . Set K = 1
.
(ii) By inspection, f (t) = u(t)u(T − t) = u(t)− u(t −T ) .
(iii) By linearity & shift theorems: F(s) = 1
s− e−Ts
s
(iv) Using mult-by-t property
G(s) = − d
dsF(s) = 1
s2 − e−sT
s2 −T e−sT
s
(v) CHECK. By DIRECT TV Calculation
g(t) = r(t)− r(t −T )−Tu(t −T )
in which case
G(s) = 1
s2 − e−Ts
s2 −T e−Ts
s
The property appears valid.
Remark: Proof by example is NOT a proof, but rather an
illustration. Proofs are logical mathematical arguments.
2. Frequency Shift Property—Dual to the Time Shift
Property: Let F(s) = L[ f (t)]. Then L[e−at f (t)] = F(s+ a) .
Proof. A simple minded proof that seems too obvious to
be true:
Step 1. By definition
L e−at f (t)⎡⎣
⎤⎦ = f (t)e−(s+a)t dt
0−
∞
∫ = f (t)e−⌢st dt
0−
∞
∫ = F(⌢s )
where
⌢s = s+ a is a new “Laplace variable”.
Step 2. Since
⌢s = s+ a , the relationship between the new
and the old is
L e−at f (t)⎡⎣
⎤⎦ = F(⌢s ) = F(s+ a)
Oh Yes. Can’t wait to use it!!!!!
Example 2. Recall: if f (t) = cos(ωt)u(t) then F(s) = s
s2 +ω 2 .
Hence if
g(t) = e−at cos(ωt)u(t) = e−at f (t)
then
G(s) = F(s+ a) = s+ a
(s+ a)2 +ω 2
Wow so cool!!!!!
Example 3. Find f (t) when F(s) = As+ B
(s+ a)2 +ω 2. Strategy:
generate a decomposition whose inverses are exponentially
decaying sines and cosines.
F(s) = As+ B
(s+ a)2 +ω 2 = A(s+ a)− Aa + B(s+ a)2 +ω 2
= A (s+ a)
(s+ a)2 +ω 2 + B − Aaω
⎛⎝⎜
⎞⎠⎟× ω
(s+ a)2 +ω 2
Strategy is complete:
f (t) = Ae−at cos(ωt)u(t)+ B − Aa
ω⎛⎝⎜
⎞⎠⎟
e−at sin(ωt)u(t)
Time Frequency Scaling Property: Let F(s) = L f (t)⎡⎣ ⎤⎦ and
a > 0 . Then L f (at)⎡⎣ ⎤⎦ =
1a
F sa
⎛⎝⎜
⎞⎠⎟
.
Proof:
L[ f (at)] = f (at)e−st dt0−
∞
∫ = f (τ )e−s τ
a⎛⎝⎜
⎞⎠⎟ dτ
a⎛⎝⎜
⎞⎠⎟
0−
∞
∫
= 1a
f (τ )e− s
a⎛⎝⎜
⎞⎠⎟τ
dτ0−
∞
∫ = 1a
F sa
⎛⎝⎜
⎞⎠⎟
Example 4. L δ (at)⎡⎣ ⎤⎦ =
1a×1= 1
a.
Example 5. Recall L sin(t)u(t)⎡⎣ ⎤⎦ =
1s2 +1
. So using the above
theorem we have, as we already knew,
L sin(ωt)u(t)⎡⎣ ⎤⎦ =1ω
1
sω
⎛⎝⎜
⎞⎠⎟
2
+1
= ωs2 +ω 2
(SUPER IMPORTANT) TIME DIFFERENTIATION PROPERTY:
L d
dtf (t)⎡
⎣⎢
⎤
⎦⎥ = sF(s)− f (0− )
Proof: This time we use integration by parts—darn!
Integration by parts
Is a leading cause
Of attacks to the heart
From students who pause
Rather than start.
Step 1. The definition—I am told that body-pump classes
cause great definition, and so by definition:
L d
dtf (t)⎡
⎣⎢
⎤
⎦⎥ =
ddt
f (t)⎡
⎣⎢
⎤
⎦⎥
0−
∞
∫ e−stdt
Step 2. The integration by parts formula. What’s “dv”? What’s “u”, not YOU?
u dv
0−
∞
∫ = uv⎤⎦0−∞
− v du0−
∞
∫
“dv” is a differential, so dv = df
dtdt = df implies v = f .
“u” is integrand leftovers, so u = e−st ⇒ du = −se−stdt
Step 3. Manipulate the definition in a non-nefarious way:
L df
dt⎡
⎣⎢
⎤
⎦⎥ =
ddt
f (t)⎡
⎣⎢
⎤
⎦⎥
0−
∞
∫ e−stdt = f (t)e−st ⎤⎦0−
∞+ s f (t)
0−
∞
∫ e−stdt
YESSS!!!!! SUCCESS!!!!!
= f (t)e−st ⎤
⎦t=∞− f (t)e−st ⎤
⎦t=0−+ sF(s) = 0− f (0− )+ sF(s)
Why a “0”? All integrated Laplace-things at t = ∞ are
AERO, I mean ZERO.
And so ends the proof whose property lived happily ever after.
Interpretation 1. differentiation in the t-world is
multiplication by s in the s-world.
Example 6. Interpretation 2—an s-domain equivalent
parallel circuit of a charged capacitor.
Part 1. Recall that iC = C
dvCdt
as per the figure below.
Part 2. Laplace transform the differential relationship:
IC (s) = CsVC (s)−CvC (0− )
Part 3. Currents to left of the equals, means currents to the
right of equals, and a sum of currents marched into the
valley of the node for a parallel equivalent circuit:
General Time Differentiation Formula.
L d n fdtn
⎡
⎣⎢⎢
⎤
⎦⎥⎥= snF(s)− sn−1 f (0− )− sn−2 f '(0− )−
....− f (n−1)(0− )
Example 7. Find the solution to the differential equation
!!f (t) = 2e−tu(t)
Step 1. Laplace transform both sides:
s2F(s)− sf (0− )− !f (0− ) = 2
s+1
Step 2. Solve for F(s) .
F(s) = 2
s2(s+1)+ f (0− )
s+!f (0− )
s2
= −2
s+ 2
s2 + 2s+1
+ f (0− )s
+!f (0− )
s2
Step 3. Invert to obtain
f (t) = −2u(t)+ 2tu(t)+ 2e−tu(t)+ f (0− )u(t)+ !f (0− )tu(t)
Integration Property:
(i) L f (τ )dτ
0−
t
∫⎡
⎣⎢⎢
⎤
⎦⎥⎥= F(s)
s
(ii) L f (τ )dτ
−∞
t
∫⎡
⎣⎢⎢
⎤
⎦⎥⎥= F(s)
s+
f (τ )dτ−∞
0−
∫s
Interpretation 1. integration in the t-world is divisioin by s
in the s-world.
Remark and a proof: A common misunderstanding of the
above theorem is that somehow we are taking the Laplace
transform of f (t) . As a result the common question is: you
told us that the Laplace transform does not look at function
values for t < 0 . And this is true. So how is the question
resolved? Actually we are finding the Laplace transform of
g(t) = f (τ )dτ
−∞
t
∫ implying g(0− ) = f (τ )dτ
−∞
0−
∫
Clearly, huh, f (t) = d
dtg(t) in which case F(s) = sG(s)− g(0− ).
Rearranging we obtain:
G(s) = F(s)
s+ g(0− )
s= F(s)
s+
f (τ )dτ−∞
0−
∫s
Example 8. Find the solution to the integro-diff eq
16 f (q)dq
0−
∞
∫ + df (t)dt
= 2u(t)
Solution. Step 1. Laplace transform both sides:
16s
F(s)+ sF(s)− f (0− ) = 2s
Equivalently
s2 +16s
⎛
⎝⎜
⎞
⎠⎟ F(s) = 2
s+ f (0− )⇒ F(s) = 2
s2 +16+ 2sf (0− )
s2 +16
Exercise. Complete the example above.
Lecture 4—Quick Test for Self-Study
1. F(s) = 1
s+ a then L tf (t −1){ } = ______________
2. F(s) = 1
s+ a then
L e−at f (t){ } = ______________
3. F(s) = 1
s+ a then L f (at){ } = ______________
4. f (t) = sin(2t)u(t) , then (i) F(s) = _______________ and
(ii) L d
dtf (t)
⎧⎨⎩
⎫⎬⎭= ______________
5. f (t) = e−a t then (i) F(s) = _______________ and
(ii) L d
dtf (t)
⎧⎨⎩
⎫⎬⎭= ______________