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)

⎧⎨⎩

⎫⎬⎭= ______________

(iii) L f (q)dq

−∞

t

∫⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= ______________

Check with each other or the TAs for solutions.