#22 (11/17/2025)

Additional Laplace transforms table


t
s
d
dt
 f(t)
s
-
f(s)
 
 − f(0)
t f(t)
− d
ds
-
f
 
 (s)
tn f(t)
(−1)n dn
dsn
-
f
 
 (s)
1
t
 f(t)
⌠
⌡ 		∞

s 
-
f
 
 (s) ds
⌠
⌡ 		t

0 
 f(t) dt
1
s
-
f
 
 (s)
H(t−a) f(t − a)
e−s a
-
f
 
 (s)
ea t f(t)
-
f
 
 (s − a)
(1)
It is roughly noted from the table above that multiplying by es (or et) has a shifting effect in the transformed domain, multiplying by t (or s) is translated into differentiation in the transformed domain and division by t (or s) is translated into integration in the transformed domain. With a proper combination of the entries above combined with the Laplace convolution theorem, it is possible to bypass the formula of the inverse Laplace transform all together.

More exercise problems

  1. Solve the following differential equation:

    ⋅⋅
    y
     
    		 (t) + y(t) = sin t,     y(0)  and
    ⋅
    y
     
    		 (0)   given.
    		(2)
    (Solution) Laplace transforms of the both sides yield:

    s2
    -
    y
     
    		 (s) − s y(0) −
    ⋅
    y
     
    		 (0) +
    -
    y
     
    		 (s) = 1
    1+s2
    		 ,
    		(3)
    which can be solved for ―y(s) as
    -
    y
     
    		 (s) = 1
    (1+s2)2
    		 + s
    1+s2
    		 y(0) + 1
    1+s2
    ⋅
    y
     
    		 (0).
    		(4)
    Hence, the inverse of the above is
    y(t)
    =
    sin t*sin t + y(0) cos t +
    ⋅
    y
     
    		 (0) sin t
    =
    ⌠
    ⌡     		t

    0 
    		 sin (t − τ) sin τ dτ1 + y(0) cos t +
    ⋅
    y
     
    		 (0) sin t
    =
    1
    2
    		 (−t cos t +sin t) + y(0) cos t +
    ⋅
    y
     
    		 (0) sin t.
    		(5)

  2. Solve the following integral equation.
    y(t) = t + ⌠
    ⌡     		t

    0 
    		 sin (t − τ) y(τ) dτ.
    		(6)
    (Solution) Laplace transform the both sides of the equation to get
    -
    y
     
    		 = 1
    s2
    		 + 1
    s2 + 1
    -
    y
     
    		 ,
    		(7)
    from which it follows
    -
    y
     
    		 = s2 + 1
    s4
    		 = 1
    s2
    		 + 1
    s4
    		 ,
    		(8)
    which can be inverted to yield
    y(t) = t + t3
    6
    		 .
    		(9)
  3. Prove

    lim
    s → ∞ 
    		 s
    -
    f
     
    		 (s) = f(0).
    (Solution) Use the definition,
    L(f  ′(t))
    =
    ⌠
    ⌡     		∞

    0 
    		 f′(t) e−s t d t
    =
    s  
    -
    f
     
    		 (s) − f(0),
    		(10)
    and let s → ∞.
  4. Prove

    lim
    s → ∞ 
    -
    f
     
    		 (s) = 0.
    (Solution)
    Use the definition,
    L(f(t))
    =
    ⌠
    ⌡     		∞

    0 
    		 f(t) e−s t d t
    =
    -
    f
     
    		 (s),
    		(11)
    and let s → ∞.
  5. Find the inverse Laplace transform for
    1
    (s + 3)4
    		 .
    (Solution)
    From
    L(t3) = 3!
    s4
    		 ,     L(ea t f(t)) =
    -
    f
     
    		 (s − a),

    L−1⎛
    ⎝     		1
    (s + 3)4
    		⎞
    ⎠     		= e−3 t t3
    3!
    		 .
  6. Find the inverse Laplace transform for
    s + 2
    s3 (s − 1)
    		 .
    (Solution) Use partial fraction to get
    s + 2
    s3 (s − 1)
    =
    3
    s − 1
    		 − 2 + 3 s + 3 s2
    s3
    =
    3
    s − 1
    		 − − 3
    s
    		 − 3
    s2
    		 − 2
    s3
    		 ,
    		(12)
    so

    L−1⎛
    ⎝     		s + 2
    s3 (s − 1)
    		⎞
    ⎠     		= 3 et −3 − 3 t − t2.
  7. Express the Laplace transform for
    t f′(t).
    (Solution)
    Note that L(t f(t)) = −[(d)/(ds)] ―f(s). Therefore,
    L(t f′(t)) = − d
    ds
    		 L(f′(t)) = − d
    ds
    		 ⎛
    ⎝     		s
    -
    f
     
    		 (s)−f(0)⎞
    ⎠     		= − ⎛
    ⎜
    ⎝
    -
    f
     
    		 (s)+ s
    d
    -
    f
     
    		 (s)
    ds
    		⎞
    ⎟
    ⎠     		.
  8. Solve the diffusion equation,
    ∂u
    ∂t
    		 = ∂2 u
    ∂x2
    		 ,
    with
    u = 0   at   t = 0,     u = 0   at   x = 0,     u = 100   at   x = 1.
    by the Laplace transform.
    (Solution) Laplace transform the original equation with respect to t to get
    s
    -
    u
     
    		 =
    d2
    -
    u
     
    d x2
    		 ,
    which can be solved for ―u as
    -
    u
     
    		 = A e√s x + B e− √s x.
    The boundary conditions are
    -
    u
     
    		 |x = 0 = A + B = 0,

    -
    u
     
    		 |x = 1 = A e√s + B e−√s = 100
    s
    		 ,
    which can be solved for A and B as
    A = 100
    s (e√s − e−√s)
    		 ,    B = − 100
    s (e√s − e−√s)
    		 .
    Therefore,
    -
    u
     
    		 = 100
    s
    		 e√sx − e−√s x
    e√s − e−√s
    The inverse of ―u needs to wait for the spring semester (ME5332).

Z-transform

If f(t) takes only discrete values, f0, f1, f2, f3 ... at t = 0, 1, 2, 3, …n as
f(t) ≡ ∞
∑
n=0 
 fn δ(t − n),
(13)
its Laplace transform is
L(f(t))
=
⌠
⌡ 		∞

0 
 f(t) e− s t dt
=
⌠
⌡ 		∞

0 
 ⎛
⎝ 		∞
∑
n=0 
 fn δ(t − n)⎞
⎠ 		e− s t dt
=
∞
∑
n=0 
 fn ⌠
⌡ 		∞

0 
 δ(t − n) e− s t dt
=
∞
∑
n=0 
 fn
(es)n
=
∞
∑
n=0 
 fn
zn
 ,
(14)
where
z ≡ es.
(15)
Equations (14) is called the Z-transform of fn and is denoted by

Z(fn) =
-
f
 
 (z) = ∞
∑
n=0 
 fn
zn
 .
(16)
The inverse Z-transform can be computed as (proof skipped)

fn = Z−1(
-
f
 
 (z)) = 1
2 πi
 ⌠
(⎜)
⌡
-
f
 
 (z) zn−1 dz,
(17)
where the integral path (in the complex plane) is a closed loop that contains all the poles of Z(an) (for details, wait until you take ME5332).
Laplace transform:

t
s
f(t)
-
f
 
 (s)=⌠
⌡ 		∞

0 
 f(t)e−st ds
(18)

Z-transform


n
z
fn
-
f
 
 (z) = ∞
∑
n=0 
 fn z−n
(19)
As the Z-transform is a special case of the Laplace transform, similar properties to those for the Laplace transform hold:

    (1) Linearity
    Z(αfn + βgn) = αZ(fn) + βZ(gn).

    (2) Shift (equivalent to derivative)
    Z(fn+1) = z
    -
    f
     
    		 (z) − z f0.
    (Proof)
    Z(fn+1)
    =
    ∞
    ∑
    n=0 
    		 fn+1
    zn
    =
    z ∞
    ∑
    n=0 
    		 fn+1
    zn+1
    =
    z ∞
    ∑
    n=1 
    		 fn
    zn
    =
    z⎛
    ⎝     		∞
    ∑
    n=0 
    		 fn
    zn
    		 − f0 ⎞
    ⎠
    =
    z
    -
    f
     
    		 (z) − z f0.
    		(20)
    Note that
    Z(fn+2)
    =
    z Z(fn+1) − z f1
    =
    z ⎛
    ⎝     		z
    -
    f
     
    		 (z)−z f0⎞
    ⎠     		−z f1
    =
    z2
    -
    f
     
    		 (z) − z2 f0 − z f1.
    		(21)

    (3) Convolution theorem

    Z(fn * gn) =
    -
    f
     
    		 (z)
    -
    g
     
    		 (z),
    where
    fn*gn ≡ n
    ∑
    m=0 
    		 fn−m gm.
Just like the Laplace transform is useful for an initial value problem in differential equations, the Z-transform is useful for difference equations.

fn
z
1
z
z−1
n
z
(z−1)2
n2
z (z+1)
(z−1)3
n3
z (z2+4 z+1)
(z−1)4
n4
z (z3+11 z2+11 z+1)
(z−1)5
an
z
z − a
sin n
z sin(1)
z2−2 z cos(1)+1
cos n
z (z−cos(1))
z2−2 z cos(1)+1
fn+1
z
-
f
 
 (z) − z f0
fn+2
z2
-
f
 
 (z) − z2 f0 − z f1
(22)
Example
(Fibonacci numbers)
davinci-fibonacci.jpg from Da Vinci Code
Find the general form of fn that satisfies

fn+2 = fn+1 + fn,     f0 = 1,   f1 = 1.
(Solution)

Apply the Z-transform on the difference equation to get

z ⎛
⎝ 		z
-
f
 
 (z) − z f0⎞
⎠ 		− z f1
=
(z
-
f
 
 (z) − z f0) +
-
f
 
 (z),
(23)
z2
-
f
 
 (z) − z2 − z
=
z
-
f
 
 (z) − z +
-
f
 
 (z),
(24)
which can be solved as

-
f
 
 (z) = z2
z2 − z − 1
 .
(25)
Noting that
z2
z2 − z − 1
 = α
α− β
 z
z − α
 − β
α− β
 z
z − β
 ,
(26)
where
α = 1 + √5
2
 ,     β = 1 − √5
2
 .
Equation (25) can be inverse Z-transformed using the table above as:

fn = α
α− β
 αn − β
α− β
 βn.
(27)
Alternatively try this:
Enter the function to be inverse z-transformed.
with respect to .


fn = −
⎛
⎝ 		1
2
 − √5
2
⎞
⎠ 		n

 
 (−5 + √5) + ⎛
⎝ 		1
2
 + √5
2
⎞
⎠ 		n

 
 (5 + √5)
10
 .
(28)

Footnotes:

1 Note that
sin (t − τ) sin τ = 1
2
 (cos (t−2 τ) − cos t)
so
⌠
⌡ 		t

0 
 cos (t−2 τ) dτ = sin t,     ⌠
⌡ 		t

0 
 cos t dτ = t cos t.


File translated from TEX by TTH, version 4.03.
On 16 Nov 2025, 11:04.