Chapter VII Wave Statistics &
Wave Spectra
Previously, the regular waves (signle frequency and amplitude)
have been studied.
However, ocean waves are almost irregular.
7.1 Introduction
1. How to use wave statistics and wave to describe (or simulate)
irregular waves.
2. How to use the previous knowledge based on (regular) linear wave
theory to calculate the properties of irregular waves.
3. Wave Statistics
4. Wave Spectra
5. FFT (decompose) and IFFT (superposition or simulation)
Regular and Irregular Waves
Ocean waves are almost always irregular or random.
Irregular waves can be viewed as the superposition of a number
of regular waves (wave components) with different frequencies
and amplitudes.
A regular wave (wave component) has a single frequency
(wavelength) and amplitude (height).
Irregular Waves: long-crested &
short-crested
All wave components are in the same direction ---- Uni-directional
irregular waves, aka, long-crested irregular waves.
Wave components are often multi-directional, ---- directional
irregular waves, aka, short-crested waves.
T
1
0 . 8
0 . 6
0 . 4
0 . 2
0
-0 . 2
-0 . 4
-0 . 6
-0 . 8
-1
0
2
4
6
8
1 0
1 2
Regular
Time t
Irregular
1 4
1 6
1 8
2 0
Wave Pattern Combining Four Regular Waves
FFT & IFFT – (Inverse) Fast Fourier Transform.
Actual (multi-directional) vs. Design (uni-directional)
Seas
7.2 Wave Height Distribution
Different from a regular wave train, in an
irregular wave train, wave heights and
wave periods are not uniform
How to count the wave heights and wave
periods in an irregular wave train.
Zero up-crossing and zero down-crossing
method.
Ocean (Irregular) Waves
Definitions of Zero-Upcrossing & Downcrossing
Root-mean-Square (RMS), Skewness and Kurtosis
Ochi (1998) Ocean Waves
7.2 Wave Height Distribution (conti.)
Probability Density Function (PDF)
Probability density function (pdf), is a
function that describes the relative
likelihood for this random variable to take
on a given value. The probability for the
random variable to fall within a particular
region is given by the integral of this
variable’s density over the region. The
probability density function is nonnegative
everywhere, and its integral over the
entire space is equal to one.
Cumulative Distribution Function (CDF)
Cumulative distribution function (CDF),
describes the probability that a realvalued random variable h with a given
probability distribution will be found at a
value less than or equal to H.
Relationship Between PDF & CDF
Let P is the cumulative distribution function
(CDF) of H and p be the related Probability
Density Function (PDF)
H
P (h  H ) 


p (u )d u
;
p (H ) 
d
dH
P (H )
Example

When measured 5 wave heights are 3.0,
3.5, 4.0, 4.2, 5.0(m), respectively

Find mean wave height

Find the rms wave height.

Find Prob[H>4.1m]
Narrow-Banded Spectra: The Rayleigh
Distribution
An irregular wave train with a narrow-banded
spectrum is that the frequencies of all its
wave components of significant energy
are concentrated near its peak frequency.
Its wave height CDF satisfies Rayleigh
distribution.

H
)  exp[(
)2 ]
P( H  H
H rms

H
)  1  exp[(
or P ( H  H
)2 ]
H rms
The related PDF is


d
H
2H


p( H ) 
[ P ( H  H )]  2 exp[(
)2 ]

H rms
H rms
dH
Examples of using Rayleigh PDF for
computations

 Hp ( H )dH
H  H1 
0


p ( H ) dH

2H 2
H 2
  2 exp[  (
) ]=
H rms
H rms
0
0

H 1/3 

Hp ( H ) dH
 1/3
H


 1.416 H rms
p ( H ) dH
 1/3
H

H 1/10 

Hp ( H ) dH
 1/10
H


 1/10
H
 1.80 H rms
p ( H ) dH

2
H rms
Example
When The root-mean square height of a
narrow-banded sea is equal to 4.24m,
using the Rayleigh distribution to
 Find Probability when [2m<η<4m]
 What is the significant wave height Hs?
 What is the probability [H<12m]
 If 600 waves are measured, how many
waves are expected to exceed H=1.2Hs?
 What is the expected maximum wave
height?
7.3 Wave Spectra
Wave Spectra
1. Discrete Spectra

Wave Energy Density Spectra
 Wave amplitude spectra
2. Continuous spectra
 Wave Energy Density Spectra**
 Wave Amplitude Spectra
See Figure 7.2 at pp194
7.3.1 Spectral Analysis
1. How to obtain an energy density spectrum
 First deriving the discrete wave amplitude
spectrum (FFT) based on measured elevation
 Secondly deriving the discrete energy density
spectrum
 Then deriving the continuous energy density
spectrum
 In simulating an irregular wave train, the
above three steps are reversed.
2. How to simulate an irregular wave train
according to a given wave spectrum (later)
7.3.2 Fourier Analysis
A wave elevation measurement for a duration
T (sec). The basic frequency or frequency
increment is
f  1 / T and   2  f .
1, cos ( n t ), and sin( n t ) are a set of orthognal
functions over [0, T ], where n  1, 2,... (integer)
The integration of the product of any two different functions
from the orthognal function set over [0, T ] is equal to zero.
7.3.2
Fourier
Analysis
T
T
 cos
 sin (n t )dt
( n t )dt  0,
0
T
 cos
 0
0
0
if n  m
T / 2 if n  m
( n t ) cos( m t )dt 
0
T
 sin (n t ) sin(m t )dt
0

0
if n  m
T / 2 if n  m
T
 cos (n t ) sin(m t )dt
 0
for all m and n
0
They can be proved using the trignometry identities.
7.3.2 Fourier Analysis (conti.)
Let  (t ) be an elevation (continuous) measurement for
the time duration [0, T ]. Then it can be decomposed as a
Fourier series.

 (t )  a0   [ an cos ( n t )  bn sin ( n t )].
1
How to obtain theose Fouier coefficients (a0 , an and bn )
T
a0 
1
 (t )dt  0,

T 0
T
T
an 
2
bn    (t ) sin( n t ) dt.
T 0
2
 (t ) cos( n t ) dt ,

T 0
7.3.2 Fourier Analysis (conti.)
a n cos ( n t )  bn sin ( n t )  An cos( n t   n )
Notice
 An cos  n cos( n t )  An sin  n sin( n t ),
By comparing the terms on the two sides of the above equ.
a n  An cos  n and bn = An sin  n
Therefore,
An 
a n2  bn2 ,
 n  tan 1 (bn , a n )
where An is the amplitude of the wave components of
the frequency and  n the related initial phase.
The related energy density of this wave component
1
 gAn2
2
Since  g is constant, the en ergy density in a discrete
En 
energy density spectrum is represneted by An2 / 2.
To compute the Resultant wave properties

Resultant wave elevation
 (t )  A0   An cos( n t   n ).
1
If we measure the elevation by setting   0
at the calm water level, then A0  0.
For practical reasons, the high frequency is truncated at a cutoff
frequency, f c . f c  Nf and hence N  f c / f  2 f c /  .

Since  (t )  A0   An cos( n t   n ) is measured at x  0,
1
For the wave elevation at other locations is given by,
N
 (t )   An cos( xk n  n t   n ).
1
When water depth is h, the resultant potential
N
 (t )  
1
An g cosh[ k n ( z  h )]
sin( xk n  n t   n ).
n
cosh( k n h )
k n is related to n t through dispersion relation, ( n ) 2  gk n tanh( k n h )
Resultant wave properties
Resultant wave velocity
N
A k g cosh[ k n ( z  h)]

u ( x, z , t ) =
cos( xk n  n t   n ).
 n n
n
cosh( k n h)
x
1
N
A k g sinh[ k n ( z  h)]

w( x , z , t ) =
sin( xk n  n t   n ).
 n n
n
cosh( k n h)
z
1
Resultant wave velocity
N
cosh[ k n ( z  h )]
du
u
a x ( x, z , t ) 
sin( xk n  n t   n ).

   An k n g
dt
cosh( k n h)
t
1
sinh[ k n ( z  h )]
dw
u N
a z ( x, z , t ) 
  An k n g
cos( xk n  n t   n ).

dt
t
cosh( k n h)
1
Wave induced dynamic pressure head
N
cosh[ k n ( z  h)]
P ( x, z , t )
1 
=  An
cos( xk n  n t   n ).
g t
cosh( k n h)
g
1
Parseval’s Theorem
Remembering that the potential energy density
T
1 1
P.E    2 (t )dt
T 02
Since potential energy density is equal to kinetic energy density
based on linear wave theory, the energy density is
T
1
E  2 P.E    2 (t )dt.
T 0
N
Substituting  (t )  a0   [ an cos ( n t )  bnsin ( n t )] into above equ,
1
T
N
1
E   {a0   [an cos (n t )  bnsin (n t )]}2 dt
T 0
1
N
an2  bn2
1 2
2
E    (t )dt  a0   [
]
T 0
2
1
T
Continuous wave spectra
There are many theoretical spectrum to
present ocean waves. The most common
ones are:
1.
JONSWAP spectrum (North sea)
2.
Pierson-Moskowitz Spectrum
Wave energy density spectra: P-M & JONSWAP Types
Pierson-Moskowitz Spectrum
  4 
g
5 f



EPM ( f ) 
exp


4 5
 fp  

4
2

f
 
   
where  --- constant depending on wind
2
JONSWAP Spectrum
  f  f 2 
p

4  exp  
2 2

 2 f p 

 
 
g
5 f

E JN ( f ) 
exp  

4 5

 4 fp  
 2  f


where  --- constant depending on wind
2
 a f  f p
 
sharp factor  =1 - 7 (average 3.3)
  b f  f p
P-M (Pierson-Moskowitz) spectrum
Fully-developed sea:
1-parameter: Vw
2-parameter: Hs & Tp

JONSWAP (JOint North-Sea WAve
Project) spectrum: storm sea
3-parameter: Hs & Tp &  (overshoot
parameter; 2-3)
JONSWAP Spectra & H1/3 and Tp
5
S ( f )   J H T f exp[ (Tp f )4 ] d
4
0.06238
where  J 
[1.094  0.01915 In ]
1
0.230  0.0336  0.185(1.9   )
2
4
1/3 p
d  exp[
1
,
fp 
Tp
5
( f / f p  1)2
2
2
]
 (sharp factor)  1 ~ 7(mean 3.3),
 0.07 f  f p 
 

 0.09 f  f p 
Goda (1987)
k-th moment of wave spectrum
mk    S ( ) d 
k
mk    k S ( ) d
Mean period T1=2π mo/m1
Mean period T2=2π √mo/m2
From a discrete energy spectrum
to a continuous energy spectrum
A n2
s ( fi ) 
, and
2 f
f c is th e c u to ff fre q .
7.4 Numerical simulation of an
irregular wave train





How to simulate an irregular wave train
according a given wave spectrum
Given significant wave height & Peak
period
Choose the type of a energy density spectrum,
for example, JONSWAP
Determine the simulation duration, say T. The
basic frequency or frequency increment f=1/T.
Discretize the related energy density spectrum
Determine the cutt-off frequency
7.4 Numerical simulation (conti.)
At each discrete frequency nf, the amplitude of
nth wave components is given by
An  2* S ( fi ) * f ,
where fi  nf and f  1/ T is the frequency increment.


The initial phase of each wave component is
randomly selected between –pi and pi.
The resultant wave elevation is hence given
by
N
 (t ) 
A
n
cos( xk n  n t   n ).
1
W hen w ater depth is h, the resultant potential
 (t ) 
N

1
An g cosh[ k n ( z  h )]
sin( xk n  n t   n ).
n
cosh( k n h )
Discretize a energy spectrum
A n2
,
S ( fi ) 
2 f
w h e re  f
is th e fre q u e n c y in c re m e n t
15.
0
s    m 2 s 
Wave amplitude of wave
component j:
7.5
Aj  2s  j  
max (cuttoff frequency)
min
0.75
  rad s 1 
1.5
Nyquist Criterion: η(t)=ΣAj cos(ωjt+ej)

Tmax=2π /∆ ω : repeated after this!
Solution: use irregular ∆ ω or perturb central
component frequency ωj

∆t < π / ωmax
Discrete spectrum to Continuous spectrum:
By using FFT, we get Aj. Then, S(ω) = Aj²/2∆ω

Relationship b/w
Hs and Tp in GOM
For H S  6 m
Tp  6.35 H S0.385
(s)
For H S  5 m
Tp  5.39 H S0.382
(s)
Revised GOM Design Condition
Wind(m/s)
Hs (m)
Current m/s
West-100
39.9
13.1
2
West-1000
49.9
16.4
2.5
WesC100
38.1
12.3
1.9
WesC1000
47.6
15.4
2.4
Cent-100
48
15.8
2.4
Cent-1000
60
19.8
3
East-100
38.4
12.2
1.9
East-1000
48
15.3
2.4
Return Period & the Probability of the Occurrence
of the Storm in A Given Year
The Return period of a storm, such as 100-year period , 50-year
period, etc, indicates the probability of the storm (with the
related strength) occurring during any given year.
mk    k S ( ) d
For example, 100-year return period means the probability of
the occurrence of the related storm is equal to 0.01 in any
given year. Similarly, 50-year return period is related the
probability of 0.02.
The probability of occurrence in a given year
= 1/(return period).
Return Period & the Probability of the Occurrence
of the Storm in the Life Span of A Platform
Assuming that the service life of a floating structure is 25
years, what is the probability of 100-year storm occurring at
least once during its service life?
mk    k S ( ) d
The probability depends on the length of service life and the
return period.
How to compute it?
Return Period & the Probability of the Occurrence
of the Storm in the Life Span of A Platform
How to compute it?
In each year, the probability for the occurrence of a storm of
100-year return period is
u  0.01
Inversely, the probability for the absense of a storm of
100-year return period is
1- u
mk    k S ( ) d
Since the probability for the occurrence or absense of each year
in the life span (say, 25 years) of a structure is independent, the
probability for the absense of a storm of 100-year storm in the
life span is
(1- u ) 25
Thus, the probability for the occurrence (once or more) is
equal to
P  1- (1- u ) 25 =1- 0.7778  0.2222
Return Period & the Probability of the Occurrence
of the Storm in the Life Span of A Platform
Binormal Distribution
p ( x)  Cnx u x v n  x
mk    k S ( ) d
where u  v  1 and x  1, 2,...., n, Cnx 
n!
x !(n  x)!
Let u stand for the probabilty of the
occurrence of a storm of given return period in a given year,
v  1  u for the probabilty of the absence of a storm of given
return period in that year, n be the years of service life,
and x the number of the occurrence of the storm during the
service life, p ( x) is the probability of that storm occurring x
times during the life span (n years).
Return Period & the Probability of the Occurrence
of the Storm in the Life Span of A Platform
p ( x)  Cnx u x v n  x
For the 100-year return period storm
mk    k S ( ) d
u  0.01 and v  1- u  0.99
the service life n  25 what is the probabilty for the 100-year
storm occurring at least once?
25
25
x 1
x 1
P   p ( x)   C25x u x (1- u ) n- x  1  (1  u ) 25
where the probability for the storm occuring once during the service life
is
1
p(1)  C25
u (1- u ) 24 .
the probability for the storm occuring exact twice during the service life
p (2)  C252 u 2 (1 - u ) 23 .