An approximate form of the Rayleigh re?ection loss and its phase:Application to reverberation calculation
Chris H.Harrison a?
NATO Undersea Research Centre,Viale San Bartolomeo400,19126La Spezia,Italy
?Received9December2009;revised26February2010;accepted3March2010?
A useful approximation to the Rayleigh re?ection coef?cient for two half-spaces composed of water
over sediment is derived.This exhibits dependence on angle that may deviate considerably from
linear in the interval between grazing and critical.It shows that the non-linearity can be expressed
as a separate function that multiplies the linear loss coef?cient.This non-linearity term depends only
on sediment density and does not depend on sediment sound speed or volume absorption.The
non-linearity term tends to unity,i.e.,the re?ection loss becomes effectively linear,when the density
ratio is about1.27.The re?ection phase in the same approximation leads to the well-known
“effective depth”and“lateral shift.”A class of closed-form reverberation?and
signal-to-reverberation?expressions has already been developed?C.H.Harrison,J.Acoust.Soc.
Am.114,2744–2756?2003?;C.H.Harrison,https://www.doczj.com/doc/1f11971752.html,put.Acoust.13,317–340?2005?;C.H.
Harrison,IEEE J.Ocean.Eng.30,660–675?2005??.The?ndings of this paper enable one to
convert these reverberation expressions from simple linear loss to more general re?ecting
environments.Correction curves are calculated in terms of sediment density.These curves are
applied to a test case taken from a recent ONR-funded Reverberation Workshop.
?2010Acoustical Society of America.?DOI:10.1121/1.3372731?
PACS number?s?:43.30.Ma,43.30.Gv,43.20.El?AIT?Pages:50–57
I.INTRODUCTION
The canonical case for shallow water propagation mod-
eling is the Pekeris waveguide where the seabed is repre-
sented as a half-space characterized by density,sound speed,
and volume absorption.In more complicated environments
variable bathymetry and refraction in the water column are
often introduced but still with a half-space seabed.A recent
ONR-sponsored pair of workshops1on reverberation model-
ing speci?ed a number of environments such as this.For this
reason the behavior of the re?ection coef?cient?amplitude,
phase,and logarithmic loss?for a half-space is of interest,
and it can be calculated by the Rayleigh re?ection coef?cient
formula,2on the assumption that the roughness does not
spoil the outward and return specular re?ections?alternative
forward scattering laws have been suggested3?.For sound speeds greater than that of water this predicts a critical angle
and a dB loss that is very nearly proportional to angle.In fact
an approximate formula for the proportionality constant has
been derived by Weston.4Consequently there is ducted
propagation within the critical angle,and the small linear
re?ection loss results in an ever narrowing Gaussian angle
distribution and hence mode-stripping.This has a profound
effect on propagation and particularly reverberation since the
propagation favors low angles whereas most scattering laws
favor high angles,or at least,reject low angles.
Combining Lambert’s law with mode-stripping,the re-
verberation behavior can be predicted by closed-form solu-
tions for range-dependent,bistatic,isovelocity,or refracting environments providing the re?ection loss is linear.5–7If the re?ection loss is not linear then these solutions are still cor-rect at very long range because the narrowness of the Gauss-ian angle distribution biases the loss to the linear part of the curve.They are also correct at short range because,regard-less of the magnitude of the re?ection loss,there are very few re?ections so there is minimal effect.For intermediate ranges there is clearly some effect,and it would be useful to be able to estimate it.This paper derives an extremely good approximation to the Rayleigh formula applicable between zero grazing and almost up to the critical angle.This clearly shows the non-linear behavior of the re?ection loss in very simple https://www.doczj.com/doc/1f11971752.html,ing the same approach one can also derive a formula for re?ection phase,and from this,one can re-derive Weston’s effective depth8,9and clarify its condi-tions of validity.
In all the closed-form reverberation calculations, whether monostatic,bistatic,range-independent or -dependent,the?nal stage with the linear re?ection loss as-sumption is to evaluate an angle integral whose integrand consists of a Gaussian times the sine of the angle.A multi-plicative correction?i.e.,a dB addition?can be tabulated for all ranges and bottom densities in the non-linear case so that the previous solutions can be easily modi?https://www.doczj.com/doc/1f11971752.html,e of these corrections is demonstrated in Sec.III by application to a test case from the recent ONR Workshop.1Solving the isoveloc-ity reverberation angle integral analytically is more dif?cult, but numerical approaches to this correction provide insight into the magnitude and importance of the non-linear re?ec-tion effect.It will be shown that the correction is a function of two parameters?i.e.,a family of curves?for all isovelocity range-independent environments.
This work was?rst reported in Ref.10.
a?Also af?liated with the Institute of Sound and Vibration Research,Univer-sity of Southampton,High?eld,Southampton SO171BJ,United Kingdom. Electronic mail:harrison@nurc.nato.int
II.THEORY A.Re?ection loss
The Rayleigh re?ection coef?cient V for a half-space is usually written 2in terms of the impedances on each side of the boundary Z 1,2,but it is more conveniently written in terms of the admittances ?the reciprocal of the impedance ?A 1,2,
V =Z 2?Z 1Z 2+Z 1=A 1?A 2
A 1+A 2
,
?1?
where
Z m =
1A m =?m c m
sin ?m
,?2?
and ?,c ,and ?are density,sound speed,and grazing angle with indices m =1,2representing water and seabed,respec-tively.In the following,we manipulate these formulae mak-ing no apologies for retaining all the steps,and only making approximations where stated.We assume that the sea is loss-less but that the seabed is an absorbing medium so that
A 2?A R +iA I ,?3?A 1?A 1+i 0,
?4?
where A 1,A R ,and A I are all real.Thus
?V ?2
=
?
A 1?A R ?iA I
A 1+A R +iA I
?
2
=
?A 1?A R ?2+A I 2?A 1+A R ?2+A I 2
=A 12+A R 2+A I 2?2A 1A R
A 12
+A R 2+A I 2+2A 1A R =1?X 1+X
,?5?
where
X =
2A 1A R A 12
+A R 2
+A I
2.
?6?
Consequently the natural-logarithmic loss ?as opposed to the loss in dB ?can be written exactly as
?log ??V ?2?=??log ?1?X ??log ?1+X ??
=????X ?X 2
/2?X 3
/3?ˉ???X ?X 2/2+X 3/3?ˉ??=2X +2/3X 3+2/5X 5+ˉ.
?7?
We now de?ne the fractional imaginary part of the wave-number in the seabed as ?,so that the complex wavenumber
is k ?1+i ??where k and ?are both real.This contrasts with Weston’s taking the imaginary part of the sound speed as the starting point.4Thus the wave function is
exp ?ik ?1+i ??r ?=exp ?ikr ?exp ??k ?r ?,
?8?
and the power decays as exp ??2k ?r ?.In terms of volume absorption in the seabed a dB/wavelength,this same power decay is written as 10?ar ?k /2??/10.Taking logs to base e we ?nd
?=
a
40?log 10?e ?=a 54.575
,
?9?
and we see that since a is typically of order or less than 1dB /?then ??1is a very good approximation since ?is always of order 0.02or less.Nevertheless,at this stage we make no assumptions about ?.
We now evaluate the admittances in terms of geoacous-tic properties and the sine of the grazing angle in the water for which we use the shorthand s .
A 1=
s ?1c 1
,?10?
A 2=??1+i ??2?Q
?2c 2
=i
??Q ?1+?2??2i ?
?2c 2
,?11?
where
Q ?
?c 2
c 1
?
2
?1?s 2?,?12?
and Q ?1always.To separate A 2into real and imaginary parts we note that it can be written as ?2c 2A 2/i =?a +ib =c +id ,where
a +i
b =?
c 2?
d 2?+i 2cd ,?13?
so
a =c 2?d 2;
b =2cd .
?14?
Solving the quadratic for c 2,stipulating that c 2must be posi-tive,we ?nd
c 2
=
a +?a 2+
b 2
2
,?15?
so from the de?nition Eq.?3?
A I =
1?2c 2?
?Q ?1+?2?+??Q ?1+?2?2+4?2
2
,
?16?
A R =
??2
?2c 2??Q ?1+?2?+??Q ?1+?2?2+4?2
.?17?
Returning to Eqs.?6?and ?7?we substitute Eqs.?10?,?16?,and ?17?to obtain
X =
2?2s ?
X o ?1c 1?2c 2??Q ?1+?2?+??Q ?1+?2?2+4?2
,?18?
where
X o =s 2
?12c 12+2?2?22c 22
??Q ?1+?2?+??Q ?1+?2?2+4?2?
+
?Q ?1+?2?+??Q ?1+?2?2+4?2
2?22c 22
.
?19?
Equations ?18?and ?19?are exact.Neglecting terms in ?2?which are extremely small,in practice ??0.02?2=4?10?4?but retaining ?and writing Q as
Q ?1=
?c 2
c 1
?
2
?s c 2?s 2?,?20?
where s c =sin ??c ?we have
X =
?2s ??1?2c 22?s c 2?s 2
??
s
2
?12c 12+
?s c 2
?s 2
?
?22c 12
?
=?2?2c 12?
?1c 22
s c
3s
?
?
1
?1?s 2/s c 2?1+???2/?1?2?1?s 2/s c 2?
.
?21?
The term in curly brackets on the right hand side of Eq.?21?is recognized as half of Weston’s linear coef?cient ?multi-plied by sine of the grazing angle ?rather than the angle it-self ?.He de?ned ?as 4
?=
4?2c 12?
?1c 22s c 3
.?22?
The remainder of Eq.?21?is the non-linear factor.This is clearly unity for small s ,tends to in?nity as s approaches s c ,and otherwise depends only on density ratio.
Returning to the re?ection loss as de?ned in Eq.?7?we see that the ?rst term in the X expansion is proportional to ?,
but the second term is proportional to ?3and is therefore always negligible since their ratio is proportional to ?2.So the re?ection loss can be written as
?log ??R ?2?=
?s
?1?s 2/s c 2?1+???2/?1?2?1?s 2/s c 2?.
?23?
It is obvious that Eq.?23?fails when s ?s c ;it is less obvious that it fails when the grazing angle is extremely close to critical.The reason can be traced to Eq.?20?where Q ?1→0as s →s c .Then the denominator of the exact Eq.?18?contains only terms in ?2,and so its neglect begins to have a serious effect.To avoid this we need
s 2?s c 2??2c 12/c 22,
?24?
but since ?2??4?10?4this is not a serious problem in
normal applications.
Some plots of this function are superimposed on the exact Rayleigh re?ection loss for various values of sound speed,density,and volume absorption in Figs.1?a ?and 1?b ?.Expanding Eq.?23?as a power series we ?nd the approxi-mation
?log ??R ?2?=?s ?1+b 1?s /s s ?2+b 2?s /s c ?4+ˉ?
??s +?s 3?1.5???2/?1?2?/s c 2,
b 1=3/2???2/?1?2;
b 2=15/8??5/2???2/?1?2
+??2/?1?4.
?25?
Although this is an approximation it suggests that the re?ec-tion loss should approach linearity when the second term is zero,i.e.,?2/?1=?3/2=1.225.
Of course,we cannot expect exact linearity since the function inevitably curves upwards just before the critical angle.However it is clear that densities lower than this value cause upward curvature ?more loss than linear ?while higher densities cause downward curvature ?less loss than linear ?.This effect is shown through the non-linear factor in Fig.2,and the closest to linear ?i.e.,?at ?can be seen to be near the density ratio of 1.25.In fact the denominator of Eq.?23?can be written as ?1+?2B ?1?v +B ?B ?2?v 2?B 2v 3?1/2with
(b)
(a)
FIG.1.The non-linear approximation,linear approximation,and exact Ray-leigh re?ection coef?cient for two half-spaces,with density ratio,sound speed,and volume absorption:?a ?1.74,1600m/s,0.2dB /?;?b ?1.88,1700m/s,0.5dB /?
.
FIG.2.?Color online ?The non-linear re?ection loss factor ??1?s 2/s c 2?1+???2/?1?2?1?s 2/s c 2???1?i.e.,Equation ?23?excluding the numerator ?vs.normalized angle with density ratio as parameter ?see key ?.
B=??2/?1?2?1and v=?s/s c?2.Its gradient is zero near v=0 if B=0.5,i.e.,when the density ratio is?1.5as before.Forc-ing the value of the denominator to be unity at v=0.25?cor-responding to approximately10°in Fig.2?leads to B =0.619and a density ratio of1.27.
B.Re?ection phase
Starting with Eq.?1?and using the same approach we can derive a formula for the phase change on re?ection. Again we write the complex re?ection coef?cient?see Eq.?5??as
V=A1?A R?iA I
A1+A R+iA I
=
?A1?A R?iA I??A1+A R?iA I?
?A1+A R?2?A I2
=A12?A R2?A I2?i2A1A I
?A1+A R?2?A I2
.?26?
Therefore the phase?is given by
tan?=
2A1A I
A12?A R2?A I2
.?27?
Substituting Eqs.?10?,?16?,and?17?we?nd the exact rela-tion
tan?=
?2s
Y o?1c1?2c2
?
??Q?1+?2?+??Q?1+?2?2+4?2,?28?
where
Y o=
s2
?12c12?
2?2
?22c22??Q?1+?2?+??Q?1+?2?2+4?2??
?Q?1+?2?+??Q?1+?2?2+4?2
2?22c22
.?29?
Note that Y o differs from X o?Eq.?19??only in two sign changes.Again neglecting terms in?2we obtain
tan?=
2s
?1c1?2c2
?Q?1?s2?
1
2c
1
2
?
Q?1
?22c22
?
=2s?2
s c?1
?1??s/s
c
?2
?1??1+??2/?1?2??s/s c?2?.?30?
The phase is understood to start in the third quadrant and move round to the fourth as s increases up to critical,i.e.,?=??+atan?2s?2s c?1?1??s/s c?2
?1??1+??2/?1?2??s/s c?2??,?31?or with a four-quadrant tangent
?=atan4Q??2s?2s c?1?1??s/s c?2,
??1??1+??2/?1?2??s/s c?2??.?32?This can be written more elegantly in terms of
sin x=s/s c?33?and
tan y=?2/?1?34?as
tan?=
sin2x sin2y
cos2x+cos2y
.?35?
The top line of Eq.?30?can be recognized as the expansion tan2?=2tan?/?1?tan2??in which?=2?,and conse-quently
tan2?=??1c1?2c2?2?Q?1?s2=???1?2?cot?
?2
??1??c1c2?2sec2??.?36?
This agrees with the formula given by Eq.?2?of Weston8?except for a typographical error?which is easily derived directly from Eq.?1?in the loss free case.It also demon-strates that the above formulae are exact for loss free seabed, and that the phase is otherwise independent of?to?rst order.
There is an interesting special case when?1=?2.Ac-cording to Eq.?34?y=?/4,so tan?=tan2x,i.e.,?=??+2asin?s/s c?.For small phase angles,phase is proportional to s,in other words the complex re?ection coef?cient rotates in a helix between zero grazing angle and the critical angle, starting at??1+i0?,passing near?0?i1?,ending at?1+i0?. Figure3shows this behavior for a number of densities.It can also be seen that the phase remains linear over a larger range of angles when?2/?1?1.25,i.e.,when the re?ection loss is linear?see Fig.2?.Equations?34?and?35?also show that as seabed density tends to in?nity?hard bottom?or zero?pres-sure release?the tangent of the phase tends to zero,indepen-dently of angle?in the former case the phase itself is zero, and in the latter it is??
.
FIG.3.?Color online?Re?ection phase?radians?for normalized grazing angle with density ratio as parameter?see key?using Eq.?31?.Note the closeness to linearity particularly near a density of about1.25.
C.Weston’s effective depth
In reality we have a single re?ection from a simple half-space boundary and we wish to replace this true re?ecting surface with a roughly equivalent,slightly deeper,pressure release boundary.This concept was originally developed by Weston8and then revisited in Refs.9,11,and12.The effect can easily be understood by inspection of typical normal mode shapes.The?rst mode?very shallow angle?tends to have a relatively small value just above the seabed,and if extrapolated downwards would approach zero at a point just below the boundary.In contrast the highest mode?at nearly the critical angle?has gradient zero at the boundary,and therefore,if extrapolated would approach zero at one quarter of its vertical wavelength below the boundary.Because this wavelength is smaller for the high order modes than for the ?rst,all modes tend to meet at approximately the same depth,an effective depth where there could have been a pres-sure release surface.This is the same phenomenon as the end-correction on musical instruments,such as a?ute or recorder.13
Mathematically we take the equation for the re?ection phase and we equate it to the phase difference?imposed on a plane wave by shifting the re?ection boundary down a distance h and re?ecting from a pressure release boundary, i.e.,
?=??+2hk o s,?37?where k o is the wavenumber in the upper medium.
From Eqs.?31?and?37?,using the shorthand of Eqs.?33?–?35?generally we have
h=
?o
4?s
atan?sin2x sin2y
cos2x+cos2y?,?38?
where?o is the wavelength in the upper medium.For small angles this reduces to
h=
?o
2?s c
?2
?1,?39?
which is identical to Eq.?5?of Ref.8.Also,the special case for any angle,where?1=?2leads to
h=
?o
2?s
asin?s/s c?.?40?
If,in addition,s is not too close to s c then
h=1/?k o s c?=?o/?2?s c?.?41?This is clearly a distance that depends on frequency and criti-cal angle but is independent of angle?or mode number?.This limiting case is obviously independent of density.
Another limiting case of interest is?2→?.From Eq.?34?,y=?/2,and so from Eq.?35?,?=0.Under these con-ditions the concept of an effective angle-independent depth to a pressure release surface certainly does not hold.In fact the surface in this extreme case is already a hard boundary. To investigate the general utility of the effective depth we plot normalized effective depth h/?o?s c?1/?2?against s/s c for various densities in Fig.4using the general formula?38?combined with Eqs.?33?and?34?.The effective depth is nearly constant when?2/?1?1.3,i.e.,when both the re?ec-tion loss and its phase are linear.
Note that there are a number of similar“effective/shift/ displacement”terms used in the literature and two genuinely different phenomena as well.Weston12distinguishes,and shows the mathematical relation between,what he calls a “wave shift”and a“beam shift.”The former results in the effective depth described here and is associated with the modal phase velocity since it is a pure phase effect.The latter results in the lateral shift?or“lateral wave”?associated with point source/receivers,spherical waves,14and the modal group velocity since it is seen,for instance,in laboratory experiments where there is a tangible arrival corresponding to the path that runs for part of its course along the boundary. Mathematically the effective depth in this paper corresponds to Eq.?4?of Ref.12and Eqs.?2?and?3?of Ref.8,as already stated.Equations?4.4.5?and?4.4.6?of Ref.14?the lateral wave?,after minor rearrangement,is the same as Eq.?A3?of Ref.9and Eq.?17?of Ref.12?except for typographical errors?.Finally note that the same effective depth term has been used with different meaning in at least two other contexts.15,16
III.REVERBERATION
A.Reverberation angle integral
The purpose of searching for an approximate form of the Rayleigh re?ection coef?cient was to be able to modify the reverberation integral under conditions when the re?ection loss was non-linear.The closed-form equations for rever-beration have been derived for various types of environments.5–7For Lambert’s law
S=?sin??1?sin??2?,?42?with isovelocity water and?at bottom a general expression
is FIG. 4.?Color online?Normalized effective depth?h/h o where h o =??o/2?s c???2/?1??against normalized grazing angle with density ratio as parameter?see key?using Eq.?38?.Note the?atness particularly when density is slightly larger than1.25.
I R =
?1rH
?
?c
sin ?exp ?
?log ??V ?2?tan ?
2H
r ?d ??
2
?r ?p ,
?43?
where r ,H ,p ,?,and ?are respectively,range,water depth,
half spatial pulse length ?1
2
ct p ?,horizontal beam width,and Lambert’s constant.
Reverberation is,of course,strictly a function of time rather than range,however,a recent paper ?Ref.17?has shown that the difference is less than 1dB for almost all practical cases,so here we retain this simple expression.For small angles with linear re?ection loss this reduces to
I R =
?1rH ?0
?c ?exp ????2
2H
r ?d ??
2
?r ?p =
??2r
3?p ?1?exp ???r ?c 2/2H ??2.?44?
Anticipating the form of the re?ection loss in Eq.?23?we put
v =??/?c ?2and de?ne a normalized range R =?rs c 2/?2H ?.The integral with non-linear re?ection loss is then of the form
I =
?
?c
?exp ??Rg ?2
?d ?=
?
1
exp ??Rg v ?d v ,?45?
where g ,written in terms of the density ratio ?,is
g =??1?v ?1+??2?1?v ???1.
?46?
For comparison,the integral from Eq.?44?is the same but with g =1,resulting in ?s c 2/2??1?exp ??R ??/R .The ratio of the non-linear to the linear integral F is then only a function of normalized range R and density ratio ?
F ??,R ?=
?0
1
exp ??Rg v ?d v
?
1
exp ??R v ?d v
.
?47?
We see that at long range ?large R ?the non-linear part of the exponential is so weak as to be ineffective,and at short range both integrands tend to unity.The result is that the ratio tends to unity at short and long range.A plot against normalized range is shown in Fig.5with density ratio as parameter through Eq.?47?.Note that because ?is a function of density ratio ?through Eq.?22??each line has a different normaliza-tion
R =r ????s c 2/2H =r ??2/?1?2c 12?/?c 22Hs c ?.
Because the function F is only dependent on ?and R it is a solution for any isovelocity range-independent environment,and one can consider it to be fully tabulated in the sense of a mathematical “special function”F ??,R ?.In this way the re-verberation solutions can still be considered closed-form since they can be written in terms of known functions.Note that although we used the small angle approximation one could still calculate this ratio without these approximations or as a conversion from Eq.?44?to any more exact formu-lation,including Eq.?43?with an arbitrary re?ection loss.
The point is that within the critical angle the ratio is still only a function of the two parameters ?,R .
B.Application to ONR reverberation workshop test CASE XI
The ONR Workshop 1Test Case XI consisted of isove-locity water ?1500m/s ?of uniform depth 100m overlaying a bottom half-space with sound speed 1700m/s,density ratio 2.0,and volume absorption 0.5dB per wavelength.The Lambert scattering coef?cient was ?=10?2.7,i.e.,?27dB.The source had a Gaussian shaded 3-dB bandwidth of one twentieth of the frequency,such that the peak amplitude of the pulse at 1m was 1?Pa,and azimuthal coverage was 2?.
The reverberation formulas used here are frequency in-dependent,and in isovelocity water they give essentially depth averaged values,so the source and receiver depths are irrelevant.The following plots show the quantity in Eq.?43??before and after correction ?with t p =1second.To convert to the ONR test case one should add ?17.26,?23.28,and ?28.72dB at frequencies 250,1000,and 3500Hz.This is the result of the differing bandwidths,and therefore pulse lengths,and a minor change because of the Gaussian shad-ing.
Figure 6?a ?shows the result of adding 20log 10F ?the correction applies to the outward and return propagation ?to the linear re?ection loss reverberation curve vs.travel time in the workshop case where density ratio was 2.0.There is a clear increase of a few dB at intermediate ranges.As could already be seen from Fig.5the worst case difference is about 3.8dB.Figure 6?b ?shows that,keeping other parameters the same a density ratio of about 1.5results in reverberation much closer to the linear case.Note that on the right there is also a slight offset in the linear re?ection loss curve com-pared with Fig.6?a ?because the density ratio also alters ?
.
FIG.5.?Color online ?The ratio of the one-way propagation integral with non-linear re?ection loss to the same integral but with linear loss ?F ??,R ?in dB from Eq.?47??against normalized range R ?R =r ????s c 2/2H =r ??2/?1?2c 12?/?c 22Hs c ??with density ratio as parameter ?see key ?.The factor F applies both to the outward and return paths implicit in reverbera-tion.
If comparisons are to be made with other models it must be remembered that the volume absorption in the water and the contribution of rays steeper than critical have been ex-cluded here.In the relevant ONR test case good agreement was found by simply adding in the absorption for the rel-evant frequency along an assumed horizontal path.
IV.CONCLUSIONS
By assuming that the imaginary part of the wavenumber in the sediment is much smaller than the real part?which is always true for wave-like propagation?an extremely good approximation to the Rayleigh re?ection coef?cient has been derived?Eq.?23??.This demonstrates explicitly the initial proportionality to angle,the onset of non-linearity,the non-linearity’s dependence on density,and its independence of other bottom parameters.The approximate form?Eq.?23??is valid for grazing angles lower than,and not too close to the critical angle.
The non-linear re?ection loss can be expressed as the linear part?obeying exactly Weston’s equation,Eq.?18?of Ref.3?multiplied by a factor which is only a function of density and https://www.doczj.com/doc/1f11971752.html,rge densities tend to?atten the re?ec-tion loss;small densities tend to steepen the re?ection loss, and the non-linear factor is shown in Fig.2.The series ex-pansion for the re?ection loss contains only odd powers of sin?,so the?rst non-linear term is in sin3?.The non-linear multiplier can be expressed in terms of a“fractional”angle ?sin?/sin?c?.
By making the same approximation formulae are de-rived for re?ection phase.These also depend only on density and fractional angle?sin?/sin?c?.The re?ection phase leads to the well-known concepts of the lateral shift and the effec-tive depth.The effective depth is re-derived in terms of the approximate phase,and it is shown that the concept should work well provided the seabed has a critical angle and all signi?cant rays re?ect at angles well inside it,as is the case with mode-stripping.The effective depth is very close to angle-independent for a density ratio of around1.27.
It was demonstrated that one can calculate a correction factor?Eq.?47?,Fig.5?for closed-form reverberation ex-pressions that converts them from linear to non-linear re?ec-tion loss.The correction is expressed as a function of nor-malized range with density ratio as parameter.The curve for a density ratio of2.0was applied to one of the test cases from the recent ONR Reverberation Workshop,and the result is shown in Fig.6.
ACKNOWLEDGMENT
Some of the test cases in the ONR Reverberation Work-shop?2006?were based on the closed-form reverberation expressions cited here.Kevin Le Page,in his role as Chair-man of the Problem De?nition Committee,happened to choose one parameter?sediment density=2?,which resulted in the re?ection loss being strongly?and rather uncharacter-istically?non-linear,thus unfortunately invalidating the closed-form solutions as strict benchmarks,but at the same time,inadvertently providing an incentive to re-establish their benchmark status.Without such stimulus this work on re?ection loss non-linearity would never have existed.
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(b)
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FIG.6.Closed-form reverberation vs.time showing the effect of non-linear re?ection loss?solid line?compared with linear?dashed??a?for the ONR Reverberation Workshop Problem XI with density ratio2.0,and?b?with density ratio1.5.
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. （初一）去绝对值常用“六招” （初一）六招”去绝对值常用“难度大，解绝对值问题要求高，绝对值是初中数学的一个重要概念，是后续学习的必备知识。不易把握，解题易陷入困境。下面就教同学们去绝对值的常用几招。一、根据定义去绝对值的值-│c│c = - 8时，求3│a│-2│b│例1、当a = -5，b = 2，负数的绝所以根据绝对值的意义即正数的绝对值是它本身，分析：这里给出的是确定的数，。代值后即可去掉绝对值。的绝对值是0对值是它的相反数，00 ＜ c = -8b =2＞0，解：因为：a = -5＜0，[ - ( - 8 ) ] = 7 2 ×2 --5）] –所以由绝对值的意义，原式= 3 [ -（”相关信息去绝对值二、从数轴上“读取c在数轴上的a、b、例2、有理数- │a│-a│+│c-b│+│a+b│位置如图所示，且│a│=│b│，化简│c的正负性，由数轴上点的位置特征，即可去绝对、a + bc - a、c-b分析：本题的关键是确定值。- a = b b 且＜c＜解：由已知及数轴上点的位置特征知：a＜0 b ) ] + 0 - ( - a ) = b –故原式= c - a + [ - ( c c - b＜0，a + b = 0 从而 c –a ＞0 ，三、由非负数性质去绝对值22的值。= 0，求-25│+ ( b –2 )ab：已知例3│a 。分析：因为绝对值、完全平方数为非负数，几个非负数的和为零，则这几个数均为“0”222 2 = 0 –由绝对值和非负数的性质：ab 解：因为│a-25 = 0 -25 │+ ( b – 2 )且= 0 ab = - 10 ab = 10或a = - 5 b = 2 故即a = 5 b = 2 或四、用分类讨论法去绝对值的值。abc≠0，求+ + 4例、若同为正号还是同为负号；两个同为正（负）号，另、c，所以只需 考虑a、b分析：因abc≠0一个为负（正）号，共八种情况。但因为两正（负）、一负（正）的 结果只有两种情况，所以其值只有四种情况。异号。b、、c、b、c有同为正号、同为负号和aa 解：由abc≠0可知，= 3 + + + = + 、c都为“+”时，b当a、= - 3 ---”时，+ + = c当a、b、都为“-+ + = 1 时，“-”、a、bc中两“+”一当+ + = - 1 “+”时，中两“-”一ca 当、b、五、用零点分段法去绝对值的最小值。2│+│x -3│-例5：求│x + 1│+│x 的值的符号也在变化。关键是把各式绝对值x -3–x 2、、在有理数范围变化，分析：xx + 1解 这类问题的基本步骤是：的取值进行分段讨论，为此要对符号去掉。x然后选取其最小值。. . 求零点、分区间、定性质、去符号。即令各绝对值代数式为零，得若干个绝对值为零的点，这些点把数轴分成几个区间，再在各区间化简求值即可。。由绝对值意义分别讨论如下：，3可确定零点为- 1，2，解：由x + 1 = 0x - 2 = 0，x - 3 = 03 + 4 = 7 ＞– 3 ) ] = -3 x + 4 -1时，原式= -( x + 1 ) + [ - ( x –2 ) ] + [ - ( x 当x＜-2 + 6 = 4 3 ) ] = - x + 6 ＞时，原式= ( x + 1 ) + [ -( x –2 ) ] + [ - ( x –当-1 ≤x ＜2 2 + 2 = 4 x + 2 ≥= –2 ) + [ - ( x –3 ) ] 当2 ≤x ＜3时，原式= ( x + 1 ) + ( x - 4 = 5 4 ≥3×3 –2 ) + ( x 3 ) = 3x –x ≥3时，原式= ( x + 1 ) + ( x –当4。故所求最小值是六、平方法去绝对值-3│、解方程│x-1│=│x例6所以对所分析：对含有绝对值的方程，用平方法是去绝对值的方法之一，但可能产生增根，求解必须进行检验，舍去增根。22 x=2是原不等式的根。x=2 x经检验，- 2x +1= x - 6x + 9 有4x =8，得解：两边平方： c在数轴上的位置、b、练习1、已知实数a │a│=│c│，化简：如图，且- b│+│a││a+c
儿童保护视力10方法 一光线须充足：光线要充足舒适，光线太弱而因字体看不清就会越看越近。 二反光要避免：书桌边应有灯光装置，其目的在减少反光以降低对眼睛的伤害。 三阅读时间勿太长：无论做功课或看电视，时间不可太长，以每三十分钟休息片刻为佳。 四坐姿要端正：不可弯腰驼背，越靠近或趴着做功课易造成睫状肌紧张过度，进而造成近视。 五看书距离应适中：书与眼睛之间的距离应以30公分为准，且桌椅的高度也应与体格相配 合，不可勉强将就。 六看电视距离勿太近：看电视时应保持与电视画面对角线六~八倍距离，每30分钟必须休息片刻 七睡眠不可太少，作息有规律：睡眠不足身体容易疲劳，易造成假性近视。 八多做户外运动：经常眺望远外放松眼肌，防止近视，向大自然多接触青山绿野，有益于眼睛的健康。 九营养摄取应无均衡：不可偏食，应特别注意维生素B类（胚芽米、麦片酵母）之摄取。 十定期做视力：凡视力不正常者应至合格眼镜公司或眼科医师处做进一步的检查。 如何保护儿童视力 孩子得了近视眼，埋怨父母，是有一定道理的。 长期以来，已有不少调查证实，近视眼的发生有一定家族性，父母亲均为近视，子女患近视的多，父母亲均无近视，子女患近视的少，父母一方有近视，子女患近视介于上述两者之间。所以儿童近视眼与父母关系十分密切。 预防近视眼，要注意以下几点： （1）不要看字迹太小或模糊的书报。 （2）教育儿童改正不合理的用眼习惯，不良习惯都会使眼睛过度疲劳，降低视力的敏锐度。 （3）加强体格锻炼，增强身体素质，可以减轻减慢近视眼的发生，尤其是室外体育运动。让孩子在空气新鲜、视野开阔的郊外进行远眺，极目欣赏祖国的山河大地，也是
去绝对值符号的几种常用方法 解含绝对值不等式的基本思路是去掉绝对值符号，使不等式变为不含绝对值符号的一般不等式，而后，其解法与一般不等式的解法相同。因此掌握去掉绝对值符号的方法和途径是解题关键。 1．利用定义法去掉绝对值符号 根据实数含绝对值的意义，即|x |=(0)(0)x x x x ≥??-????≤?； |x |>c (0)0(0)(0)x c x c c x c x R c <->>???≠=??∈c (c >0)来解，如|ax b +|>c (c >0)可为ax b +>c 或ax b +<－c ；|ax b +| 独立同分布随机变量序列的顺序统计方法 设有限长度离散随机变量序列12,,...,n x x x ，对其按从小到大的顺序排列，得到新的随机序列12,,...,n y y y ，满足：12...n y y y ≤≤≤；假设12,,...,n x x x 是独立同分布的连续取值型随机变量，每个变量的概率分布函数及概率密度分布函数分别为(),()F x f x 。 (1)求(1)k y k n ≤≤的概率密度分布函数()k y f y 解：k y 在y 处无穷小邻域取值的概率()k y f y dy 可以等效为这样一些事件发生的概率之 和：12,,...,n x x x 这n 个随机变量中有任意一个在y 处无穷小邻域取值，而剩余的n -1个随机变量中有任意k -1个的取值小于等于y ，对应的另外n -k 个变量的取值大于等于y 事件的个数(变量的组合数)为111n n k -???? ???-???? ，每个事件的概率为1[()]()[1()]k n k f y dy F y F y ---，则 11()()()[1()]11k k n k y n n f y dy f y dyF y F y k ---????=- ???-???? => 1!()()[1()]() (1)(1)!()! k k n k y n f y F y F y f y k n k n k --= -≤≤-- (2)求随机变量,(1)k l y y k l n ≤<≤的联合概率密度分布函数(,)k l y y f u v 解：(,) ()k l y y k l <在平面上的点(,) ()u v v u ≥处无穷小邻域取值的概率 随机变量及其分布知识点汇总 知识点一 离散型随机变量及其分布列 （一）、离散型随机变量的分布列 一般地，设离散型随机变量X 可能取的值为12,,,,,i n x x x x ??????，X 取每一个值 (1,2,,)i x i n =???的概率()i i P X x p ==，则称以下表格 为随机变量X 的概率分布列，简称X 的分布列. 离散型随机变量的分布列具有下述两个性质： （1）0,1,2,,i P i n =???≥ （2）121n p p p ++???+= 1．两点分布 如果随机变量X 的分布列为 则称X 服从两点分布，并称=P(X=1)p 为成功概率. 2．超几何分布 一般地，在含有M 件次品的N 件产品中，任取n 件，其中恰有X 件次品，则事件{}X k =发生的概率为： (),0,1,2,3,...,k n k M N M n N C C P X k k m C --=== 则随机变量X 的概率分布列如下： {}*min ,,,,,,m M n n N M N n M N N =≤≤∈其中且。 注意：超几何分布的模型是不放回抽样 知识点二 条件概率与事件的独立性 （一）、条件概率 一般地，设A,B 为两个事件,且()0P A >,称() (|)() P AB P B A P A =为在事件A 发生的条件下,事件B 发生的条件概率. 0(|)1P B A ≤≤ 如果B 和C 互斥，那么[()|](|)(|)P B C A P B A P C A =+ （二）、相互独立事件 设A ，B 两个事件，如果事件A 是否发生对事件B 发生的概率没有影响(即 ()()()P AB P A P B =),则称事件A 与事件B 相互独立。 ()()()A B P AB P A P B ?=即、相互独立 一般地，如果事件A 1,A 2,…,A n 两两相互独立，那么这n 个事件同时发生的概率，等于每个事件发生的概率的积，即1212(...)()()...()n n P A A A P A P A P A =. 注意：(1)互斥事件：指同一次试验中的两个事件不可能同时发生； (2)相互独立事件：指在不同试验下的两个事件互不影响. （三）、n 次独立重复试验 1．一般地，在相同条件下，重复做的n 次试验称为n 次独立重复试验. 在n 次独立重复试验中，记i A 是“第i 次试验的结果”，显然， 1212()()()()n n P A A A P A P A P A ???=??? “相同条件下”等价于各次试验的结果不会受其他试验的影响 注意: 独立重复试验模型满足以下三方面特征 第一：每次试验是在同样条件下进行； 第二：各次试验中的事件是相互独立的； 第三：每次试验都只有两种结果，即事件要么发生，要么不发生. 2．n 次独立重复试验的公式： n A X A p n A k 一般地，在次独立重复试验中，设事件发生的次数为，在每次试验中事件发生的概率为，那么在次独立重复试验中，事件恰好发生次的概率为 ()(1),0,1,2,...,.(1)k k n k k k n k n n P X k C p p C p q k n q p --==-===-其中，而称p 为成功 保温层厚度的计算与校核 1 已知条件 保温棉内侧对流换热系数h1=70w/(m2·k)，温度分别为0℃、-60℃、-138℃。铝片的厚度∝1为5mm，传热系数λ1=236w/(m2·k)。保温棉的传热系数λ2=0.022 w/(m2·k)。保温棉外侧的空气温度为35℃，其表面温度查空气焓湿图，取35℃、65%相对湿度情况下的露点温度。保温棉外侧的对流换热系数h2=8 w/(m2·k)。 2 保温棉厚度计算 2.1 露点温度 空气温度T a=35℃，相对湿度为65%时，查空气焓湿图得到露点温度T d=27.57℃。2.2最大允许冷损失量的计算 根据《工业设备及管道绝热工程设计规范（GB50264-97）》，最大允许冷损失量应按以下公式进行计算： 当T a-T d≤4.5时： [Q]=-(T a-T d)αs; 当T a-T d >4.5时： [Q]=-4.5αs 其中αs绝热层外表面向周围环境的放热系数。 T a-T d=（35-27.57）℃=7.43℃，故最大允许冷损失量 [Q]=-4.5αs=-4.5×8=-36w。 2.3 保温棉厚度的计算 由传热公式知： [Q]= (T i-T a)/ (1 ?1+∝1 λ1 +∝2 λ2 +1 ?2 ) 其中∝2为保温层的厚度。 由此得到∝2=λ2×（T i?T a Q ?1 ?1 ?1 ?2 ?∝1 λ1 ） 1 保温层内侧温度为0℃时 保温层厚度∝2= λ2×T i?T a Q ?1 ?1 ?1 ?2 ?∝1 λ1 =0.022×0?35 ?36 ?1 70 ?1 8 ?0.005 236 =0.018m 2 保温层内侧温度为-60℃时 保温层厚度∝2= λ2×T i?T a Q ?1 ?1 ?1 ?2 ?∝1 λ1 =0.022×?60?35 ?36 ?1 70 ?1 8 ?0.005 236 =0.054m 3 保温层内侧温度为-138℃时 保温层厚度∝2= λ2×T i?T a Q ?1 ?1 ?1 ?2 ?∝1 λ1 =0.022×?138?35 ?36 ?1 70 ?1 8 ?0.005 236 =0.103m 3 保温层厚度的校核 设保温层外侧表面的温度为T f 1 保温层内侧温度为0℃时 取保温层厚度∝2=0.025m 传热量[Q] = (T i-T a)/ (1 ?1+∝1 λ1 +∝2 λ2 +1 ?2 )= (0-35)/ (1 70 +0.005 236 +0.025 0.022 +1 8 )=-27.44w T f=T a+Q ?2=35?27.44 8 =31.57℃>T d=27.57℃故符合要求。 数学学科自习卷（二） 一、选择题 1．将三颗骰子各掷一次，记事件A ＝“三个点数都不同”，B ＝“至少出现一个６点”，则条件概率()P A B ，() P B A 分别是（ ） A.6091，12 B.12，6091 C.518，6091 D.91216，12 2．设随机变量ξ服从正态分布()3,4N ，若()()232P a P a ξξ<-=>+，则a 的值为 A ．73 B ．53 C ．5 D ．3 3．已知随机变量ξ～)2,3(2N ，若23ξη=+,则D η= A ． 0 B ． 1 C ． 2 D ． 4 4．同时拋掷5枚均匀的硬币80次，设5枚硬币正好出现2枚正面向上，3枚反面向上的次数为ξ，则ξ的数学期望是（ ） A ．20 B ．25 C. 30 D ．40 5． 甲乙两人进行乒乓球比赛, 约定每局胜者得1分, 负者得0分, 比赛进行到有一人比对方多2分或打满6局时停止, 设甲在每局中获胜的概率为 23,乙在每局中获胜的概率为13 ,且各局胜负相互独立, 则比赛停止时已打局数ξ的期望()E ξ为（ ） A ．24181 B ．26681 C ．27481 D ．670243 6．现在有10奖券，82元的，25元的，某人从中随机无放回地抽取3奖券,则此人得奖金额的数学期望为（ ） A ．6 B ．395 C ．415 D ．9 7．一个篮球运动员投篮一次得3分的概率为a ，得2分的概率为b ，不得分的概率为c ，,,(0,1)a b c ∈，且无其它得分情况，已知他投篮一次得分的数学期望为1，则ab 的最大值为 （ ） A ．148 B ．124 C ．112 D ．16 8．位于数轴原点的一只电子兔沿着数轴按下列规则移动：电子兔每次移动一个单位，移动的方向向左或向右，并且向左移动的概率为 23，向右移动的概率为13，则电子兔移动五次后位于点(1,0)-的概率是 （ ） A ．4243 B ．8243 C ．40243 D ．80243 第二章 随机变量及其分布 复习 一、随机变量. 1. 随机试验的结构应该是不确定的.试验如果满足下述条件： ①试验可以在相同的情形下重复进行；②试验的所有可能结果是明确可知的，并且不止一个；③每次试验总是恰好出现这些结果中的一个，但在一次试验之前却不能肯定这次试验会出现哪一个结果. 它就被称为一个随机试验. 2. 离散型随机变量：如果对于随机变量可能取的值，可以按一定次序一一列出，这样的随机变量叫做离散型随机变量.若ξ是一个随机变量，a ，b 是常数.则b a +=ξη也是一个随机变量.一般地，若ξ是随机变量，)(x f 是连续函数或单调函数，则)(ξf 也是随机变量.也就是说，随机变量的某些函数也是随机变量. 3、分布列：设离散型随机变量ξ可能取的值为：ΛΛ,,,,21i x x x ξ取每一个值),2,1(Λ=i x 的概率p x P ==)(，则表称为随机变量ξ的概率分布，简称ξ的分布列. 121i 注意：若随机变量可以取某一区间内的一切值，这样的变量叫做连续型随机变量.例如：]5,0[∈ξ即ξ可以取0～5之间的一切数，包括整数、小数、无理数. 典型例题： 1、随机变量ξ的分布列为(),1,2,3(1) c P k k k k ξ== =+……，则P(13)____ξ≤≤= 2、袋中装有黑球和白球共7个，从中任取两个球都是白球的概率为1 7 ，现在甲乙两人从袋中轮流摸去一 球，甲先取，乙后取，然后甲再取……，取后不放回，直到两人中有一人取到白球时终止，用ξ表示取球的次数。（1）求ξ的分布列（2）求甲取到白球的的概率 3、5封不同的信，放入三个不同的信箱，且每封信投入每个信箱的机会均等，X 表示三哥信箱中放有信件树木的最大值，求X 的分布列。 4 已知在全部50人中随机抽取1人抽到喜爱打篮球的学生的概率为5 ． （1）请将上面的列联表补充完整； （2）是否有99.5％的把握认为喜爱打篮球与性别有关？说明你的理由； （3）已知喜爱打篮球的10位女生中，12345,,A A A A A ，，还喜欢打羽毛球，123B B B ，，还喜欢打乒乓球，12C C ，还喜欢踢足球，现再从喜欢打羽毛球、喜欢打乒乓球、喜欢踢足球的女生中各选出1名进行其他方面的调查，求1B 和1C 不全被选中的概率． (参考公式：2 ()()()()() n ad bc K a b c d a c b d -=++++，其中n a b c d =+++) 保温层厚度计算(2021新版) Safety management is an important part of enterprise production management. The object is the state management and control of all people, objects and environments in production. ( 安全管理 ) 单位：______________________ 姓名：______________________ 日期：______________________ 编号：AQ-SN-0646 保温层厚度计算(2021新版) 保温层厚度计算有A种方法，选择介绍四种方法：经济厚度法；直埋管道保温热力法；多层绝热层法；允许降温法。将计算结果经对比分析后选定厚度。 1．保温层经济厚度法 (1)厚度公式 式中δ——保温层厚度，m； Do ——保温层外径，m； Di ——保温层内径，取0．125m； A1 ——单位换算系数，A1 =1．9×10-3 ； λ——保温材料制品导热系数，取0．028W／(m·℃)； τ——年运行时间，取5840h； fn ——热价，现取7元／106kJ； t——设备及管道外壁温度，不计玻璃钢管酌保温性能，取介质温度55℃； ta ——保温结构周围环境的空气温度，取极端土壤地温5℃； Pi ——保温结构单位造价， Pl ——保温层单位造价，硬质聚氨酯泡沫塑料造价1700元／m3 ； P2 ——保护层单位造价，玻璃钢保护层取135元／m2 ； S——保温工程投资贷款年分摊率，按复利率计息， n——计算年限，取15年； i——年利率(复利率)，取7％； a——保温层外表向外散热系数，取11．63W／(cm2 ·℃)。 用试差法，经计算δ=22．5mm。 (2)管道保温层表面散热损失 式中q——单位表面散热损失，W／m。经计算q=42．2W／m，满足国标GB4272—84《设备及管道保温技术通则》要求。 (3)温降计算 式中△t ——卑位长度温降，℃／km； Q——流量，kg／h； 去绝对值常用“六招”（初一） 去绝对值常用“六招” （初一） 绝对值是初中数学的一个重要概念，是后续学习的必备知识。解绝对值问题要求高，难度大，不易把握，解题易陷入困境。下面就教同学们去绝对值的常用几招。 一、根据定义去绝对值 例1、当a = -5，b = 2， c = - 8时，求3│a│-2│b│- │c│的值 分析：这里给出的是确定的数，所以根据绝对值的意义即正数的绝对值是它本身，负数的绝对值是它的相反数，0的绝对值是0。代值后即可去掉绝对值。 解：因为：a = -5＜0，b =2＞0，c = -8＜0 所以由绝对值的意义，原式= 3 [ -（-5）] – 2 ×2 - [ - ( - 8 ) ] = 7 二、从数轴上“读取”相关信息去绝对值 例2、有理数a、b、c在数轴上的 位置如图所示，且│a│=│b│，化简│c-a│+│c-b│+│a+b│-│a│ 分析：本题的关键是确定c - a、c-b、a + b的正负性，由数轴上点的位置特征，即可去绝对值。 解：由已知及数轴上点的位置特征知：a＜0＜c＜b 且- a = b 从而 c – a ＞0 ， c - b＜0， a + b = 0 故原式= c - a + [ - ( c – b ) ] + 0 - ( - a ) = b 三、由非负数性质去绝对值 例3：已知│a2-25│+ ( b – 2 )2 = 0，求ab的值。 分析：因为绝对值、完全平方数为非负数，几个非负数的和为零，则这几个数均为“0”。解：因为│a2-25│+ ( b – 2 )2 = 0 由绝对值和非负数的性质：a2-25 = 0 且b – 2 = 0 即a = 5 b = 2 或a = - 5 b = 2 故ab = 10或ab = - 10 四、用分类讨论法去绝对值 例4、若abc≠0，求+ + 的值。 分析：因abc≠0，所以只需考虑a、b、c同为正号还是同为负号；两个同为正（负）号，另一个为负（正）号，共八种情况。但因为两正（负）、一负（正）的结果只有两种情况，所以其值只有四种情况。 解：由abc≠0可知，a、b、c有同为正号、同为负号和a、b、c异号。 当a、b、c都为“+”时，+ + = + + = 3 当a、b、c都为“-”时，+ + = - - - = - 3 当a、b、c中两“+”一“-”时，+ + = 1 当a、b、c中两“-”一“+”时，+ + = - 1 五、用零点分段法去绝对值 例5：求│x + 1│+│x - 2│+│x -3│的最小值。 “让心灵之窗更明亮”爱眼护眼主题班会教案 一、班会教学目的： 1、初步了解眼睛构造，知道视力对学习、生活的重要性。 2、培养科学用眼的意识，养成正确用眼、认真做眼操的好习惯。 3、学会爱护自己的眼睛，保护视力，同时爱护自己。 二、班会准备： 让学生分小组从家长及课外书中去查找一些关于眼睛的各方面的资料，做眼睛保健操的正确的动作，并获得一些有关眼睛保健方面的知识。 拍摄三位家长有关近视的录像，一位医生解说近视成因的录像 三、班会过程： 1、导入——观看有关家长视力不好的视频，谈感受： 敬爱的老师，亲爱的同学：大家好！今天的班会课由我们俩为大家主持。 先请大家观看几段录像。（a说说自己得近视的原因；b一位家长说说从小的理想是当一名飞行员，但因为近视无法实现自己的梦想；c在生活中近视给自己带来诸多的不便，如雨天，镜片遇雨水模糊。吃热腾腾的食物时，镜片也糊了等。）看了这几段录像，你有什么感受吗？生谈感受 预设：我觉得如果近视了，对我们的生活很有影响；我想我得好好爱护我的眼睛了。…… 主持人点评，小结：听了爸爸妈妈们的心声，我们更加明白了视力的重要。 2、宣布主题，班会开始： 同学们，我们之所以能看到如此五彩缤纷的世界，正是因为我们每个人都拥有一双明亮的眼睛。我们交流时，眼睛是表情达意的好帮手，因此它也被人称做是心灵之窗。可是，如果我们视力不好，就会看不清身边的一切，必须戴上眼镜，严重的还会失明，多可怕啊！所以今天我们一起来召开一个主题班会，说说心灵之窗——我们的眼睛。 《让心灵之窗更明亮》主题班会，现在开始！ 3、了解眼睛构造： 我们天天都用到眼睛，看到身边的一切，那眼睛到底是什么样子的？ 下面我们先来听听第一小组（4人）的介绍。他们介绍的是“眼睛的构造”点击出示眼睛结构图 A、这是眼睛的结构图，这是结膜、虹膜、角膜，这是房水、晶状体、视网膜、巩膜，这是视神经。眼睛是一个复杂器官，我们要掌握的三个重要概念是角膜，晶状体和视网膜。（分别点指） B、我来介绍角膜，角膜就像是相机的镜头，是位于眼球前壁的一层透明膜，它将光线传递并集中到眼睛里。 4、了解视力受损的严重性 原来我们的眼睛是个如此复杂的器官，要想清晰地看到事物，看到美好的大自然，还真要注意保护了。要是视力受损，，后果可真是不堪设想。刚才爸爸妈妈们已经谈了自己的感受，同学们，你们知道视力不好，有哪些后果吗？ 预设： 1、老师写在黑板上的板书看不见 2、我长大想当警察，要是视力不好，就不能当警察了 3、广告牌上的字都看不清…… 鼓励学生大胆表达所了解的视力受损的严重性，发散性；主持人要灵活地点评。 同学们刚才说得很好，但视力不好的后果还不止这些，瞧（点击出示六条危害）赶快自己读读吧。 请六位同学读，每人读一条。 5、了解视力受损的原因 视力不好，给我们的生活和学习带来很多的不便和麻烦，那么到底是什么原因造成了视力受损呢？让我们来听一听医生是怎么说的吧？（播放解说一般情况下视力受损的原因DV） 听了医生的解说，你知道了近视有哪些原因呢？随机点评。 同学们可以对照着改变一下自己用眼的坏习惯，对我们的眼睛肯定有帮助。听了专家的介绍，再来让我们听一听同学的心声吧。 第七周多维随机变量，独立性 7.4独立随机变量期望和方差的性质 独立随机变量乘积的期望的性质： Y X ,独立，则()()() Y E X E XY E =以离散型随机变量为例，设二元随机变量(),X Y 的联合分布列() ,i j P X x Y y ==已知，则()()(),i j i j P X x Y y P X x P Y y ====?=， () 1,2,,; 1,2,,i m j n == ()() 11,m n i j i j i j E XY x y P X x Y y =====∑∑()() 11 m n i j i j i j x y P X x P Y y =====∑∑()() 1 1 m n i i j j i j x P X x y P Y y =====∑∑()() E X E Y =***********************************************************************独立随机变量和的方差的性质： Y X ,独立，则()()() Y Var X Var Y X Var +=+()()() 2 2 Var X Y E X Y E X Y ??+=+-+?? ()222E X XY Y =++()()()()22 2E X E X E Y E Y ??-++? ? ()()()()2 2 22E X E X E Y E Y =-+-()()()22E XY E X E Y +-()()()() 2 2 22E X E X E Y E Y =-+-()() Var X Var Y =+若12,,,n X X X 相互独立，且都存在方差，则()() 121 n m k k Var X X X Var X =+++=∑ ***********************************************************************利用独立的0-1分布求和计算二项分布随机变量()~,X b n p 期望和方差 我们在推导二项分布随机变量的方差时，已经利用了独立随机变量和的方差等于方差 “随机变量及其分布”简介 北京师范大学数学科学院李勇 随机变量是研究随机现象的重要工具之一，他建立了连接随机现象和实数空间的一座桥梁，使得我们可以借助于有关实数的数学工具来研究随机现象的本质，从而可以建立起应用到不同领域的概率模型，如二项分布模型、超几何分布模型、正态分布模型等。 在本章中将通过具体实例，帮助学生理解取有限值的离散型随机变量及其分布列、均值、方差的概念，理解超几何分布和二项分布的模型并能解决简单的实际问题，使学生认识分布列对于刻画随机现象的重要性，认识正态分布曲线的特点及曲线所表示的意义。 一、内容与要求 1. 随机变量及其分布的概念。 通过具体实例使学生理解随机变量及其分布列的概念，认识随机变量及其分布对于刻画随机现象的重要性。要求学生会用随机变量表达简单的随机事件，并会用分布列来计算这类事件的概率。 2.超几何分布模型及其应用。 通过实例，理解超几何分布及其导出过程，并能进行简单的应用。 3. 二项分布模型及其应用。 通过具体实例使学生了解条件概率和两个事件相互独立的概念，理解n次独立重复试验和二项分布模型，并能解决一些简单的实际问题。 4.离散随机变量的均值与方差。 通过实例使学生理解离散型随机变量均值、方差的概念，能计算简单离散型随机变量的均值、方差，并能解决一些实际问题。 5.正态分布模型。 借助直观使学生认识正态分布曲线的特点及含义。 二、内容安排及说明 1.全章共安排了4个小节，教学约需12课时，具体内容和课时分配如下（仅供参考）： 2．1 离散型随机变量及其分布列约3课时 2．2 二项分布及其应用约4课时 2．3 离散型随机变量的均值与方差约3课时 2．4 正态分布约1课时 小结约1课时 2. 本章知识框图 3.对内容安排的说明。 研究一个随机现象，可以借助于随机变量，而分布描述了随机变量取值的概率分布规律。二项分布和超几何分布是两个应用广泛的概率模型．为了使学生能够更好地理解它们，并能用来解决一些实际问题，教科书在内容安排上作了如下考虑： (1) 为学生把注意力集中在随机变量的基本概念和方法的理解上，通过取有限个不同 值的随机变量为载体介绍这些概念，以便他们能更好的应用这些概念解决实际问 题。例如，如何定义随机变量来描述所感兴趣的随机事件；一个具体的随机变量都 能表达什么样的事件，如何表达这些事件；如何用分布列来表达随机事件发生的概 率等。 (2) 介绍超几何分布模型及其应用，其目的是 i. 让学生了解它的广泛应用背景，并使学生能够应用该分布设计一些能够丰富学生课外 去绝对值符号的几种常用方法 解含绝对值不等式的基本思路是去掉绝对值符号，使不等式变为不含绝对值符号的一般不等式，而后，其解法与一般不等式的解法相同。因此掌握去掉绝对值符号的方法和途径是解题关键。 1．利用定义法去掉绝对值符号 根据实数含绝对值的意义，即|x |=(0)(0)x x x x ≥??-????≤? ；|x |>c (0)0(0)(0)x c x c c x c x R c <->>???≠=??∈c (c >0)来解，如|ax b +|>c (c >0)可为ax b +>c 或ax b +<－c ；|ax b +| 随机变量及其分布知识点整理 一、离散型随机变量的分布列 一般地，设离散型随机变量X 可能取的值为12,,,,,i n x x x x ??????，X 取每一个值(1,2,,)i x i n =???的概率()i i P X x p ==，则称以下表格 为随机变量X 的概率分布列，简称X 的分布列. 离散型随机变量的分布列具有下述两个性质： （1）0,1 ,2,,i P i n =???≥ （2）121n p p p ++???+= 1．两点分布 如果随机变量X 的分布列为 则称X 服从两点分布，并称=P(X=1)p 为成功概率. 2．超几何分布 一般地，在含有M 件次品的N 件产品中，任取n 件，其中恰有X 件次品，则事件{}X k =发生的概率为： (),0,1,2,3,...,k n k M N M n N C C P X k k m C --=== {}*min ,,,,,,m M n n N M N n M N N =≤≤∈其中且。 注：超几何分布的模型是不放回抽样 二、条件概率 一般地，设A,B 为两个事件,且()0P A >,称()(|)() P AB P B A P A =为在事件A 发生的条件下,事件B 发生的条件概率. 0(|)1P B A ≤≤ 如果B 和C 互斥，那么[()|](|)(|)P B C A P B A P C A =+ 三、相互独立事件 设A ，B 两个事件，如果事件A 是否发生对事件B 发生的概率没有影响(即()()()P AB P A P B =),则称事件A 与事件B 相互独立。()()()A B P AB P A P B ?=即、相互独立 一般地，如果事件A 1,A 2,…,A n 两两相互独立，那么这n 个事件同时发生的概率，等于每个事件发生的概 圆梦教育中心 随机变量及其分布知识点整理 一、离散型随机变量的分布列 一般地，设离散型随机变量X 可能取的值为12,,,,,i n x x x x ??????，X 取每一个值(1,2,,)i x i n =???的概率 ()i i P X x p ==，则称以下表格 为随机变量X 的概率分布列，简称X 的分布列. 离散型随机变量的分布列具有下述两个性质： （1）0,1,2,,i P i n =???≥ （2）121n p p p ++???+= 1．两点分布 则称X 服从两点分布，并称=P(X=1)p 为成功概率. 2．超几何分布 一般地，在含有M 件次品的N 件产品中，任取n 件，其中恰有X 件次品，则事件{}X k =发生的概率为： (),0,1,2,3,...,k n k M N M n N C C P X k k m C --=== {}*min ,,,,,,m M n n N M N n M N N =≤≤∈其中且。 注：超几何分布的模型是不放回抽样 二、条件概率 一般地，设A,B 为两个事件,且()0P A >,称() (|)() P AB P B A P A = 为在事件A 发生的条件下,事件B 发生的条件概率. 0(|)1P B A ≤≤ 如果B 和C 互斥，那么[()|](|)(|)P B C A P B A P C A =+U 三、相互独立事件 设A ，B 两个事件，如果事件A 是否发生对事件B 发生的概率没有影响(即()()()P AB P A P B =),则称事件A 与事件B 相互独立。()()()A B P AB P A P B ?=即、相互独立 一般地，如果事件A 1,A 2,…,A n 两两相互独立，那么这n 个事件同时发生的概率，等于每个事件发生的概率的积，即1212(...)()()...()n n P A A A P A P A P A =. 风管保温层要工程量计算方 法 标准化文件发布号：（9312-EUATWW-MWUB-WUNN-INNUL-DDQTY-KII 风管保温层要工程量计算方法 1、矩形按矩形单边长度加一个保暖厚度作为边长计算； 2、圆形按园半径加一个保温厚度作为半径； 3、其中：保温厚度=设计要求的保温厚度+规范规定的允许超厚系数%（即保温厚度*）。 4、通风空调风管橡塑板保温体积计算公式： （1）矩形风管=（长+宽+保温厚度*）*2*长度*保温厚度* （2）圆形风管=（直径+保温厚度**2）**长度*保温厚度* 5、通风空调风管橡塑板保温面积计算公式： （1）矩形风管=（长+宽+保温厚度*）*2*长度=保温面积 （2）圆形风管=（直径+保温厚度**2）**长度=保温面积 6、风管保温层厚度计算方法 1、可以用风管面积乘以一个系数来确定，系数一般取15%左右，视风管大小、施工方法确定。 2、公式：(a+b+4d)*2*L（a、b分别为风管长宽、L为风管长度） 3、公式这样算出来还是要乘以一个损耗及包法兰边的系数 4、直接用风管面积乘以15%左右最方便，也比较准确。（参考方法） 如果你自己弄不明白，或没时间计算，建议找代算，根据情况不同，费用不等。 套定额 套用保温定额中有关于风管保温的定额 一、其他方法 1、你可以搜索下小蚂蚁算量，能做工程量计算、预算，高质、高效 2、你可以在网上搜下预算造价单位，有一些单位做的比较好 3、你可以去第三方平台委托别人做，平台上注意防骗，你可以找单位、也可以找个人来做。 二、注意点 1、计算工程量应按照工程所在地的定额或规定标准计算； 2、工程量计算熟悉定额、规定是基础； 3、计算工程量前看清楚图纸是前提，应注意小的注释，以免看漏看错是计算结果出现错误； 4、工程量计算原则上是不允许错误的，希望不要抱侥幸态度去计算工程量。 保护视力方法大全 保护视力的运动 一、转眼法： 选一安静场所，或坐或站，全身放松，清除杂念，二目睁开，头颈不动，独转眼球。先将眼睛凝视正下方，缓慢转至左方，再转至凝视正上方，至右方，最后回到凝视正下方，这样，先顺时针转9圈。再让眼睛由凝视下方，转至右方，至上方，至左方，再回到下方，这样，再逆时针方向转6圈。总共做4次。每次转动，眼球都应尽可能地达到极限。这种转眼法可以锻炼眼肌，改善营养，使眼灵活自如，炯炯有神。（注意：头颈不动，极力，慢,转到后颈发酸时，立即按摩颈部，直至酸感消失。） 二、眼呼吸凝神法： 选空气清新处，或坐或立，全身放松，二目平视前方，徐徐将气吸足，眼睛随之睁大，稍停片刻，然后将气徐徐呼出，眼睛也随之慢慢微闭，连续做9次。 三、熨眼法： 此法最好坐着做，全身放松，闭上双眼，然后快速相互摩擦两掌，使之生热，趁热用双手捂住双眼，热散后两手猛然拿开，两眼也同时用劲一睁，如此3～5次，能促进眼睛血液循环，增进新陈代谢。 坚持天天做，保护好眼睛。 看不见不要眯眼看，那样最容易近视。 四、提倡望远训练。少年眼球处于生长发育阶段，调节力很强，每天可进行一定时间的望远训练。如清晨眺望远处的建筑物或树木，或在夜晚辨认天空的星斗，还可以在日常休息时对远处某一目标进行辨认，认真对待望远训练并持之以恒，对预防近视眼的发生和发展是很有收益的。 眼疲劳食疗验方 1.黑豆粉1匙，核桃仁泥1匙，牛奶1标，蜂蜜1匙。制法：黑豆500克，炒熟后待冷，磨成粉。核桃仁500克，炒微焦去衣，待冷后捣如泥。取以上两种食品各1匙，冲入煮沸过的牛奶1杯后加入蜂蜜1匙。吃法：早晨或早餐后服。或当早餐，另加早点。说明：黑豆含有丰富蛋白质与维生素B1等，营养价值高，又因黑色食物入肾，配合核桃仁，可增加补肾力量，再加上牛奶和蜂蜜，这些食物含有较多的维生素B1、钙、磷等，能增强眼内肌力、加强调节功能，改善眼疲劳的症状。 2.枸杞子10克，桑椹子10克，山药10克，红枣10个。制法：将上述四种药物水煎两次独立同分布随机变量序列的顺序统计方法(2019)
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