Mixing and separating method for a plurality signals

Information

  • Patent Grant
  • 6442224
  • Patent Number
    6,442,224
  • Date Filed
    Friday, February 5, 1999
    25 years ago
  • Date Issued
    Tuesday, August 27, 2002
    22 years ago
Abstract
A method and device for mixing and separating a plurality signals. In the mixing method, each of the m signal Si(t) to be mixexd within a time perid [T0, T1] is sampled for n samples Si(tj), j=1,2 . . . n, wherein tε[T0, T1], T0, T1 εR, t is time variable. Each sample is multiplied by a coefficient function iaj(t) which is a linear independent set (i=1,2 . . . m, j=1,2 . . . n), thus obtaining m transformed signals for Si(t): Si0⁡(t)=∑j=1n⁢ ⁢[i⁢aj⁢(t)⁢Si⁢(tj)].By summing above m transformed signals obtains the mixed transformed signals are obtained: SM⁢(t)=∑i=1m⁢ ⁢[Si0⁢(t)].As to the separating method, the coefficient function iaj(t) is a linear independent set (i=1,2 . . . m, j=1,2 . . . n), therefore Si(tj) are the unknown of m×n linear equation set and can be solved by known linear algebraic method.
Description




FIELD OF THE INVENTION




This invention relates to a mixing and separating method for a plurality signals, more particularly, to achieve control circuit for signal processing.




BACKGROUND OF THE INVENTION




The electronic communication has grown more and more prosperous and the problem of available of communication channel becomes even serious. The bandwidth of a specific communication medium is limited, therefore how to most exploit the avail bandwidth is essential. It is the object of the invention to provide a method for transmitting a plurality of signals in a pair of transmission line or a single channel.




SUMMARY OF THE INVENTION




Principles of the Invention I




The present invention is based on the principle of unique solution condition for a set of N linearly equations, i.e., linearly independence. Based on this principle, each of the m signal S


i


(t) within period [T


0


, T


1


] is sampled for n samples S


i


(t


j


), j=1,2 . . . n, wherein tε[T


0


, T


1


], T


0


, T


1


εR, t is time variable. Each sample is multiplied by a coefficient function


i


a


j


(t) which is a linear independent set (i=1,2 . . . m, j=1,2 . . . n), thus obtaining m transformed signals for S


i


(t):








S
i
0



(
t
)


=




j
=
1

n







[



a
j


i




(
t
)





S
i



(

t
j

)



]












summing above m transformed signals obtains the mixed transformed signals:







SM


(
t
)


=




i
=
1

m







[


S
i
0



(
t
)


]












The mixed transformed signals have m×n variables S


i


(t


j


) with coefficient


i


a


j


(t). If party A transmits SM(t) during time [T


0


, T


1


] to party B, party B will obtain message of m×n S


i


(t


j


), (i=1,2 . . . m, j=1,2 . . . n), wherein the bandwidth depends on the max bandwidth of


i


a


j


(t). More particularly, party A can m messages S


i


(t) (i=1,2 . . . m) to party B during time [T


0


, T


1


], if the samples (unknowns) S


i


(t


1


), S


i


(t


2


), S


i


(t


3


) . . . S


i


(t


n


), ((i=1,2 . . . m) are sufficient to represent S


i


(t) (i=1,2 . . . m) during time [T


0


, T


1


].




The party B can resolve S


i


(t


i


) upon receiving SM(t) if m×n−1 differential means are provided to obtain differential signals SM′(t), SM″(t) . . . SM


m×n−1


(t):














i
=
1

m










j
=
1

n







[



a
j


i




(
t
)





S
i



(

t
j

)



]



=

SM


(
t
)













i
=
1

m










j
=
1

n







[



a
j


i




(
t
)





S
i



(

t
j

)



]



=


SM




(
t
)



















i
=
1

m










j
=
1

n







[



a
j

(


m
×
n

-
1

)


i




(
t
)





S
i



(

t
j

)



]



=


SM


m
×
n

-
1




(
t
)







(
1
)













Eq( 1) is an m×n equation set, wherein functions


i


a


j


(t) (i=1,2 . . . m, j=1,2 . . . n) are linear independent. Therefore, S


i


(t


i


) in Eq(1) has unique solution because the Wronskin (determinant)of functions


i


a


j


(t) is not equal to zero.




Therefore, the S


i


(t


i


) can be calculated by choosing a specific time t


0


within [T


0


, T


1


] and obtain


i


a


j




(u)


(t


0


) and SM


(u)


(t


0


). Moreover, each S


i


(t) can be calculated (i=1,2 . . . m,j=1,2 . . . n, u=0, 1,2 . . . m×n−1).




Principles of the Invention II




The above solving procedure requires m×n−1 differential means to solve S


i


(t), the hardware structure is bulky. However, party B can also take m×n samples after receiving SM(t):














i
=
1

m










j
=
1

n







[



a
j


i




(

t
1

)





S
i



(

t
j

)



]



=

SM


(

t
1

)













i
=
1

m










j
=
1

n







[



a
j


i




(

t
2

)





S
i



(

t
j

)



]



=

SM


(

t
2

)



















i
=
1

m










j
=
1

n







[



a
j


i




(

t

m
×
n


)





S
i



(

t
j

)



]



=

SM


(

t

m
×
n


)







(
2
)













wherein t


1


, t


2


. . . t


m×n


are all within [T


0


, T


1


] and t


u


≠t


v


if u≠v, (u, v=0,1,2 . . . m×n). S


i


(t) has unique solution because


i


a


j


(t) (i=1,2 . . . m, j=1,2 . . . n) are linear independent in [T


0


, T


1


].




Principles of the Invention III




In above scheme, party requires to take m×n samples within [T


0


, T


1


] even thought the differential means can be saved. The sample frequency will increase when the number of signal (m) increases. Therefore, the sampling rate of the A/D should be considered to determine the number of signal m.




To increase m and keep hardware compact, a compromise is to use m−1 differential means to get m differential signals (including original SM(t)), and to take n samples for each signal within [T


0


, T


1


] thus obtaining following equation set:














i
=
1

m










j
=
1

n







[



a
j


i




(

t
1

)





S
i



(

t
j

)



]



=

SM


(

t
1

)













i
=
1

m










j
=
1

n







[



a
j


i




(

t
2

)





S
i



(

t
j

)



]



=

SM


(

t
2

)



















i
=
1

m










j
=
1

n







[



a
j


i




(

t
n

)





S
i



(

t
j

)



]



=

SM


(

t
n

)













i
=
1

m










j
=
1

n







[



a
j


i




(

t
1

)





S
i



(

t
j

)



]



=


SM




(

t
1

)













i
=
1

m










j
=
1

n







[



a
j


i




(

t
2

)





S
i



(

t
j

)



]



=


SM




(

t
2

)



















i
=
1

m










j
=
1

n







[



a
j


i




(

t
n

)





S
i



(

t
j

)



]



=


SM




(

t
n

)



















i
=
1

m










j
=
1

n







[



a
j

(

m
-
1

)


i




(

t
n

)





S
i



(

t
j

)



]



=


SM

(

m
-
1

)




(

t
n

)







(
3
)















i


a


j


(t) (i=1,2 . . . m, j=1,2 . . . n) are linear independent. Therefore, S


i


(t


i


) in Eq(3) has unique solution because the Wronskin (determinant) of functions


i


a


j


(t) is not equal to zero.




Party B has a plurality of ways to create m×n linear independent equation set form SM(t) as will be described below.














i
=
1

m










j
=
1

n







[



a
j


i




(

t
1

)





S
i



(

t
j

)



]



=

SM


(

t
1

)













i
=
1

m










j
=
1

n







[



a
j


i




(

t
2

)





S
i



(

t
j

)



]



=

SM


(

t
2

)



















i
=
1

m










j
=
1

n







[



a
j


i




(

t
n

)





S
i



(

t
j

)



]



=

SM


(

t
n

)













i
=
1

m










j
=
1

n







[


1
D




a
j


i




(

t
1

)





S
i



(

t
j

)



]



=


1
D



SM


(

t
1

)














i
=
1

m










j
=
1

n







[


1
D




a
j


i




(

t
2

)





S
i



(

t
j

)



]



=


1
D



SM


(

t
2

)














i
=
1

m










j
=
1

n







[


1
D




a
j


i




(

t
n

)





S
i



(

t
j

)



]



=


1
D



SM


(

t
n

)




















i
=
1

m










j
=
1

n







[


1

D

m
-
1






a
j


i




(

t
n

)





S
i



(

t
j

)



]



=


1

D

m
-
1





SM


(

t
n

)








(
4
)













wherein







1

D
u





a
j


i




(

t
v

)












 and







1

D
u




SM


(

t
v

)












 is uth integration of


i


a


j


(t) and SM(t) from 0 to t


v


., u=1,2 . . . m−1, v=1,2 . . . n.




Another alternative is:














i
=
1

m










j
=
1

n







[



a
j


i




(

t
1

)





S
i



(

t
j

)



]



=

SM


(

t
1

)













i
=
1

m










j
=
1

n







[



a
j


i




(

t
2

)





S
i



(

t
j

)



]



=

SM


(

t
2

)



















i
=
1

m










j
=
1

n







[



a
j


i




(

t
n

)





S
i



(

t
j

)



]



=

SM


(

t
n

)













i
=
1

m










j
=
1

n







[

Δ



a
j


i




(

t
1

)





S
i



(

t
j

)



]



=

Δ






SM


(

t
1

)














i
=
1

m










j
=
1

n







[

Δ



a
j


i




(

t
2

)





S
i



(

t
j

)



]



=

Δ






SM


(

t
2

)














i
=
1

m










j
=
1

n







[

Δ



a
j


i




(

t
n

)





S
i



(

t
j

)



]



=

Δ






SM


(

t
n

)




















i
=
1

m










j
=
1

n







[


Δ

m
-
1





a
j


i




(

t
n

)





S
i



(

t
j

)



]



=


Δ

m
-
1




SM


(

t
n

)








(
5
)













wherein Δ


u




i


a


j


(t


v


) and Δ


u


SM(t


v


) is uth differential of


i


a


j


(t) and SM(t) at t


v


, u=1,2 . . . m−1, v=1,2 . . . n.




Still another alternative is:














i
=
1

m










j
=
1

n







[



a
j


i




(

t
1

)





S
i



(

t
j

)



]



=

SM


(

t
1

)













i
=
1

m










j
=
1

n







[



a
j


i




(

t
2

)





S
i



(

t
j

)



]



=

SM


(

t
2

)



















i
=
1

m










j
=
1

n







[



a
j


i




(

t
n

)





S
i



(

t
j

)



]



=

SM


(

t
n

)













i
=
1

m










j
=
1

n







[





a
j


i




(

t
1

)






S
i



(

t
j

)



]



=



SM


(

t
1

)














i
=
1

m










j
=
1

n







[





a
j


i




(

t
2

)






S
i



(

t
j

)



]



=



SM


(

t
2

)














i
=
1

m










j
=
1

n







[





a
j


i




(

t
n

)






S
i



(

t
j

)



]



=



SM


(

t
n

)




















i
=
1

m










j
=
1

n







[





m
-
1





a
j


i




(

t
n

)






S
i



(

t
j

)



]



=




m
-
1




SM


(

t
n

)








(
6
)













wherein ∇


u




i


a


j


(t


v


) and ∇


u


SM(t


v


) is uth summation of


i


a


j


(t) and SM(t) from 0 to t


v


. u=1,2 . . . m−1, v=1,2 . . . n.




The determinant in each matrix of Eqs(4)-(6) is not zero because


i


a


j


(t) (i=1,2 . . . m, j=1,2 . . . n) are linear independent in [T


0


, T


1


]. Therefore, S


1


(t) has unique solution for Eqs (4)-(6).




Moreover, party B can mix the operations of differential, integration, difference, summation and sampling to create m×n linear independent equation set. For example, taking differential, integration, difference, summation for number of r1, r2,r3 and r4, and taking sample number of h, such that (r1+r2+r3+r4)h=m×n, party B can create m×n linear independent equation set for solving S


i


(t


j


). However, the other methods are not described here for clarity.




Principles of the Invention IV




A particular choice of


i


a


j


(t) is described below, wherein


i


a


j


(t) thus selected are orthonormal for t within period [T


0


, T


1


]










T
0


T
1






a
l


k




(
x
)





a
j


i




(
x
)









x



=

{




1
;






when





k

=


i





and





l

=
j








0
;






when





k



i





or





l


j
















wherein i,k=1,2 . . . m, 1,j=1,2 . . . n.




in this situation










T
0


T
1





&AutoLeftMatch;

SM


(
x
)


&AutoRightMatch;




a
l


k




(
x
)









x



=





T
0


T
1





[




i
=
1

m










J
=
1

n









a
j


i




(
x
)





s
i



(

t
j

)





]




a
l


k




(
x
)









x



=


S
k



(

t
l

)













Therefore, when the transformed and mixed signal SM(t) is sent to the receiver during time period [T


0


, T


1


] the receiver party multiplies the received signal with


k


a


l


(t) (k=1,2 . . . m, 1=1,2 . . . n) and integrates the result between time period [T


0


, T


1


] to obtain S


k


(t


j


) (the Ith sample for the kth signal). This indicates that each sample value for each signal S


i


(t


j


) can be calculated without the step of solving the linear algebraic equation set.




Below describes the way to orthonormalize the function group {


i


a


j


(t) (i=1,2 . . . m, j=1,2 . . . n)} within time period [T


0


, T


1


].




First, m×n functions G


1


(t), G


2


(t) . . . G


m×n


(t) linearly independent within time period [T


0


, T


1


] are selected and let









h


(

r
,
s

)


=




T
0


T
1






G
r



(
x
)





G
s



(
x
)









x




;




r

,

s
=
1

,

2











m
×
n






A
0

=
1





A
v

=



&LeftBracketingBar;




h


(

1
,
1

)





h


(

1
,
2

)





h


(

1
,
3

)








h


(

1
,
v

)







h


(

2
,
1

)





h


(

2
,
2

)





h


(

2
,
3

)








h


(

2
,
v

)







h


(

3
,
1

)





h


(

3
,
2

)





h


(

3
,
3

)








h


(

3
,
v

)
























h


(

v
,
1

)





h


(

v
,
2

)





h


(

v
,
3

)








h


(

v
,
v

)





&RightBracketingBar;






v

=

1.2











m
×
n












then establishing the function







A
v

=

&LeftBracketingBar;




h


(

1
,
1

)





h


(

1
,
2

)





h


(

1
,
3

)








h


(

1
,
u

)







h


(

2
,
1

)





h


(

2
,
2

)





h


(

2
,
3

)








h


(

2
,
u

)
























h


(


u
-
1

,
1

)





h


(


u
-
1

,
2

)





h


(

u
-
1.3

)








h


(


u
-
1

,
u

)








G
1



(
t
)






G
2



(
t
)






G
3



(
t
)










G
u



(
t
)


)




&RightBracketingBar;





u
=

1.2











m
×
n











then the function










l
u



(
t
)


=


1



A

u
-
1




A
u







P
u



(
t
)




;





u
=
1


,

2











m
×
n











is orthonormal within time period [T


0


, T


1


].




Moreover, assume


i


a


j


(t)=Q(I−1)n+j(t) (i=1,2 . . . m, j=1,2 . . . n), that is






1


a


2


(t)=Q


1


(t)






1


a


2


(t)=Q


2


(t)





















1


a


n


(t)=Q


n


(t)






2


a


1


(t)=Q


1+n


(t)






2


a


2


(t)=Q


2+n


(t)





















2


a


n


(t)=Q


2n


(t)






3


a


1


(t)=Q


1+2n


(t)






3


a


2


(t)=Q


2+2n


(t)





















3


a


2


(t)=Q


3n


(t)





















m


a


n


(t)=Q


m×n


(t)




Apparently, function group {


i


a


j


(t) (i=1,2 . . . m, j=1,2 . . . n)} are orthonormal within time period [T


0


, T


1


]. In above procedure for receiver party to restore S


k


(t


1


), the step of solving linear algebraic equation is eliminated. However, m×n integrals are required, this will make the hardware complicated. The present invention provide following approach.




By choosing suitable function group {G


u


(t)|u=1,2 . . . m×n},


k


a


1


(t) can be expressed into Power series as following:











a
l


k




(
t
)


=







a
l


k




(
b
)


+







a
l


k




(
b
)








t


(

t
-
b

)



+






1

2
!









a
l


k




(
b
)









(

t
-
b

)

2


+














1

3
!





a
l
′′′

k




(
b
)





(

t
-
b

)

3












;





b





R















Because the high frequency components of


k


a


l


(t) are limited, the first several terms are sufficiently to represent


k


a


l


(t). Assuming that first M terms are considered here,


k


a


l


(t) can be expressed as:









a
l


k




(
t
)







q
=
0


M
-
1









[


1

q
!





a
l

(
q
)


k




(
b
)





(

t
-
b

)

q


]






b




R










then














S
k



(

t
l

)


=








T
0


T
1






SM


(
x
)


k




a
l



(
x
)









x









=








q
=
0


M
-
1








[


1

q
!





a
l

(
q
)


k




(
b
)







T
0


T
1





SM


(
x
)





(

x
-
b

)

q








x




]









(
7
)













The receiver party only need to store m×n×M data








1

q
!





a
l

(
q
)


k




(
b
)








(


k
=
1

,

2











m

,

l
=
1

,

2











n

,





q
=
0

,


1











M

;

b





R



)


,










then calculate M integrals









T
0


T
1





SM


(
x
)





(

x
-
b

)

q








x












upon receiving the SM(t). The m×n samples S


k


(t


1


) are restored. In other word, receiver party only requires to prepare M integrators other than m×n.




Moreover, above approach is also applicable to the data compress technology. The m×n samples S


k


(t


1


) can be approximately represented by M data:











T
0


T
1





SM


(
x
)





(

x
-
b

)

q








x



;





q
=
0


,
1
,



2











M

-
1

;





b

R












In other word, the function group {


i


a


j


(t) (i=1,2 . . . m, j=1,2 . . . n)} are orthonormalized within time period [T


0


, T


1


] by above procedure. The M data;











T
0


T
1





SM


(
x
)





(

x
-
b

)

q








x



;





q
=
0


,
1
,



2











M

-
1

;





b

R












are calculated by using known m×n samples S


k


(t


1


). Afterward, Eq(7) is employed to restore m×n samples S


k


(t


1


), wherein









T
0


T
1





SM


(
x
)





(

x
-
b

)

q








x












are the data after compression and







1

q
!





a
l

(
q
)


k




(
b
)












refers to the restoring parameter. It is apparent that the invention is applicable both to transmission of mass data or the compression technology of signal




Principles of the Invention V




From above description, party A sends message to party B segment by segment with time duration [T


0


, T


1


]. However, party A need to send a synchronous signal before sending signal containing message.




Therefore, the duration [T


0


, T


1


] is divided into first synchronous period [T


0


, T


1


′] for sending synchronous signal, and second information period [T


1


′, T


1


] for sending information (T


0


<T


1


′<T


1


). The decrease of information period due to the incorporation of synchronous period will not influence bandwidth because the maximum bandwidth depends on


i


a


j


(t)











The various objects and advantages of the present invention will be more readily understood from the following detailed description when read in conjunction with the appended drawing, in which:




BRIEF DESCRIPTION OF THE DRAWINGS





FIG. 1

is an analog adding circuit in the invention.





FIG. 2

show an inverted amplifier scheme in the invention.





FIG. 3

shows the generating circuit for mixed signal SM(t).





FIG. 4

is the schematic diagram of the control circuit


12


which outputs the mixing signal and the synchronous signal alternatively.





FIG. 5

is the control circuit


17


which make the analog signal be transferred into the digital signal, then be storage into the memory.





FIG. 6

, is the whole hardware of party A.





FIG. 7

is a figure of a differential circuit comprising four differential means.





FIG. 8

is the separated circuit


24


which separates the several kind of the original type of the mixing signal.





FIG. 9

is the control circuit 22-i which can resolve the parallel equation, and store the resolve result into the memory.




FIG.


10


. is the whole hardware of party A.





FIG. 11

is the schematic diagram of the all embodiment according by this invention, the number


19


is the party A shown in

FIG. 6

, the number


26


is the party B shown in FIG.


10


.





FIG. 12

is the timing schematic diagram of the R


1


, R


2


and CK which shown in the block


10


of FIG.


4


.











DETAILED DESCRIPTION OF THE INVENTION




Embodiment




Hereinafter, a preferred embodiment is used to substantially explain the present invention, wherein the communication medium is assumed to be an ideal (distortionless) medium.




In this embodiment, the Si(t) in Eq. (3) is solved and below list some important issues.




1. The choice of [T


0


, T


1


′], [T


1


′, T


1


], m, n and


i


a


j


(t):




Provided that




[T


0


, T


1


′]=[0, ε];




[T


1


′, T


1


]=[ε, 5ε],ε={fraction (1/1000)} sec;




m=5




n=8






i


a


j


(t)=G(j+40I−40,t), i=1,2 . . . 5; j=1,2, . . . ,40;




wherein G(1,t)=cos [2(400+151)πt]; 1=1,2 . . . 200.




and taking 40 samples Si(t1), . . . , Si(t40) for Si(t) (i=1, 2, 3, 4, 5) in [T


0


, T


1


′]=[0, ε]




and further assuming




Sa(1)=S


1


(t


1


)




Sa(2)=S


1


(t


2


)














Sa(40)=S


1


(t


40


)




Sa(41)=S


2


(t


1


)




Sa(42)=S


2


(t


2


)














Sa(80)=S


2


(t


40


)














Sa(200)=S


5


(t


40


).




Then the party A (sender) create a mixed transformed signal SM(t)










SM


(
t
)


=




u
=
1

200







[


G


(

u
,
t

)





S
a



(
u
)



]






(
7
)













the SM(t) in above equation is transformed from party A to party B within [T


1


′, T


1


]=[ε, 5ε] and the maximum transmitting frequency is 3.4 kHz which depends on the function group cos [2(400+151)πt], not on the maximum frequency of Si(t).




2. The parameter setting in party B




The function group cos [2(400+151)πt] is linear independent in tε[ε, 5ε], therefore the below determinant is not zero.






W
=

&LeftBracketingBar;

&AutoLeftMatch;








G


(

1
,

X
1


)





G


(

2
,

X
1


)








G


(

200
,

X
1


)







G


(

1
,

X
2


)





G


(

2
,

X
2


)








G


(

200
,

X
2


)





















G


(

1
,

X
40


)





G


(

2
,

X
40


)








G


(

200
,

X
40


)








G




(

1
,

X
1


)






G




(

2
,

X
1


)









G




(

200
,

X
1


)








G




(

1
,

X
2


)






G




(

2
,

X
2


)









G




(

200
,

X
2


)






















G




(

1
,

X
40


)






G




(

2
,

X
40


)









G




(

200
,

X
40


)








G




(

1
,

X
1


)






G




(

2
,

X
1


)









G




(

200
,

X
1


)






















G

(
4
)




(

1
,

X
40


)






G

(
4
)




(

2
,

X
40


)









G

(
4
)




(

200
,

X
40


)









&AutoRightMatch;

&RightBracketingBar;











wherein X1,X2 . . . X40 are corresponding to the sampling points within tε[ε, 5ε].




For convenience's sake, let




D(1,j)=G(jX1)




D(2,j)=G(j,X2)














D(40,j)=G(j,X40)




D(41,j)=G′(j,X1)




D(42,j)=G′(j,X2)














D(80,j)=G′(j,X40)




D(81,j)=G″(j,X1)














D(200,j )=G


(4)


(j,X40)




W can be written as






W
=

&LeftBracketingBar;




D


(

1
,
1

)





D


(

1
,
2

)








D


(

1
,
200

)







D


(

2
,
1

)





D


(

2
,
2

)








D


(

2
,
200

)





















D


(

200
,
1

)





D


(

200
,
2

)








D


(

200
,
200

)





&RightBracketingBar;











let







z


(

i
,
j

)


=

&LeftBracketingBar;




D


(

1
,
1

)





D


(

1
,
2

)








D


(

1
,

j
-
1


)





D


(

1
,

j
+
1


)








D


(

1
,
200

)







D


(

2
,
1

)





D


(

2
,
2

)








D


(

2
,

j
-
1


)





D


(

2
,

j
+
1


)








D


(

2
,
200

)






























D


(


i
-
1

,
1

)





D


(


i
-
1

,
2

)








D


(


i
-
1

,

j
-
1


)





D


(


i
-
1

,

j
+
1


)








D


(


i
-
1

,
200

)







D


(


i
+
1

,
1

)





D


(


i
+
1

,
2

)








D


(


i
+
1

,

j
-
1


)





D


(


i
+
1

,

j
+
1


)








D


(


i
+
1

,
200

)






























D


(

200
,
1

)





D


(

200
,
2

)








D


(

200
,

j
-
1


)





D


(

200
,

j
+
1


)








D


(

200
,
200

)





&RightBracketingBar;











wherein i≧2, 199≧j




let







Z


(

1
,
j

)


=

&LeftBracketingBar;




D


(

2
,
1

)





D


(

2
,
2

)








D


(

2
,

j
-
1


)





D


(

2
,

j
+
1


)








D


(

2
,
200

)







D


(

3
,
1

)





D


(

3
,
2

)








D


(

3
,

j
-
1


)





D


(

3
,

j
+
1


)








D


(

3
,
200

)






























D


(

200
,
1

)





D


(

200
,
2

)








D


(

200
,

j
-
1


)





D


(

200
,

j
+
1


)








D


(

200
,
200

)





&RightBracketingBar;






Z


(

200
,
j

)


=

&LeftBracketingBar;




D


(

1
,
1

)





D


(

1
,
2

)








D


(

1
,

j
-
1


)





D


(

1
,

j
+
1


)








D


(

1
,
200

)







D


(

2
,
1

)





D


(

2
,
2

)








D


(

2
,

j
-
1


)





D


(

2
,

j
+
1


)








D


(

2
,
200

)






























D


(

199
,
1

)





D


(

199
,
2

)








D


(

199
,

j
-
1


)





D


(

199
,

j
+
1


)








D


(

199
,
200

)





&RightBracketingBar;






Z


(

i
,
1

)


=

&LeftBracketingBar;




D


(

1
,
2

)





D


(

1
,
3

)








D


(

1
,
200

)







D


(

2
,
2

)





D


(

2
,
3

)








D


(

2
,
200

)





















D


(


i
-
1

,
2

)





D


(


i
-
1

,
3

)








D


(


i
-
1

,
200

)







D


(


i
+
1

,
2

)





D


(


i
+
1

,
3

)








D


(


i
+
1

,
200

)







D


(

200
,
2

)





D


(

200
,
3

)








D


(

200
,
200

)





&RightBracketingBar;






Z


(

i
,
200

)


=

&LeftBracketingBar;




D


(

1
,
1

)





D


(

1
,
2

)








D


(

1
,
199

)







D


(

2
,
1

)





D


(

2
,
2

)








D


(

2
,
199

)





















D


(


i
-
1

,
1

)





D


(


i
-
1

,
2

)








D


(


i
-
1

,
199

)







D


(


i
+
1

,
1

)





D


(


i
+
1

,
2

)








D


(


i
+
1

,
199

)







D


(

200
,
1

)





D


(

200
,
2

)








D


(

200
,
199

)





&RightBracketingBar;











moreover let








R


(


u,v


)=(−1


)u+v




Z


(


u,v


)


/W


  (8)






Wherein u,v=1, 2, . . . ,200




The R(u,v) in Eq(8) is the reverse-transform parameter, the value thereof are calculated by computer and then save in memory.




After party B receiving the signal as (7) from party A, party B uses differential means to obtain SM(t), SM′(t), SM″(t), SM′″(t) and SM


(4)


(t), and then takes sample to get 200 data including SM(X1), SM(X2), . . . SM(X40), SM′(X1), SM′(X2), . . . SM′(X40), . . . SM


(4)


(X1), . . .




SM


(4)


(X40)




For convenience, let




α(1)=SMG(X1)




α(2)=SM(X2)














α(40)=SM(X40)




α(41)=SM′(X1)




α(42)=SM′(X2)














α(80)=G′(X40)




α(81)=Gα(X1)









α(161)=G


(4)


(X1)














α(200)=G


(4)


(X40)




Sa(j) can be calculated by below equation











S
a



(
j
)


=




i
=
1

200







[


α


(
i
)




R


(

i
,
j

)



]






(
9
)













wherein i=1,2, . . . 200




In (9), Sa(1), Sa(2)..Sa(40) are the samples in S


1


(t) taken by party A, Sa(41), Sa(42) . . . Sa(80) are the samples in S


2


(t) taken by party A, . . . Sa(161), Sa(162) . . . Sa(200) are the samples in S


5


(t) taken by party A. Eq (9) is apparently a reverse-transform formula.




3. The hardware of party A (sender)





FIG. 1

is an analog adding circuit wherein


010


is a high-gain amplifier,


020


is feedback resistor,


021


-


1


˜


021


-


200


are input resistors. If all input resistors have resistance same as that of the feedback resistor, then








e




0


=−(


e




1




+e




2




+ . . . +e




200


)






wherein e


1


,e


2


. . . e


200


are input voltages, and e


0


is output voltage.





FIG. 2

show an inverted amplifier scheme wherein output voltage e


0


equal to e


in


multiplied by sample data Sa(i), (i=1,2 . . . ,200) and then inverted.




Below are features of FIG.


2


.




1.


011


is operational amplifier




2.


022


-


0


˜


022


-


7


are eight serially-connected feedback resistors and the action thereof depend on the on-off state of electronic switch


040


-


0


˜


040


-


7


.


022


-j is shorted and has no action when


040


-j is on, and has action when


040


-j is off. The on-off state of


040


-j is controlled by b


j


.


040


-j is on when b


j


is low, (j=0, 1,2 . . . 7).




3. Setting Sa(i) (i=1,2 . . . 200) as binary data in byte base, and has value of b


7


X2


7


+b


6


X2


6


+ . . . b


1


X


2


+b


0


X2


0


.




4. Let resistance of


022


-


0


˜


022


-


7


are r, 2r, 2


2


r, . . . 2


7


r, respectively, the resistance of input resistor


023


is r.




5. The resistance of feedback resistor will be controlled by Sa(i) (i=1,2 . . . 200) and becomes b


7


X2


7


r+b


6


X2


6


r+ . . . b


1


X


2


r+b


0


Xr=Sa(i)r.




 Therefore, input/output voltage has below relationship






e


0




=−Sz


(


I


)


e




in









FIG. 3

shows the generating circuit for mixed signal SM(t) which has following features.




1.


06


-l is the generating circuit for functions cos [2(400+15l)πt] (l=1,2, . . . 200, l is time veriable) which generate those functions when signal in RESET off.




2.


05


-


1


˜


05


-


200


is the circuit shown in FIG.


2


.




3.


03


is the circuit shown in FIG.


1


.




4.


041


-


1


,


041


-


2


are electronic switch,


070


is inverter,


041


-


1


OFF and


041


-


2


ON when SW HIGH.




5. When RS signal disappears and SW HIGH, ST generates signal SM(t) with below form:







SM


(
t
)


=




i
=
1

200







[


Sa


(
i
)




cos


[

2


(

400
+
ε

)


π





t

]



]













FIG. 4

has following features




1. DATA BUS DS


1


˜DS


5


come from DS


1


˜DS


5


in

FIG. 5

which send Sa(1)˜Sa(200) to data register


11


, wherein DS


1


sends Sa(1)˜Sa(40), DS


2


sends Sa(41)˜Sa(80) . . . DS


5


sends Sa(161)˜Sa(200), RS


1


is the timing control terminal for data.




2.


08


is circuit in

FIG. 3

,


09


is synchronous signal generator and sends synchronous signal when RESET signal is over,


012


is operational amplifier,


024


is feedback resistor of


012


,


025


-


1


and


025


-


2


are input resistors with same resistance,


042


-


1


and


042


-


2


are electronic switch,


042


-


1


is off and


042


-


2


is on when R


2


is HIGH,


071


is inverter.




3.


10


is a timing control circuit which begin to function when RS is excited and generate timing as shown in FIG.


12


.




4. Therefore, output S


0


sends synchronous signal within [T


0


, T


1


′] and mixed transformed signal SM(t) within [T


1


′, T


1


]





FIG. 5

has following features




1. The box


14


enclosed by dashed line has 5 A/D converters


13


-


1


˜


13


-


5


to convert the analog signals from D


0


˜D


4


to digital signals, wherein CL is CLOCK control for sampling.






2


. The box


16


enclosed by dashed line has 5 memory means


15


-


1


˜


15


-


5


to store the digital data from A/D converters


13


-


1


˜


13


-


5


, wherein AD


1


˜AD


5


are address bus, R and W are read and write control.




3. From the control of W end, the digital data from ND converters


13


-


1


˜


13


-


5


are stored in 5 memory means


15


-


1


˜


15


-


5


, from the control of R end, the stored digital data are read and output from DS


1


˜DS


5







FIG. 6

is the whole hardware of party A, wherein


12


is circuit in

FIG. 4

,


17


is circuit in FIG.


5


. Moreover,


18


is a well-know timing circuit and the description is omitted.




4. The hardware of party B





FIGS. 7-10

is block diagram of party B.

FIG. 7

is a figure of a differential circuit comprising four differential means


20


-


1


,


20


-


2


,


20


-


3


and


20


-


4


, and the description is omitted for they are well known art.





FIGS. 8 and 9

have following features




1.


22


-i(i=1,2 . . . 5) in

FIG. 8

is circuit in

FIG. 9








2


.


220


in

FIG. 9

is a circuit for solving linear equation set and has parametric memory for storing reverse-transform parameter R(u,v) (u,v=1,2 . . . 200), CE is a timing control end.




3.


221


in

FIG. 9

is memory for storing the result of


220


, wherein Asi(i=1,2 . . . 5) are address bus, DRi(i=1,2 . . . 5) are data bus and R and W are read and write control ends.




4. In

FIG. 8

,


23


-


1


˜


23


-


5


are D/A converters, EN is chip enable, R, W, CK are the same as R, W, CK in

FIG. 9

, data bus DS


1


˜DS


5


are DS


1


˜DS


5


in

FIG. 9







FIG. 10

is the whole hardware of party A and has below features.




1.


21


is the differential circuit in

FIG. 7

,


17


is that shown in

FIG. 5

,


21


execute first, second, third, fourth order differential operation on SM(t) to get SM′(T), SM″(T), SM′″(T), SM


(4)


(T), and send those signal with SM(t) to D


0


˜D


4


in


17


, those signals are digitized by 5 A/D converters in


17


and stored in memory


15


-


1


˜


15


-


5


in


17


.




2.


24


is same as that shown in

FIG. 8

, the terminals around the dashed box are the same as those in FIG.


8


.




3.


25


is a timing control circuit, which initial


21


to execute first, second, third, fourth order differential operation on Si and store those signals with Si into


17


, then 5 A/D converters


13


-


1


˜


13


-


5


in


17


digitize those signal and store the result in


15


-


1


˜


15


-


5


, the data in


15


-


1


˜


15


-


5


are fetched to


220


in


22


-i (i=1,2 . . . 5) for solving linear equation set, the solution are stored in


221


, the data in


221


are sent to


23


-


1


˜


23


-


5


to convert into analog signal S


1


(5)-S


5


(t), thus reverse transforming the data.




4. Circuit


25


is well known and the description thereof is omitted for clarity.




Although the present invention has been described with reference to the preferred embodiment thereof, it will be understood that the invention is not limited to the details thereof. Various substitutions and modifications have suggested in the foregoing description, and other will occur to those of ordinary skill in the art. Therefore, all such substitutions and modifications are intended to be embraced within the scope of the invention as defined in the appended claims.



Claims
  • 1. A method for mixing a plurality of signals, comprising the steps of:(a) using a plurality of data bus for receiving a plurality of analog signals within a time period, each signal being continuous and being able to be mathematically represented by an equation of Si(t) within the time period [T0, T1], wherein m is integer, i=1,2, . . . , m, t is time variable, T0, T1 εR, tε[T0, T1]; (b) using a timing control circuit to sample the signals Si(t) within the time period [T0, T1] and obtain n samples for each signal, said samples being mathematically represented by Si(tj), wherein n is integer, tjε[T0, T1], j=1, 2, . . . n; (c) using a generating circuit to generate a predetermined time dependent function group which is a linear independent set and can be mathematically represented by iaj(t), wherein i=1,2 . . . m, j=1,2 . . . n; (d) using a multiplying circuit to multiply each sample Si(tj) with a corresponding time dependent function iaj(t) and generating a transformed signal which can be mathematically represented as Si0⁢(t)=∑j=1n⁢ ⁢[i⁢aj⁢(t)⁢Si⁢(tj)];(e) using an adding circuit to sum up all transformed signals Si0(t) to form a mixed transformed signal which be mathematically represented as SM(t), wherein SM⁢(t)=∑i=1m⁢ ⁢[Si0⁢(t)].
  • 2. The method as in claim 1, wherein the time dependent function group iaj(t) (i=1 . . . m, j=1,2 . . . n) are such selected that every function of iaj(t) is linear independent for tε[T0, T1].
  • 3. The method as in claim 1, wherein the time dependent function group is mathematically represented as kal(t) (k=1,2 . . . m, l=1,2 . . . n) within the time period [T0, T1] and is mathematically determined by the following functions:selecting a set of m×n functions G1(t), G2(t) . . . Gm×n (t) which are linearly independent within the time period [T0, T1]; letting h⁡(r,s)=∫T0T1⁢Gr⁡(x)⁢Gs⁡(x)⁢ ⁢ⅆx; ⁢rs=1,2⁢ ⁢…⁢ ⁢m×n;A0=1;Av=&LeftBracketingBar;h⁡(1,1)h⁡(1,2)h⁡(1,3)…h⁡(1,v)h⁡(2,1)h⁡(2,2)h⁡(2,3)…h⁡(2,v)h⁡(3,1)h⁡(3,2)h⁡(3,3)…h⁡(3,v)……………h⁡(v,1)h⁡(v,2)h⁡(v,3)…h⁡(v,v)&RightBracketingBar;; ⁢v=1.2⁢ ⁢…⁢ ⁢m×n;then establishing the function Av=&LeftBracketingBar;h⁡(1,1)h⁡(1,2)h⁡(1,3)…h⁡(1,u)h⁡(2,1)h⁡(2,2)h⁡(2,3)…h⁡(2,u)……………h⁡(u-1,1)h⁡(u-1,2)h⁡(u-1,3)…h⁡(u-1,u)G1⁡(t)G2⁡(t)G3⁡(t)…Gu⁡(t))&RightBracketingBar;; ⁢u=1.2⁢ ⁢…⁢ ⁢m×n;then the function Qu⁡(t)=1Au-1⁢Au⁢Pu⁡(t); ⁢u=1,2⁢ ⁢…⁢ ⁢m×n; ⁢and then lettingiaj(t)=Q(i−1)n+j(t) (i=1,2 . . . m, l=1,2 . . . n).
  • 4. The method as in claim 3, wherein kal(t) (k=1,2 . . . m, l=1,2 . . . n) are orthonormal within the time period [T0, T1]).
  • 5. The method as in claim 1, further comprising a step for solving m×n Si(tj) (i=1,2 . . . m, j=1,2 . . . n), said step taking m×n samples to establish a m×n linear equation set with unknowns S1(t1), S1(t2), S1(t3). . . S1(tn), S2(t1), S2(t2) . . . Sm(t1), Sm(t2), . . . Sm(tn) and using a hardware to solve this linear equation set with linear algebraic method to get Si(tj) (i=1, . . . m, j=1,2 . . . n).
  • 6. The method as in claim 1, further comprising a step for solving m×n Si(tj) (i=1,2 . . . m, j=1,2 . . . n) said step establishing m differential equations for SM(t) with respect to t and with each term having different order including zero, then taking n samples to m differential equations within tε[T0, T1], and establishing a m×n linear equation set with unknowns S1(t1), S1(t2), S1(t3). . . S1(tn), S2(t1), S2(t2) . . . Sm(t1), Sm(t2), . . . Sm(tn) and solving this linear equation set with linear algebraic method to get Si(tj) (i=1, . . . m, j=1,2 . . . n).
  • 7. The method as in claim 1, further comprising a step for solving m×n Si(tj) (i=1,2 . . . m, j=1,2 . . . n), said step performing m integration operation to the transformed signal from timed=0 to time=t, and with each term having different order including zero, then taking n samples to m integral equations within tε[T0, T1], and establishing a m×n linear equation set with unknowns S1(t1), S1(t2), S1(t3). . . S1(tn), S2(t1), S2(t2) . . . Sm(t1), Sm(t2), . . . Sm(tn) and solving this linear equation set with linear algebraic method to get Si(tj) (i=1, . . . m, j=1,2 . . . n).
  • 8. The method as in claim 1, further comprising a step for solving m×n Si(tj) (i=1,2 . . . m, j=1,2 . . . n), said step establishing r differential equations for SM(t) (r<m) with respect to t and with each term having different order including-zero, then performing m−r integration operation to SM(t) from time=0 to time=t, and with each term having different order including zero, taking n samples to r differential equations and m−r integral equation within tε[T0, T1], and establishing a m×n linear equation set with unknowns S1(t1), S1(t2), S1(t3). . . S1(tn), S2(t1), S2(t2) . . . Sm(t1), Sm(t2), . . . Sm(tn) and solving this linear equation set with linear algebraic method to get Si(tj) (i=1, . . . m, j=1,2 . . . n).
  • 9. The method as in claim 1, further comprising a step for solving m×n Si(tj) (i=1,2 . . . m, j=1,2 . . . n), said step establishing m difference equations for SM(t) with respect to t and with each term having different order including zero, then taking n samples to m difference equations within tε[T0, T1], and establishing a m×n linear equation set with unknowns S1(t1), S1(t2), S1(t3). . . S1(tn), S2(t1), S2(t2) . . . Sm(t1), Sm(t2), . . . Sm(tn) and solving this linear equation set with linear algebraic method to get Si(tj) (i=1, . . . m, j=1,2 . . . n).
  • 10. The method as in claim 1, further comprising a step for solving m×n Si(tj) (i=1,2 . . . m, j=1,2 . . . n), said step performing m summation operation to SM(t) from time=0 to time=t, and with each term having different order including zero, then taking n samples to m summation equations within tε[T0, T1], and establishing a m×n linear equation set with unknowns S1(t2), S1(t3). . . S1(tn), S2(t1), S2(t2) . . . Sm(t1), Sm(t2), . . . Sm(tn) and solving this linear equation set with linear algebraic method to get Si(tj) (i=1, . . . m, j=1,2 . . . n).
  • 11. The method as in claim 1, further comprising a step for solving m×n Si(tj) (i=1,2 . . . m, j=1,2 . . . n), said step establishing r difference equations for SM(t) (r<m) with respect to t and with each term having different order including zero, then perform m−r summation operation to SM(t) from time=0 to time=t, and with each term having different order including zero, taking n samples to r difference equations and m−r summation equation within tε[T0, T1], and establishing a m×n linear equation set with unknowns S1(t1), S1(t2), S1(t3) . . . S1(tn), S2(t1), S2(t2) . . . Sm(t1), Sm(t2), . . . Sm(tn) and solving this linear equation set with linear algebraic method to get Si(tj) (i=1, . . . m, j=1,2 . . . n).
  • 12. The method as in claim 1, further comprising a step for solving m×n Si(tj) (i=1,2 . . . m, j=1,2 . . . n), said step establishing r1 differential equations for SM(t) (r1=0 or <=m) with respect to t and with each term having different order including zero, establishing r2 difference equations for SM(t) (r2=0 or <=m) with respect to t and with each term having different order including zero, then performing r3 integration operation to SM(t) (r3=0 or <=m) from time=0 to time=t, and with each term having different order including zero, then performing r4 summation operation to SM(t) (r4=0 or <=m) from time=0 to time=t, and with each term having different order including zero, wherein r1+r2+r3+r4=m, then taking n samples to r1 differential equations, r2 difference equations, r3 integral equation and r4 summation equations within tε[T0, T1]: and establishing a m×n linear equation set with unknowns S1(t2), S1(t3) . . . S1(tn), S2(t1), S2(t2) . . . Sm(t1), Sm(t2), . . . Sm(tn) and solving this linear equation set with linear algebraic method to get Si(tj) (i=1, . . . m, j=1,2 . . . n).
  • 13. The method as in claim 1, wherein the predetermined time dependent function group iaj(t) (i=1, . . . m, j=1,2 . . . n) for SM(t) are linear independent such that the Si(t) within the time period [T0, T1] can be solved by linear algebra technique.
  • 14. The method as in claim 4, further comprising a step for restoring m×n Si(tj) (i=1,2 . . . m, j=1,2 . . . n), from SM(t), wherein {iaj(t) (i=1,2 . . . m, j=1,2 . . . n)} are orthonormal within the time period [T0, T1], said step comprising: multiplying iaj(t) to SM(t) and integrating the result within time period [T0, T1] to get Si(tj) (i=1,2 . . . m, j=1,2 . . . n).
  • 15. The method as in claim 4, further comprising a step for restoring m×n Sk(tl) (k=1,2 . . . m, l=1,2 . . . n) from SM(t), wherein {iaj(t) (i=1, 2 . . . m, j=1,2 . . . n)} are orthonormal within the time period [T0, T1], said step comprising:preparing m×n×M restoring parameters 1q!⁢al(q)k⁢(b)⁢ ⁢(k=1,2⁢ ⁢…⁢ ⁢m,l=1,2⁢ ⁢…⁢ ⁢n,q=0,1⁢ ⁢…⁢ ⁢M;b∈R), calculating M data, ∫T0T1⁢SM⁢(x)⁢(x-b)q⁢ ⁢ⅆx;q=0, 1, 2 . . . M−1; bεR calculating Sk(tl) using following equation Sk⁢(tl)=∑q=0M-1⁢ ⁢[1q!⁢al(q)k⁢(b)⁢∫T0T1⁢SM⁢(x)⁢(x-b)⁢ q⁢ⅆx].
  • 16. The method as in claim 4, wherein after obtaining SM(t) by using functions Si0⁢(t)=∑j=1n⁢ ⁢[i⁢aj⁢(t)⁢Si⁢(tj)]and SM⁢(t)=∑i=1m⁢ ⁢[Si0⁢(t)],further comprising the following step:choose {iaj(t) (i=1,2 . . . m, j=1,2 . . . n)} by method in claim 3; calculate SM(t) using Si0⁢(t)=∑j=1n⁢ ⁢[i⁢aj⁢(t)⁢Si⁢(tj)]and SM⁢(t)=∑i=1m⁢ ⁢[Si0⁢(t)] calculating M compression data by using the function of ∫T0T1⁢SM⁢(x)⁢(x-b)q⁢ ⁢ⅆx;q=0, 1 ,2 . . . M−1; bεR using SM(t), wherein t within period [T0, T1].
  • 17. The method according to claim 1, further comprising the following steps after step (d):using the timing control circuit to divid the time period [T0, T1] into a synchronous time period and an information time period; sending a synchronous signal in the synchronous period; sending mixed transformed signal SM(t) in the information period.
  • 18. A hardware device comprising:m A/D converters for sampling Si(t) and digitizing the samples; m×n signal generators for generating iaj(t) signals; m×n multipliers for calculating iaj(t) and Si(t), wherein Si(tj) is the j-th sampling of Si(t); an adder for calculating ∑j=1n⁢ ⁢[i⁢aj⁢(t)⁢Si⁢(tj)]a synchronous signal generator for generating synchronous signal within the time period [T0, T1]; and a timing controller for regulating the timing control of above components.
  • 19. A hardware device for separating a mixed transformed signal SM(t), comprising:a first circuit for detecting a synchronous signal; (m×n)2 memory device for use in solving linear algebra equations; at least one logic circuit for pre-handling the mixed transformed signal SM(t) to obtain a output result; a plurality of A/D converters for sampling said output result to obtain a plurality of sampling results and digitizing the sampling results; a plurality of adders and multipliers; a calculating means for calculating original samples Si(tj) of the Si(t) according to the data stored in original samples Si(tj) of the Si(t) and then outputting from the A/D converters; and a timing controller for regulating the timing control of above components.
  • 20. The device as in claim 19, further comprising m D/A converters for transforming the original samples Si(tj) (i=1,2 . . . m, j=1,2 . . . n) to analog signals.
  • 21. The device as in claim 19, wherein said logic circuit comprises a plurality of differential circuits for taking differential to SM(t).
  • 22. The device as in claim 19, wherein said logic circuit comprises a plurality of difference circuits for taking difference to SM(t).
  • 23. The device as in claim 19, wherein said logic circuit comprises a plurality of integration circuits for taking integration to SM(t) from 0 to t.
  • 24. The device as in claim 19, wherein said logic circuit comprises a plurality of summation circuits for taking summation to SM(t) from 0 to t.
US Referenced Citations (2)
Number Name Date Kind
6137848 Ho et al. Oct 2000 A
6148024 Ho et al. Nov 2000 A