General formal for complex symmetric part of sequence x[n]











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0
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Given:



(1) Complex Symmetric Sequence has the property:



$$x[n]=x^{*}[-n]$$



What’s the general formula for the Complex Symmetric Component of the sequence x[n], in terms of x[n] and x*[n]?



I'm guessing it's this:



$$x_{complex symetric}[n] =frac{1}{2}left(xleft[nright]+x^*left[-nright]right)$$



How to prove this equation is true or false?






sequences-and-series complex-numbers







share|cite|improve this question
























  • What does $;x^*;$ mean here? And what is $;[n];$ ?
    – DonAntonio
    Nov 13 at 19:08












  • x*[n] means the complex conjugate of x[n]. if x[n]=a[n] + i b[n], then x*[n] = a[n] - i b[n].
    – Bill Moore
    Nov 13 at 19:13










  • x[n] means that its a discrete function, that is, its only defined at integer values of n. This is in contrast to a continuous function: x(t) that can be defined at any real number t.
    – Bill Moore
    Nov 13 at 19:24












  • But then $;-x^*[n]=-a[n]+ib[n] neq a[n]+ib[n];$ ...! By the way, with $;x[n];$, do you mean $;x_n;$ ?
    – DonAntonio
    Nov 13 at 19:24












  • And if $;x;$ is a function defined on integers or naturals, what is the problem to denote its values by $;x(n);$ instead of the confusing $;x[n];$ ?
    – DonAntonio
    Nov 13 at 19:26

















up vote
0
down vote

favorite












Given:



(1) Complex Symmetric Sequence has the property:



$$x[n]=x^{*}[-n]$$



What’s the general formula for the Complex Symmetric Component of the sequence x[n], in terms of x[n] and x*[n]?



I'm guessing it's this:



$$x_{complex symetric}[n] =frac{1}{2}left(xleft[nright]+x^*left[-nright]right)$$



How to prove this equation is true or false?






sequences-and-series complex-numbers







share|cite|improve this question
























  • What does $;x^*;$ mean here? And what is $;[n];$ ?
    – DonAntonio
    Nov 13 at 19:08












  • x*[n] means the complex conjugate of x[n]. if x[n]=a[n] + i b[n], then x*[n] = a[n] - i b[n].
    – Bill Moore
    Nov 13 at 19:13










  • x[n] means that its a discrete function, that is, its only defined at integer values of n. This is in contrast to a continuous function: x(t) that can be defined at any real number t.
    – Bill Moore
    Nov 13 at 19:24












  • But then $;-x^*[n]=-a[n]+ib[n] neq a[n]+ib[n];$ ...! By the way, with $;x[n];$, do you mean $;x_n;$ ?
    – DonAntonio
    Nov 13 at 19:24












  • And if $;x;$ is a function defined on integers or naturals, what is the problem to denote its values by $;x(n);$ instead of the confusing $;x[n];$ ?
    – DonAntonio
    Nov 13 at 19:26















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given:



(1) Complex Symmetric Sequence has the property:



$$x[n]=x^{*}[-n]$$



What’s the general formula for the Complex Symmetric Component of the sequence x[n], in terms of x[n] and x*[n]?



I'm guessing it's this:



$$x_{complex symetric}[n] =frac{1}{2}left(xleft[nright]+x^*left[-nright]right)$$



How to prove this equation is true or false?






sequences-and-series complex-numbers







share|cite|improve this question















Given:



(1) Complex Symmetric Sequence has the property:



$$x[n]=x^{*}[-n]$$



What’s the general formula for the Complex Symmetric Component of the sequence x[n], in terms of x[n] and x*[n]?



I'm guessing it's this:



$$x_{complex symetric}[n] =frac{1}{2}left(xleft[nright]+x^*left[-nright]right)$$



How to prove this equation is true or false?






sequences-and-series complex-numbers




sequences-and-series complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 16:56

























asked Nov 13 at 19:05









Bill Moore

1176




1176












  • What does $;x^*;$ mean here? And what is $;[n];$ ?
    – DonAntonio
    Nov 13 at 19:08












  • x*[n] means the complex conjugate of x[n]. if x[n]=a[n] + i b[n], then x*[n] = a[n] - i b[n].
    – Bill Moore
    Nov 13 at 19:13










  • x[n] means that its a discrete function, that is, its only defined at integer values of n. This is in contrast to a continuous function: x(t) that can be defined at any real number t.
    – Bill Moore
    Nov 13 at 19:24












  • But then $;-x^*[n]=-a[n]+ib[n] neq a[n]+ib[n];$ ...! By the way, with $;x[n];$, do you mean $;x_n;$ ?
    – DonAntonio
    Nov 13 at 19:24












  • And if $;x;$ is a function defined on integers or naturals, what is the problem to denote its values by $;x(n);$ instead of the confusing $;x[n];$ ?
    – DonAntonio
    Nov 13 at 19:26




















  • What does $;x^*;$ mean here? And what is $;[n];$ ?
    – DonAntonio
    Nov 13 at 19:08












  • x*[n] means the complex conjugate of x[n]. if x[n]=a[n] + i b[n], then x*[n] = a[n] - i b[n].
    – Bill Moore
    Nov 13 at 19:13










  • x[n] means that its a discrete function, that is, its only defined at integer values of n. This is in contrast to a continuous function: x(t) that can be defined at any real number t.
    – Bill Moore
    Nov 13 at 19:24












  • But then $;-x^*[n]=-a[n]+ib[n] neq a[n]+ib[n];$ ...! By the way, with $;x[n];$, do you mean $;x_n;$ ?
    – DonAntonio
    Nov 13 at 19:24












  • And if $;x;$ is a function defined on integers or naturals, what is the problem to denote its values by $;x(n);$ instead of the confusing $;x[n];$ ?
    – DonAntonio
    Nov 13 at 19:26


















What does $;x^*;$ mean here? And what is $;[n];$ ?
– DonAntonio
Nov 13 at 19:08






What does $;x^*;$ mean here? And what is $;[n];$ ?
– DonAntonio
Nov 13 at 19:08














x*[n] means the complex conjugate of x[n]. if x[n]=a[n] + i b[n], then x*[n] = a[n] - i b[n].
– Bill Moore
Nov 13 at 19:13




x*[n] means the complex conjugate of x[n]. if x[n]=a[n] + i b[n], then x*[n] = a[n] - i b[n].
– Bill Moore
Nov 13 at 19:13












x[n] means that its a discrete function, that is, its only defined at integer values of n. This is in contrast to a continuous function: x(t) that can be defined at any real number t.
– Bill Moore
Nov 13 at 19:24






x[n] means that its a discrete function, that is, its only defined at integer values of n. This is in contrast to a continuous function: x(t) that can be defined at any real number t.
– Bill Moore
Nov 13 at 19:24














But then $;-x^*[n]=-a[n]+ib[n] neq a[n]+ib[n];$ ...! By the way, with $;x[n];$, do you mean $;x_n;$ ?
– DonAntonio
Nov 13 at 19:24






But then $;-x^*[n]=-a[n]+ib[n] neq a[n]+ib[n];$ ...! By the way, with $;x[n];$, do you mean $;x_n;$ ?
– DonAntonio
Nov 13 at 19:24














And if $;x;$ is a function defined on integers or naturals, what is the problem to denote its values by $;x(n);$ instead of the confusing $;x[n];$ ?
– DonAntonio
Nov 13 at 19:26






And if $;x;$ is a function defined on integers or naturals, what is the problem to denote its values by $;x(n);$ instead of the confusing $;x[n];$ ?
– DonAntonio
Nov 13 at 19:26












1 Answer
1






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A Conjugate Symmetric Sequence (CSS) is defined as:



$$x[n] = x^{*}[-n]$$



A Anti-Conjugate Symmetric Sequence (CAS) is defined as:



$$x[n] = - x^{*}[-n]$$



It can be shown that any sequence can be written as the sum of an CSS sequence and an CAS sequence as follows:



$$x[n] = x_{css}[n] + x_{cas}[n] $$



$$x[n] = (0.5)(x[n] + X^{*}[-n]) + (0.5)(x[n] - X^{*}[-n])$$



$$x[n] = x[n] $$



we can see that CSS Sequence is equal to:



$$x_{css}[n] = (0.5) (x[n] + x^{*}[n])$$






share|cite|improve this answer





















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    up vote
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    down vote













    A Conjugate Symmetric Sequence (CSS) is defined as:



    $$x[n] = x^{*}[-n]$$



    A Anti-Conjugate Symmetric Sequence (CAS) is defined as:



    $$x[n] = - x^{*}[-n]$$



    It can be shown that any sequence can be written as the sum of an CSS sequence and an CAS sequence as follows:



    $$x[n] = x_{css}[n] + x_{cas}[n] $$



    $$x[n] = (0.5)(x[n] + X^{*}[-n]) + (0.5)(x[n] - X^{*}[-n])$$



    $$x[n] = x[n] $$



    we can see that CSS Sequence is equal to:



    $$x_{css}[n] = (0.5) (x[n] + x^{*}[n])$$






    share|cite|improve this answer

























      up vote
      0
      down vote













      A Conjugate Symmetric Sequence (CSS) is defined as:



      $$x[n] = x^{*}[-n]$$



      A Anti-Conjugate Symmetric Sequence (CAS) is defined as:



      $$x[n] = - x^{*}[-n]$$



      It can be shown that any sequence can be written as the sum of an CSS sequence and an CAS sequence as follows:



      $$x[n] = x_{css}[n] + x_{cas}[n] $$



      $$x[n] = (0.5)(x[n] + X^{*}[-n]) + (0.5)(x[n] - X^{*}[-n])$$



      $$x[n] = x[n] $$



      we can see that CSS Sequence is equal to:



      $$x_{css}[n] = (0.5) (x[n] + x^{*}[n])$$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        A Conjugate Symmetric Sequence (CSS) is defined as:



        $$x[n] = x^{*}[-n]$$



        A Anti-Conjugate Symmetric Sequence (CAS) is defined as:



        $$x[n] = - x^{*}[-n]$$



        It can be shown that any sequence can be written as the sum of an CSS sequence and an CAS sequence as follows:



        $$x[n] = x_{css}[n] + x_{cas}[n] $$



        $$x[n] = (0.5)(x[n] + X^{*}[-n]) + (0.5)(x[n] - X^{*}[-n])$$



        $$x[n] = x[n] $$



        we can see that CSS Sequence is equal to:



        $$x_{css}[n] = (0.5) (x[n] + x^{*}[n])$$






        share|cite|improve this answer












        A Conjugate Symmetric Sequence (CSS) is defined as:



        $$x[n] = x^{*}[-n]$$



        A Anti-Conjugate Symmetric Sequence (CAS) is defined as:



        $$x[n] = - x^{*}[-n]$$



        It can be shown that any sequence can be written as the sum of an CSS sequence and an CAS sequence as follows:



        $$x[n] = x_{css}[n] + x_{cas}[n] $$



        $$x[n] = (0.5)(x[n] + X^{*}[-n]) + (0.5)(x[n] - X^{*}[-n])$$



        $$x[n] = x[n] $$



        we can see that CSS Sequence is equal to:



        $$x_{css}[n] = (0.5) (x[n] + x^{*}[n])$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 at 17:06









        Bill Moore

        1176




        1176






























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