General formal for complex symmetric part of sequence x[n]
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
|
show 9 more comments
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
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
|
show 9 more comments
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
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
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
|
show 9 more comments
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
|
show 9 more comments
1 Answer
1
active
oldest
votes
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:
$$x_{css}[n] = (0.5) (x[n] + x^{*}[n])$$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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:
$$x_{css}[n] = (0.5) (x[n] + x^{*}[n])$$
add a comment |
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:
$$x_{css}[n] = (0.5) (x[n] + x^{*}[n])$$
add a comment |
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:
$$x_{css}[n] = (0.5) (x[n] + x^{*}[n])$$
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:
$$x_{css}[n] = (0.5) (x[n] + x^{*}[n])$$
answered Nov 21 at 17:06
Bill Moore
1176
1176
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997168%2fgeneral-formal-for-complex-symmetric-part-of-sequence-xn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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