# In OFDM, do we need to send all the outputs of the IFFt block?

Started by January 7, 2014
```In OFDM Transmitter, when you do the IFFT of input samples
x(0),x(1),x(2)....x(N-1), then at the output of the IFFT block you get a
sequence X[0],X[1],....X[N-1].

My question is, that do we then transmit X[0],X[1],....X[N-1] one by one
over the channel (and hence the need for parallel to serial blockafter IFF
block)?

If that is so, don't you think we only need to transmit only one element
from the set {X[0],X[1],....X[N-1] } ??? because if you consider, lets say,
X[0] then this particular element has information about all the samples of
x(0),x(1),x(2)....x(N-1) (according to the formula of DFT) and we need only
send only this element and not the complete set {X[0],X[1],....X[N-1] }?

_____________________________
Posted through www.DSPRelated.com
```
```On Tuesday, January 7, 2014 9:19:59 AM UTC-5, engrMunna wrote:
> If that is so, don't you think we only need to transmit only one element
> from the set {X[0],X[1],....X[N-1] } ??? because if you consider, lets say,
> X[0] then this particular element has information about all the samples of
> x(0),x(1),x(2)....x(N-1) (according to the formula of DFT) and we need only
> send only this element and not the complete set {X[0],X[1],....X[N-1] }?
>

You're wrong. X[0] is a function of all the elements x[0...N-1], but in the general case you cannot recover all of x[0...N-1] just from X[0].

If c=a+b, can you compute both a and b just from observing c?
```
```On Tue, 7 Jan 2014 07:01:41 -0800 (PST), julius <juliusk@gmail.com>
wrote:

>On Tuesday, January 7, 2014 9:19:59 AM UTC-5, engrMunna wrote:
>> If that is so, don't you think we only need to transmit only one element
>> from the set {X[0],X[1],....X[N-1] } ??? because if you consider, lets say,
>> X[0] then this particular element has information about all the samples of
>> x(0),x(1),x(2)....x(N-1) (according to the formula of DFT) and we need only
>> send only this element and not the complete set {X[0],X[1],....X[N-1] }?
>>
>
>You're wrong. X[0] is a function of all the elements x[0...N-1], but in the general case you cannot recover all of x[0...N-1] just from X[0].
>
>If c=a+b, can you compute both a and b just from observing c?

It'd be a very popular compression scheme if it worked.   ;)

Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com
```
```On Tuesday, January 7, 2014 1:01:01 PM UTC-5, Eric Jacobsen wrote:
> On Tue, 7 Jan 2014 07:01:41 -0800 (PST), julius <juliusk@gmail.com>
>
> wrote:
>
>
>
> >On Tuesday, January 7, 2014 9:19:59 AM UTC-5, engrMunna wrote:
>
> >> If that is so, don't you think we only need to transmit only one element
>
> >> from the set {X[0],X[1],....X[N-1] } ??? because if you consider, lets say,
>
> >> X[0] then this particular element has information about all the samples of
>
> >> x(0),x(1),x(2)....x(N-1) (according to the formula of DFT) and we need only
>
> >> send only this element and not the complete set {X[0],X[1],....X[N-1] }?
>
> >>
>
> >
>
> >You're wrong. X[0] is a function of all the elements x[0...N-1], but in the general case you cannot recover all of x[0...N-1] just from X[0].
>
> >
>
> >If c=a+b, can you compute both a and b just from observing c?
>
>
>
> It'd be a very popular compression scheme if it worked.   ;)
>
>
>
>
>
> Eric Jacobsen
>
> Anchor Hill Communications
>
> http://www.anchorhill.com

Don't give them any ideas, now ... !
```
```On 1/7/14 1:03 PM, julius wrote:
> On Tuesday, January 7, 2014 1:01:01 PM UTC-5, Eric Jacobsen wrote:
>> On Tue, 7 Jan 2014 07:01:41 -0800 (PST), julius<juliusk@gmail.com>
>> wrote:

>>
>>> If c=a+b, can you compute both a and b just from observing c?
>>
>> It'd be a very popular compression scheme if it worked.   ;)
>>
>
> Don't give them any ideas, now ... !

remember that Zeosync farce from around 2000?

--

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

```
```On Tuesday, January 7, 2014 1:22:10 PM UTC-5, robert bristow-johnson wrote:
>
> remember that Zeosync farce from around 2000?
>

That idea is re-invented every couple of years. We can write a patent on this, and I'll even let you be the first author.
```