Sirs,

I have some questions on OFDM method

1) what is meant by orthogonality in frequency domain. I understand
orthogonality in time domain
2) what is the significance of keeping the inter carrier seperation equal
to 1/symbol time

Thanks, Manish

>Sirs,
>
>I have some questions on OFDM method
>
>1) what is meant by orthogonality in frequency domain. I understand
>orthogonality in time domain
>2) what is the significance of keeping the inter carrier seperation equal
>to 1/symbol time
>
>Thanks, Manish
>

I am not "sure" I know the answer but my simple perspective is that similar
to fft leakage from bin to bin. If you have a sine wave of exact number of
cycles and of a frequency that exactly hits a bin centre then no leakage
occurs and you get one single line for your frequency.

At the receiver of an ofdm you can't get full cycles or exact frequency for
the fft of a symbol frame so leakage will occur but when the frequency
separation is = n/(fft period) then each frequency will occupy bins such
that it does not cross talk with next frequency bins.

The separation is n/symbol time and minimum value for n is 1.

Kadhiem

On 10/20/2012 1:33 PM, manishp wrote:
> Sirs,
>
> I have some questions on OFDM method
>
> 1) what is meant by orthogonality in frequency domain. I understand
> orthogonality in time domain

Frequency elements a and b are orthogonal if

  +infinity
   INTEGRAL(f(a)*f(b))dt=0
  -infinity

Thus, any two frequencies not the same, or two signals of the same 
frequency that are in quadrature are orthogonal. That is the basis of 
calculating the Fourier transform. the FFT algorithm is a mathematical 
shortcut for that calculation.

> 2) what is the significance of keeping the inter carrier seperation equal
> to 1/symbol time

Reasonable guard band.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

>Frequency elements a and b are orthogonal if
>
>  +infinity
>   INTEGRAL(f(a)*f(b))dt=0
>  -infinity

Hello Jerry,

Since convolution includes intgration already, does the above equation mean
there is a double integration. First due to convolution and then over the
resultant signal. Can you please clarify?

Thanks, Manish ...

On Mon, 22 Oct 2012 09:44:34 -0500, "manishp" <58525@dsprelated>
wrote:

>>Frequency elements a and b are orthogonal if
>>
>>  +infinity
>>   INTEGRAL(f(a)*f(b))dt=0
>>  -infinity
>
>Hello Jerry,
>
>Since convolution includes intgration already, does the above equation mean
>there is a double integration. First due to convolution and then over the
>resultant signal. Can you please clarify?
>
>Thanks, Manish ...

Stated more simply, if the dot product of two vectors is zero, then
they are orthogonal to each other.


Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com

e...@ieee.org (Eric Jacobsen) writes:

> On Mon, 22 Oct 2012 09:44:34 -0500, "manishp" <58525@dsprelated>
> wrote:
>
>>>Frequency elements a and b are orthogonal if
>>>
>>>  +infinity
>>>   INTEGRAL(f(a)*f(b))dt=0
>>>  -infinity
>>
>>Hello Jerry,
>>
>>Since convolution includes intgration already, does the above equation mean
>>there is a double integration. First due to convolution and then over the
>>resultant signal. Can you please clarify?
>>
>>Thanks, Manish ...
>
> Stated more simply, if the dot product of two vectors is zero, then
> they are orthogonal to each other.

That is one definition. There are others, e.g., two "continuous-time"
signals f(t) and g(t) are said to be orthogonal over some interval T if

  \int_{\tau}^{\tau + T} f(t) \cdot g(t) dt = 0, 

where \tau is a given point in time (usually either 0 or -T/2).
-- 
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com

On Mon, 22 Oct 2012 13:33:08 -0400, Randy Yates
<y...@digitalsignallabs.com> wrote:

>e...@ieee.org (Eric Jacobsen) writes:
>
>> On Mon, 22 Oct 2012 09:44:34 -0500, "manishp" <58525@dsprelated>
>> wrote:
>>
>>>>Frequency elements a and b are orthogonal if
>>>>
>>>>  +infinity
>>>>   INTEGRAL(f(a)*f(b))dt=0
>>>>  -infinity
>>>
>>>Hello Jerry,
>>>
>>>Since convolution includes intgration already, does the above equation mean
>>>there is a double integration. First due to convolution and then over the
>>>resultant signal. Can you please clarify?
>>>
>>>Thanks, Manish ...
>>
>> Stated more simply, if the dot product of two vectors is zero, then
>> they are orthogonal to each other.
>
>That is one definition. There are others, e.g., two "continuous-time"
>signals f(t) and g(t) are said to be orthogonal over some interval T if
>
>  \int_{\tau}^{\tau + T} f(t) \cdot g(t) dt = 0, 
>
>where \tau is a given point in time (usually either 0 or -T/2).
>-- 

I don't know what script language that is or what it really says, but
it looks to me like it's either a dot product or inner product.   If
so, it's essentially the same definition of orthogonality.


Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com

On 10/22/12 2:09 PM, Eric Jacobsen wrote:
> On Mon, 22 Oct 2012 13:33:08 -0400, Randy Yates
> <y...@digitalsignallabs.com>  wrote:
>
...
>>
>>   \int_{\tau}^{\tau + T} f(t) \cdot g(t) dt = 0,
>>
>> where \tau is a given point in time (usually either 0 or -T/2).
>> --
>
> I don't know what script language that is

it's LaTeX.  looks like it to me.  we use it at Wikipedia and also on 
the physicsforums.com site.

-- 

r b-j                  r...@audioimagination.com

"Imagination is more important than knowledge."

Randy Yates <y...@digitalsignallabs.com> wrote:
> e...@ieee.org (Eric Jacobsen) writes:


(snip on orthogonality)

>> Stated more simply, if the dot product of two vectors is zero, then
>> they are orthogonal to each other.

> That is one definition. There are others, e.g., two "continuous-time"
> signals f(t) and g(t) are said to be orthogonal over some interval T if

>  \int_{\tau}^{\tau + T} f(t) \cdot g(t) dt = 0, 

> where \tau is a given point in time (usually either 0 or -T/2).

Well, as I understand the mathematics, first you extend the vector to
infinity and call it a function. (It might still have finite bounds,
but defined at an infinite number of points.)

The same extension gets you from the Fourier series to the Fourier
transform. 

I still remember NOT learning the difference between Fourier series
and transform, as taught by my physics TA at 9:00 AM. That was the
first time I had to extend the idea of a vector to infinity, and
was too much to learn so quickly.

-- glen

robert bristow-johnson <r...@audioimagination.com> wrote:
> On 10/22/12 2:09 PM, Eric Jacobsen wrote:
>> On Mon, 22 Oct 2012 13:33:08 -0400, Randy Yates
>> <y...@digitalsignallabs.com>  wrote:
>>
> ...
>>>
>>>   \int_{\tau}^{\tau + T} f(t) \cdot g(t) dt = 0,
>>>
>>> where \tau is a given point in time (usually either 0 or -T/2).
>>> --
>>
>> I don't know what script language that is

> it's LaTeX.  looks like it to me.  we use it at Wikipedia and also on 
> the physicsforums.com site.

It would be LaTeX (or plain TeX) if you put $ or $$ around it.

-- glen