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
OFDM
Started by ●October 20, 2012
Reply by ●October 20, 20122012-10-20
>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
Reply by ●October 21, 20122012-10-21
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 domainFrequency 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 timeReasonable guard band. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●October 22, 20122012-10-22
>Frequency elements a and b are orthogonal if > > +infinity > INTEGRAL(f(a)*f(b))dt=0 > -infinityHello 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 ...
Reply by ●October 22, 20122012-10-22
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
Reply by ●October 22, 20122012-10-22
eric.jacobsen@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
Reply by ●October 22, 20122012-10-22
On Mon, 22 Oct 2012 13:33:08 -0400, Randy Yates <yates@digitalsignallabs.com> wrote:>eric.jacobsen@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
Reply by ●October 22, 20122012-10-22
On 10/22/12 2:09 PM, Eric Jacobsen wrote:> On Mon, 22 Oct 2012 13:33:08 -0400, Randy Yates > <yates@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 isit's LaTeX. looks like it to me. we use it at Wikipedia and also on the physicsforums.com site. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●October 22, 20122012-10-22
Randy Yates <yates@digitalsignallabs.com> wrote:> eric.jacobsen@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
Reply by ●October 22, 20122012-10-22
robert bristow-johnson <rbj@audioimagination.com> wrote:> On 10/22/12 2:09 PM, Eric Jacobsen wrote: >> On Mon, 22 Oct 2012 13:33:08 -0400, Randy Yates >> <yates@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