On 10/23/2012 3:02 PM, robert bristow-johnson wrote:
>
> consider spending some time at Wikipedia either contributing new
> material to technical pages or standing vigil against errors and crap.
> then you'll develop a TeX interpreter in your brain. (and they have a
> page explaining stuff at http://en.wikipedia.org/wiki/Wikipedia:Math .
> looks like they added a few symbols, \circledast and \triangleq . wish
> they had those before. the first is what *i* think should be used for
> convolution and the latter for a defined equality.)
And how do these equations get shown on the web page, as a graphic?
Rick
Reply by robert bristow-johnson●October 23, 20122012-10-23
On 10/23/12 1:48 PM, Tim Wescott wrote:
> On Tue, 23 Oct 2012 11:12:09 -0400, robert bristow-johnson wrote:
>
>> On 10/23/12 4:31 AM, manishp wrote:
>>>
>>> Yes. I assumed asterisk operator above to be convolution function. My
>>> mistake indeed ...
>>
>> it's the problem with ASCII math (maybe one reason Randy used TeX
>> script, but then you gotta know what "\cdot" means)
>>
>> in my opinion, the naked asterisk is an unfortunate choice for
>> convolution in the textbooks and other lit. i would have put that
>> asterisk in a little circle to make it look like is "more" than
>> multiplication. in ASCII math, i surround it with parenths to make it
>> look like something else:
>>
>> y(t) = h(t) (*) x(t)
>>
>> i am now trying to get out of the habit of using * for multiplication on
>> these pages but sometimes it's unavoidable.
>
> For multiplication I try to use something like (a(t))(b(t)). For
> convolution, I try to say "convolve".
>
> Oh, if we'd all just install TeX interpreters in our heads...
>
consider spending some time at Wikipedia either contributing new
material to technical pages or standing vigil against errors and crap.
then you'll develop a TeX interpreter in your brain. (and they have a
page explaining stuff at http://en.wikipedia.org/wiki/Wikipedia:Math .
looks like they added a few symbols, \circledast and \triangleq . wish
they had those before. the first is what *i* think should be used for
convolution and the latter for a defined equality.)
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge."
Reply by Eric Jacobsen●October 23, 20122012-10-23
On Tue, 23 Oct 2012 11:29:33 -0400, Randy Yates
<yates@digitalsignallabs.com> wrote:
>robert bristow-johnson <rbj@audioimagination.com> writes:
>
>> On 10/23/12 4:31 AM, manishp wrote:
>>>
>>> Yes. I assumed asterisk operator above to be convolution function.
>>> My mistake indeed ...
>>
>> it's the problem with ASCII math (maybe one reason Randy used TeX
>> script, but then you gotta know what "\cdot" means)
>>
>> in my opinion, the naked asterisk is an unfortunate choice for
>> convolution in the textbooks and other lit. i would have put that
>> asterisk in a little circle to make it look like is "more" than
>> multiplication. in ASCII math, i surround it with parenths to make it
>> look like something else:
>>
>> y(t) = h(t) (*) x(t)
>>
>> i am now trying to get out of the habit of using * for multiplication
>> on these pages but sometimes it's unavoidable.
>>
>>>
>>> But coming back to main question. I do understand that sub-carriers are
>>> orthogonal, but question is in what way this property is useful?
>>> That is, orthogonality in sub-carriers ...
>>
>> you can put different and unrelated information of similar bandwidth
>> on the two different carriers if they are orthogonal.
>
>Exactly. It's why I once asked in (information theory?) class why anyone
>would use BPSK instead of QPSK - twice the data rate for the same
>bandwidth. I suppose it's done due to other concerns, e.g., simpler
>receiver.
There's a 3dB difference in link margin. That's huge. ;)
Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com
Reply by Randy Yates●October 23, 20122012-10-23
Tim Wescott <tim@seemywebsite.com> writes:
> On Tue, 23 Oct 2012 11:29:33 -0400, Randy Yates wrote:
>
>> robert bristow-johnson <rbj@audioimagination.com> writes:
>>
>>> On 10/23/12 4:31 AM, manishp wrote:
>>>>
>>>> Yes. I assumed asterisk operator above to be convolution function. My
>>>> mistake indeed ...
>>>
>>> it's the problem with ASCII math (maybe one reason Randy used TeX
>>> script, but then you gotta know what "\cdot" means)
>>>
>>> in my opinion, the naked asterisk is an unfortunate choice for
>>> convolution in the textbooks and other lit. i would have put that
>>> asterisk in a little circle to make it look like is "more" than
>>> multiplication. in ASCII math, i surround it with parenths to make it
>>> look like something else:
>>>
>>> y(t) = h(t) (*) x(t)
>>>
>>> i am now trying to get out of the habit of using * for multiplication
>>> on these pages but sometimes it's unavoidable.
>>>
>>>
>>>> But coming back to main question. I do understand that sub-carriers
>>>> are orthogonal, but question is in what way this property is useful?
>>>> That is, orthogonality in sub-carriers ...
>>>
>>> you can put different and unrelated information of similar bandwidth on
>>> the two different carriers if they are orthogonal.
>>
>> Exactly. It's why I once asked in (information theory?) class why anyone
>> would use BPSK instead of QPSK - twice the data rate for the same
>> bandwidth. I suppose it's done due to other concerns, e.g., simpler
>> receiver.
>
> BPSK has a larger distance between decision points (dang, I can't
> remember the real term).
yeah, me either - but i know what you mean.
> QPSK requires twice the power at the same
> bandwidth (in essence, it requires the _same_ energy/symbol) to get the
> same bit error rate.
On Tue, 23 Oct 2012 03:31:46 -0500, manishp wrote:
>>Tim Wescott <tim@seemywebsite.com> wrote:
>>> On Mon, 22 Oct 2012 09:44:34 -0500, manishp 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?
>>
>>> How did convolution get into this? You were asking about
>>> orthogonality
>
>>> in the frequency domain -- that has no direct relationship to
> convolution.
>>
>>Jerry used *, the symbol for multiply in most computer languages, but
>>also similar to the symbol some use for convolution.
>>
>>-- glen
>
> Yes. I assumed asterisk operator above to be convolution function. My
> mistake indeed ...
>
> But coming back to main question. I do understand that sub-carriers are
> orthogonal, but question is in what way this property is useful? That
> is, orthogonality in sub-carriers ...
At its most basic, because no sub-carrier interferes with any other. So
regardless of what you put in subcarrier n, it doesn't affect the
information in subcarrier m so long as n != m.
--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?
Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply by Tim Wescott●October 23, 20122012-10-23
On Tue, 23 Oct 2012 11:29:33 -0400, Randy Yates wrote:
> robert bristow-johnson <rbj@audioimagination.com> writes:
>
>> On 10/23/12 4:31 AM, manishp wrote:
>>>
>>> Yes. I assumed asterisk operator above to be convolution function. My
>>> mistake indeed ...
>>
>> it's the problem with ASCII math (maybe one reason Randy used TeX
>> script, but then you gotta know what "\cdot" means)
>>
>> in my opinion, the naked asterisk is an unfortunate choice for
>> convolution in the textbooks and other lit. i would have put that
>> asterisk in a little circle to make it look like is "more" than
>> multiplication. in ASCII math, i surround it with parenths to make it
>> look like something else:
>>
>> y(t) = h(t) (*) x(t)
>>
>> i am now trying to get out of the habit of using * for multiplication
>> on these pages but sometimes it's unavoidable.
>>
>>
>>> But coming back to main question. I do understand that sub-carriers
>>> are orthogonal, but question is in what way this property is useful?
>>> That is, orthogonality in sub-carriers ...
>>
>> you can put different and unrelated information of similar bandwidth on
>> the two different carriers if they are orthogonal.
>
> Exactly. It's why I once asked in (information theory?) class why anyone
> would use BPSK instead of QPSK - twice the data rate for the same
> bandwidth. I suppose it's done due to other concerns, e.g., simpler
> receiver.
BPSK has a larger distance between decision points (dang, I can't
remember the real term). QPSK requires twice the power at the same
bandwidth (in essence, it requires the _same_ energy/symbol) to get the
same bit error rate.
At least, if the caffeine is working yet...
And yes, both transmit and receive are simpler with BPSK.
--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?
Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply by Tim Wescott●October 23, 20122012-10-23
On Tue, 23 Oct 2012 11:12:09 -0400, robert bristow-johnson wrote:
> On 10/23/12 4:31 AM, manishp wrote:
>>
>> Yes. I assumed asterisk operator above to be convolution function. My
>> mistake indeed ...
>
> it's the problem with ASCII math (maybe one reason Randy used TeX
> script, but then you gotta know what "\cdot" means)
>
> in my opinion, the naked asterisk is an unfortunate choice for
> convolution in the textbooks and other lit. i would have put that
> asterisk in a little circle to make it look like is "more" than
> multiplication. in ASCII math, i surround it with parenths to make it
> look like something else:
>
> y(t) = h(t) (*) x(t)
>
> i am now trying to get out of the habit of using * for multiplication on
> these pages but sometimes it's unavoidable.
For multiplication I try to use something like (a(t))(b(t)). For
convolution, I try to say "convolve".
Oh, if we'd all just install TeX interpreters in our heads...
--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?
Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply by Randy Yates●October 23, 20122012-10-23
robert bristow-johnson <rbj@audioimagination.com> writes:
> On 10/23/12 4:31 AM, manishp wrote:
>>
>> Yes. I assumed asterisk operator above to be convolution function.
>> My mistake indeed ...
>
> it's the problem with ASCII math (maybe one reason Randy used TeX
> script, but then you gotta know what "\cdot" means)
>
> in my opinion, the naked asterisk is an unfortunate choice for
> convolution in the textbooks and other lit. i would have put that
> asterisk in a little circle to make it look like is "more" than
> multiplication. in ASCII math, i surround it with parenths to make it
> look like something else:
>
> y(t) = h(t) (*) x(t)
>
> i am now trying to get out of the habit of using * for multiplication
> on these pages but sometimes it's unavoidable.
>
>>
>> But coming back to main question. I do understand that sub-carriers are
>> orthogonal, but question is in what way this property is useful?
>> That is, orthogonality in sub-carriers ...
>
> you can put different and unrelated information of similar bandwidth
> on the two different carriers if they are orthogonal.
Exactly. It's why I once asked in (information theory?) class why anyone
would use BPSK instead of QPSK - twice the data rate for the same
bandwidth. I suppose it's done due to other concerns, e.g., simpler
receiver.
--
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com
Reply by robert bristow-johnson●October 23, 20122012-10-23
On 10/23/12 4:31 AM, manishp wrote:
>
> Yes. I assumed asterisk operator above to be convolution function.
> My mistake indeed ...
it's the problem with ASCII math (maybe one reason Randy used TeX
script, but then you gotta know what "\cdot" means)
in my opinion, the naked asterisk is an unfortunate choice for
convolution in the textbooks and other lit. i would have put that
asterisk in a little circle to make it look like is "more" than
multiplication. in ASCII math, i surround it with parenths to make it
look like something else:
y(t) = h(t) (*) x(t)
i am now trying to get out of the habit of using * for multiplication on
these pages but sometimes it's unavoidable.
>
> But coming back to main question. I do understand that sub-carriers are
> orthogonal, but question is in what way this property is useful?
> That is, orthogonality in sub-carriers ...
you can put different and unrelated information of similar bandwidth on
the two different carriers if they are orthogonal.
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge."
Reply by Randy Yates●October 23, 20122012-10-23
Randy Yates <yates@digitalsignallabs.com> writes:
> eric.jacobsen@ieee.org (Eric Jacobsen) writes:
>
>> 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.
>
> It's (La)TeX. Punch it in here to see:
>
> http://www.codecogs.com/latex/eqneditor.php
>
> Is summing a countable number of values the same as integration? If not,
> then this is a different definition. Same concept at some level, but at
> the base operations level it's different.
From [herstein, p.193]:
DEFINITION The vector space V over F is said to be an _inner product
space_ if there is DEFINED for any two vectors u, v \in V an element
(u, v) in F such that [...].
So they're both inner products but each one has to be defined to satisfy
these properties depending on the space.
--Randy
@book{herstein,
title = "Topics in Algebra",
author = "I.N. Herstein",
publisher = "Wiley",
edition = "second",
year = "1975"}
--
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com