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# Discussion Groups | Comp.DSP | a question about the effect of channel estimation and equalization in ofdm

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# a question about the effect of channel estimation and equalization in ofdm - MrJiang - 2009-06-24 09:39:00

```in OFDM systems, the LS channel estimation result should be

H'=P'/P

if we compensate the received data symbols with one tap scheme, the
equalization result should be

D''=D'/H' = P*D'/P';

one the right of the above equation, D'/P' looks like the differential
decoding result

so the coherent demodulation with LS estimation and onetap equalization
sees like the differential demodulation.

but why the coherent modulation is so popular in OFDM system.

```
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# Re: a question about the effect of channel estimation and equalization in ofdm - julius - 2009-06-24 10:28:00

```On Jun 24, 9:39 am, "MrJiang" <misterji...@126.com> wrote:
> in OFDM systems, the LS channel estimation result should be
>
> H'=P'/P
>
> if we compensate the received data symbols with one tap scheme, the
> equalization result should be
>
> D''=D'/H' = P*D'/P';
>
> one the right of the above equation, D'/P' looks like the differential
> decoding result
>
> so the coherent demodulation with LS estimation and onetap equalization
> sees like the differential demodulation.
>
> but why the coherent modulation is so popular in OFDM system.

It's because if you really think about it, A=A'/B, so therefore B''=B'/
C.
```
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# Re: a question about the effect of channel estimation and equalization in ofdm - MrJiang - 2009-06-25 22:10:00

```>On Jun 24, 9:39 am, "MrJiang" <misterji...@126.com> wrote:
>> in OFDM systems, the LS channel estimation result should be
>>
>> H'=P'/P
>>
>> if we compensate the received data symbols with one tap scheme, the
>> equalization result should be
>>
>> D''=D'/H' = P*D'/P';
>>
>> one the right of the above equation, D'/P' looks like the differential
>> decoding result
>>
>> so the coherent demodulation with LS estimation and onetap
equalization
>> sees like the differential demodulation.
>>
>> but why the coherent modulation is so popular in OFDM system.
>
>It's because if you really think about it, A=A'/B, so therefore
B''=B'/
>C.
>

sorry, I can't get your meaning.
could you explain it?
```
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# Re: a question about the effect of channel estimation and equalization in ofdm - MrJiang - 2009-06-26 01:13:00

```>On Jun 24, 9:39 am, "MrJiang" <misterji...@126.com> wrote:
>> in OFDM systems, the LS channel estimation result should be
>>
>> H'=P'/P
>>
>> if we compensate the received data symbols with one tap scheme, the
>> equalization result should be
>>
>> D''=D'/H' = P*D'/P';
>>
>> one the right of the above equation, D'/P' looks like the differential
>> decoding result
>>
>> so the coherent demodulation with LS estimation and onetap
equalization
>> sees like the differential demodulation.
>>
>> but why the coherent modulation is so popular in OFDM system.
>
>It's because if you really think about it, A=A'/B, so therefore
B''=B'/
>C.
>

sorry, I can't get your meaning.
could you explain it?
```
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# Re: a question about the effect of channel estimation and equalization in ofdm - sasuke - 2009-06-26 08:55:00

```>>On Jun 24, 9:39 am, "MrJiang" <misterji...@126.com> wrote:
>>> in OFDM systems, the LS channel estimation result should be
>>>
>>> H'=P'/P
>>>
>>> if we compensate the received data symbols with one tap scheme, the
>>> equalization result should be
>>>
>>> D''=D'/H' = P*D'/P';
>>>
>>> one the right of the above equation, D'/P' looks like the
differential
>>> decoding result
>>>
>>> so the coherent demodulation with LS estimation and onetap
>equalization
>>> sees like the differential demodulation.
>>>
>>> but why the coherent modulation is so popular in OFDM system.
>>
>>It's because if you really think about it, A=A'/B, so therefore
>B''=B'/
>>C.
>>
>
>sorry, I can't get your meaning.
>could you explain it?
>

Please specify what H, D and P are.
```
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