Dear Friends.

Thank many of you for your help. After a night of think, I got myself the following code to solve some close bottom min data set and eliminated the near by repeat one. Still，

Many thanks for you tips and inspirations!

Xiaxing

BOTTS
NUMB = Dimensions[BOTTS][[1]]
For[i = NUMB - 1, i > 0, i--, 
 BOTTS = If[BOTTS[[i + 1]] - BOTTS[[i]] < 5, Drop[BOTTS, {i}], BOTTS];]
BOTTS

The BOTTS before the code gives

{2, 262, 520, 522, 775, 1028, 1274, 1517, 1757, 1993, 2225, 2455, \
2684, 2905, 3125, 3344, 3558, 3560, 3769, 3978}

The BOTTS after the code gives

{2, 262, 522, 775, 1028, 1274, 1517, 1757, 1993, 2225, 2455, 2684, \
2905, 3125, 3344, 3560, 3769, 3978}

The code eliminated of the the nearby peak@520，@3558

A simple peak finder

findPeaks[list_] := (list // BlockMap[(#[[1]] < #[[2]] > #[[3]] &), #, 3, 1] & // Position[True]) + 1

On Mike's sample data

peaks = findPeaks@data;
values = Extract[data, peaks];

ListPlot[{data, Transpose[{Flatten@peaks, values}]}, 
 PlotStyle -> {Automatic, Red}]

Without a better understanding of your requirements, I think something like this might work:

data = {100, 3, 2, 103, 102, 5, 4, 2, 105, 104, 103, 3};

findPeaks[ data_List, tolerance_List] := Module[
   {
    peaks = FindPeaks[data]
    , lastPeak = {-Infinity, -Infinity}
    , forward
    , reverse
    },
   forward = Map[
     With[
       {
        retVal = If[
          And[
           First[#1] -  First[lastPeak] <= tolerance[[1]],
           Last[#1] - Last[lastPeak] <= tolerance[[2]]
           ]
          ,
           Nothing
          ,
          #1
          ]
        }
       ,
       lastPeak = #1
       ] &
     ,
     peaks
     ];


   reverse = Map[
     With[
       {
        retVal = If[
          And[
           First[#1] -  First[lastPeak] <= tolerance[[1]],
           Last[#1] - Last[lastPeak] <= tolerance[[2]]
           ]
          ,
           Nothing
          ,
          #1
          ]
        }
       ,
       lastPeak = #1
       ] &
     ,
     Reverse[forward]
     ];

   Reverse[reverse]

   ];

findPeaks[data, {3, 3}]

Hi Rohit

Sorry the data is positive, and I am looking for the bottom position. So I multiply it with (-1) and looking for the peaks instead

Thanks for looking into my problem.

Xiaxing

Please post the code you used to generate the plot. There are no negative values in Example.txt but the plot in OSC-Signal.jpg has negative values.