Hi, please, how can i find respectively the position of first 3 suppose i have:
A = [3 3 3 3 0 0 3 3 3 3 0 0 0 3 3 3 3 0 3 3 3 3]; [pks,locs]= find(A==3); % but only each first position of 3
Thanks to the person who can help me 🙂
MATLAB: The first Position of the same values find
1
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Let MM - your 'required colum'.
MM =[3,3,0,0,3,3,0,0,0,3,3,0,0,3,3,0,0,3,0]'; m = MM;m(m == 0) = nan;%lo - define the indices of the values that need to be changed:
lo = fillmissing(m,'previous') == 3 & fillmissing(m,'next') == 3 & isnan(m); % ii - assign numbers to intervals:
ii = cumsum([0;diff(lo)>0]).*lo; %{
p - count the amount of each value of vector ii:1 element - the number of zeros (0's);2 element the number of values in the first interval.3 element the number of values in the second interval, etc.:%}
p = accumarray(ii(:)+1,1);% Let a - the values for to be changed (vector with length equal max(ii), in our case - 4):
a = [3,4,4,3]'; % HERE fixed..
% Condition for intervals with length equal 1
a(p == 1) = 3;% Replacement:
MM(lo) = a(ii(lo));Hallo Herr Müller
1. my preferred way to generate signal 'trains' is with command repmat
clear all fs = 150000; ts = 1/fs; tpuls = 10e-3;% chirp_with
f0 = 94; f2 = 9000; N= 20;% number of my chirp
t=0:ts:tpuls; % ungerade zahl endet auf null/t=0:ts:tpuls;
y = chirp(t, f0, t(end), f2);figure(1);plot(t, y); grid on;t2=repmat(t,1,2);t_guard=zeros(1,300);y0=[y t_guard]; % daß ist deine y
y2=repmat(y0,1,3);figure(2);plot(y2);grid on
2.
the function pulstran is also suggested by Mathworks support in
but
- pulstran only has 3 predefined functions, if one needs a custom function, as it happens in most of the cases, a handle to the custom function, or the same anonymously defined function has to be passed to repmat instead.
- you need 2 time vectors, one for the base chirp, and an extended one for the resulting signal.
Following an example with pulstran and custom function:
fs = 150000; ts = 1/fs; tpuls = 10e-3;% chirp_with f0 = 94; f2 = 9000; N= 20;% number of my chirpt=[0:ts:tpuls]; % ungerade zahl endet auf null/t=0:ts:tpuls; y = chirp(t, f0, t(end), f2);figure(1);plot(t, y); grid on;T = 0:1/fs:3-1/fs;prf = 1; pw = 0.1;D = 0:1/prf:3-1/prf;Y = pulstran(T,D,@(t)chirp(t,f0,t(end),f2).*(t>=0).*(t<pw));figure(2);plot(T,Y);grid on
you may also want to modify f0 and f2 passed to pulstran by a factor N
T = 0:1/fs:3-1/fs;prf = 1; pw = 0.4;D = 0:1/prf:3-1/prf;N=12; % decimation needed to keep same frequencies
Y = pulstran(T,D,@(t)chirp(t,f0/N,t(end),f2/N).*(t>=0).*(t<pw));figure(3);plot(T,Y);grid on
.
note that there is a slight variation in the top frequency of the last pulse compared to the first pulse.
Function pulstran is useful but requires a more elaborate configuration than repmat as shown above, and the simplicity of using only one time vector may also pay off avoiding complications.
3.
you can vary the intervals between chirped pulses either modifying the guard interval from the simple sequence of zeros to diverse lengths when using repmat, or with the parameter D in pulstran
So Herrn Müller
if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?
To any other reader, please if you find this answer of any help solving your question,
please click on the thumbs-up vote link,
Servus und thanks in advance
John BG
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