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extracting elements from rows and columns

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ABDUL
ABDUL 2018 年 2 月 7 日
閉鎖済み: MATLAB Answer Bot 2021 年 8 月 20 日
i have a input matrix of size
ofdm_symbol=randn(8,10,2)+(1j*randn(8,10,2))
the size of this matrix is 8x10x2
i want to access specific rows and columns and pad with zeros in the code
time_domain_signal=abs(ifft([ofdm_symbol(:,1:(end/2)-1,ii) zeros(N,(L-1)*N,ii(1:k1))
ofdm_symbol(:,5:8,ii(1:k1))])
and it gives the error as
Dimensions of matrices being concatenated are not consistent.
can you help me in debugging the code
  9 件のコメント
ABDUL
ABDUL 2018 年 2 月 8 日
i am getting the result for the code segment . but i want to modify the same code for obtaining in the 3d matrix.
Walter Roberson
Walter Roberson 2018 年 2 月 8 日
Your code fragment
zeros(N,(L-1)*N,ii(1:k1))
does not work with a 3D matrix unless k1 happens to be exactly 1: if k1 is greater than 1, then 1:k1 is a vector and ii(1:k1) would be a vector of sizes. If any of those other than the first was something other than 1, then you would be working with an array of at least 4 dimensions.
In the phrase
[ofdm_symbol(:,1:(end/2)-1,ii) zeros(N,(L-1)*N,ii(1:k1))
ofdm_symbol(:,5:8,ii(1:k1))]
the number of rows needs to match (which it appears to do). But the number of pages (third dimension) also needs to match, so size(ofdm_symbol(:,1:(end/2)-1,ii),3) needs to match size(zeros(N,(L-1)*N,ii(1:k1)),3) and size(zeros(N,(L-1)*N,ii(1:k1)),4) upwards need to be 1. size(ofdm_symbol(:,1:(end/2)-1,ii),3) is length(ii) . size(zeros(N,(L-1)*N,ii(1:k1)),3) is ii(1). size(zeros(N,(L-1)*N,ii(1:k1)),4) upwards are ii(2) through ii(k1)

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