# Shifting signals to the right

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Curious Mind on 18 Jun 2020
Commented: Adam Danz on 19 Jun 2020
Hi, I have a matrix, M which is 10*2000 double (10 rows of signals and each signal contains 2000 variables). I want to shift each of the signal in this matrix to the right by say 1. I have a code (below) that can shift a single signal (that is if M contains a single signal, 1*2000). How do I modify this to shift each signal in M at the same time?
The code code for shifting a single signal is:
p = M;
Shift = 1;
Shifted_M = circshift(p,Shift);
In summary, I want a code that can shift each independent row signal in a matrix and output the shifted matrix as Shifted_M. After that I would like to plot this shifted matrix with the original data to visualize it.
Thanks

#### 1 Comment

Tanveer ul haq on 18 Jun 2020
% consider your matrix is M then:
M=[1 2 3; 4 5 6; 7 8 9];
Shift = 1;
Shifted_M = circshift(M',Shift);
rotated_matrix = Shifted_M'

Adam Danz on 18 Jun 2020
If you'd like to circularly shift the data "A" by "K" positions along the "dim" dimension,
So, you'll need to specify the dimension.

Adam Danz on 19 Jun 2020
"The code is just adding 1 to all the variables and not shifting their positions."
That's incorrect. When you plot two variable using plot(r,q) the x-coordinates are defined by r. To shift them one unit to the right, you would add 1 to the x-coordinates. plot(r+1,q).
It makes sense, right? If r=3 and you want to shift it by 1 unit to the right, then r0=3+1.
Here's an example.
r = 1:10;
q = sort(randn(1,10));
cla()
plot(r, q, 'r-o')
hold on
plot(r+1, q, 'b-o') It's still not clear how your matrix values should be visualized. For example, for matrix M,
m = [ 2 4 1 0 8
8 2 1 9 0
1 9 2 0 3];
if the x coordinates are x=[1 2 3 4 5] and the y coordinates are y=[1 2 3], then at coordinate (1,1) the value of m is 2. Does that mean you want a 3D plot were m contains the z-values?
Curious Mind on 19 Jun 2020
You are correct! I was adding the 1 to the wrong thing. It works now. Also I figured out the plot part. I appreciate your patience and help! Thank you.
Adam Danz on 19 Jun 2020
Ah, good! Sometimes describing the problem is more difficult than finding the solution. Glad it all worked out!

R2019b

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