how to plot the vectors having different length

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hi all, i have two vectors one is time and other is amplitude. length of time is 140 whereas length of amplitude is double of it 279, i hv to plot time v/s amplitude but as i use plot(t,y),it gives an error that vectors should be of same length. what should i do....

Answers (5)

Jos (10584)
Jos (10584) on 24 Feb 2016
You can use interpolation
X_known = [1 2 3 6 8 10] ;
Y_known = [2 5 6 11 18 23] ;
X_desired = 1:10
Y_desired = interp1(X_known, Y_known, X_desired)
plot(X_desired, Y_desired,'rs-', X_known, Y_known,'bo')

Matt Fig
Matt Fig on 16 Oct 2012
Edited: Matt Fig on 16 Oct 2012
If you know the starting time and ending time for the amplitude, you can just make your own time vector of the correct length.
t = linspace(beginningtime,endingtime,length(y));
plot(t,y)
This assumes the y data was sampled at equal time spacings. If that is not the case, you should give more information about how you ended up with two vectors of different lengths and the relationship between t(idx) and y(idx) for a given idx.
  1 Comment
kavya saxena
kavya saxena on 16 Oct 2012
hello,thanx for ur reply. actually my problem is that i hv 2 signals with same time nd i m doing cross correlation using xcorr()function. it makes my signal length double for the same time. i m not getting the cross correlation plot which having one peak value against the time interval. i m sending u what i did
t=startnode(:,1); ux1=startnode(:,2); ux2=endnode(:,2); y=xcorr(ux1,ux2); n=length(t); plot(t,y); it gives error that vector must be same length. when i did according to u it again add the two values of ux1 and ux2 and gives a signal sort of plot, not any peak value as cross correlation function gives. m i using this function correctly? plz help me..

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Flavio Baduini
Flavio Baduini on 3 Apr 2016
Edited: Flavio Baduini on 3 Apr 2016
The question is of primary importance, I'm facing the same while plotting raw and averaged sensors data.
.
ex:
raw data: y (a vector of length 1000)
averaged data: z (a vector of length 100)
.
You can define a common x-axis and plot your vectors with respect to that. I chose time and I made an average of time values for each 10 samples (10 because I'm averaging 10 values each):
.
time: t (a vector of length 1000)
time averaged: ta (a vector of length 100)
.
Now just plot the values with the same length and you'll get two graphs with the raw and averaged data, in the same plot and with the correct spacing
.
plot(t,y)
plot(ta,z)

kavya k
kavya k on 16 May 2017
i have to plot the graph between the lumens(different ND filters) and pixel intensity. pixel intensity of an image is of size 240*320 .like that i have to plot for 4 images. i am getting an error as vectors are of different sizes. can any one help me. i have attached the file what the code i have written. plz help me out of this.....
  2 Comments
Jan
Jan on 16 May 2017
Please do not append a new question in the section for answers of another thread. Delete this message and open a new thread instead. Thank you.
Jan
Jan on 16 May 2017
Edited: Jan on 16 May 2017
Please do not append a new question in the section for answers of another thread. Delete this message and open a new thread instead. Include a complete copy of the error message there. Thank you.

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SABRINA SABRINA
SABRINA SABRINA on 23 Apr 2019
hi
I have a file.txt which has 20 points following x and 17 points following y with combinations E = 20 * 17 points I want to draw E in the xy plane BUT it displays error because they do not have the same dimensions

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