Area Under Curve and above X axis only, and between x axis limits.

I need to calculate the area under the curve. The curve may go down the x axis, but I dont' want to calculate that area. And then it may come up the axis again. Also, I need to calculate area in the limits of 1Hz to 6Hz, as shown in the image attached.

Please help out. I went through the trapz() function, but I don't know, how I can measure the area in this case. Thanks in advance.

[Looky]

The easiest solution would be to make a copy of your data and just set everything below zero to zero. The use trapz:

y=yourYData(yourXData>1&yourXData<6);
x=yourXData(yourXData>1&yourXData<6);

y(y<0)=0;
area=trapz(x, y )

[Bhanu Pratap Singh]

Yeah, that would work. Thanks a lot Looky.

Answer 1

[Allen]

If you are trying to determine the area under the curve between the frequencies of 1 Hz to 6 Hz for all y-values greater than 0, then Looky's suggestion will do the trick. However, if you are trying to determine the area under the curve between 1 Hz to 6 Hz and for all y-values greater than some threshold value other than zero, you will need to some additional adjustments before running trapz(). From the image you provided it looks like 1 V/H is your target threshold.

% Assign the y-value threshold to a variable. Makes is clearer to see how and where it is used,
% as well as makes is easy to change if needed.
thres = 1;

% Extract x- and y-data for frequencies in the desired range of 1-6 Hz.
x = yourXData(yourXData>1 & yourXData<6); % x-values in the range
y = yourYData(yourXData>1 & yourXData<6)-thres; % y-values in range & with a threshold adjustment

% Set all values that are <0 (these were originally all values less than the threshold) to zero.
y(y<0) = 0;

% Use trapz function to calculate the area under the curve.
area = trapz(x, y )

[Bhanu Pratap Singh]

Yes, I need the threshold to be 1. I think I can do just the following, to make all values less than the threshold to be zero. Thanks a lot for helping me out.

% Set all values that are < 1 (these were originally all values less than the threshold) to zero.
y(y<1) = 0;

[Looky]

Hey Bhanu,

no, this is not the same! Just take @Allen's Solution, it has a variable threshold via the thres variable, that is what you want.

Your solution will give you the area under the curve but enclosed with x-axis and not your threshhold at 1 !