# Peak Detection Of Every Pulse

I have this signal as the following figure:

Now, I need to know each beginning index of each pulse as described in the following figure:

I have tried to detect the peaks through obtaining the cross correlation function with the sequence that I am sending and the Signal that I am receiving (which is exactly as if am doing auto-correlation). But they are not aligned as the following figure:

Can you suggest a solution to this problem, please? I am kinda stuck and this is my last step in my project.

I can show the codes if needed.

• This looks like you simply need to use a threshold to compare your signal to. What's wrong with that (very straight-forward) method? Commented May 7, 2018 at 17:08
• Can you provide the code and data? Commented May 7, 2018 at 18:00
• @MarcusMüller I tried to use threshold before, but my data is not at the same level, so I cannot take one threshold to all the snaps at a time. I only can take for each snap, which will cost me a lot of time. It is not efficient also to take threshold for every batch of data alone. Commented May 8, 2018 at 10:44
• @Seth I can provide it though email. Please send me your email. Commented May 8, 2018 at 10:44
• How about thresholding with the average of each batch? that wouldn't be a big loss in efficiency, would it? also, how about using a median filter and a derivative to detect the edges? Commented May 10, 2018 at 6:57

filtering -> thresholding -> edge detection

apply low pass filter:

averageFilterCoefficient = 0.01
for each sample of x:
y[i] = y[i-1]*(1-averageFilterCoefficient) + x[i]*averageFilterCoefficient


this simple code should remove all high frequencies and produce more smooth signal. Now apply thresholding:

threshold = 100
for each sample of y:
if y[i] > threshold then
z[i] = 1
else
z[i] = 0


Now you have nice logical signal, lets find rising edges:

for each sample of z:
if z[i-1] = 0 and z[i] = 1 then
print i


You just need to find the right values for exponential average filter coefficient and thresholding level.

• You don't have a moving-average filter, you have an exponential average filter.
– Ben
Commented Jun 6, 2018 at 20:33