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I'm trying to understand the cross wavelet function in R, but can't figure out how to convert the phase lag arrows to a time lag with the biwavelet package. For example:

require(gamair)
data(cairo)
data_1 <- within(cairo, Date <- as.Date(paste(year, month, day.of.month, sep = "-")))
data_1 <- data_1[,c('Date','temp')]
data_2 <- data_1

# add a lag
n <- nrow(data_1)
nn <- n - 49
data_1 <- data_1[1:nn,]
data_2 <- data_2[50:nrow(data_2),]
data_2[,1] <- data_1[,1]

require(biwavelet)
d1 <- data_1[,c('Date','temp')]
d2 <- data_2[,c('Date','temp')]
xt1 <- xwt(d1,d2)
plot(xt1, plot.phase = TRUE)

enter image description here enter image description here

These are my two time series. Both are identical but one is lagging the other. The arrows suggest a phase angle of 45 degrees - apparently pointing down or up means 90 degrees (in or out of phase) so my interpretation is that I'm looking at a lag of 45 degrees.

How would I now convert this to a time lag i.e. how would I calculate the time lag between these signals?

I've read online that this can only be done for a specific wavelength (which I presume means for a certain period?). So, given that we're interested in a period of 365, and the time step between the signals is one day, how would one alculate the time lag?

MichaelChirico
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Emma Tebbs
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2 Answers2

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So I believe you're asking how you can determine what the lag time is given two time series (in this case you artificially added in a lag of 49 days).

I'm not aware of any packages that make this a one-step process, but since we are essentially dealing with sin waves, one option would be to "zero out" the waves and then find the zero crossing points. You could then calculate the average distance between zero crossing points of wave 1 and wave 2. If you know the time step between measurements, you can easy calculate the lag time (in this case the time between measurement steps is one day).

Here is the code I used to accomplish this: 

#smooth the data to get rid of the noise that would introduce excess zero crossings)
#subtracted 70 from the temp to introduce a "zero" approximately in the middle of the wave
spline1 <- smooth.spline(data_1$Date, y = (data_1$temp - 70), df = 30)
plot(spline1)
#add the smoothed y back into the original data just in case you need it
data_1$temp_smoothed <- spline1$y

#do the same for wave 2
spline2 <- smooth.spline(data_2$Date, y = (data_2$temp - 70), df = 30)
plot(spline2)
data_2$temp_smoothed <- spline2$y

#function for finding zero crossing points, borrowed from the msProcess package
zeroCross <- function(x, slope="positive")
{
  checkVectorType(x,"numeric")
  checkScalarType(slope,"character")
  slope <- match.arg(slope,c("positive","negative"))
  slope <- match.arg(lowerCase(slope), c("positive","negative"))

  ipost  <- ifelse1(slope == "negative", sort(which(c(x, 0) < 0 & c(0, x) > 0)),
  sort(which(c(x, 0) > 0 & c(0, x) < 0)))
  offset <- apply(matrix(abs(x[c(ipost-1, ipost)]), nrow=2, byrow=TRUE), MARGIN=2, order)[1,] - 2
  ipost + offset
}

#find zero crossing points for the two waves
zcross1 <- zeroCross(data_1$temp_smoothed, slope = 'positive')
length(zcross1)
[1] 10

zcross2 <- zeroCross(data_2$temp_smoothed, slope = 'positive')
length(zcross2)
[1] 11

#join the two vectors as a data.frame (using only the first 10 crossing points for wave2 to avoid any issues of mismatched lengths)
zcrossings <- as.data.frame(cbind(zcross1, zcross2[1:10]))

#calculate the mean of the crossing point differences
mean(zcrossings$zcross1 - zcrossings$V2)
[1] 49

I'm sure there are more eloquent ways of going about this, but it should get you the information that you need.

scribbles
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0

In my case, for the tidal wave in semidiurnal, 90 degree equal to 3 hours (90*12.5 hours/360 = 3.125 hours). 12.5 hours is the period of semidiurnal. So, for 45 degree equal to -> 45*12.5/360 = 1.56 hours.

Thus in your case: 90 degree -> 90*365/360 = 91.25 hours. 45 degree -> 45*365/360= 45.625 hours.

meddy
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