In some DSP slides, I've noted that when we try to choose a window for truncating a signal, sometimes a window is chosen that minimizes the processing delay (in my example there were a rectangular, Hamming and Bartlet)...but what is exactly this processing delay?

The minimum process delay I would have to have with Rectangular window?

Sorry, but why is Bartlet? I thought that the minimum phase was proportional at the width of the main lobe (I've resolved thanks, I did not remember that energy end amplitude is related.)


1 Answer 1


One interpretation of processing delay is that the causal window, $w[n]$ has more energy close to $n=0$ than any other window, $w_o[n]$ with the same frequency response (see section 5.8.4).

So that $$ \sum_{n=0}^M |w_o[n]|^2 \le \sum_{n=0}^M |w[n]|^2 $$ for all integer $M \ge 0$ .

So if I plot these sums for the three windows, you can see that the Hamming window (barely) gives the least processing delay compared with the Bartlett, and the rectangular is worst.

enter image description here

R Code Only Below


N <- 128

r <- rep(1,N)
h <- hamming(N)
b <- bartlett(N)

r <- r/sqrt(sum(r*r))
h <- h/sqrt(sum(h*h))
b <- b/sqrt(sum(b*b))

#r <- r/sum(r)
#h <- h/sum(h)
#b <- b/sum(b)

rd <- rep(0,N)
hd <- rep(0,N)
bd <- rep(0,N)

for (idx in seq(1,N))
  if (idx != 1)
    rd[idx] <- rd[idx-1] + r[idx]*r[idx]
    hd[idx] <- hd[idx-1] + h[idx]*h[idx]
    bd[idx] <- bd[idx-1] + b[idx]*b[idx]
    rd[idx] <- r[idx]*r[idx]
    hd[idx] <- h[idx]*h[idx]
    bd[idx] <- b[idx]*b[idx]

plot(rd, col="blue", type ="l", ylab="delay", ylim=c(0,max(c(rd,hd,bd))))
lines(hd, col="red")
lines(bd, col="black")

legend(0, 1.0, c("rectangular", "hamming", "bartlett"), col = c("blue", "red", "black"), lty = c(1, 1, 1))
  • $\begingroup$ sorry...but in that case than what exactly would be better between rettangular, hamming and bartlet for minimizing the processing delay?...because if i look this property..i could think that all windows could be suitable... $\endgroup$
    – alb084
    May 17, 2017 at 20:01
  • $\begingroup$ @alb084 Really? Calculate the cumulative sum for different values of $M$ and see. $\endgroup$
    – Peter K.
    May 17, 2017 at 20:06

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