Processing delay window-function

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.)

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.

R Code Only Below

#41046

require(signal)
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]
}
else
{
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))

• 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... – alb084 May 17 '17 at 20:01
• @alb084 Really? Calculate the cumulative sum for different values of $M$ and see. – Peter K. May 17 '17 at 20:06