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According to Discrete Time Signal Processing by Al Oppenheim 3rd Edition, the impulse response of a moving average filter is as below
My question is that shouldn't h[n] be having the dirac delta function instead of the unit function u[n] (See Eq 2.9).
This is mentioned in example 2.3 of the same book
Taking Equation $2.74$ from Example $3$ and setting $M_1=0$ gives:
$$ h[n] = \frac{1}{M_2+1} \sum_{k=0}^{M_2} \delta[n-k] $$
Now lets take a closer look at this: $h[n]$ is non-zero only for certain values of $n$. Its a good a starting point as any, so lets look at $n=0$. We have:
$$ h[0] = \frac{1}{M_2+1} \bigg( \delta[0] + \delta[-1] + ... + \delta[-M_2] \bigg) = \frac{1}{M_2+1}$$
For $n=1$, we have:
$$ h[1] = \frac{1}{M_2+1} \bigg( \delta[1] + \delta[0] + ... + \delta[1-M_2] \bigg) = \frac{1}{M_2+1}$$
For $n=2$, we have:
$$ h[2] = \frac{1}{M_2+1} \bigg( \delta[-2] + \delta[-1] + \delta[0]+ ... + \delta[2-M_2] \bigg) = \frac{1}{M_2+1}$$
And this pattern continues, the only non-zero value, $\delta[0]$, percolates through until $n=M_2$, then we have:
$$h[M_2] = \frac{1}{M_2+1} \bigg( \delta[-M_2] + \delta[-M_2+1] + ... + \delta[0] \bigg) = \frac{1}{M_2+1}$$
And for $n = M_2 + 1$:
$$ h[M_2+1] = \frac{1}{M_2+1} \bigg( \delta[-M_2-1] + \delta[-M_2] + ... + \delta[-1] \bigg) = 0$$
And you can see, for all $n > M_2$, $h[n]=0$. So, altogether we can write $h[n]$ as equal to $\frac{1}{M_2+1}$ for $n \in [0, M_2]$ and zero otherwise.
All Equation $2.90$ does is treats the summation of impulses (from Equation $2.74$) as unit step functions, a very useful trick! Since $h[n]$ starts at $n=0$ and ends at $n=M_2$ we do: $u[n]-u[n-M_2-1]$ to get a length $M_2$ box car, then scale it by $\frac{1}{M_2+1}$ to get the right amplitude. Hope this helps!
