# Find a band-pass filter

The question is: how can I define $$h_1[n]$$ in such a way that $$h [ n ] = \delta [n - 1 ] + 2 \delta [n -2 ] + h_ 1 [n]$$ is a band-pass filter. My thought was the following.

Firstable, I wrote the $$Z$$-transform as $$\displaystyle{H (z) = H_ 1 ( z) + z^ {-1} +2z^ {-2} = \frac{z^ 2 H_1 ( z) + z + 2 }{z ^ 2 }}$$. Then I thought, since the sistem basically has two poles in 0, to add two poles, in $$z = 1, -1$$ (in order to have the amplitude of $$H( e^ {i\omega})$$ vanishing for $$\omega = 0, \pi$$). I decided to take a signal of the form $$h _1 [n] = A\delta [n + 1 ]+ B \delta [n+2 ] \implies H _1(z) = A z + B {z ^2 }$$ and I got, by the previous conditions, a system of equations: $$\begin{equation*} \begin{cases} H ( -1 ) = H_1( -1 )+( -1)^ {-1} + 2 (-1 ) ^ 2 = -A+ B +1= 0 \\ H (1) = H _1(1) +(1)^ {-1} +2(1 ) ^ 2 = A + B +3 = 0 \end{cases} \end{equation*}$$ so that $$A= -1, B = -2$$. Therefore I concluded that $$h[n] =- \delta [n+1 ] - 2\delta [n+2] + \delta [n-1] + 2 \delta[n-2]$$ is a band-pass filter. I draw the graph of the amplitude and it seems to be a kind of absolute value of a damping sinusoidal function (with zeros for $$\omega = 0, \omega _ 1, \pi$$, where $$\omega _ 1$$ is a value between $$\frac \pi 2$$ and $$\frac 2 3 \pi$$).

I was wondering if my argument is correct. Any help would be appreciated, thanks in advance.

You seem to making this more complicated than it needs to be. If you want $$h[n] = h_0[n] + h_1[n]$$ to be a bandpass, than $$h_1[n]$$ is simply
$$h_1[n] = h[n] - h_0[n]$$
In other words: design any bandpass $$h[n]$$ you like, subtract 1 from the first coefficient and 2 from the second coefficient, and you have your $$h_1[n]$$