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First, you can rewrite the thing as

$$h[n] = h_1[n] + h_2[n] \ast (h_3[n] + h_4[n]) = h_1[n] + h_2[n] \ast h_3[n] + h_2[n] \ast h_4[n]$$

So you basically have three parallel impulse responses that you need to add up. The first two are really simple:

$$ h_1[n] = \begin{bmatrix} 1 & .5 & 0 &0 & ... \end{bmatrix} $$ $$ h_2[n] \ast h_3[n] = \begin{bmatrix} -1 & .5 & 0 &0 & ... \end{bmatrix} $$

$h_2[n] \ast h_4[n]$ is a bit more tricky since $h_4[n]$ is actually an IIR filter. This easiest done in the Z-domain. $h_4$ has a pole at z = 0.5 and $h_2$ has corresponding zero at z = 0.5 as well. The simply cancel each other $$H_2(z) \cdot H_4(z) = \frac{1-0.5\cdot z^{-1}}{1} \cdot \frac{1}{1-0.5\cdot z^{-1}} = 1$$

$$H_2(z) \cdot H_4(z) = \frac{1-0.5\cdot z^{-1}}{1} \cdot \frac{-1}{1-0.5\cdot z^{-1}} = -1$$

and we get

$$ h_2[n] \ast h_4[n] = \begin{bmatrix} 1 & 0 & 0 &0 & ... \end{bmatrix} $$$$ h_2[n] \ast h_4[n] = \begin{bmatrix} -1 & 0 & 0 &0 & ... \end{bmatrix} $$

Summing to all up yields

$$ h[n] = \begin{bmatrix} 1 & 1 & 0 &0 & ... \end{bmatrix} $$$$ h[n] = \begin{bmatrix} -1 & 1 & 0 &0 & ... \end{bmatrix} $$

First, you can rewrite the thing as

$$h[n] = h_1[n] + h_2[n] \ast (h_3[n] + h_4[n]) = h_1[n] + h_2[n] \ast h_3[n] + h_2[n] \ast h_4[n]$$

So you basically have three parallel impulse responses that you need to add up. The first two are really simple:

$$ h_1[n] = \begin{bmatrix} 1 & .5 & 0 &0 & ... \end{bmatrix} $$ $$ h_2[n] \ast h_3[n] = \begin{bmatrix} -1 & .5 & 0 &0 & ... \end{bmatrix} $$

$h_2[n] \ast h_4[n]$ is a bit more tricky since $h_4[n]$ is actually an IIR filter. This easiest done in the Z-domain. $h_4$ has a pole at z = 0.5 and $h_2$ has corresponding zero at z = 0.5 as well. The simply cancel each other $$H_2(z) \cdot H_4(z) = \frac{1-0.5\cdot z^{-1}}{1} \cdot \frac{1}{1-0.5\cdot z^{-1}} = 1$$

and we get

$$ h_2[n] \ast h_4[n] = \begin{bmatrix} 1 & 0 & 0 &0 & ... \end{bmatrix} $$

Summing to all up yields

$$ h[n] = \begin{bmatrix} 1 & 1 & 0 &0 & ... \end{bmatrix} $$

First, you can rewrite the thing as

$$h[n] = h_1[n] + h_2[n] \ast (h_3[n] + h_4[n]) = h_1[n] + h_2[n] \ast h_3[n] + h_2[n] \ast h_4[n]$$

So you basically have three parallel impulse responses that you need to add up. The first two are really simple:

$$ h_1[n] = \begin{bmatrix} 1 & .5 & 0 &0 & ... \end{bmatrix} $$ $$ h_2[n] \ast h_3[n] = \begin{bmatrix} -1 & .5 & 0 &0 & ... \end{bmatrix} $$

$h_2[n] \ast h_4[n]$ is a bit more tricky since $h_4[n]$ is actually an IIR filter. This easiest done in the Z-domain. $h_4$ has a pole at z = 0.5 and $h_2$ has corresponding zero at z = 0.5 as well. The simply cancel each other

$$H_2(z) \cdot H_4(z) = \frac{1-0.5\cdot z^{-1}}{1} \cdot \frac{-1}{1-0.5\cdot z^{-1}} = -1$$

and we get

$$ h_2[n] \ast h_4[n] = \begin{bmatrix} -1 & 0 & 0 &0 & ... \end{bmatrix} $$

Summing to all up yields

$$ h[n] = \begin{bmatrix} -1 & 1 & 0 &0 & ... \end{bmatrix} $$

    Post Undeleted by Hilmar
    Post Deleted by Hilmar
1
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First, you can rewrite the thing as

$$h[n] = h_1[n] + h_2[n] \ast (h_3[n] + h_4[n]) = h_1[n] + h_2[n] \ast h_3[n] + h_2[n] \ast h_4[n]$$

So you basically have three parallel impulse responses that you need to add up. The first two are really simple:

$$ h_1[n] = \begin{bmatrix} 1 & .5 & 0 &0 & ... \end{bmatrix} $$ $$ h_2[n] \ast h_3[n] = \begin{bmatrix} -1 & .5 & 0 &0 & ... \end{bmatrix} $$

$h_2[n] \ast h_4[n]$ is a bit more tricky since $h_4[n]$ is actually an IIR filter. This easiest done in the Z-domain. $h_4$ has a pole at z = 0.5 and $h_2$ has corresponding zero at z = 0.5 as well. The simply cancel each other $$H_2(z) \cdot H_4(z) = \frac{1-0.5\cdot z^{-1}}{1} \cdot \frac{1}{1-0.5\cdot z^{-1}} = 1$$

and we get

$$ h_2[n] \ast h_4[n] = \begin{bmatrix} 1 & 0 & 0 &0 & ... \end{bmatrix} $$

Summing to all up yields

$$ h[n] = \begin{bmatrix} 1 & 1 & 0 &0 & ... \end{bmatrix} $$