2 added 5 characters in body edited Apr 23 '18 at 17:06 Hilmar 11.4k1212 silver badges1818 bronze badges 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 occurred Apr 23 '18 at 17:04 Post Deleted by Hilmar occurred Apr 23 '18 at 16:24 1 answered Apr 23 '18 at 16:19 Hilmar 11.4k1212 silver badges1818 bronze badges 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 $$h[n] = \begin{bmatrix} 1 & 1 & 0 &0 & ... \end{bmatrix}$$