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Given is the illustrated circuit diagram of a linear, time-invariant, time-discrete system:

enter image description here

How do I show that the total system has the impulse response $h[n] = aδ[n] + bδ [n - 1] + cδ [n - 2]$ and determine the constants a,b and c?

with the following impulse responses of the subsystems:

$h_1[n] = δ[n] + (1/2)δ[n − 1]$

$h_2[n] = δ[n] − (1/2)δ[n − 1]$

$h_3[n] = −δ[n]$

$h4[n] = − (1/2)^n σ[n]$

So what I have done:

$h_{3||4}=−δ[n]−(1/2)^n σ[n]$

and than:

$h_{3||4+2}=\sum_{k=-\infty}^{\infty}(δ[n-k] − (1/2)δ[n-k − 1])(−δ[n]−(1/2)^n σ[n])$

$h_{3||4+2}=−δ[n]−(1/2)^n σ[n]-\frac{1}{2}(−δ[n-1]−(1/2)^{n-1} σ[n-1])=−δ[n]-\frac{1}{2}^nδ[n]+\frac{1}{2}δ[n-1]$

$h_{3||4+2+1}=(\frac{1}{2})^nδ[n]-δ[n-1]$

But the solution is $a=1$ and $b=-1$. How is $a=1$?

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  • $\begingroup$ $((h_3+h_4) \circledast h2) + h_1$? $\endgroup$ – A_A Apr 21 '18 at 13:15
  • $\begingroup$ What is $\sigma [n]$ ? $\endgroup$ – Hilmar Apr 21 '18 at 13:18
  • $\begingroup$ @Hilmar it's unit step function $\endgroup$ – Alena Apr 21 '18 at 14:41
  • $\begingroup$ @A_A Sum of h3 and h4 convolution with h2 and than sum with h1 :D $\endgroup$ – Alena Apr 21 '18 at 14:41
  • $\begingroup$ Indeed, just double checking that that's what your intention is because the last bit there seems to imply "3,4 parallel, 2 serial, 1 serial" (?) $\endgroup$ – A_A Apr 21 '18 at 14:49
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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

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

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Working in z-domain makes the problem easy

$$ H_{1}(z) = 1 + \frac{1}{2}z^{-1} = \frac{2z+1}{2z}$$ $$ H_{2}(z) = 1 - \frac{1}{2}z^{-1} = \frac{2z-1}{2z}$$ $$ H_{3}(z) = -1$$ $$ H_{4}(z) = \frac{-z}{z - \frac{1}{2}} = \frac{-2z}{2z-1}$$

$$ H_{3}(z) + H_{4}(z) = \frac{-4z + 1}{2z + 1} $$ $$ H_{2}(H_{3} + H_{4}) = (\frac{2z-1}{2z})(\frac{1-4z}{2z-1}) = \frac{1-4z}{2z}$$

$$ H(z) = H_{1} || H_{2}(H_{3} + H_{4}) = \frac{2z+1}{2z} + \frac{1-4z}{2z} =\frac{1-z}{z} $$ $$ H(z) = z^{-1} - 1 $$ $$ \implies h[n] = -\delta[n] + \delta[n-1] $$ Hence, $a = -1, b = 1, c = 0$

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