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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:

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

But problem is that I don't know how to calculate this and than to do it in parellel with $h_1$ to get this form $h[n] = aδ[n] + bδ [n - 1] + cδ [n - 2]$ ?

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Since this is a homework type problem I will not present the complete solution, but I'll help you to solve it yourself.

First of all, don't write out the convolution as an infinite sum, but simply use the fact that for any sequence $x[n]$ the following holds:

$$x[n]\star\delta[n-k]=x[n-k]\tag{1}$$

The lower path of the given system has the impulse response

$$h_2[n]\star(h_3[n]+h_4[n])=h_2[n]\star h_3[n]+h_2[n]\star h_4[n]=h_a[n]+h_b[n]\tag{2}$$

and if you use $(1)$ to evaluate $h_a[n]$ and $h_b[n]$ in $(2)$ you'll quickly see that both are (very simple) finite impulse response.

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The magic of the impulse response is that you can find the response... with an impulse. So first you can input a delta function as $x$, and see manually what happen.

But you can also observe that the superposition principle applies: you can redraw the diagram with three parallel branches whose respective filters are:

  • $h_1$
  • $h_2 \star h_3$
  • $h_2 \star h_4$

with the same input (a delta), that all add at the end:

$$ h = h_1 + h_2 \star h_3 + h_2 \star h_4\,.$$

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