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Consider the filter which equation can be represented by $y[n] = \frac{1}{3}x[n] + \frac{1}{3}x[n-1] + \frac{1}{3}x[n-2]$, in $x[n]$ and $y[n]$ are sequence of input and output of the system respectively.

Mark the correct option about the sequence ${h[n]}$ filter response with respect the unit pulse ($\delta_0[n] = 1$) in $n=0$.

A) h[1] = 2/3

B) h[1] + h[5] = 1/3

C) h[0] = 0

D) h[2] = 3

E) h[2] + h[3] = 1

My attempt:

$Y(z) = \frac{1}{3}\cdot X(z)(1+z^{-1} + z^{-2})$

$1+z^{-1} + z^{-2}$ its a Finite Filter Response right?

$X(z) = 1$

$Y(z) = H(z) \Rightarrow H(z) = \frac{1}{3}(1+z^{-1} +z^{-2})$

I stuck here because and really cant realize to envolve the discrete domain with this expression on Z domain.

Any hints?

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As the name implies, you get the impulse response by using the an impulse as the input signal. Hence $$h[n] = \frac{1}{3} \cdot \delta[n] + \frac{1}{3} \cdot \delta[n-1] + \frac{1}{3} \cdot \delta[n-2]$$

This means $h[n]$ is only non zero for $n=0,1,2$ and specifcally $$h[0] = \frac{1}{3}, h[1] = \frac{1}{3}, h[2] = \frac{1}{3} $$

With that, you work through the answers one by one and B is indeed the only correct one. $h[1] = \frac{1}{3}$ and $h[5] = 0$ so the sum $h[1]+h[5]$ is indeed $\frac{1}{3}$

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  • $\begingroup$ Omg. Thank you so much. I wish you would give me some books or links about that specific topic. It would be very useful to me. One more time you've saved my day! $\endgroup$ – miguel747 Jun 20 '18 at 1:45
  • $\begingroup$ ccrma.stanford.edu/~jos/filters, dspguide.com are both free and very good introductions. $\endgroup$ – Hilmar Jun 20 '18 at 13:38
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You can see by definition that your system is LTI System.
Moreover it is casual as it is only depends on input from the past.

The quaestion asks for response on the Unit Pulse signal on time 0.
In LTI Casual System it is the coefficient which responds to to input at time $ n $ -> Its value is $ \frac{1}{3} $.

I'm not sure where $ h \left[ 1 \right] + h \left[ 5 \right] $ is from though.
As clearly this system can be represented by 3 Coefficients which you found in Z Domain (Though they are also given from the finite differences equation).

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