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We have $Y(k)=x(k-1)+ kx(k-5)+x(k)^4$ .I have to find the impulse response for the function

So I know that $G(z) = \frac{Y(z)}{X(z)}$ but how do I relate that to this?

$Y(z)=(z^{-1}) + k(z^{-5})+ z^4$

mathreadler
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Xizi
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1 Answers1

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The system with input $x(k)$ and output $y(k) = x(k-1)+kx(k-5)+x(k)^4$ is not a linear system, so the usual transfer function methods don't apply.

However, we can still find the impulse response, which is the output $y(k)$ when the input is $x(k) = \delta(k) = \begin{cases}1 & \text{if} \ k = 0 \\ 0 & \text{if} \ k \neq 0\end{cases}$.

Since $x(k) = 0$ for every value of $k$ except $0$, $y(k) = x(k-1)+kx(k-5)+x(k)^4 = 0$ for every value of $k$ except $1$, $5$, and $0$.

Can you evaluate $y(1)$, $y(5)$, and $y(0)$?

JimmyK4542
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  • y(1)=0,Y(5)=0 and Y(0)=1? – Xizi Sep 01 '15 at 13:10
  • ^No that's not quite right. Plug $k = 1$ into $y(k) = x(k-1)+kx(k-5)+x(k)^4$ and evaluate the left side. Remember that $x(0) = 1$ and $x(k) = 0$ for any number $k \neq 0$. Then repeat this for $k = 5$ and $k = 0$. – JimmyK4542 Sep 01 '15 at 13:12
  • so y(5) for example is 1 ,y(o) is 1 and y(1) is 1 – Xizi Sep 01 '15 at 13:20
  • I found these, but whats the impulse response? :( – Xizi Sep 01 '15 at 13:22
  • That's almost right. Check $y(5)$ again. The impulse response is the output of the system when a unit impulse $\delta(k)$ is applied as the input. In this case, if we apply $x(k) = \delta(k)$ to the input, then the output $y(k)$ is the impulse response. From what I said in my post, we know that the impulse response is zero at all times $k$ except $1$, $5$, and $0$. Now, you just need to find the impulse response at these times. – JimmyK4542 Sep 01 '15 at 16:18