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I am given the following discrete time transfer function : $$G_d(z^{-1})=z^{-d}\frac{b_0+b_1z^{-1}}{1+a_1z^{-1}}$$ which has the following impulse response $$g_d[n]=\{0,1,-0.1,-0.05,...\}$$

How can I find the transfer function?($b_0$, $b_1$ etc)

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  • $\begingroup$ are you asking, how do I find $b_0$ $b_1$, and $a_1$? from the given impulse response? $\endgroup$ – Stanley Pawlukiewicz Jun 23 '17 at 21:56
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$$\begin{align} G_d(z^{-1})&=z^{-d}\frac{b_0+b_1z^{-1}}{1+a_1z^{-1}}\\ &=z^{-d}\frac{b_0}{1+a_1z^{-1}}+z^{-(d+1)}\frac{b_1}{1+a_1z^{-1}} \end{align}$$ Hence, considering

$$\mathcal{Z}^{-1}\{\frac{1}{1-az^{-1}}\}=a^nu[n]$$

and

$$\mathcal{Z}^{-1}\{z^{-n_0}X(z)\}=x[n-n_0]$$

we have $$g_d[n]=b_0a_1^{(n-d)}u[n-d]+b_1a_1^{n-(d+1)}u[n-(d+1)]$$ Now the unknowns can be found by equating the terms apropriately.

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