A digital filter with input x[k] and output y[k] is described by the difference equation: 

$$y[k]=1/6(3x[k]+2x[k-1]+x[k-2])$$

The discrete-time transfer function of the filter H(z) is:

A.   $$(6z^2)/(3z^2+2z+1)$$
B.   $$(6z)/(3z^2+2z+1)$$
C.   $$3z^2+2z+1$$
D.   $$1/6[(3z^2+2z+1)/z^2]$$

 

In order to find the discrete-time transfer function from the difference equation, we must convert our function into the Z-domain using Z-transforms. Let's first move our $\frac{1}{6}$ factor from the right side of the equal sign to the left before transforming our function:

$$\begin{gathered} y\left[k\right] = \frac{1}{6}\left(3x\right[k]+2x[k-1]+x[k-2\left]\right) \\ 6y\left[k\right]=3x\left[k\right]+2x[k-1]+x[k-2] \end{gathered}$$

Now we are ready to convert to our Z domain. We will use the following three properties to do so:

$$\begin{gathered} Property \#1: Z \left[y\right[n\left]\right] = Y\left(z\right) ORZ\left[x\right[n\left]\right] = X\left(z\right) \\ Property \#2: Z\left[ax\right[n]+b[y\left[n\right]] = aX(z) + bY(z) \\ Property \#3: Z[x[n-n_0]] = z^{-n_0}X(z) \end{gathered}$$

Using these three properties, let's now convert to the Z domain:

$$\begin{gathered} Z\left[6y\right[k\left]\right] = Z\left[3x\right[k]+2x[k-1]+x[k-2\left]\right] \\ 6Y\left(z\right) = 3X\left(z\right) +2z^{-1}X\left(z\right)+z^{-2}X\left(z\right) \end{gathered}$$

Since our transfer function, $H\left(z\right)$ is as follows:

$H\left(z\right) = \frac{Y\left(z\right)}{X\left(z\right)}$

Let's put our function in the Z domain into this form:

$$\begin{gathered} 6Y\left(z\right) = 3X\left(z\right) +2z^{-1}X\left(z\right)+z^{-2}X\left(z\right) \\ 6Y\left(z\right) = X\left(z\right)(3+2z^{-1}+z^{-2}) \\ \frac{Y\left(z\right)}{X\left(z\right)}=\frac{3+2z^{-1}+z^{-2}}{6} \\ H\left(z\right) = \frac{3+2z^{-1}+z^{-2}}{6} \end{gathered}$$

We can now multiply each side by $\frac{z^2}{z^2}$ to eliminate our negative exponents and get our answer in its simplified form:

$$\begin{gathered} \left(\frac{z^2}{z^2}\right)H\left(z\right) = \frac{3+2z^{-1}+z^{-2}}{6}\left(\frac{z^2}{z^2}\right) \\ H\left(z\right) = \frac{3z^2+2z+1}{6z^2} \end{gathered}$$

This corresponds to answer choice D.