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

$$y[n]=1/2x[n]+1/3y[n-1]$$

The impulse response for this filter is most nearly:

A.   $$h[n]=2delta[n]-2/3delta[n-1]$$
B.   $$h[n]=1/2delta[n]-1/3delta[n-1]$$
C.   $$h[n]=1/2(1/3)^n, n>=0$$
D.   $$h[n]=1/2(-1/3)^n, n>=0$$

Let's first start off solving for our solution by placing all y terms on one side and x terms on the other:

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

Let's now transform our function from the time to frequency domain, using the properties of Z transforms as shown below:

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

Using these properties:

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

We know that our transfer function is H(z) as represented in the equation below, so let's convert our function into this form to then solve for the transfer function:

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

Now that we have our transfer function, we want to convert our transform function back to the time domain in order to arrive at our impulse response function. We can see we have a common transform in our expression, as shown in this table:

Using this transform pair and converting back to the time domain:

$$\begin{gathered} H\left(z\right) = \frac{1}{2}\frac{1}{1-{\displaystyle \frac{1}{3}}{z}^{-1}} \\ h\left[n\right] =\frac{1}{2}\left(\frac{1}{3}u\right[n\left]\right), n\ge 0 \end{gathered}$$

Therefore our final answer is C.