Impulse response of recursive DT system

[dionysian]

Homework Statement

$$y[n] - 1.8\cos (\frac{\pi }{{16}})y[n - 1] + 0.81y[n - 2] = x[n] + 0.5x[n - 1]$$
Determine the impulse response $$h[n]$$ by calculating the zero-state response with $$x[n] = \delta [n]$$

Homework Equations

$$y[n] - 1.8\cos (\frac{\pi }{{16}})y[n - 1] + 0.81y[n - 2] = x[n] + 0.5x[n - 1]$$

The Attempt at a Solution

$$y[n] - 1.8\cos (\frac{\pi }{{16}})y[n - 1] + 0.81y[n - 2] = x[n] + 0.5x[n - 1]$$
$$y[n] = 1.8\cos (\frac{\pi }{{16}})y[n - 1] - 0.81y[n - 2] + x[n] + 0.5x[n - 1]$$
$$h[n] = 1.8\cos (\frac{\pi }{{16}})y[n - 1] - 0.81y[n - 2] + \delta [n] + 0.5\delta [n - 1]$$
The zero-state response means
$$\begin{array}{l} y[- 1] = 0 \\ y[- 2] = 0 \\ \end{array}$$
So my question is how can I write out the impulse response explicitly in one expression. I think I could calculate it by $$h[0],h[1],h[2]...$$ but I want to write it out as just one expression but the recursive terms$$y[n - 1]$$ and $$y[n - 2]$$ are kind of throwing me off.

 

[KataKoniK]

To write the impulse response explicitly in one expression, you can use the z-transform. The z-transform of the given equation is:

Y(z) - 1.8cos(\frac{\pi}{16})z^{-1}Y(z) + 0.81z^{-2}Y(z) = X(z) + 0.5z^{-1}X(z)

Using the property of the z-transform that relates the output to the input for a linear time-invariant system, we get:

H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1-1.8cos(\frac{\pi}{16})z^{-1} + 0.81z^{-2}} + \frac{0.5z^{-1}}{1-1.8cos(\frac{\pi}{16})z^{-1} + 0.81z^{-2}}

Using partial fraction decomposition, we can write the above expression as:

H(z) = \frac{A}{1-\alpha z^{-1}} + \frac{B}{1-\beta z^{-1}}

where A and B are constants and \alpha and \beta are the roots of the denominator polynomial. In this case, \alpha = 0.9 and \beta = 0.9. Solving for A and B, we get:

A = \frac{1}{1-0.9} = 10
B = \frac{0.5}{1-0.9} = 5

Therefore, the impulse response in one expression is:

h[n] = 10(0.9)^n + 5(0.9)^n u[n-1]