# Missing delay in heavyside step function

I found the following task that was inspired by an example in the book A. V. Oppenheim and R. W. Schafer, "Discrete-Time Signal Processing", 3rd Edition, 2014.

Consider the 2nd-order IIR filter $$H(z) = \frac{1}{1 - 2\cos(\theta)z^{-1} + z^{-2}}\,.$$

Show that the impulse response $$h[n] = \frac{\sin(\theta(n+1))}{\sin(\theta)} u[n] \, ,$$ with $$u[n]$$ the heavyside step function.

## My solution:

I started from $$\sin(\theta n) u[n] \leftrightarrow \frac{\sin(\theta)z^{-1}}{1-2\cos(\theta) z^{-1} + z^{-2}} \, ,$$ which I looked up in a table and divided both sides by $$\sin(\theta)$$, i.e. $$\frac{\sin(\theta n)}{\sin(\theta)} u[n] \leftrightarrow \frac{z^{-1}}{1-2\cos(\theta) z^{-1} + z^{-2}} \, .$$ Now I just had to get rid of $$z^{-1}$$, hence I multiplied with $$z$$ in z-domain, which corresponds to a convolution with $$\delta[n+1]$$ in time-domain, i.e. $$\frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1] \leftrightarrow \frac{1}{1-2\cos(\theta) z^{-1} + z^{-2}} \, .$$

Comparing my solution with $$h[n]$$ in the task above, the only difference is the delayed function $$u[n]$$.

Does anybody see what I might did wrong or what properties can be exploited here?

We know that $$\displaystyle \frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1]$$ has value $$0$$ for $$n < -1$$ since $$u[n+1]=0$$ for $$n < -1$$. At $$n=-1$$, $$u[n+1]$$ jumps to value $$1$$, but $$\sin(\theta (n+1))\bigr|_{n=-1} = \sin(\theta (-1+1))$$ has value $$0$$, and so $$\displaystyle \frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1]$$ just happens to equal $$\displaystyle \frac{\sin(\theta (n+1))}{\sin(\theta)} u[n]$$.
No need to convolve with $$\delta[n+1]$$ (or subtract $$\delta[n+1]$$ as Ch. Siva Ram Kishore suggests, why I don't know).