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?


1 Answer 1


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).

  • $\begingroup$ Beautiful! I did not recognize this coincidence, thank you. $\endgroup$
    – Gsslng
    Commented Jun 6, 2019 at 16:19

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