How to calculate impulse response from frequency response whose denominator has an imaginary roots

Question

It is quite easy to solve when the denominator of the frequency response has no imaginary values..... But I do I solve something like this?

$$H(\omega)=\frac{1+2e^{-2j\omega}}{1-\frac{1}{15}e^{-j\omega}+\frac{1}{5}e^{-2j\omega}}$$

Answer 1

$$H(\omega)=\frac{1+2e^{-2j\omega}}{1-\frac{1}{15}e^{-j\omega}+\frac{1}{5}e^{-2j\omega}}$$

Assuming this is the frequency response of a discrete-time system, which is the transfer function with $z=e^{-j\omega}$ for $\omega \in [0, 2\pi)$, then for that case specifically we can deduce the transfer function to be:

$$H(z) = \frac{Y(z)}{X(z)} =\frac{1+2z^{-2}}{1-\frac{1}{15}z^{-1}+ \frac{1}{5}z^{-2}}$$

From which we get:

$$Y(z)(1-\frac{1}{15}z^{-1}+ \frac{1}{5}z^{-2}) = X(z)(1+2z^{-2})$$

And from this with the inverse z-transform we can get the impulse response (unit sample response).

Answer 2

@Dan_Boschen has already given an answer to recover the difference equation that includes feedback terms, which is all one really needs for a practical implementation of the filter.

However, if one wants $h[n]$ explicitly, then it is easiest to put the expression in a form where tables of Z-Transforms can be used to find the inverse Z-transform.

So let's use this table: https://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html

We can see from the given $H(e^{j\omega})$ and the Z-Transform table, that some combination of the Z-Transforms of $\cos(k\omega_0T)$ and $\sin(k\omega_0T)$ is likely what we'll need to come up with.

$$\begin{align} H(z) &= \dfrac{1+2z^{-2}}{1-\dfrac{1}{15}z^{-1}+\dfrac{1}{5}z^{-2}}\\ \\ &=5 \cdot \dfrac{1+2z^{-2}}{1-2z^{-1}\dfrac{1}{6}+z^{-2}}\\ \\ &=5 \cdot \dfrac{z^2+2}{z^2-2z\dfrac{1}{6}+1}\\ \\ \text{Let} \cos\left(\omega_0T\right) &= \dfrac{1}{6} \quad\text{and}\quad \sin\left(\omega_0T\right) = +\dfrac{\sqrt{35}}{6}\\ \\ H(z) &= 5 \cdot \dfrac{z^2+2}{z^2-2z\cos\left(\omega_0T\right)+1}\\ \\ H(z) &= 5 \cdot \dfrac{z^2-z\cos\left(\omega_0T\right)}{z^2-2z\cos\left(\omega_0T\right)+1} + 5 \cdot \dfrac{z\cos\left(\omega_0T\right)+2}{z^2-2z\cos\left(\omega_0T\right)+1}\\ \\ &= 5 \cdot \dfrac{z\left(z-\cos\left(\omega_0T\right)\right)}{z^2-2z\cos\left(\omega_0T\right)+1} + 5 \cdot \dfrac{z\cos\left(\omega_0T\right)}{z^2-2z\cos\left(\omega_0T\right)+1}+5 \cdot \dfrac{2}{z^2-2z\cos\left(\omega_0T\right)+1}\\ \\ &= 5 \cdot \dfrac{z\left(z-\cos\left(\omega_0T\right)\right)}{z^2-2z\cos\left(\omega_0T\right)+1} + 5 \dfrac{\cos\left(\omega_0T\right)}{\sin\left(\omega_0T\right)}\cdot \dfrac{z\sin\left(\omega_0T\right)}{z^2-2z\cos\left(\omega_0T\right)+1}+5 \cdot \dfrac{2}{z^2-2z\cos\left(\omega_0T\right)+1}\\ \\ &= 5 \cdot \dfrac{z\left(z-\cos\left(\omega_0T\right)\right)}{z^2-2z\cos\left(\omega_0T\right)+1} + 5 \dfrac{\cos\left(\omega_0T\right)}{\sin\left(\omega_0T\right)}\cdot \dfrac{z\sin\left(\omega_0T\right)}{z^2-2z\cos\left(\omega_0T\right)+1}+\dfrac{10}{\sin\left(\omega_0T\right)} \cdot\dfrac{1}{z} \dfrac{z\sin\left(\omega_0T\right)}{z^2-2z\cos\left(\omega_0T\right)+1}\\ \end{align}$$

So, taking inverse Z-Transforms, the explicit expression for the impulse response is

$$h[k] = 5 \cos\left(k\cdot\omega_0T\right) + 5\cdot\dfrac{1}{\sqrt{35}}\sin\left(k\cdot\omega_0T\right) +10\cdot\dfrac{6}{\sqrt{35}}\sin\left([k-1]\cdot\omega_0T\right) $$

where $\omega_0T = \mathrm{Arccos}\dfrac{1}{6}$.