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So I have the following equation $$ y[n]=0.9y[n-1]+0.1x[n] $$

We can find easily find the transfer function, which is

$$ H(z)=\frac{0.1}{1-0.9z^{-1}} $$

and from that, the frequency response, which is

$$ H(e^{j \omega})=\frac{0.1}{1-0.9 e^{-j \omega}} $$

To find the cutoff frequency we just make the magnitude of the above equal to $\frac{1}{\sqrt{2}}$. Then, $\omega_0 = 0.0675\Rightarrow f_0 = 0.0107$.

Okay, so how can we say that given $f_0 = f_c$ the filter is a LPF or HPF?

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    $\begingroup$ Does the frequency response increase or decrease as the frequency increases? $\endgroup$ – MBaz Nov 17 '17 at 2:30
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Types of filters, such as LPF, HPF, BPF or BSF, are not described by their cutoff frequencies $f_c$ but through their passband or stopband frequency $\omega$ region.

For example, in your case of the filter $H(z) = 1 / (1 - 0.9 z^{-1})$, the passband is in the low frequencies (where the magnitude of the frequency response $H(e^{j\omega})$ is greater than $1/\sqrt{2}$. And by your definition that region corresponds to $\omega \in (-f_c , f_c)$ which indicates a lowpass range of frequencies.

Other filters like HPF and BPF follow in a similar manner. Note that specific filters such as comb or notch types are defined in a different style than this.

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It's actually the Z Transform

\begin{equation} H(z) = \frac{0.1}{1-0.9z^{-1}} \end{equation}

Now get the frequency response by evaluating \begin{equation} H(e^{j\omega}) = \frac{0.1}{1-0.9z^{-1}} \Bigr|_{z = e^{j\omega}} \end{equation}

You can now plot magnitude and phase for a given frequency $\omega$

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  • $\begingroup$ you're correct mitch. i have enough rep that i can edit his question. the only thing i need to do to fix yours (which is not the same as completing it) is to change $w$ to $\omega$. $\endgroup$ – robert bristow-johnson Nov 17 '17 at 4:55

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