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Given LCCDE system. Is it possible to calculate the frequency response using Fourier transform?

(y[n] - 1/15y[n-1] + 1/5y[n-2] = x[n]+2x[n-1])

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  • $\begingroup$ What makes you think that this system is unstable? $\endgroup$
    – Matt L.
    Mar 30, 2022 at 19:13

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It is slightly more intuitive to solve this problem using the Z-domain, which is a more generalized version of the Fourier transform. The Z-transform of a discrete-time signal $x[n]$ is

$$X(z)=\sum_{n=-\infty}^{\infty}{x[n]z^{-n}}.$$

Applying this transform to both $x$ and $y$ in your equation,

$$y[n]-\frac{1}{15}y[n-1]+\frac{1}{5}y[n-2]=x[n]+2x[n-1],$$

becomes

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

Collecting and solving for the definition of the transfer function:

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

Then the frequency response function is simply the transfer function evaluated at $z=e^{j\omega}$. Recall that the Fourier transform is evaluated on the unit circle.

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

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    $\begingroup$ If you replace $z$ by $e^{j\omega}$ you imply something about the region of convergence of $H(z)$. In such a case it's always a good idea to make this assumption explicit. $\endgroup$
    – Matt L.
    Mar 30, 2022 at 19:15

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