How to get the difference equation from impulse response?

I'm trying to solve a problem about filter design. We want to design a equalizer filter which have the following frequency response :

How can i get the difference equation from the impulse responce which is defined by:

$$\frac{2 f_1}{F_s} \left ( {\rm sinc}(2 \pi f_1 n T_s) (g_1-g_2) \right )+ \frac{2 f_2}{F_s} \left ( {\rm sinc}(2 \pi f_2 n T_s) (g_2-g_3) \right )+ \frac{2}{F_s} \left ( f_3 g_3 {\rm sinc}(2 \pi f_3 n T_s)-f_0\cdot g_1 {\rm sinc}(2 \pi f_0 n T_s)\right )$$

• Your sinc() impulse response is missing a window function to truncate its length. Otherwise it will be of infinite length; the difference equation will be impractical; ie. infinite order. For infinitely long impulse reponses (IIR), only recursive finite-order difference equations can be practically computed. Nov 12 '21 at 21:54
• What should I do in order to get a finite length ? I will multiply this h(n) by w(n) ? After that how can I find the difference equation ? Nov 13 '21 at 10:14

$$y[k] = \sum_{n=-\infty}^\infty x[k-n] \cdot \left[ \frac{2 f_1}{F_s} \left ( {\rm sinc}(2 \pi f_1 n T_s) (g_1-g_2) \right )+ \frac{2 f_2}{F_s} \left ( {\rm sinc}(2 \pi f_2 n T_s) (g_2-g_3) \right )+ \frac{2}{F_s} \left ( f_3 g_3 {\rm sinc}(2 \pi f_3 n T_s)-f_0\cdot g_1 {\rm sinc}(2 \pi f_0 n T_s)\right) \right]$$