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I'm learning about DSP and I've doubts.

The resampling process can be do it with only one filter.

$L$: Increment Sample Rate Integer

$M$: Decrement Sample Rate Integer

$f_S$: Frequency Sampling

For Low Pass Filter:

The frequencies allowed are in the range: from $0$ until $f_{pass}$,

The frequencies denied are in the range: from $f_{stop}$ until $f_S/2$

$f_{pass}$: Highest pass frequency

$f_{stop}$: Lowest cut frequency

I'm simplifying my calculations as well:

I'm defining: $ 0 < \delta < 1$

$$f_{stop} = f_S/2\\ f_{pass} = \delta f_{stop} = \delta f_S/2$$

According to Known formula:

$$\Delta F = (f_{stop}-f_{pass})/f_S\\ f_c = (f_{stop}+f_{pass})/2 $$

For Hamming Window:

$$N = 3.3/\Delta F$$


Questions

Assuming $f_S = 24000$, $\delta = 0.97$ and the most demanding requirements for FIR filter calculus.

  1. If $L=3$ and $M=2$; How can I to calculate (step by step) $N=?$ and $f_c = ?$

  2. If $L=2$ and $M=3$; How can I to calculate (step by step) $N=?$ and $f_c = ?$

EDIT:

  1. Case 1

$$f_{S1} = 24000\\f_{stop1} = f_{S1}/2 = 12000\\f_{pass1} = 0.97*f_{S1}/2 = 11640$$

$L=3$ $$f_{S2} = 72000\\f_{stop2} = f_{S2}/2 = 36000\\f_{pass2} = 0.97*f_{S2}/2 = 34920$$

$M=2$ $$f_{S3} = 36000\\f_{stop3} = f_{S3}/2 = 18000\\f_{pass3} = 0.97*f_{S3}/2 = 17460$$

  1. Case 2

$$f_{S1} = 24000\\f_{stop1} = f_{S1}/2 = 12000\\f_{pass1} = 0.97*f_{S1}/2 = 11640$$

$L=2$ $$f_{S2} = 48000\\f_{stop2} = f_{S2}/2 = 24000\\f_{pass2} = 0.97*f_{S2}/2 = 23280$$

$M=3$ $$f_{S3} = 16000\\f_{stop3} = f_{S3}/2 = 8000\\f_{pass3} = 0.97*f_{S3}/2 = 5173$$

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