Can someone give my a link to study to solve this problem. I can not find any formula for this problem...
Suppose we filter singals with the below pre-emphasis filter:
y(n)=x(n)−ax(n−1)
How can I compute a if I know that:
- fs=10.000Hz
- length 100 in which 98 is zero.
- Frequency response at 2100Hz is 1.3429 dB
I don't want the solution only a hint I found this in my notes: 10log10(a^2*w^2+1)=db but with this formula i don't use some information which given.
Thanks
Answer
Hope this is right... h[n]=δ[n]−aδ[n−1] H(ejw)=1−ae−jw=(1−acos(ω))+jasin(ω) |H(ejω)|=√1+a2−2acos(ω) DBω≜ 20\log((1 + a^2 -2a\cos(\omega))^{1/2})= DB_\omega 1 + a^2 -2a\cos(\omega)= 10^{DB_\omega/10} a^2 -2a\cos(\omega) +1-10^{DB_\omega/10}=0
with \omega = 2 \pi \frac{2100}{10000}, \ DB_\omega=1.343 \ \Rightarrow \ a \in \{0.9, -0.4026\}
a = -0.4026 results in a low-pass pre-emphasis whereas a = 0.9 results in a high-pass pre-emphasis.
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