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≜20log(|H(ejω)|)
20log((1+a2−2acos(ω))1/2)=DBω
1+a2−2acos(ω)=10DBω/10
a2−2acos(ω)+1−10DBω/10=0
with ω=2π210010000, DBω=1.343 ⇒ a∈{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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