Monday, January 14, 2019

impulse response - pre-emphasis filter - parameter a


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(n1)


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[n1]

H(ejw)=1aejw=(1acos(ω))+jasin(ω)
|H(ejω)|=1+a22acos(ω)
DBω20log(|H(ejω)|)
20log((1+a22acos(ω))1/2)=DBω
1+a22acos(ω)=10DBω/10
a22acos(ω)+110DBω/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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