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) 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.


No comments:

Post a Comment

digital communications - Understanding the Matched Filter

I have a question about matched filtering. Does the matched filter maximise the SNR at the moment of decision only? As far as I understand, ...