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.


  • $\begingroup$ "Impulse response at 2100Hz is 1.3429 dB" i really dunno what that means. did you mean "Frequency response at 2100Hz is 1.3429 dB"? $\endgroup$ – robert bristow-johnson Jan 12 '15 at 21:22
  • $\begingroup$ @robertbristow-johnson Yes, I am sorry. Do you know any formula for this problem ? $\endgroup$ – Aggeliki P Jan 12 '15 at 21:28
  • $\begingroup$ What do you mean by "length 100 in which 98 is zero"? $\endgroup$ – Matt L. Jan 13 '15 at 7:55

Hope this is right... $$h[n]=\delta[n]-a\delta[n-1]$$ $$H(e^{jw})= 1- ae^{-jw} =(1-a\cos(\omega))+ja\sin(\omega)$$ $$|H(e^{j\omega})|= \sqrt{1 + a^2 -2a\cos(\omega)}$$ $$DB_\omega \triangleq 20\log(|H(e^{j\omega})|)$$ $$20\log((1 + a^2 -2a\cos(\omega))^{1/2})= DB_\omega$$ $$1 + a^2 -2a\cos(\omega)= 10^{DB_\omega/10}$$ $$a^2 -2a\cos(\omega) +1-10^{DB_\omega/10}=0$$

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.

  • $\begingroup$ Pre-emphasis usually emphasizes the high frequencies, so I'd go for the $a=0.9$ solution. $\endgroup$ – Matt L. Jan 13 '15 at 7:54
  • $\begingroup$ @Bulent S. Could you please explain me this line: H(ejw)=1−ae−jw=(1−acos(ω))+jasin(ω) Which formula did you use? $\endgroup$ – Aggeliki P Jan 16 '15 at 22:40
  • $\begingroup$ use $e^{j\theta} = cos(\theta) + j\sin(\theta)$ This is the Euler' Relation. $\endgroup$ – Fat32 Jan 16 '15 at 22:50
  • $\begingroup$ @BulentS. in this solution you don't use the information that the singal has length n=100(98 is zero), right? $\endgroup$ – Aggeliki P Jan 16 '15 at 23:00
  • $\begingroup$ I dont know how to use that information? :) What do you mean by signal length n=100 (98) zero? (as Matt-L already asked it) $\endgroup$ – Fat32 Jan 16 '15 at 23:02

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