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"? $\endgroup$ Jan 12, 2015 at 21:22
  • $\begingroup$ @robertbristow-johnson Yes, I am sorry. Do you know any formula for this problem ? $\endgroup$ Jan 12, 2015 at 21:28
  • $\begingroup$ What do you mean by "length 100 in which 98 is zero"? $\endgroup$
    – Matt L.
    Jan 13, 2015 at 7:55

1 Answer 1


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, 2015 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? $\endgroup$ Jan 16, 2015 at 22:40
  • $\begingroup$ use $e^{j\theta} = cos(\theta) + j\sin(\theta)$ This is the Euler' Relation. $\endgroup$
    – Fat32
    Jan 16, 2015 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? $\endgroup$ Jan 16, 2015 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, 2015 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.