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:

$y(n)=x(n) - ax(n-1)$

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

• "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"? Jan 12 '15 at 21:22
• @robertbristow-johnson Yes, I am sorry. Do you know any formula for this problem ? Jan 12 '15 at 21:28
• What do you mean by "length 100 in which 98 is zero"? 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.
• Pre-emphasis usually emphasizes the high frequencies, so I'd go for the $a=0.9$ solution. Jan 13 '15 at 7:54
• use $e^{j\theta} = cos(\theta) + j\sin(\theta)$ This is the Euler' Relation. Jan 16 '15 at 22:50