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Can someone help me with this problem:

Suppose we filter signals with the below pre-emphasis filter:

$$y[n]=x[n]-0.8x[n-1]$$

I have to calculate the impulse response of the filter (ok easy) and then I have to find the magnitude of Frequency Response (dB) at $F = 2000 Hz$ with $Fs = 16000 Hz$

Can someone tell me the formula? I am confused :(

Thanks

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  • $\begingroup$ I am also confused... :( What is the frequency response "meter"? You mean value? $\endgroup$
    – jojeck ♦
    Jan 19, 2015 at 22:51
  • $\begingroup$ @jojek |H(e^{j\omega})| $\endgroup$
    – Ewan Terry
    Jan 19, 2015 at 22:54
  • $\begingroup$ This is the magnitude, not a "meter" then. $\endgroup$
    – jojeck ♦
    Jan 19, 2015 at 22:55
  • $\begingroup$ @jojek I am sorry, I change it $\endgroup$
    – Ewan Terry
    Jan 19, 2015 at 22:57
  • $\begingroup$ Answer posted, you might want to accept it. $\endgroup$
    – jojeck ♦
    Jan 19, 2015 at 23:36

1 Answer 1

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Your filter transfer function in the Z-transform domain is given by:

$$H(z)=1-0.8z^{-1} $$

Substituting $z=e^{j\omega} $:

$$H(j\omega)=1-0.8e^{-j\omega} $$

Knowing you the sampling frequency and frequency of interest you can calculate the angular frequency: $\omega_0= 2\pi\frac{2000}{16000}=\dfrac{\pi}{4}$

Substituting you get:

$$H(j\omega_0)=1-0.8e^{-j\dfrac{\pi}{4}}$$

Taking the magnitude:

$$|H(j\omega_0)| = \left|1-0.8\dfrac{1-i}{\sqrt{2}}\right|\approx 0.7132 $$

Converting to decibel scale:

$$|H_{dB}(j\omega_0)|=20\log_{10}0.7132=-2.93\mathrm{dB} $$

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  • $\begingroup$ Firstly thank you so much I found this: dsp.stackexchange.com/questions/19992/… look Bulent S. answer is similar problem (he want tha a parameter) is his solution right? $\endgroup$
    – Ewan Terry
    Jan 19, 2015 at 23:37
  • $\begingroup$ Sure thing, that is correct as well. $\endgroup$
    – jojeck ♦
    Jan 19, 2015 at 23:39

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