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Pre–emphasis - Signal Processing

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sparrow Jr
sparrow Jr 2018-8-11
评论： hanif omar 2021-2-14
I am trying to compute the Pre-emphasis of a signal and the formular is below:
y[n] = x[n] - 0.95 x[n-1]
Let:
x=[1 3 5 9 4 3 2 1]
First..
x[1] = 1
x[n - 1] =0
y[n] = 1 - (0.95 * 0) = 1
Second..
x[2] = 3
x[n - 1] = 1
y[n] = 3 - (0.95 * 1) = 2.05
Therefore:
y[n] = 3 - (0.95 * 1) = 2.05
But I don't understand how this has increased the energy of the signal at a higher frequency? Because 3 is greater than 2.05 etc..
How this filter work?

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回答（1 个）

Dimitris Kalogiros
Dimitris Kalogiros 2018-8-11
编辑：Dimitris Kalogiros 2018-8-11
Your question regards frequency response of FIR filters. I suggest to study (and run) the following script.
clear; clc; close all;
%frequency response of a filter with differencial equation y[n]=x[n]-0.95*x[x-1]
h=[1 -0.95];
freqz(h); zoom on; grid on;
title('frequency response')
%attenuation of a LOW frequency signal
t=0:1:500;
f1=0.01;
x=sin(2*pi*f1*t);
y=filter(h,1,x);
figure;
plot(x,'-b'); hold on;
plot(y, '-r'); zoom on; grid on;
legend('input', 'output');
title('attenuation of a LOW frequency signal')
%enhancement of a HIGH frequency signal
t=0:1:100;
f1=0.3;
x=sin(2*pi*f1*t);
y=filter(h,1,x);
figure;
plot(x,'-b'); hold on;
plot(y, '-r'); zoom on; grid on;
legend('input', 'output');
title('enhancement of a HIGH frequency signal');
Which illustrates how your filter works. How it emphasizes high frequencies and attenuates lowers
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Dimitris Kalogiros
Dimitris Kalogiros 2019-9-30
I used filter h=[1 -0.95] twice.
Each time, x is the input of the filter and y is the output.
At both examples, x was a sinusoidal signal, with frequency f1.
I use f1, in order to show what frequences are going to be attenuated and what frequencies are going to be enchaned.
t=0:1:500;
f1=0.01;
x=sin(2*pi*f1*t);
The above piece of code is just to generate filter's input.
You can just replace x with your audio signal
hanif omar
hanif omar 2021-2-14
if x is input.can i put x is analogread ?

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