Pre–emphasis - Signal Processing 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[n] = 0.34
 x[n - 1] = 0.24

Therefore:
 y[n] = 0.34 - (0.95 * 0.24) = 0.112 

But I don't understand how this has increased the energy of the signal at a higher frequency? Because 0.34 is greater than 0.112.. Am I not calculating something right here?
 A: The transformation you are describing is essentially a high-pass filter. Here is an intuitive explanation: suppose you feed a signal of magnitude 1 and of zero frequency in this filter. Then this signal is equal to $x[n]=1$. Hence the output of your filter will be $y[n]=0.05$. Observe the the energy of the output signal is much smaller than the energy of the input signal. Next, suppose that you feed a signal of small frequency in the filter. Because the frequency is small, adjacent samples will tend to be similar in magnitude and so the difference $x[n]-0.95 x[n-1]$ will tend to be very small. The numerical experiment that you are presenting, does not give you any information, because if $x[n]=0.34, x[n-1]=0.24$, then this does not say anything about the frequency of $x$. How about the other values of $x$, i.e. $x[n-2], x[n-3], e.t.c.$?
A: This attenuates slow-changeable components of a signal.
Imagine that undesirable constant C was added to your signal.
This constant would be 20 times attenuated on output of your filter.
High frequency components would pass this filter without such attenuation.
A: This is a convolution of $x$ with a two point signal $z$ with $z[0] = 1$ and $z[1] = \alpha = 0.95$.  The spectrum of $x$ is therefore multiplied by the spectrum of $z$.  The spectrum of $z$ is given by $\hat{z}[x] = 1 - \alpha e^{-2 \pi i x}$.  The square norm of this spectrum is
$$
\left|\hat{z}[x]\right|^2 = (1 - \alpha)^2 + 4 \alpha \sin^2(\pi x).
$$
So for $\alpha \in [0, 1]$ the norm $\left| \hat{z}(x) \right|$ has a minimum $1-\alpha$ at $x = 0$ and a maximum $1+\alpha$ at $x = \tfrac{1}{2}$.  For values of $\alpha$ close to one it acts as a low pass filter as was already observed in the other answers and emphasises high frequencies.
