로우패스필터 알고리즘 - loupaeseupilteo algolijeum

Looking at your first line of C-code:

y += a*(x-y)

It can be converted to an algebraic relation between the samples of the filter output sequence $y[n]$ and the filter input $x[n]$ as:

$$y[n+1]-(1-a)y[n]=ax[n]$$ which is also equivalent to $$y[n]-(1-a)y[n-1]=ax[n-1]$$

And the associated transfer function of this filter is $$H(z) = \frac{a z^{-1}}{1 - (1-a)z^{-1}} = \frac{a z^{-1}}{1 + (a-1)z^{-1}}$$

Now your application will probably require a causal filter, (which means only current and past input is available and should be used to produce current output) in which case then the poles of the filter should reside inside the unit circle , i.e., $|z_p| < 1$.

Since this filter has a single pole which is at $z_p=(1-a)$ then we have; $|1-a| < 1$ and hence; $$0 < a <2$$ is the allowed range of real $a$ for which you can have a stable and causal filter.

Now to have a casual and stable lowpass filter you require that the pole is along the positive side of the real line, i.e., $0 < z_p < 1$ which means that we require $0 < a < 1$.

Otherwise when $ 1 < a < 2$ the filter becomes a highpass filter (actually it will be some other form of high-boost or shelving type filter rather than a strict high-pass filter which should block low frequencies, which these filters won't), as the pole will become negative for that range of $a$.

Note also that $a=1$ produces output equal to input shifted by 1 samples: $y[n] = x[n-1]$

Below are a few frequency response plots for different values of valid $a$ range:

Let's start with a very basic example of the generic problem at hand: understanding the effect of a digital filter on the spectrum of a digital signal. The purpose of this example is to provide motivation for the general theory discussed in later chapters.

Our example is the simplest possible low-pass filter. A low-pass filter is one which does not affect low frequencies and rejects high frequencies. The function giving the gain of a filter at every frequency is called the amplitude response (or magnitude frequency response). The amplitude response of the ideal lowpass filter is shown in Fig.1.1. Its gain is 1 in the passband, which spans frequencies from 0 Hz to the cut-off frequency

Definition of the Simplest Low-Pass

The simplest (and by no means ideal) low-pass filter is given by the following difference equation:

(2.1)

where

It is important when working with spectra to be able to convert time from sample-numbers, as in Eq.

where

To further our appreciation of this example, let's write a computer subroutine to implement Eq.

Figure: Implementation of the simple low-pass filter of Eq.

/* C function implementing the simplest lowpass:
 *
 *      y(n) = x(n) + x(n-1)
 *
 */
double simplp (double *x, double *y,
               int M, double xm1)
{
  int n;
  y[0] = x[0] + xm1;
  for (n=1; n < M ; n++) {
    y[n] =  x[n]  + x[n-1];
  }
  return x[M-1];
}

In this implementation, the first instance of

We may call xm1 the filter's state. It is the current ``memory'' of the filter upon calling simplp. Since this filter has only one sample of state, it is a first order filter. When a filter is applied to successive blocks of a signal, it is necessary to save the filter state after processing each block. The filter state after processing block

Figure 1.4 illustrates a simple main program which calls simplp. The length 10 input signal x is processed in two blocks of length 5.

Figure 1.4: C main program for calling the simple low-pass filter simplp

/* C main program for testing simplp */
main() {
  double x[10] = {1,2,3,4,5,6,7,8,9,10};
  double y[10];

  int i;
  int N=10;
  int M=N/2; /* block size */
  double xm1 = 0;

  xm1 = simplp(x, y, M, xm1);
  xm1 = simplp(&x[M], &y[M], M, xm1);

  for (i=0;i<N;i++) {
    printf("x[%d]=%f\ty[%d]=%f\n",i,x[i],i,y[i]);
  }
  exit(0);
}
/* Output:
 *    x[0]=1.000000     y[0]=1.000000
 *    x[1]=2.000000     y[1]=3.000000
 *    x[2]=3.000000     y[2]=5.000000
 *    x[3]=4.000000     y[3]=7.000000
 *    x[4]=5.000000     y[4]=9.000000
 *    x[5]=6.000000     y[5]=11.000000
 *    x[6]=7.000000     y[6]=13.000000
 *    x[7]=8.000000     y[7]=15.000000
 *    x[8]=9.000000     y[8]=17.000000
 *    x[9]=10.000000    y[9]=19.000000
 */

You might suspect that since Eq.

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Finding the Frequency Response
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Introduction