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The Simplest Lowpass Filter

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.

Figure 1.1: Amplitude response (gain versus frequency) specification for the ideal low-pass filter.
\begin{figure}\input fig/kfigtpone.pstex_t \end{figure}

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 $ f_c$ Hz, and its gain is 0 in the stopband (all frequencies above $ f_c$). The output spectrum is obtained by multiplying the input spectrum by the amplitude response of the filter. In this way, signal components are eliminated (``stopped'') at all frequencies above the cut-off frequency, while lower-frequency components are ``passed'' unchanged to the output.

Definition of the Simplest Low-Pass

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

$\displaystyle y(n) = x(n) + x(n - 1) \protect$ (2.1)

where $ x(n)$ is the filter input amplitude at time (or sample) $ n$, and $ y(n)$ is the output amplitude at time $ n$. The signal flow graph (or simulation diagram) for this little filter is given in Fig.1.2. The symbol ``$ z^{-1}$'' means a delay of one sample, i.e., $ z^{-1}x(n) = x(n - 1)$.

Figure 1.2: System diagram for the filter $ y(n) = x(n) + x(n - 1)$.
\begin{figure}\input fig/kfig2p2.pstex_t \end{figure}

It is important when working with spectra to be able to convert time from sample-numbers, as in Eq.$ \,$(1.1) above, to seconds. A more ``physical'' way of writing the filter equation is

$\displaystyle y(nT) = x(nT) + x[(n - 1)T]\qquad n = 0, 1, 2, \ldots\,, $

where $ T$ is the sampling interval in seconds. It is customary in digital signal processing to omit $ T$ (set it to 1), but anytime you see an $ n$ you can translate to seconds by thinking $ nT$. Be careful with integer expressions, however, such as $ (n - k)$, which would be $ (n-k)T$ seconds, not $ (nT - k)$. Further discussion of signal representation and notation appears in §A.1.

To further our appreciation of this example, let's write a computer subroutine to implement Eq.$ \,$(1.1). In the computer, $ x(n)$ and $ y(n)$ are data arrays and $ n$ is an array index. Since sound files may be larger than what the computer can hold in memory all at once, we typically process the data in blocks of some reasonable size. Therefore, the complete filtering operation consists of two loops, one within the other. The outer loop fills the input array $ x$ and empties the output array $ y$, while the inner loop does the actual filtering of the $ x$ array to produce $ y$. Let $ M$ denote the block size (i.e., the number of samples to be processed on each iteration of the outer loop). In the C programming language, the inner loop of the subroutine might appear as shown in Fig.1.3. The outer loop might read something like ``fill $ x$ from the input file,'' ``call simplp,'' and ``write out $ y$.''

Figure: Implementation of the simple low-pass filter of Eq.$ \,$(1.1) in the C programming language.
  
/* 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 $ x(n-1)$ is provided as the procedure argument xm1. That way, both $ x$ and $ y$ can have the same array bounds ( $ 0,\dots,M-1$). For convenience, the value of xm1 appropriate for the next call to simplp is returned as the procedure's value.

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 $ m$ is then the starting state for block $ m+1$.

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.$ \,$(1.1) is the simplest possible low-pass filter, it is also somehow the worst possible low-pass filter. How bad is it? In what sense is it bad? How do we even know it is a low-pass at all? The answers to these and related questions will become apparent when we find the frequency response of this filter.

Next Section:
Finding the Frequency Response
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Introduction