The simplest non trivial FIR filter is the filter that computes the running average of two contiguous samples, and the corresponding convolution can be expressed as

If we put a sinusoidal signal into the filter, the output will still be a sinusoidal signal scaled in amplitude and delayed in phase according to the frequency response , which is

Figure 1

Magnitude and phase response for the averaging filter

The one just presented is a first-order (or lenght 22) filter, because it uses only one sample of the past sequence of input. From Figure 1 we see that the frequency response is of the low-pass kind, because the high frequencies are attenuated as compared to the low frequencies. Attenuating high frequencies means smoothing the rapid signal variations. If one wants a steeper frequency response from an FIR filter, the order must be increased or, in other words, more samples of the input signal have to be processed by the convolution operator to give one sample of output.

A symmetric second-order FIR filter has an impulse response whose form is a 0 a 1 a 0 a 0 a 1 a 0 , and the frequency response turns out to be H⁢ω=( a 1 +2⁢ a 0 ⁢cosω)⁢e−(i⁢ω) H ω a 1 2 a 0 ω ω . Convolution can be expressed as

Figure 2

Magnitude and phase response of a second-order FIR filter

Given the simple low-pass filters that we have just seen, it is sufficient to change sign to a coefficient to obtain a high-pass kind of response, i.e. to emphasize high frequencies as compared to low frequencies. For example, the Figure 3 displays the magnitude of the frequency responses of high-pass FIR filters of the first and second order, whose impulse responses are, respectively 0.5−0.5 0.5 0.5 and 0.17654−0.646930.17654 0.17654 0.64693 0.17654 .

Figure 3

Frequency response (magnitude) of first- (left) and second- (right) order high-pass FIR filter.
(a) (b)

To emphasize high frequencies means to make rapid variations of signal more evident, being those variations time transients in the case of sounds, or contours in the case of images.

In 2D, the impulse response of an FIR filter is a convolution mask with a finite number of elements, i.e. a matrix. In particular, the averaging filter can be represented, for example, by the convolution matrix 19⁢( 111 111 111 ) 1 9 1 1 1 1 1 1 1 1 1 .

Example 1: Noise cleaning

The low-pass filters (and, in particular, the smoothing filters) perform some sort of smoothing of the input signal, in the sense that the resulting signal has a smoother design, where abrupt discontinuities are less evident. This can serve the purpose of reducing the perceptual effect of noises added to audio signals or images. For example, the code reported below loads an image, it corrupts with white noise, and then it filters half of it with an averaging filter, thus obtaining Figure 4.

Figure 4

Smoothing

// smoothed_glass
// smoothing filter, adapted from REAS:
// http://processing.org/learning/topics/blur.html

size(210, 170); 
PImage a;  // Declare variable "a" of type PImage 
a = loadImage("vetro.jpg"); // Load the images into the program 
image(a, 0, 0); // Displays the image from point (0,0) 
 
// corrupt the central strip of the image with random noise 
float noiseAmp = 0.2;
loadPixels(); 
for(int i=0; i<height; i++) { 
  for(int j=width/4; j<width*3/4; j++) { 
    int rdm = constrain((int)(noiseAmp*random(-255, 255) + 
	      red(pixels[i*width + j])), 0, 255); 
    pixels[i*width + j] = color(rdm, rdm, rdm);
  } 
} 
updatePixels(); 
  
int n2 = 3/2; 
int m2 = 3/2; 
float  val = 1.0/9.0;
int[][] output = new int[width][height]; 
float[][] kernel = { {val, val, val}, 
                     {val, val, val}, 
                     {val, val, val} }; 
  
// Convolve the image 
for(int y=0; y<height; y++) { 
  for(int x=0; x<width/2; x++) { 
    float sum = 0; 
    for(int k=-n2; k<=n2; k++) { 
      for(int j=-m2; j<=m2; j++) { 
        // Reflect x-j to not exceed array boundary 
        int xp = x-j; 
        int yp = y-k; 
        if (xp < 0) { 
          xp = xp + width; 
        } else if (x-j >= width) { 
          xp = xp - width; 
        } 
        // Reflect y-k to not exceed array boundary 
        if (yp < 0) { 
          yp = yp + height; 
        } else if (yp >= height) { 
          yp = yp - height; 
        } 
        sum = sum + kernel[j+m2][k+n2] * red(get(xp, yp)); 
      } 
    } 
    output[x][y] = int(sum);  
  } 
} 
 
// Display the result of the convolution 
// by copying new data into the pixel buffer 
loadPixels(); 
for(int i=0; i<height; i++) { 
  for(int j=0; j<width/2; j++) { 
    pixels[i*width + j] = 
	      color(output[j][i], output[j][i], output[j][i]);
  } 
} 
updatePixels(); 
         
	

For the purpose of smoothing, it is common to create a convolution mask by reading the values of a Gaussian bell in two variables. A property of gaussian functions is that their Fourier transform is itself gaussian. Therefore, impulse response and frequency response have the same shape. However, the transform of a thin bell is a large bell, and vice versa. The larger the bell, the more evident the smoothing effect will be, with consequential loss of details. In visual terms, a gaussian filter produces an effect similar to that of an opalescent glass superimposed over the image. An example of Gaussian bell is 1273⁢( 14741 41626164 72641267 41626164 14741 ) 1 273 1 4 7 4 1 4 16 26 16 4 7 26 41 26 7 4 16 26 16 4 1 4 7 4 1 .

Conversely, if the purpose is to make the contours and salient tracts of an image more evident (edge crispening or sharpening), we have to perform a high-pass filtering. Similarly to what we saw in Section 4 this can be done with a convolution matrix whose central value has opposite sign as compared to surrounding values. For instance, the convolution matrix ( -1-1-1 -19-1 -1-1-1 ) -1 -1 -1 -1 9 -1 -1 -1 -1 produces the effect of Figure 5.

Figure 5

Edge crispening

In the audio field, second-order IIR filters are particularly important, because they realize an elementary resonator. Given the difference equation

Figure 7

Magnitude and phase response of the second-order IIR filter