A MULT I- CHANNEL TELEVISION APPARATUS 37 



set of three is placed a prism Fi which diverts the normally incident 

 light upward; the second hole is left clear; the third is covered by a 

 prism P2 turned to divert the light downward. If wc imagine the 

 prisms removed and a single channel used instead of the three that are 

 proposed, it is clear that the holes would have to be spaced three times 

 as far apart so that no more than one would be included in the frame/ 

 at one time. The diameters of the holes, and the radial separation of 

 the first and last in the spiral would be unchanged. Quite apart, 

 therefore, from the smaller frequency bands which are sufficient to 

 carry each of the three sets of signals, which is the principal objective 

 sought, there is realized in this arrangement a reduced size of apparatus 

 for the same size of disc holes. 



Studying more closely the division of the light into three sets of 

 beams, it is important to note that the signals transmitted by any one 

 of the three sets of holes are continuous — as one hole of a given prism 

 series passes out of the frame the next of the same series comes in. 

 The signals generated in each photoelectric cell are accordingly exactly 

 like those of a single-channel system. 



Before describing the details of the apparatus, the general relation- 

 ship between the number of image elements, band width, number of 

 channels, and shape of picture may be developed. For this purpose, 

 let the following symbols be used. 



B = frequency band available in one channel, in cycles per second. 



F = repetition frequency of images, per second. 



C = number of communication channels. 



n = total number of scanning holes. 

 ajb = ratio of tangential to radial dimensions of frame. 



a — angular opening of frame. 



We shall assume that the picture elements into which the frame is 

 imagined divided are symmetrical in shape, i.e. either circles or squares. 

 We then have that 

 the number of picture elements in the radial direction = number of 



holes = n\ 

 the number of picture elements in the tangential direction = {alb)-n. 

 Now the number of signal cycles corresponding to this number of 



elements is (1/2) • {a[b)n. 

 The number of cycles per second in one transit along the frame 



= {\l2)-{alb)-n'F- 

 over the whole picture it is (1/2)' (ajb)-n- F-n = {l/2)ia/b)Fn-; 



