 Today we want to see the topic on Interlace Scanning. The learning outcome of this topic is at the end of this session student will be able to explain the concept of Interlace Scanning. The contents of this topic are introduction, horizontal scanning, vertical scanning, number of scanning lines, interlace scanning and scanning periods. Scanning means the division of picture into many horizontal lines is called scanning. The flickering effect can be minimized with the help of interlace scanning. And to avoid this flickering we use interlace scanning and minimum picture frames per second requirement is 50. The horizontal line scanning frequency is required for 625 line system with a period of 64 microsecond divided by 2 is equal to 32 microsecond. Now before going to the interlace scanning we want to understand first what is the means of horizontal scanning and vertical scanning. Now this figure shows the horizontal scanning of path of scanning beam in covering the picture area. This diagram is drawn width versus height and here shows start of a line is from top left hand side to right hand side. This will dash line is shows and this coil is a trace line and this will dotted line shows the retrace path. This is a trace path and this is a retrace path. And this all these processes happen in the raster. Now that images is shown here. This graph is shown current versus time. This is a horizontal current. Now in this diagram this diagram shows waveform of current in the horizontal deflection coil producing linear scanning in the horizontal direction. Here for the first line it will start from the zero and it will reach to the maximum IH maximum and after when it will reach to the maximum it will suddenly drop down to the zero current zero and again it will start and again it will come to zero. This starting from zero to maximum this path is called as a trace path and from the maximum point to the zero it is called as a retrace path. Now this diagram is again shown here in the raster of 625 lines. For the interlayer scanning we require minimum 625 lines and that is explained with the help of this diagram. This diagram shows it will start from the most left hand side it will start from the zero it will reach up to this IMAX and that will be shown here and when it will reach to the IMAX it will suddenly drop down to the zero and again it will start. This will called as a trace and this will called as a retrace. This one cycle of a deflection current is equal to trace plus retrace. Now vertical scanning same the horizontal scanning the vertical scanning is a start from the top side left hand side top left hand side and it will reach to the right hand side. This will called as a trace path and it again it will dotted line shows the retrace path this dash line it shows the trace and this will be the retrace path. This linear vertical retrace and the retrace is done with the help of this vertical direction this will be the show the vertical retrace. Now this the previous diagram is explained with the help of this one here the vertical current versus time withdrawn the vertical deflection current. Here again it will start from the zeros it will reach up to this maximum current and after when it will reach to the maximum current it will drop down to the zero. This is from zero to maximum current whichever showing this is the vertical trace and which is the maximum to zero this will shows the retrace path. That's this is a first frame second frame in this way this up to 625 frames we have to show here and this is a raster this is a W means width and H means height and this is a trace period that's the height total see the trace period is equal to the height of that raster that will be the shown here it will be start from this zero and reach to the maximum and after that one it will again goes to the zero this is a path of a trace and this is a path of a retrace this is the current waveforms in vertical deflection coils. Now number of scanning lines how many numbers of a scanning lines we require here for the number of a scanning lines we require here the dark white dark white these are the strips dark white and this is a beam spot if we want the correctly the picture transmission in the TV then the beam spot is correctly mentioned in between this dark and white strips this is a beam path and this diagram shows about the scanning this spot perfectly along with the black and white lines. Now alternate lines are a black and white the electrical information corresponding to the brightness of each bar will be correctly reproduced during the scanning process if it will satisfy these two conditions which are these two conditions one is a thickness of a scanning beam is equal to width of each white and black bar and second one is number of scanning lines equal to number of bars. Now this diagram shows critical weaving distance as determined by the ability of the eye to resolve the two space pictures. Now in this diagram this diagram shows D is equal to 4H this eye of the observer this is a two-distance black picture element just resolved is the nth bar n plus one bar and this is what eye angle and we see here we calculate the eye here we calculate the distance how much distance we require to two adjacent black picture element just resolved we have to calculate here. Now the maximum number of alternate light and dark elements which can be resolved by the eye is given by the formula nv is equal to 1 by alpha p. Now what is this terminology where nv is total number of fill lines elements to be resolved in the vertical direction alpha means minimum resolving angle of the eye expressed in radiance. Now this alpha we want to measure in the radiance that we calculated with the help of pi by 180 into 1 by 60 this is a standard value. Now rho is equal to D by H that is a weaving distance divided by picture height is equal to standard value we use here 4 substituting the standard values into equation 1 we get now here pause the video for two minutes and you put these standard values into that equation and calculate the what is the value. Okay after the putting the value of that equation nv is equal to 1 by pi by 80 180 into 1 by 60 into 4 we get 860 means a distinct pickup of the picture information results total number of a scanning lines here 860 and the scanning beam just pass over the each bar. Now the the most important part that is the interlayer scanning in television pictures an effective rate of 50 vertical scans per second is utilized to reduce the flicker this is accomplished by increasing the downward rate of travel of the scanning electron beam so that every alternate line gets scan instead of every successive line thus the total number of lines are divided into two graphs two groups called the fields each fill is scanned alternatively this method of scanning is known as interlay scanning. Now this diagram shows the principle of interlay scanning here note that the vertical retest time has been assumed to be 0 now here it will start from the beginning from the first fill that is first it will start from the top left hand side and it will reach to the right side and after that when it will goes to the left again right this will be the trace path and when is if you observe this diagram it will start from the left hand side at the 0th position and it will reach for the end of the first fill it will be the at the middle of this line and after when we reach the here then it again retrace path it will start the middle of this one and again it will start before the second field and in this way this one is a first fill and in this way in the second field now this will process will be done if you observe in this diagram it will be the scanning is for different for all lines and scanning for different for the even lines now for successful interlay scanning the 625 lines of each frame or picture are divided into sets of 312.5 lines to achieve this the horizontal sweep oscillator is made to work at a frequency of 15,625 hertz that is how we calculate this one with the help of 312.5 into 50 is equal to 1500 is 15,625 now scanning periods the most important concept here are two types of a scanning periods one is a horizontal deflection current and second one is vertical deflection current now horizontal deflection current it will diagram shows the horizontal deflection current it will again the total time period is equal to the trace plus retrace now total time period is equal to trace period is 52 microsecond and for the retrace it will require 12 microsecond therefore total time become 52 microsecond plus 12 microsecond that is 64 microsecond this will is equal to 64 microsecond and if we calculate the frequency it will require 15,625 hertz for this horizontal deflection current for this vertical deflection current here the total time required for vertical deflection current is for the trace it will require 18.720 millisecond and for the retrace it will require 1.280 millisecond therefore the total time required for trace and retrace we have to add that one therefore total time required is trace plus retrace period that is 18.720 millisecond plus 1.280 millisecond it becomes 20 millisecond and from this time period we calculate the frequency that is a 50 hertz for the vertical deflection current the horizontal and vertical sweep oscillator operate continuously to achieve the first sequence of internal scanning that is 20 horizontal lines now 24 the even field and 24 the odd field therefore it becomes as a 40 lines therefore this leaves the active number of lines NA for scanning the picture details equal to 625 minus 40 that is equal to 585 instead of 625 lines we use only 585 lines actually scanned per frame this is a scanning sequence as we told that the for the scanning sequence we require the scanning for this first field and for the second field is a different that is for the first vertical trace it will show it will start from the zero position from the left hand side from A to the B and it will reach to the middle of the B that is a first vertical trace that is equal to 292.5 lines that is 1 to 292.5 and after when it will reach to the B it will again goes to the retrace path that requires 292.5 to 312.5 is equal to 20 lines therefore it for the first field it will require 312.5 lines and the same sequence is applied for this second vertical trace and here we get the for second field is a 312.5 lines that is if we add this one we get the one frame picture is equal to 625 lines my advantage advantage for this internal scanning is to avoid the flicker it is better than sequential scanning and it conserves the bandwidth now difference between the progressive and internal scanning the progressive scanning and internal scanning is shown here first in in the progressive scanning every successive line is being scanned but for the internal scanning the first scans the odd lines from the top to bottom and then it scans the lines those are scripted in the in the previous scanning the effective number of pictures scanned per seconds are 25 frames per second in progressive scanning and in the internal scanning it requires 50 frames per second flicker problems will occur in the progressive scanning but with in the internal scanning the flicker problem is our total number of a line scanned at a time from top to bottom is a 620 lines in the progressive scanning total number of a line scanned at a time from top to bottom are 312.5 lines for the internal scanning references for these topics thank you