 Today we want to see reception of Waste Agile Sideband and signals. The learning outcome of this topic is, at the end of this session, students will be able to explain the concept of Waste Agile Sideband reception. The contents of these topics are introduction, transmission efficiency and complete channel bandwidth. Now, recall what is the means of SSB? SSB means transmission of signals with the help of single sideband. An SSB signal with carrier cannot be demodulated by an envelope detector. To produce effective multiplication of the carrier with the sideband, there are two types of A for transmission that synchronous demodulation and a square law device. Envelope detection can be used in television receivers if the sideband amplitude is small as compared to the carrier. The envelope of the SSB with carrier signal equal to the modulating signal. If the picture carriers were successfully modulated equal to depth by a series of frequency throughout the video frequency range employed by the system. Now figure 1 shows intermediate frequency i.e. IF receiver response curve. This figure relative video detector output voltage versus frequency in MHz. Now if you observe this frequency, the Waste Agile Sideband is of a 2-voltage is up to 0.75 MHz. And beyond the 0.75 MHz this Waste Agile Sideband linearly reduced to 1V up to 1.25 MHz. And after that 1.25 MHz it is a constantly up to 5 MHz it is a 1V. Means from 2 to 1V means at the frequency 0.75 MHz to 1.25 MHz it will be reduced linearly. Now if you observe the voltage present across 1.20V to 5V it is a double across 0 to 0.75 the voltage is double as compared to the voltage across 1.25 MHz to the 5 MHz. Now this figure 2 shows transmitted output characteristic for Waste Agile Sideband signal. This figure is drawn relative transmitter output versus the frequency. Now here this will show of a total 7 MHz and P means is a picture carrier and S means sound carrier and total effective bandwidth we use here 5 MHz out of the 7 MHz. And if you observe this relative transmitted output will be linearly increased from 0 to some extent from frequency at the range 0 to 0.75. After that 1 it will linearly to 5 MHz. After that 1 it will linearly reduce and it will be 0 again at the 5.5 MHz. Now this figure shows the desired receiver characteristics for correct reproduction of produce signals. This figure drawn relative receiver output versus the frequency. Now if you observe here it will linearly increase up to 1V after that it will be the constant. Now this Waste Agile Sideband can we prove how we say that this will be the linear. This will be linear we said that mathematical equation if A plus B is equal to 1 and A dash plus B dash equal to 1. Now observe the 1 by 1. Now if you observe the frequency at 0.5 MHz what is the relative output voltage at the 0.5 MHz it will present at both the upper sideband. This will part is called as a upper sideband means above the 0 it will called as upper sideband and below the 0 it will called as a lower sideband. Now at the 0.5 voltage the frequency is 0.2 that is a lower sideband and for the upper sideband for the 0.5 MHz it will be the 0.8 when you add this 0.2 plus 0.8 it will become as 1. Therefore 0.2 is A and B is a 0.8 then it will become as a 1. Now similarly if you observe for the 0.7 MHz if you observe the 0.7 MHz for that 0.7 MHz it will present at the upper sideband as well as for the lower sideband. For the upper sideband for 0.7 MHz it will be the if you observe it will be the 0.92 and for this A dash will be present at the 0.0 not 8. Now if you add this 0.92 plus 0.0 it will become 1 therefore this condition will be satisfied that is E dash plus B dash equal to 1. Now similarly after that 0.5 and 0.7 MHz if you observe for the 0.75 MHz at that 0.75 MHz the lower sideband is 0 and upper sideband it will be represented as 0.75 it will be represented at 1 therefore lower sideband is 0 and upper sideband for the 0.75 MHz it will be the 1 therefore 0 plus 1 it become as a 1. Now for this characteristics A plus B for the vestigial sideband will be satisfied. Now D merits of a vestigial sideband transmission the transmitted power is wasted in the vestigial sideband filters. Second a loss of about 6 decibels in the signal to noise voltage ratio at the transmitter phase and amplitude distortion of the picture signal occurs. It is very difficult to tune IF stage means intermediate stage intermediate frequency stage of the receiver to correspond exactly with the ideal desired response. The main important thing is FM channel bandwidth that is a frequency modulation channel bandwidth. The channel bandwidth required for the frequency modulation is the formula is bandwidth is equal to 2 into N into FM. Now N means where FM means frequency of the modulating wave N means number of the significant side frequency components and the value of N is determined from the modulation index. The higher frequency in speech or music have much less amplitude as compared to the lower audio frequency. Now the maximum frequency deviation is equal to 75 kHz and the modulating frequency is range between 25 Hz to 15 kHz. Now for the frequency modulation bandwidth we require two things. The first one things is maximum frequency deviation and second thing is modulating frequency and if we know these two things and we can calculate easily the bandwidth. Now we calculate the bandwidth. How we calculate the bandwidth? See if 15 kHz we taken an example if a 15 kHz tone has a unit amplitude equal to the maximum allowed amplitude then MF modulation index of that frequency is equal to 75 divided by 15 that is equal to 5 and for that 5 we calculate the value of JN with the help of this Bessel function. From the Bessel function table for the MF is equal to 5 the significant value of JN is equal to 7. Now for this N is equal to 7 therefore we put these all these values into the bandwidth equation that is bandwidth equal to 2 into N into MF that is 2 into 7 into 15 that become 210 kHz. Now from that example we conclude that bandwidth depends on in the frequency modulation that is depends on the tone amplitude and in the amplitude modulation the bandwidth depends on tone frequency that's the two different parts. Now take a similar example for the in the 625b television if the maximum deviation that is equal to plus or minus 50 kHz and modulating frequency equal to 15 kHz then modulation index for that frequency carrier is equal to 15 by 15 that equal to 3 and from the Bessel function if the value of MF is equal to 3 then significant value of for that JN is equal to 5. Now put that values into that bandwidth equation then bandwidth equal to 2 into 5 into 15 that will become as 150 kHz. Now Kershaw's rule now before going to the Kershaw's rule what is the use for that one see Kershaw's rule is also used for the calculation for bandwidth now whatever we calculate the bandwidth with the help of the previous frequency modulation equation that will give the exact value but that we use the Bessel function. But here without using the Bessel function we can calculate the bandwidth but with the help of this Kershaw's rule that the bandwidth is near approximate that Kershaw's rule now what is that one Kershaw's rule we see one by one. Kershaw's rule states that to a good approximation the bandwidth requirement to pass and frequency modulation wave is equal to twice the sum of the deviation and the highest modulating frequency. Now as already said that this Kershaw's rule is used for the bandwidth calculation. Now we calculate the bandwidth calculation whichever we see in the example the before two examples we calculate the bandwidth equation with that formula we calculate the bandwidth. Now with the help of the Kershaw's rules we can now calculate the bandwidth. The required bandwidth is equal to 2 into 75 plus 15 that becomes 180 kHz and for the 620 file-in system the required bandwidth is equal to 2 into 50 plus 15 is equal to 130 kHz and now if you observe that is a near approximate to the previous calculation. Now this is the channel bandwidth for color transmission. Now here this is the total 7 MHz channel is required that is the video signal spectrum and this is the color signal spectrum. Now out of this video spectrum it is the 7 MHz and out of this the one 4.433 MHz is required for video signal spectrum plus half of the color signal spectrum and for the video signal spectrum and color spectrum and sound carrier and color and picture carrier require the 5.5 MHz. Now allocation of a frequency bands for television signal transmission for effective amplitude modulation and better selectivity at an RF and IF tune amplifier in the receiver that is carrier frequency equal to 10 into highest modulating frequency. Now consider carrier frequency equal to 10 MHz then highest video modulating frequency equal to 5 MHz that deviation is equal to 50%. Now television standards the picture and sound signal standards as specified by the International Radio Consolidative Committee that is CCI for the 625B monochrome system we want to see in the table. This is the table here number of lines for picture frame request 625 field frequency that is 50 interlaced ratio is 2 is to 1 picture is required 25 picture frame frequency line frequency is 15625 aspect ratio 4 by 3 scanning sequence is left to right and field talk to bottom. The references for this topic is as follows. Thank you.