 Today we want to see video signal transmission for monochrome TV. Learning outcome of this topic is at the end of this section student will be able to explain the concept of video signal transmission for monochrome TV. The contents of these topics are first we want to see introduction. In most of the television system before the transmission we require two modulation signal one is amplitude modulation and second one is frequency modulation. Amplitude modulation is used for transmission of a picture signal and the frequency modulation is used to transmit the sound signal. The channel bandwidth is determined by the highest video frequency required for proper picture reception and maximum sound carrier frequency deviation permitted in a TV system means total channel bandwidth is calculated on video frequency and the sound carrier frequency. The bandwidth of the channel is summation of highest video frequency and maximum sound carrier frequency. Now need for modulation. We want to recall here what is the means of a modulation. You pause the video and think on what is the means of modulation. Okay modulation means message signal whichever we transmit the signal that contains the message that is called as modulation. It is impossible to transmit a signal by itself that's why we require modulation. Difficulty in the use of unmodulated wave is the need for long antennas for efficient radiation and reception. For example a quarter wavelength antenna for the transmitting frequency of 15 kHz would be 5000 meters long. Now can you imagine for transmission of a 15 kHz we require 5000 meters long vertical antenna means near approximately 5 kms this is unthinkable and in fact impracticable. That's why without modulation we cannot transmit the signal for that one we require modulation. Both picture and sound signals from different stations are connected within the same range of frequencies therefore radiation from different stations would be hopelessly and inextricably mixed up and it would be impossible to separate one from the other at the receiving end. That's why this is another reason why we require the modulation. In order to able to separate it is necessary to translate them all to different portions of the electromagnetic spectrum depending on the carrier frequency assigned to each station. Now we see amplitude modulation. Now the first equation that is E c is equal to E c cos omega ct is the equation for carrier wave modulation and for second equation that is modulating signal that is E m cos omega mt is the equation for modulating signal. Now the equation of the modulation wave is E is equal to A cos omega ct where A is equal to E c plus k E m cos omega mt. E c means carrier wave and E m means modulation wave where k is a constant of the modulator. Now the figure one shows the unmodulated carrier wave means here the message signal is not present here only the carrier signal is present this is a graph is drawn E c versus time carrier wave versus time and this carrier is shown with the help of E c equal to E c cos omega ct. Now figure two shows the modulation signal. Modulation signal is drawn E m E m means modulation is graph is drawn E m versus time. This signal is called as message signal and this is represented by with the help of equation E m is equal to E m cos omega mt. Now the figure three shows the modulated wave means here figure three is a combination of a figure one and figure two that is unmodulating signal and the modulated signal this is shown as this is the message signal this message signal and this one is inner bracket is shown as the carrier signal and this is a combination of a both message signal and the carrier signal and when this both the common the equation become as E c plus E m cos omega mt and minus E c plus E m cos omega mt this graph is drawn carrier versus the time this is a modulation wave. Now the frequency spectrum of this E m amplitude wave is the FC is the carrier wave at the right hand side is shown as the fm and this is upper side band frequency is located at FC plus fm for the left hand side that is a LSB lower side band and it is located at the FC minus fm frequency and this is a graph is of frequency spectrum of A m wave. Now E is equal to E c plus K E m cos omega mt into cos omega ct. Now we separate this above equation simplified with the above equation they become as E is equal to E c 1 plus E m cos omega mt into cos omega ct where M is a modulation index is equal to K into E m by E c at K E m is equal to E c then at that time modulation index become 1 and the corresponding depth of modulation is then termed as 100 percent. Therefore the above equation becomes now E is equal to E c cos omega ct plus M E c by 2 cos omega c minus omega mt minus M E c by 2 cos omega c plus omega m into t. Now channel bandwidth in the 625 line TV system the frequency component present in the video signal from DC to 5 MHz a double sideband A m transmission would occupy a total bandwidth of 5 MHz for upper band bandwidth and for the 5 MHz for the lower bandwidth therefore total becomes 10 MHz. The actual band space allocated to the television channel would have to be still greater because with the practical filter characteristics it is not possible to terminate the bandwidth of the signal abruptly at the edge of the sidebands therefore an attenuation slope of 0.5 MHz is provided at each edge of the two sidebands therefore 0.5 MHz for the right hand side and 0.5 MHz for the left hand side means 0.5 MHz for upper sideband and 0.5 MHz for the lower sideband then total becomes 1 MHz is required for the total band space. Television channel it has its associated with first is a frequency modulated FM signal that carries the sound signal the carrier frequency require 5.5 MHz of the picture signal the small guard band if we require 0.25 MHz so that total practical figure of this channel bandwidth would be 11.25 MHz means 5.5 MHz for the upper sideband 5.5 MHz for the lower sideband that becomes 11 MHz and the remaining guard band we require for that one is 0.25 MHz then total bandwidth channel bandwidth would require 11.25 MHz. Now this is a diagram shows total channel bandwidth using double sideband picture signal here the graph is drawn amplitude versus frequency relative to picture carrier now if you observe this diagram P indicates the picture and S indicates the sound carrier now in this graph if you know the graph upper sideband USB is from 0 to 5.5 and the lower sideband is from 0 to 5.5 then it become total 11 MHz and the remaining 5.5 to 5.75 that is 0.25 MHz that is the guard band this total bandwidth is required for the 11 MHz such but such a bandwidth is too large that limits the number of channels in a given high frequency spectrum allocated for 3B transmission therefore to ensure spectrum conservation some saving in the bandwidth allotted to each channel is desirable for the transmission of a that signal it require 11 MHz but that is a very huge that's why for minimization of that bandwidth we go for next modulation that is single sideband transmission now SSB now equation 2 which we derived in the previous slide that reveals that the carrier component conveys no information because the information is present only at the message signal carrier is contains no message in that equation 2 because its amplitude and frequency remain constant no matter carrier frequency is necessary at the receiver for recovering the modulating frequency FM now upper sideband equation is FC plus FM minus FC and the lower sideband equation is FC minus FC minus FM in transmission the carrier frequency is radiated along with the sideband component in all radio broadcast and TV system results in simpler transmitting equipment needs diode detector at the receiver for recovering the modulation components without an undue distortion from the equation 2 we conclude the three things first two sidebands are images of each other second one is each one is equally affected by changes in the modulating voltage amplitude via the component MEC by 2 now third one is any changes in the frequency of the modulating signal results in identical changes in the band spread of the two sidebands therefore we require from this statement only one sideband therefore all information can be conveyed by the use of one sideband only and it's result in saving of 5 megahertz per channel because we are not using the second channel and it saves the 5 megahertz channel the magnitude of the detector signal in the receiver will be just half of that obtained when both the sideband are transmitted the references for this topic is thank you