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4. SIGNAL TRANSMISSION AND CHANNEL BANDWIDTH. Prepared by Sam Kollannore U. Lecturer, Department of Electronics M.E.S.College, Marampally, Aluva-7. Picture – amplitude modulated Sound – frequency modulated
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4. SIGNAL TRANSMISSION AND CHANNEL BANDWIDTH Prepared by Sam Kollannore U. Lecturer, Department of Electronics M.E.S.College, Marampally, Aluva-7
Picture – amplitude modulated • Sound – frequency modulated • Channel bandwidth – determined by the highest video frequency required for proper picture reception and the maximum sound carrier frequency deviation permitted in a TV system • Need for modulation - antenna size - interference - poor radiation at low frequency
Amplitude Modulation • In amplitude modulation the intelligence to be conveyed is used to vary the amplitude of the carrier wave. • ec = Ec cos ωct is the carrier wave and • em = Em cos ωmt is the modulating signal. • The equation of the modulated wave is :e = A cos ωct • where A = (Ec + kEm cos ωmt) when k is a constant of the modulator. On substituting the value of A we get : e = (Ec + kEm cos ωmt) cos ωct = Ec (1 + m cos ωmt) cos ωct where m = kEm/Ec is the modulation index At kEm = Ec , m = 1 and the corresponding depth of modulation is then termed as 100%.
If the modulating signal consists of more than a single frequency, as it would be for a video signal, the equation can be extended to include the sum and difference of the carrier and all frequency components of the modulating signal. Therefore if the modulated wave is to be transmitted without distortion by this method, the transmission channel must be atleast of width 2fm centred on fc.
Channel Bandwidth • Frequency components present in the video signal extends from 0 to 5 MHz • Therefore Bandwidth required = 2× 5MHz = 10MHz • Attenuation slope of 0.5MHz is provided at the edge of the two side bands (i.e. 2 × 0.5MHz = 1MHz) • Each channel has its associated FM sound signal whose carrier frequency located at 5.5MHz (in the upper limit) • A small guard band of 0.25MHz for each channel • So total Channel Bandwidth = 10 + 1 + 0.25 = 11.25MHz
Channel Bandwidth …contd • Such a bandwidth is too large - limits the number of channels in a given high frequency spectrum • Two side bands are identical - only one is necessary - thus saving 5MHz per channel - SSB • Carrier conveys no information - but its presence is necessary at the receiver for recovering the modulating frequency fm from the USB or from LSB. Therefore it is transmitted • Thus results in simpler transmitting equipment • Only needs an inexpensive diode detector at the receiver for demodulation • In Television transmission we use VSB - Vestigial Side Band transmission
Vestigial Side Band Transmission • In the video signal, very low frequency modulating components exist along with rest of the signal • These components give rise to sidebands very close to the carrier frequency – difficult to remove by physically realizable filters • Again the low video frequencies contain the most important information of the picture • Complete suppression of the lower sideband would result in phase distortion at these frequencies • Therefore we cannot fully suppress one complete sideband • As a compromise only a part of the LSB is suppressed • Radiated signal consist of : Full USB + Carrier + Vestige of the partially suppressed LSB • This pattern of transmission is known as Vestigial Side Band Transmission or A5C transmission
Vestigial Side Band Transmission …contd • Frequencies up to 0.75 MHz of the LSB are fully radiated • Attenuation slope of 0.5 MHz at either end • FM sound signal occupies a frequency spectrum of about ±75 KHz around the sound carrier • Guard band of 0.25 MHz – allowed on the sound carrier side – for interchannel separation
For a 625 line system, bandwidth requirement = 2(50+15) = 130KHz - close to the value calculated earlier
Television Signal standards contd …
