1INDEX1. TELEVISION SIGNALS BROADCASTING IN U.S.A .2. MODULATION AND DEMODULATION SCHEMES.3. BROADCASTING OF INTENSITY , SOUND ANDCOLOR4. ERRORS IN TRANSMISSION.5. MOVING PICTURE EXPERT GROUP (MPEG).2There are mainly three standards in use across the world . They are1. National television systems committee standards ( NTSC)2. Phase altered line (PAL)3. Sequential with memory(SECAM)Television signals in U.S.A are broadcast using the NTSC-M standards. According toNTSC-M standards used in U.S.ANumber of horizontal scans per frame = 525number of frames per second = 59.94(~ 60). FIGURE 1. TWO FIELDS INTERLACE TO FORM ONE FRAME3Usually the image is scanned along the lines shown above until it completes scanning theimage. This type of scanning is called progressive scanning. But to reduce the effectof flickering the frame shown above is divided into two fields and each field is used toshow consecutive images . The odd fields represent the first image and even fieldsrepresent the image that follows. This type of scanning is called interlaced scanning.This type of scanning reduces the effect of flicker to a considerable extent.So there are ,525 Horizontal scan paths per frame.262.5 Horizontal scan paths per field.30 frames (complete pictures per second).60 fields (half pictures) per second.So the horizontal scan frequency = 525 x 30 = 15,750 .Therefore Time for each scan = 63.5 x 10 - 6 FIGURE 2. SCANNING PATTERN FOR A FIELD4FIGURE 3. TV VIDEO SIGNAL.PROBLEM : A 40 cm x 30 cm television is scanned on 525 scan lines , each of which contains 650samples. We wish to consider one frame of the video as a two - dimensional signal,(i.e.,ignore the time variable and assume that there is no interlacing). Determine the sampling matrix, V , that defines this sampling lattice .SOLUTION :To determine V , We need two sampling vectors that define the lattice. This choice is notunique, But one convenient choice is to let V1 point down the scan line and let V2 point tothe first point on the next scan line.5 q 40/650 cm 0 V1 = V2 = 30/525x525 cm 30/525 cm .0615 cm 0 cm V = .000088 cm .0571 cm 2p x 650/40 -2p /40U = rad/cm 0 2 p6 MODULATION : In TV Transmission the use of FM is made for Audio transmissionand AM for Video transmission.Vestigial Sideband modulation (VSB) is used for the following reasons :1. Video signal exhibits a large bandwidth and significant low-frequency content whichsuggests the use of VSB2. The circuitry for demodulation in the receiver should be simple and therefore cheap.VSB demodulation uses a simple envelope detection.But the practical TV signals are not exactly VSB modulated due to the followingreasons :I) The power at the transmitter is very high and it would be expensive to rigidly control thefiltering of sidebands. Instead, a VSB filter is inserted in the receiver where the powers arelow.FIGURE 4. SPECTRA OF MODULATING SIGNAL AND CORRESPONDINGDSB , SSB, VSB SIGNALS .For SSB signals the output is given by fssb (w) = m(t) cos wct + mn(t) sin wctfvsb(w) = [ M(w + wc) + M(w-wc)]H(w) ------ (1)H(w) is the Vestigial Shaping filter7We require that m(t) is recoverable from fvsb(t)So, ed(t) = 2*fvsb(t) coswct <---> [fvsb(w+wc) + fvsb(w-wc)] -------- (2)For distortionless reception we haveeo(t) <---> C *M(w) where C is a constant.Choosing C = 1 we haveH(w + wc) + H(w - wc) = 1 |w| <= 2*p*BFor any real filter H(-w) = H*(w)So H(wc+w) + H*(wc-w) = 1 |w| <= 2*p*Bor H(wc+x) + H*(wc+x) = 1 |x| <=2*p*BIf we construct a filter of the form H(w) e-jwtd the term e-jwtd represents a puredelay.Hence only H(w) need satisfy the above equation as |H(w)| is realSo |H(wc+x)| + |H(wc-x) = 1 |x| <= 2*p*BSince fvsb(w) is a band pass Spectrum we can express,fvsb(t) = mc(t) coswc(t) + ms(t) sinwct -------- (1)If fvsb(t) is multiplied by 2*coswct, we get2*fvsb(t)*Coswct = mc(t) + mc(t) cos 2wct + ms(t) sin 2wct -------------------------------------eliminated by lpfSo, mc(t) = m(t)To determine ms(t), (1) is multiplied by 2Sinwct & LPF, the output is ms(t).Thus VSB suppresses the transmitted Sideband and compensates it with thegradual roll off filter. Thus it has both the cost and increased bandwidth advantage overDSB and SSB.The above reasons justify the use of VSB for TV broadcasting.8FIGURE 5.(a) IDEALIZED AMPLITUDE RESPONSE SPECTRUM OF A TRANSMITTED TVSIGNAL.(b) AMPLITUDE RESPONSE OF VSB SHAPING FILTER IN THE RECEIVER.9The demodulation of VSB transmitted waves can be done with the use of a simpleenvelope detector at the receiver (a)(b)(c) FIGURE 6. (a) AMPLITUDE DEMODULATOR ( ENVELOPE DETECTOR). (b) AMPLITUDE MODULATED WAVE . (c) OUTPUT OF THE ENVELOPE DETECTOR .10RBG - Y-U-V (intensity - luminance - chrominance)or Y-I-R (luminance, chrominance(hue, saturation))All colors are synthesized by mixing the 3 primary colors-Red , Blue and YellowIn TV, Red, Blue and Green (blue + yellow) are used instead, due to the reason thatPhosphors that glow with these colors are available when excited by an electron beam.In TV cameras the optical system resolves the image into 3 primary colors. A set ofcamera tubes can be used to produce these 3 primary color images. But there are somedifficulties at the reception end. It requires 3 times as much bandwidth as monochrome TVand secondly it is not compatible with the existing system because monochrome TVreceives only one color.This problem is solved by signal matrixing. The information about mr(t), mg(t) andmb(t) can be transmitted by 3 signals, each of which is a linear combination of the aboveand provided they are independent of each other.my(t) = 0.3mr(t) + 0.59mg(t) + 0.11mb(t)mi(t) = 0.60mr(t) - 0.28mg(t) - 0.32mb(t)mq(t) = 0.21mr(t) - 0.52mg(t) + 0.31mb(t)Signals mr(t), mg(t) and mb(t) are normalized to a maximum value of 1 so that eachamplitude range lies between 0 to 1my(t) is always > 0mi(t)
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