Television Engineering — Guided Practice

1. Map position to time. Each row is one scan line, and the 8 columns are 8 equal slices of the active time (the blanking interval is ignored in this sketch).
2. Read each row left to right. Row 3: W W B B W W W W. Row 6: W W B B B B B W. Row 8: all W.
3. Convert to levels. White → high; black → low. A change of colour is a step in the voltage.

Interpretation. The vertical bar is the same on every row, so every row crosses it at the same time offset; rows 6 and 7 also cross the foot, so their low interval is longer (5 cells instead of 2). A wider black feature produces a longer low interval, and a feature that is the same on consecutive rows produces identical consecutive waveforms.

1. (a) \[T_H = \frac{1}{f_H} = \frac{1}{15\,734\ \text{Hz}} = 63.56\times10^{-6}\ \text{s} = 63.56\ \mu\text{s}\].
2. (b) \[T_{\text{active}} = T_H - T_{\text{blank}} = 63.56-10.9 = 52.66\ \mu\text{s}\approx 52.7\ \mu\text{s}\]. Fraction \(=52.66/63.56 = 0.829\), about 83 %.
3. (c) During the active interval the beam moves left to right and its brightness follows the video voltage: this is the visible picture. During blanking (10.9 µs = front porch 1.5 + sync 4.7 + back porch 4.7) the signal is below black, the beam is off, and the beam flies back (retrace) and the sweep circuit resets.

Interpretation. About 17 % of every line carries no picture. The textbook rounds to a 63.6-µs line, ≈ 10 µs of blanking and 53.5 µs active; the difference changes later bandwidth estimates by about 2 %. Retrace is a physical motion; blanking is a signal level that hides it and also carries the sync pulse and, in colour, the burst.

1. (a) Field rate \(=2\times25 = 50\) fields/s.
2. (b) Lines per field \(=625/2 = 312.5\).
3. (c) Line frequency:\[625\ \text{lines/frame}\times25\ \text{frames/s}=15\,625\ \text{Hz}\]Check: \(312.5\times 50 = 15\,625\) Hz. Line period \(=64\ \mu\)s.
4. (d) One frame is a complete picture. It is built from two fields, each containing every second line: field 1 the odd lines, field 2 the even lines. Lines per frame \(=\) lines per field \(\times 2\); line rate \(=\) lines per frame \(\times\) frame rate.

Interpretation. For an odd line count the number of lines per field is a half-integer (312½ here, 262½ in NTSC), so each field ends or starts in the middle of a line. This is deliberate: both fields have the same duration, so one master oscillator and dividers drive all the timing. For NTSC colour, \(525\times29.97=15\,734\) Hz and \(2\times29.97=59.94\) fields/s.

1. (a) One black + one white stripe is one cycle, so there are \(N/2=120\) cycles in \(52.7\ \mu\)s. \[ f=\frac{N}{2T_{\text{active}}}=\frac{240}{2\times52.7\ \mu\text{s}}=2.277\ \text{MHz} \] Units: elements \(/\ \mu\)s \(=\) MHz.
2. (b) \(2.28\ \text{MHz} < 4.2\ \text{MHz}\): yes, the fundamental passes.
3. (c) A square wave also contains odd harmonics: \(3f=6.83\) MHz and \(5f=11.4\) MHz. Both lie above 4.2 MHz and are removed. Only the fundamental survives, so the stripes appear as soft, sinusoidal bars, not crisp black-and-white edges.

Interpretation. Bandwidth limits detail in two ways: the finest pattern whose fundamental fits (here \(N_{\max}=2T_{\text{active}}B\approx443\)), and the sharpness of edges, which needs harmonics. Try 240 stripes in the Video Bandwidth explorer.

1. (a) \[B=\frac{N}{2T_{\text{active}}}=\frac{600}{2\times51.2\ \mu\text{s}}=5.86\ \text{MHz}\].
2. (b) 5.86 MHz exceeds 5 MHz, so a 5-MHz band cannot fully resolve 600 elements: it resolves about \(2\times51.2\times5=512\), the textbook’s figure.
3. (c) Per picture height: \(600\times\tfrac34 = 450\) TV lines.

Interpretation. Bandwidth, active line time and horizontal resolution are tied together by one relation. To raise detail you must raise bandwidth (a wider channel) or lengthen the active time (a lower line rate, hence less vertical detail or a lower frame rate). Always say whether a figure is per full width or per picture height.

1. (a) \[Y'=0.299(0.8)+0.587(0.4)+0.114(0.2)=0.2392+0.2348+0.0228=0.497\].
2. (b) \(R'-Y'=0.8-0.497=0.303\); \(B'-Y'=0.2-0.497=-0.297\).
3. (c) \(U=0.492(-0.297)=-0.146\), \(V=0.877(0.303)=0.266\). Amplitude: \[M=\sqrt{U^2+V^2}=\sqrt{0.0213+0.0708}=0.303\] Phase: \[\varphi=\operatorname{atan2}(V,U)=\operatorname{atan2}(0.266,-0.146)\approx 119^\circ\] which is close to the I axis at 123° (orange).
4. (d) Brightness already occupies the video band as \(Y'\); two further colour signals must fit in the same 4.2 MHz. A cosine and a sine at one frequency are orthogonal, so each colour signal can modulate one of them (DSB-SC) and still be recovered separately by a synchronous detector. Two signals therefore share one subcarrier frequency instead of needing two.

Interpretation. The phase 119° says which colour (orange-ish) and the amplitude 0.30 says how saturated. A grey pixel has \(R'=G'=B'\), so \(U=V=0\) and no chrominance is sent. The receiver must regenerate the subcarrier in the right phase, which is why every line carries a colour burst. Check the numbers in the Chrominance explorer.

1. (a) Picture carrier \(=210+1.25=211.25\) MHz. Sound carrier \(=211.25+4.5=215.75\) MHz. Colour subcarrier \(=214.829545\) MHz (\(211.25+3.579545\)).
2. (b) \(216-215.75=0.25\) MHz: a guard region above the FM sound.
3. (c) DSB-AM of 4.2-MHz video needs \(2B=8.4\) MHz for the picture sidebands alone, which does not fit in a 6-MHz channel together with the sound. SSB would need only 4.2 MHz, but video has major energy at and near 0 Hz, immediately next to the carrier, so the filter would need an impractically abrupt cut there. VSB keeps the whole upper sideband and a controlled vestige (about 0.75 MHz, rolling off by the channel edge) of the lower one, so the picture needs about \(4.2+0.75\approx5\) MHz and the filter slope can be gentle, provided it has odd symmetry about the carrier.

Interpretation. The 1.25 MHz below the picture carrier holds the vestige; the 4.2 MHz above carries the video; 0.3 MHz up to the sound carrier and 0.25 MHz above it complete the 6 MHz. Note the distinction between the 4.2-MHz baseband video limit and the 6-MHz RF channel. This is the historical analog allocation; present-day digital broadcasting in the same channels is a different system.