Television Engineering — From Images to RF Signals

That requirement forces four engineering tools: scanning (serialize), synchronization (re-assemble), modulation (reach the channel), and frequency multiplexing (add colour and sound).

Part B

Raster Scanning and Video Signal Generation

Image → scan line → brightness waveform → sequence of lines → reconstructed image.

A camera sensor (a vidicon tube historically, a CCD today) examines the picture one narrow strip at a time. Along each strip it outputs a voltage proportional to brightness.

• Horizontal position \(x\) becomes time within a line.
• Vertical position \(y\) becomes which line — i.e. time within a field.

At the receiver the process is reversed: the same waveform drives the brightness of a beam that sweeps the screen in step with the camera. Image → line → waveform → lines → image.

Guided solution →

• The receiver has its own sweep oscillators. Left alone, analog circuits drift, so lines would start at slightly different positions and the picture would skew sideways.
• The transmitter therefore adds a sync pulse to every line (and a longer pattern at the end of every field). The receiver’s sync separator finds those pulses because they go below the black level, a range the picture never uses.
• Sync is also what tells the receiver which line is the first line of a field.

Guided solution →

Part C

Interlaced Scanning

Fields, frames, half lines, and the NTSC timing numbers.

The price: fast-moving edges are in different places in the two fields. They look jagged (“combing”) when the two fields are shown together, and thin horizontal detail can flicker (“line twitter”).

The pulses use a double-rate pattern (equalising pulses) so that field 1 and field 2, which begin half a line apart, look the same to the receiver’s sync circuits.

Guided solution →

Part D

Video Signal Bandwidth

From picture detail to hertz.

1. Let the active scan time be \(T_{\text{active}}\), and let \(N\) be the number of alternating black and white elements along one line.
2. One cycle = one black + one white element, so the line holds \(N/2\) cycles.
3. The period of one cycle is \(T_0 = T_{\text{active}} / (N/2) = 2T_{\text{active}}/N\).
4. The fundamental frequency of the square wave is its reciprocal:

Doubling the detail (\(N\)) doubles the frequency. Halving the active time (scanning twice as fast) also doubles it. Higher picture detail or faster scanning demands more bandwidth.

So the result is an estimate of the highest useful video frequency, good to a few per cent, not a precise standard.

1. Cycle period at the limit: \(1/4.2\text{ MHz} = 0.238\ \mu\text{s}\). One element (half a cycle): \(0.119\ \mu\text{s}\).
2. Textbook active time: about \(53.5\ \mu\text{s}\) (a 63.6-µs line less roughly 10 µs of horizontal blanking).
3. Elements along the line: \(N = 53.5/0.119 \approx 449.5\), so \(R_H \approx 450\).

Only some of the visible lines are independently resolvable, since lines and picture details are not perfectly aligned. Notice that the ≈ 337 TVL horizontal figure and the ≈ 340-line vertical figure are close: NTSC was designed so that horizontal and vertical detail are about balanced.

Always say which one you mean: across the full width, or per picture height.

Given: 625 lines, 25 frames/s, \(f_H = 15\,625\) Hz, 80 % of each line active, \(R_H \approx 512\), about 580 displayed lines. Find the bandwidth and vertical resolution.

1. \(T_H = 1/15\,625 = 64\ \mu\text{s}\); \(T_{\text{active}} = 0.8 \times 64 = 51.2\ \mu\text{s}\).
2. One element: \(51.2/512 = 0.1\ \mu\text{s}\); one cycle: \(0.2\ \mu\text{s}\).
3. \(B \approx 1/(0.2\ \mu\text{s}) = 5\ \text{MHz}\) — the same as \(f=N/(2T)=512/(2\times51.2\ \mu\text{s})\).
4. \(R_V = 0.8 N_L = 0.8 \times 580 \approx 464\) lines.

Guided solution →

Guided solution →

Part E

Adding Colour Without Tripling the Bandwidth

RGB, luminance, chrominance, and quadrature multiplexing.

Any colour can be made by mixing red, green and blue light, so a colour camera is really three cameras behind colour filters, producing \(R'\), \(G'\), \(B'\) signals (the prime marks that each has been gamma-corrected for the display).

• Sending three full-bandwidth signals would need three times the spectrum.
• It would also be incompatible: millions of black-and-white receivers already existed in the 1950s.
• The eye resolves fine brightness detail far better than fine colour detail, so colour can be sent with much less bandwidth than brightness.

• The weights reflect how bright the eye finds each primary: green most, blue least. The textbook writes them as 30 %, 59 % and 11 %.
• Notation: the textbook calls this \(Y\). Strictly it is \(Y'\) (“luma”), computed from gamma-corrected \(R'G'B'\), not true photometric luminance. This lesson writes \(Y'\) for the signal and \(Y\) only where quoting the textbook.
• White (\(R'=G'=B'=1\)) gives \(Y'=1\); any grey gives \(Y'\) equal to that grey level; pure blue gives only \(0.114\).

Compatibility works in both directions: a colour broadcast shows a normal grey picture on an old set, and a monochrome broadcast has no colour, so the colour set shows a normal grey picture.

Brightness is already sent as \(Y'\), so only two more numbers per point are needed. Take colour differences, which are zero for any grey:

NTSC rotates this pair by 33° into I (in-phase) and Q (quadrature), chosen to match how sharply the eye sees orange–cyan versus green–magenta detail:

The textbook rounds these to I = 60 % R − 28 % G − 32 % B and Q = 21 % R − 52 % G + 31 % B. Because I carries the detail the eye resolves best it is allowed about 1.5 MHz; Q is limited to about 0.5 MHz. (Many later receivers demodulate with equal narrow bandwidths; the I/Q split is the historical design.)

This is the same idea as Week 4’s quadrature branches. A cosine and a sine at the same frequency are orthogonal, so two independent signals can be sent and separated:

A synchronous detector multiplies by \(2\sin(\omega_{sc}t)\) or \(2\cos(\omega_{sc}t)\) and low-pass filters; the cross terms average to zero:

Each modulator is a DSB suppressed-carrier AM stage: no subcarrier is transmitted, only sidebands. That is why the receiver must regenerate the subcarrier, with the right phase.

Phase error rotates every hue by the same angle: a 20° error shifts all colours around the colour wheel, which is what the old “tint” control adjusted.

Guided solution →

Part F

Vestigial-Sideband Modulation

Getting 4.2 MHz of video onto an RF carrier.

Week 3: amplitude-modulating a carrier with a message of maximum frequency \(B\) creates an upper and a lower sideband, each \(B\) wide:

• That is for the picture sidebands alone. Adding the FM sound carrier 4.5 MHz above the picture carrier makes the total even wider.
• The whole TV channel is 6 MHz. Ordinary DSB-AM does not fit, even before any guard band.
• The lower sideband duplicates the upper one. It is the redundant half.

• Video has important energy at and near 0 Hz: the average brightness of a whole picture is a DC component, and slowly changing areas of the picture are very low frequencies. In RF terms that is right next to the carrier.
• SSB would need a filter that passes everything above \(f_c\) and kills everything below it, with a vanishingly narrow transition. That cannot be built with practical filters at VHF/UHF.
• Any gradual cut near the carrier damages the very frequencies carrying picture brightness and large-area tone, producing smeared, uneven pictures.

Let the transmit signal be the DSB spectrum shaped by a filter \(H(f)\). Coherent detection (multiply by the carrier, low-pass filter) returns

The picture is recovered without distortion if the bracket is constant for all \(|f|\le B\):

• Above the vestige width \(f_v\): \(H(f_c+f)=1\) and \(H(f_c-f)=0\), sum = 1.
• Inside \(\pm f_v\): the slope must have odd symmetry about the carrier, passing through 50 %: what the lower side lacks the upper side contributes, and vice versa.
• Vestige abruptly kept (no slope): the sum is 2 near DC, so low frequencies are doubled.

• The transmitter removes most of the lower sideband; the receiver’s IF filter (a surface-acoustic-wave, SAW, filter in sets of the textbook’s era) supplies the final precise Nyquist slope.
• The slope is arranged so the picture carrier sits at the 50 % (−6 dB) point of the response, giving \(H(f_c+f)+H(f_c-f)=\text{const.}\) automatically.
• The same filter strongly attenuates the sound IF, so the sound does not leak into the video detector.
• US standard IFs: picture 45.75 MHz, sound 41.25 MHz. Example: channel 4, picture 67.25 MHz, sound 71.75 MHz, local oscillator 113 MHz, so \(113-67.25 = 45.75\) MHz and \(113-71.75 = 41.25\) MHz. Because the oscillator is above the signal, the spectrum is mirrored at the IF.

An envelope detector works because the picture carrier is large; later receivers use synchronous detection, which reduces the distortion that VSB envelope detection can introduce.

Part G

The Complete Analog Television RF Channel

NTSC System M: picture, colour, sound and a 6-MHz allocation.

Budget from the lower channel edge \(L\): 1.25 MHz up to the picture carrier (this region holds the vestige) + 4.2 MHz of upper video sideband + 0.3 MHz to the sound carrier + 0.25 MHz above the sound carrier = 6 MHz.

Guided solution (including the VSB-versus-DSB explanation) →

• Everything above describes analog NTSC System M. Full-power analog broadcasting in the United States ended in 2009, replaced by the digital ATSC system, which also fits in 6-MHz channels.
• Digital broadcasts compress the picture and send bits, not a scanned voltage. ATSC 1.0 uses an 8-level vestigial-sideband modulation (8-VSB), so VSB survived even though analog video did not.
• Interlacing, however, did not disappear: formats such as 1080i still scan two fields per frame.
• The textbook chapter you used is historical for its analog TV sections: use it for the engineering principles, not as a description of what a modern receiver does.

The five demos need only a browser. This optional activity repeats the chain numerically:

1. Turn rows of a test image into a sampled brightness waveform (with blanking and sync).
2. Plot that waveform and its FFT, and find the stripe frequency \(N/(2T)\).
3. Amplitude-modulate it onto a carrier and compare DSB with VSB filter responses.

Download week07_tv_video_chain.mRead the notes →

Plain MATLAB; no toolbox or Simulink model is required, and none is provided this week. See the study article for the exact verification statement.

EE 351 · Week 7

From a picture to a 6-MHz RF channel

• Scanning turns \(I(x,y,t)\) into one voltage; sync keeps the receiver in step.
• NTSC: 525 lines, 29.97 frames/s (two fields, 59.94 Hz); \(T_H=63.56\) µs.
• Detail costs bandwidth: \(f=N/(2T_{\text{active}})\); about 4.2 MHz for NTSC.
• Colour: \(Y'\) plus quadrature I/Q on 3.579545 MHz, locked by the burst.
• VSB: one sideband plus a vestige, with \(H(f_c+f)+H(f_c-f)\) constant.
• Channel: picture \(L+1.25\) MHz, FM sound 4.5 MHz above it, all in 6 MHz.