EE 351 · Week 7 · Study article

Television Engineering — From Images to RF Signals

A stand-alone reading companion to the Week 7 slides: one picture, followed all the way from a camera to a 6-MHz radio channel.

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1. Television as a communication problem

Radio transmits sound, which is one number at each instant: the pressure \(p(t)\). A picture is a number at every point, the brightness \(I(x,y)\), and a moving picture adds time, \(I(x,y,t)\). Yet one antenna and one radio channel can only carry one value at a time.

The solution is to serialize the picture into a one-dimensional voltage \(v(t)\) by scanning; to add synchronization so the receiver knows where each instant belongs; to modulate the voltage onto a radio-frequency carrier; and to multiplex colour and sound with it in frequency. The whole chain is:

It reuses ideas from earlier weeks: the sidebands of DSB-AM (Week 3), the single-sideband idea and quadrature branches (Week 4), frequency-division multiplexing of a stereo subcarrier (Week 5), and FM for the sound (Week 6).

2. Raster scanning and the video signal

A camera examines the picture in horizontal strips, left to right and top to bottom. Horizontal position becomes time within a line, and vertical position becomes the line number. The standard picture has a 4:3 aspect ratio.

As a worked example, take a black letter F on a white background. Each scan line gives a waveform that is high where the beam sees white and low where it crosses the black letter:

Four ideas must be kept apart: visible scanning (the trace, which is the picture); horizontal retrace (the physical flyback of the beam); horizontal blanking (a signal level below black so that the beam is off, containing the retrace); and synchronization (a pulse inside the blanking interval that marks each line start).

The line period is \(T_H = 1/f_H = 63.56\ \mu\text{s}\), of which about 10.9 µs is blanking and about 52.7 µs is active picture. (The textbook uses the rounder figures of about 10 µs and 53.5 µs.) In the plots here white is high, but the RF picture carrier uses negative modulation: sync tips give the largest carrier amplitude.

3. Interlaced scanning and the NTSC numbers

Flicker is hidden only if the screen is refreshed roughly 50–60 times per second, but sending 60 complete frames per second would double the video bandwidth. Interlacing compromises: field 1 draws the odd lines and field 2 the even lines in between. The screen is refreshed 60 times a second while full detail is sent only about 30 times a second. The cost is combing on fast motion and line twitter on thin horizontal detail.

When colour was added, the line frequency was tied to the 4.5-MHz sound spacing, which lowered the original 15 750 Hz, 60 Hz and 30 Hz by a factor of 1000/1001.

The half line makes both fields equally long and gives them the same retrace distance, so a single master oscillator and dividers can generate all the timing. After each field about 20 line periods are blanked for the vertical retrace; they contain equalising and broad pulses that let the receiver recognise a field boundary. That is why only about 485 of the 525 lines can carry picture.

Line standard, colour encoding and RF channel plan are separate choices. PAL-M pairs 525/60 lines with PAL colour in a 6-MHz channel; PAL B/G uses 7- or 8-MHz channels; other 625-line systems have other sound spacings. Only System M is analysed here.

4. Video bandwidth

Scan across alternating black and white vertical stripes. The voltage is a square wave. If the active scan time is \(T_{\text{active}}\) and the line holds \(N\) alternating elements, there are \(N/2\) cycles, each lasting \(2T_{\text{active}}/N\). The fundamental frequency is therefore

Units: elements per (2 × seconds) is hertz; with \(T_{\text{active}}\) in microseconds the answer is in megahertz. Assumptions: the worst-case alternating pattern, only the fundamental counted, and \(T_{\text{active}}\) excluding blanking. Sharp edges need harmonics beyond the fundamental, so the formula estimates the highest useful video frequency; it does not give an exact standard.

The 4.2-MHz NTSC case

At 4.2 MHz the cycle period is 0.238 µs and one element is 0.119 µs. With about 53.5 µs of active time the line holds \(N=53.5/0.119\approx 449.5\) elements, so \(R_H\approx 450\). (With 52.7 µs the result is about 443.) This is horizontal resolution across the full picture width; per picture height, a 4:3 picture reduces it to about \(450\times\frac34\approx 337\) TV lines. The vertical resolution is about \(R_V\approx 0.7\,N_L\), roughly \(0.7\times485\approx340\) lines, so horizontal and vertical detail are balanced.

A PAL-based example (textbook Example 23-2)

625 lines, \(f_H=15\,625\) Hz, 80 % active, \(R_H\approx512\), about 580 displayed lines. \(T_H=64\ \mu\text{s}\), \(T_{\text{active}}=51.2\ \mu\text{s}\); one element lasts \(51.2/512=0.1\ \mu\text{s}\), a cycle \(0.2\ \mu\text{s}\), so \(B\approx5\) MHz; \(R_V=0.8\times580\approx464\) lines.

5. Colour without tripling the bandwidth

A colour camera has three sensors, producing \(R'\), \(G'\) and \(B'\) (the primes mean gamma-corrected). Sending three full-bandwidth signals would triple the spectrum and break compatibility with the black-and-white sets already in homes. NTSC instead sends the brightness exactly as before and adds a narrow colour signal.

The textbook calls this signal \(Y\); strictly it is the luma \(Y'\). A monochrome receiver uses only \(Y'\) and the sync. The colour information is the pair of colour differences, rotated by 33° into the I and Q signals:

I is allowed about 1.5 MHz and Q about 0.5 MHz, because the eye resolves orange–cyan detail better than green–magenta detail. I and Q amplitude-modulate two copies of a 3.579545-MHz subcarrier that are 90° apart; each modulator is DSB suppressed-carrier AM. (The chapter’s opening paragraph says 3.85 MHz, but its detailed treatment and the standard use 3.579545 MHz, which is used here.)

Sine and cosine at one frequency are orthogonal, so a synchronous detector recovers them separately: \(\text{LPF}\{2C\sin\omega_{sc}t\}=U\), \(\text{LPF}\{2C\cos\omega_{sc}t\}=V\). Writing \(C(t)=M\cos(\omega_{sc}t+\varphi)\), the amplitude \(M=\sqrt{U^2+V^2}\) is the saturation and the phase \(\varphi\), measured against the burst, is the hue.

A picture’s spectrum clusters at multiples of \(f_H\). Placing the subcarrier at an odd multiple of \(f_H/2\) interleaves the chrominance lines between the luminance lines, so both can share the upper part of the 4.2-MHz video band with limited mutual interference.

Because the subcarrier is suppressed, the receiver’s oscillator has no phase information. About 8–12 cycles (nominally 9) of the unmodulated subcarrier are therefore placed on the back porch of every line. A phase-locked loop compares to this colour burst so demodulation uses the transmitter’s phase reference; a phase error rotates every hue by the same angle (the old “tint” control).

6. Vestigial-sideband modulation

Ordinary DSB-AM of a video signal with \(B=4.2\) MHz needs \(2B=8.4\) MHz for the picture sidebands alone, more than the 6-MHz channel before the sound is added. The lower sideband duplicates the upper one. Single-sideband (SSB) would need only \(B\), but video has important energy at and near 0 Hz (average brightness and large areas), which in RF terms lies right next to the carrier. An SSB filter would need a vanishingly narrow transition there.

Vestigial sideband keeps one full sideband and a controlled remnant, the vestige, of the other. Let the DSB spectrum be shaped by a filter \(H(f)\). Coherent detection returns

so the picture is recovered without distortion if \(H(f_c+f)+H(f_c-f)\) is constant for \(|f|\le B\). Above the vestige width \(f_v\) one side passes completely and the other is blocked (sum = 1). Within \(\pm f_v\) the slope must have odd symmetry about the carrier and pass through 50 % at \(f_c\): what the lower side lacks the upper side supplies. If the vestige were simply kept flat and then cut off, the sum would be 2 near DC and the low frequencies would be doubled. The Week 7 sideband explorer shows both cases.

In practice the transmitter removes most of the lower sideband and the receiver’s IF filter (a SAW filter in the textbook’s era) supplies the final Nyquist slope, with the picture carrier at its 50 % point. US receivers use picture and sound IFs of 45.75 MHz and 41.25 MHz; for channel 4 (picture 67.25 MHz, sound 71.75 MHz) a 113-MHz oscillator gives 113 − 67.25 = 45.75 MHz and 113 − 71.75 = 41.25 MHz, mirroring the spectrum.

7. The 6-MHz analog channel

With \(L\) the lower channel edge, the picture carrier is at \(L+1.25\) MHz (AM, vestigial sideband), the colour subcarrier is 3.579545 MHz above it, and the FM sound carrier is 4.5 MHz above the picture carrier, at \(L+5.75\) MHz, with a maximum deviation of ±25 kHz. For example channel 13 occupies 210–216 MHz: picture \(210+1.25=211.25\) MHz, colour subcarrier 214.83 MHz, sound \(211.25+4.5=215.75\) MHz.

4.2 MHz versus 6 MHz. 4.2 MHz is the highest frequency of the baseband video; 6 MHz is the width of the allocated RF channel. From the lower edge: 1.25 MHz to the picture carrier (holding the vestige), 4.2 MHz of upper video sideband, 0.3 MHz to the sound carrier and 0.25 MHz above it. The figure is idealised; the real transmitted spectrum is bounded by a regulatory emission mask and by the receiver’s filtering.

8. Historical analog versus present practice

Full-power analog broadcasting in the United States ended in 2009 and was replaced by the digital ATSC system, which also fits in 6-MHz channels and uses an 8-level vestigial-sideband modulation (8-VSB). Digital broadcasts send compressed bits rather than a scanned voltage. Interlacing survives in formats such as 1080i. Treat the analog sections of the textbook chapter as a statement of engineering principles, not of what a modern receiver does.

9. Optional MATLAB activity

week07_tv_video_chain.m is a self-contained script in plain MATLAB (no toolboxes, no Simulink). It (1) turns an image row into a sampled video line with sync, porches and active video; (2) takes the FFT of a 100-stripe line and compares the peak with \(N/(2T_{\text{active}})\); and (3) amplitude-modulates a multi-line video signal (the RF carrier is scaled down to 8 MHz) and compares DSB, VSB with an odd-symmetric slope, and VSB with an abrupt vestige.

Verification statement. The script was run in MATLAB R2024b in batch mode (figures hidden) on 20 October 2026 and completed without error. Its built-in assertions reported: FFT peak 0.9496 MHz versus \(N/(2T_{\text{active}})=0.9496\) MHz; \(N_{\max}=2T_{\text{active}}B\approx442\); RMS recovery error 0.0000 (DSB), 0.0001 (VSB, Nyquist slope) and 1.03 (VSB, abrupt vestige); and \(H(f_c+f)+H(f_c-f)=1\) for the Nyquist slope. It was not run on other MATLAB releases, and the figure windows were not inspected visually. No Simulink model is provided this week. The HTML demonstrations do not need MATLAB.

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