4. ANALOG TV SIGNALS MEASUREMENT

Transcription

1 Goals of measurement 4. ANALOG TV SIGNALS MEASUREMENT 1) Measure the amplitudes of spectral components in the spectrum of frequency modulated signal of Δf = 50 khz and f mod = 10 khz (relatively to unmodulated carrier amplitude). Check Bessel zeros. 2) Display TV test signals 1 to 8 (except 6 and 7) on oscilloscope and compare one line of video signal with the picture on the monitor. 3) Measure the amplitudes of the signal No. 3 and frequencies of the signal No. 8. Measuring equipments TV signals generator GTS 11 Signal generator Agilent 33220A Oscilloscope Tektronix TDS 3012B Measurement block diagram 75 Ohm generator FM signal generator OSC (FFT) TV generator (75 Ohm) TV monitor (75 Ohm) OSC Introduction Fig. 1: Block diagram left for 1), right for 2) and 3) The TV camera that is on the beginning of the TV chain (camera, transmitter, receiver and cathode ray tube - CRT) converts a visual picture or scene into an electrical signal. The picture is divided into thousands of individual elements (points) and the scanning electron beam scans across the picture line by line from the top to bottom and produces a video signal that is proportional to the light intensity at the individual elements being scanned. Maximum amplitude of video signal corresponds to white points and minimum amplitude to the black ones. At the receiver, the CRT converts the electrical signal back to optical picture. TV camera and CRT output display scanning must be perfectly synchronized. A picture is "drawn" on a TV screen by sweeping an electrical signal horizontally across the screen one line at a time. The amplitude of this signal versus time represents the instantaneous brightness at that physical point on the screen. Figure 2 shows the signal amplitude relationship to the brightness on the screen. At the end of each line, there is a horizontal blanking interval with horizontal synchronization (sync) pulse that tells the scanning circuit to retrace to the left edge of the screen and then start scanning the next line. Starting at the top, all of the lines on the screen are scanned in this way. One complete set of lines makes a picture. This is called a frame. Once the first complete picture is scanned, there is vertical blanking interval with vertical sync pulse that tells the scanning circuit to retrace to the top of the screen and start scanning the next frame, or picture. This sequence is repeated at a fast enough rate so that the displayed images are perceived to have continuous motion. 1/8 ZS ( )

2 Fig.2: Horizontal scan versus screen brightness TV standard D is used in Czech Republic. The screen is divided to 625 lines which repeats with frequency Hz so that each line is 64 μs in duration. The frame frequency is 25 Hz. An interlaced scanning is used, when "painting" the picture on the screen. Each picture, referred to as a frame, is divided into two separate sub-pictures and referred to as fields. Two fields make up a frame. An interlaced picture is painted on the screen in two passes, by first scanning the horizontal lines of the first field and then retracing to the top of the screen and scanning the horizontal lines for the second field in-between the first set. Field 1 consists of odd-numbered lines and field 2 consists of even-numbered lines. The field frequency is thus 50 Hz. As a result of these data the maximum frequency of video signal 6,5 MHz (and TV bandwidth as well) can be determined. TV transmission is usually realized by radio channel. The video signal is amplitude modulated onto a carrier. As the usage of the double side band amplitude modulation would demand a RF bandwidth of 13 MHz a vestigial sideband modulation (v.s.b.) is used. The audio-signal is transmitted separately by means of frequency modulation. A distance between video and audio carriers is 6,5 MHz or 5,5 MHz according to standard used, the total bandwidth (include FM audio signal) is 8 MHz. Color TV principle By monochromatic TV transmission, each pixel of screen gets information about brightness of scanned scene. The color TV has to carry information not only about brightness but also color information must be transmitted. Any system for color TV has to be compatible with black-and-white TV. That is, a monochrome receiver can reproduce a color transmission in black-and-white shades and a monochrome transmission is reproduced black and white by a color receiver. At the color TV transmitter the scene to be televised is actually scanned by three separate cameras, each camera being sensitive only to one of the three colors of red, blue, and green. Since various combinations of these three colors can be mixed to form any color to which the human eye is sensitive, an electrical representation of a complete color scene is possible. The three-color cameras scan the scene in unison, with the red (R), green (G), and 2/8 ZS ( )

3 blue (B) color content separated into three different signals. These signals are fed into the transmitter signal processing circuits and create the luminance signal Y Y = 0,30 R + 0,59 G + 0,11 B or U Y = 0,30 U R + 0,59 U G + 0,11 U B and the chrominance, or color signals (R - Y); (B - Y) or (U R U Y ); (U B U Y ). The Y signal contains just the right proportion of red, blue, and green such that it creates a normal black-and-white picture. It modulates the video carrier just as does the signal from a single black-and-white camera with 6 MHz bandwidth. The color signals are used to modulate the color subcarrier. Quadrature double-sideband suppressed-carrier amplitude modulation (QAM) is used. QAM is a system in which two signals (B Y), (R - Y) modulate two carriers that are at the same frequency but are 90 out of phase with one another. To remain compatible, the same 8-MHz total bandwidth must be used, but more information must be transmitted. A form of multiplexing overcomes this problem. It turns out, that the video signal information is clustered at Hz intervals throughout its 6-MHz bandwidth. Midway between these Hz clusters of information are unused spaces. By generating the color information around just the right color subcarrier (4, MHz) frequency it becomes centered in clusters exactly between black-and-white signals. This is known as interleaving. At the receiver, a monochrome set will detect only the Y signal while the color set is able to detect chrominance signals too and figure out three separate signals R, G, B. They are made to illuminate groups of red, green, and blue phosphor dots on the screen and the original scene is reproduced in color. Sound transmission As we said before, TV transmitter consists of two separate transmitters. The sound transmitter is actually a frequency modulated (FM) system. Let us consider a modulating signal is a pure tone um () t = U m cosω mt. Then a formula for frequency modulated signal takes the form u FM ( t) = U c J n ( β )cos( ω c + nω m ) t. n= Amplitudes of FM frequency components are determined by Bessel functions J n (β ), where Δf β = f m is the modulation index and Δf is the frequency deviation (maximum frequency deviation from the frequency of the unmodulated carrier), which is proportional to the modulating signal amplitude U m. Spectrum FM contains an infinite number of side components (bands) spaced at multiples of the modulating frequency f m = ω m/ 2π above and below the carrier. Fortunately the amplitude of these side bands approaches a negligible level the farther away they are from the carrier, which allows transmission within finite bandwidth. The theoretical infinite spectrum can be limited in an empirical way. For this purposes, the Carson s formula can be used BFM = 2( f m max + Δf max ), where f m max is the maximum modulating frequency and Δf max is the maximum deviation. 3/8 ZS ( )

4 Measurement notes Task 1) As the source of frequency modulated signal use the generator Agilent (Fig.1). Choose the carrier frequency 1 MHz, level 1 V RMS and modulating frequency 10 khz. Measure the amplitudes of sideband frequencies in db in regard to the amplitude of unmodulated carrier (0 dbv) for frequency deviation 50 khz. The FM spectrum will include all spectral components with amplitude over 40 db. Amplitudes of spectral components enter into the table 1, compare them with home prepared results and sign differences. Outline the measured spectrum into the measurement conclusion. Tab. 1: Amplitudes of spectral components for Δ f = 50 khz and f m = 10 khz Values of Bessel functions Measured values Difference n A/A c [ - ] A/A c [ db ] n A/A c [ db ] Bessel func. measured value [ db ] As we can see from Fig. 5, Bessel functions J n (β ) are zero for definite values of β, listed in the table 2. Choose either of β from the table, calculate β f m = Δ f and set it on the generator. Prove at least for two cases that spectral components, accordant with chosen n, have zero value. Tab. 2: Zero values of Bessel functions β Order of root n , , , , , , , , , , , , , , , , , , , ,6160 Task 2) Connect the oscilloscope to the generator of TV test signals GT 11 (Fig. 1). Test signals allow controlling parameters of TV chain as frequency, transmission characteristics and nonlinear distortion. Adjust step-by-step test signals 1 till 8 (except the 6 and 7) on the generator, display them on the oscilloscope and draw picture from monitor in the Table 3. 4/8 ZS ( )

5 Tab. 3: Test TV signals on the oscilloscope and matched picture on the monitor Oscilloscope Monitor 5/8 ZS ( )

6 Oscilloscope Monitor Task 3) For test signal 3 (Fig.4), measure the white, black, blanking and sync. levels and mark the values into the table 4. The test signal 8, which includes reference impulse with amplitude 0,42 V and six frequency beams of the same amplitude (Fig. 5), allows controlling transmission characteristic of the TV chain. Display always only one from six beams (on the screen is only harmonic signal), measure it s frequency and mark the values into the table 5. Keep eye on the structure of the picture on the monitor. 6/8 ZS ( )

7 level [%] [V] white level f 1 f 2 f 3 f 4 f 5 f t black level blanking level sync. level Fig. 4: Test signal No. 3 Fig. 5: Test signal No. 8 Tab.4: Measured amplitude levels Level Standard Level [V] White level 1 Black level 0,35 Blanking level 0,30 Sync. level 0 Measured Level [V] Tab. 5: Frequency of frequency beams f f f f f f H omework shall be done before the measurement starts Find the relative amplitudes of spectral components from graph of Bessel functions (Fig. 6) for modulating frequency 10 khz and frequency deviation 50 khz, convert the levels to db and write them down into table 1. Help: Δf β = = 5. f m Carrier frequency (n = 0) has relative amplitude -0,178 (from Fig. 6), then A/A c [ - ] = 0,178 and 0,178 A/A c [ db ] = 20 log = -14,99 db. 1 Continue for other spectral components (n = 1, 2 8). 7/8 ZS ( )

8 A /Ac [-] carrier, n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 n = 6 n = 7 n = 8 n = 9 Ac is the amplitude of unmodulated carrier (n = 0, β = 0) modulation index β Fig. 6: Bessel functions - relative amplitudes A/A c of the spectral components 8/8 ZS ( )