hmwk 3 sol - EECS 314 Fall 2006 HW 03 Problem 1 Student's...

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EECS 314 Fall 2006 HW 03 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) © 2006 Alexander Ganago Page 1 of 2 The big picture Behind each television screen (traditional, not a flat one) sits a CRT = Cathode Ray Tube. In a CRT, beams of electrons ( cathode ray is an old term for a beam of electrons) are produced by electron guns (one for each color) and deflected by electric fields, which are generated by high-voltage power supplies connected to the pairs of vertical and horizontal plates so that each beam sweeps the entire surface of the screen. To each pair of plates, a high-voltage saw- tooth (ramp) signal is applied, which makes the beam sweep the whole the length (or height) of the screen at a constant speed, and ensures nearly instantaneous return of the beam to the original position (during the return motion the beam is blanked, its brightness made “blacker than black”). A simplified sequence of scans is shown on the sketch (the return motion is shown with dashed lines). Note that the beam moves slowly from top to bottom, and moves fast from left to right, because the frequency of the saw-tooth waveform for vertical deflection is low, while the frequency of the saw-tooth waveform for horizontal deflection is high. Problem (see the sample solution on the next page) The US standard for television involves 30 frames (called fields ) per second; each frame consists of 525 lines (out of which only 483 are visible; others – at the edges of the screen – are blanked). For the ramp signal used for the horizontal (line) sweep, calculate: a. The frequency in Hz b. The period in μ sec c. The Rise Time in μ sec.
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EECS 314 Fall 2006 HW 03 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) © 2006 Alexander Ganago Page 2 of 2 Solution by Alexander Ganago The frequency of the ramp used for horizontal scan equals (30 Hz) (525) = 15 750 Hz. The period equals 1 / (15 750 Hz) = 63.49 10 -6 sec = 63.49 μ sec The Rise Time equals 0.8 (period) = 50.79 μ sec
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Problem 2 Solution (Barry Mullins) f = 30kHz A = V ppk = 3000V The frequencies of the n th harmonic will be n*f. The amplitude (in V ppk ) of the n th harmonic of a saw-tooth waveform is given by: A n = 2 A π n To convert this to V pk and V RMS , use V pk = V ppk 2 = A n 2 V RMS = V 2 è!!! 2 = A n 2 è!!! 2 The table can then be filled out. Component Frequency (kHz) Vpk (V) Vrms (V) Fundamental 30 954.96 675.26 2nd Harmonic 60 477.48 337.63 3rd Harmonic 90 318.32 225.09 4th Harmonic 120 238.74 168.81 5th Harmonic 150 190.99 135.05 Calculation for third harmonic: f 3 = 3f = 90Hz V = A n 2 = 2 H 3000 V L 2 H 3 π = 318.32 V L V = A n 2 è!!! 2 = 2 H V L 2 H 3 π L è!!! 2 = 225.09 V Problem 2 Solution Page 1 of 1
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EECS 314 Fall 2006 HW 03 Problem 3 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) The big picture The spectrum of a saw-tooth wave at the frequency f and peak amplitude A (which has zero average) has both odd-numbered and even-numbered components: () 1 n = integer Saw-tooth wave of peak amplitude A = 2 1s i n ( 2 2 =s i n ( 2 ) 2 sin(2 2 180 ) 2 2 sin(2 3 ) 3 2 sin(2 4 180 ) . ..
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hmwk 3 sol - EECS 314 Fall 2006 HW 03 Problem 1 Student's...

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