b V 2 V c t p 02 ms d Amplitude 8 V 2 V 6 V e tilt 1 2 100 V V V V 8V 75 V 2

B v 2 v c t p 02 ms d amplitude 8 v 2 v 6 v e tilt 1

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b V = 2 V c. t p = 0.2 ms d. Amplitude = 8 V 2 V = 6 V e. % tilt = 1 2 100% V V V × V = 8V + 7.5 V 2 = 7.75 V % tilt = 8 V 7.5 V 7.75 V × 100% = 6.5% 2. a. negative-going b. +7 mV c. 3 μ s d. 8 mV ( from base line level) e. V = 8 mV 7 mV 15 mV = 2 2 = 7.5 mV % Tilt = 1 2 8 mV ( 7 mV) 100% 100% 7.5 mV V V = V − − × × = 1mV 100% 7.5 mV × = 13.3% f. T = 15 μ s 7 μ s = 8 μ s prf = 1 T = 1 8 s μ = 125 kHz g. Duty cycle = p t T × 100% = 3 s 8 s μ μ × 100% = 37.5% 3. a. positive-going b. b V = 10 mV c. t p = 8 10 4 ms = 3.2 ms d. Amplitude = (30 10)mV = 20 mV e. % tilt = 1 2 100% V V V × V = 30 mV + 28 mV 2 = 29 mV % tilt = 30 mV 28 mV 29 mV × 100% 6.9%
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334 CHAPTER 24 4. t r (0.2 div.)(2 ms/div.) = 0.4 ms t f (0.4 div.)(2 ms/div.) = 0.8 ms 5. tilt = 1 2 V V V = 0.1 with V = 1 2 2 V V + Substituting V into top equation, 1 2 1 2 2 V V V V + = 0.1 leading to V 2 = 1 0.95 1.05 V or V 2 = 0.905(15 mV) = 13.58 mV 6. a. t r = 80% of straight line segment = 0.8(2 μ s) = 1.6 μ s b. t f = 80% of 4 μ s interval = 0.8(4 μ s) = 3.2 μ s c. At 50% level (10 mV) t p = (8 1) μ s = 7 μ s d. prf = 1 1 20 s = T μ = 50 kHz 7. a. T = (4.8 2.4)div. [ ] 50 s/div. μ = 120 μ s b. f = 1 1 120 s T μ = = 8.33 kHz c. Maximum Amplitude: (2.2 div.)(0.2 V/div.) = 0.44 V = 440 mV Minimum Amplitude: (0.4 div.)(0.2 V/div.) = 0.08 V = 80 mV 8. T = (3.6 2.0)ms = 1.6 ms prf = 1 1 1.6 ms = T = 625 Hz Duty cycle = 0.2 ms 100% 1.6 ms p t = T × × 100% = 12.5%
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CHAPTER 24 335 9. T = (15 7) μ s = 8 μ s prf = 1 1 8 s T μ = = 125 kHz Duty cycle = (20 15) 5 100% 100% = 100% 8 8 p t s T s μ μ × = × × = 62.5% 10. T = (3.6 div.)(2 ms/div.) = 7.2 ms prf = 1 1 7.2 ms = T = 138.89 Hz Duty cycle = 1.6 div. 100% 3.6 div. p t = T × × 100% = 44.4% 11. a. T = (9 1) μ s = 8 μ s b. t p = (3 1) μ s = 2 μ s c. prf = 1 1 8 s T μ = = 125 kHz d. V av = (Duty cycle)(Peak value) + (1 Duty cycle)( V b ) Duty cycle = 2 s 100% 100% 8 s p t T μ μ × = × = 25% V av = (0.25)(6 mV) + (1 0.25)( 2 mV) = 1.5 mV 1.5 mV = 0 V or V av = (2 s)(6 mV) (2 s)(6 mV) 8 s μ μ μ = 0 V e. V eff = 6 6 (36 10 )(2 s) (4 10 )(6 s) 8 s μ μ μ × + × = 3.46 mV 12. Eq. 24.5 cannot be applied due to tilt in the waveform. (Method of Section 13.6) Between 2 and 3.6 ms V av = 1 (3.4 ms 2 ms)(2 V) + (3.6 ms 3.4 ms)(7.5 V) + (3.6 ms 3.4 ms)(0.5 V) 2 3.6 ms 2 ms = 1 (1.4 ms)(2 V) + (0.2 ms)(7.5 V) + (0.2 ms)(0.5 V) 2 1.6 ms = 2.8 V + 1.5 V + 0.05 V 1.6 = 2.719 V
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336 CHAPTER 24 13. Ignoring tilt and using 20 mV level to define t p t p = (2.8 div. 1.2 div.)(2 ms/div.) = 3.2 ms T = (at 10 mV level) = (4.6 div. 1 div.)(2 ms/div.) = 7.2 ms Duty cycle = 3.2 ms 100% 100% 7.2 ms p t T × = × = 44.4% V av = (Duty cycle)(peak value) + (1 Duty cycle)( V b ) = (0.444)(30 mV) + (1 0.444)(10 mV) = 13.320 mV + 5.560 mV = 18.88 mV 14. V av = (Duty cycle)(Peak value) + (1 Duty cycle)( V b ) Duty cycle = p t T (decimal form) = (8 1) s 20 s μ μ = 0.35 V av = (0.35)(20 mV) + (1 0.35)(0) = 7 mV + 0 = 7 mV 15. Using methods of Section 13.8: A 1 = b 1 h 1 = [(0.2 div.)(50 μ s/div.)][(2 div.)(0.2 V/div.)] = 4 μ sV A 2 = b 2 h 2 = [(0.2 div.)(50 μ s/div.)][(2.2 div.)(0.2 V/div.)] = 4.4 μ sV A 3 = b 3 h 3 = [(0.2 div.)(50 μ s/div.)][(1.4 div.)(0.2 V/div.)] = 2.8 μ sV A 4 = b 4 h 4 = [(0.2 div.)(50 μ s/div.)][(1 div.)(0.2 V/div.)] = 2.0 μ sV A 5 = b 5 h 5 = [(0.2 div.)(50 μ s/div.)][(0.4 div.)(0.2 V/div.)] = 0.8 μ sV V av = (4 4.4 2.8 2.0 0.8) sV 120 s μ μ + + + + = 117 mV 16. Using the defined polarity of Fig. 24.57 for υ C , V i = 5 V, V f = +20 V and τ = RC = (10 k Ω )(0.02 μ F) = 0.2 ms a. υ C = V i + ( V f V i )(1 e t / τ ) = 5 + (20 ( 5))(1 e t /0.2 ms ) = 5 + 25(1 e t /0.2 ms ) = 5 + 25 25 e t /0.2 ms υ C = 20 V 25 V e t /0.2 ms
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CHAPTER 24 337 b.
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