Lesson+6 - The re-engineering of history 1 Lesson 6 Lesson...

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Lesson 6 1 The re-engineering of history.
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Sinusoids, Impedances, and Steady State Response Chapters 9-10 Lesson 6 Lesson 6 2
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Challenge 5 Lamp resistance R x = 80 . Lamp fires when lamp voltage is 50 V x 70 V. Discharge circuit time constant =1ms (given) Peak power dissipation P R1 100 mW (packaging limitation). Lesson 6 + - V 1 =60V R 1 Charging Circuit R x + - V x Discharge Circuit I 1 I x C Xenon Lamp 3
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Lesson 6 Charging Circuit Peak Power Constraint: P(t)= I 1 2 R 1 0.1W I 1 (peak) = max( V 1 - V C )/R 1 . Peak power dissipated occurs when V C (0)=0 (fully discharged). P 1 (peak) = I 1 (peak) 2 R 1 = {(60/R 1 ) 2 }R 1 = (60) 2 /R 1 0.1W Requires that R 1 36 k (say 36 k ) - given. + - V 1 =60V R 1 Charging Circuit R x + - V x Discharge Circuit I 1 I x C Xenon Lamp 4
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Lesson 6 Discharge Circuit Time Constant Constraint: Given R x = 80 , the discharge time constant is: = R x C = 80C = 10 -3 Therefore C = 12.5 F. - given + - V 1 =60V R 1 Charging Circuit R x + - V x Discharge Circuit I 1 I x C Xenon Lamp 5
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Lesson 6 Returning to the charging circuit : charging = R 1 C = 36k*12.5 = 0.45 s (5 =2.25 s) V C(charging) (0-)= V C(charging) (0+)= 0.0 V (begins fully discharged) V C(charging) ( )= 60.0 V Therefore: V C (charging) (t)= 60 + (0 - 60)e -t/ =60(1- e -t/0.45 ) + - V 1 =60V R 1 Charging Circuit R x + - V x Discharge Circuit I 1 I x C Xenon Lamp 6
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Lesson 6 V C (charging) (t)=60(1- e -t/0.45 ) Require V C (charging) (t’)=(60 -60)e -t’/ = 50 (lamp fires) or -60e -t’/t = -10. Solving for -t’/ = ln(-10/-60) = -1.8 t’ = 1.8*t = 1.8*0.45 = 0.81 s ~ 1.0s 60 50 0 0 t’ t Threshold firing voltage (50 V) v C (t) t’~1 sec. 7
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Lesson 6 Sinusoids Lesson 6 Professor Leonhard Euler 8
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x(t) = A cos( (t)) What is the signal’s frequency? Lesson 6 (t) = d (t)/dt ex: (t) = 0 t + (t) = d (t)/dt = 0 9
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0 10 20 30 40 50 60 70 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Lesson 6 Sinusoids (a.k.a. harmonic oscillation) x(t) = sin( 0 t) y(t) = cos( 0 t) 0 = 2 f 0 , f 0 in Hz T 0 =1/f 0 (sec.) f 0 = 1/64 (shown) 10
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0 10 20 30 40 50 60 70 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Lesson 6 x(t) shows a first zero crossing at t=0. y(t) shows a first shown zero crossing at 8 seconds or /4 rad. later. We say that y(t) lags x(t) by 8 seconds or /4 = 45 . We can also say that x(t) leads y(t) by 8 seconds or /4 = 45 . Why do we do this? x(t) = sin( 0 t) y(t) = sin( 0 t- /4) f 0 = 1/64 f 0 = 1/64 t 11
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Lesson 6 Frequency: Given x(t) = sin( 0 (t)), the signal’s instantaneous frequency is: 0 (t)= d 0 (t)/dt Familiar cases: 0 (t) = 0 t+ 0 , (t)= d 0 (t)/dt = 0 (constant frequency case) 0 (t) = a x(t) ; 0 (t)= d 0 (t)/dt = a dx(t)/dt (called phase modulation or PM) 0 (t) = a x(t)dt; 0 (t)= d 0 (t)/dt = a x(t) (called frequency modulation or FM) How does one perform the differentiation and integration? OpAmps 12
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Lesson 6 Some important sinusoidal identies (also see formula supplement): cos(a) cos(b) = ½ cos(a+b) + ½ cos(a-b) (modulaiton) Euler’s equation cos( 0 t) = (e j 0 t + e -j 0 t )/2 sin( 0 t) = (e j 0 t - e -j 0 t )/2j e j 0 t = cos( 0 t) + jsin( 0 t) 13
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Lesson 6 Euler e j0 Imaginary projection Real projection 14
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Lesson 6 Euler e j30 Imaginary projection Real projection 15
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Lesson 6 Euler e j60 Imaginary projection Real projection 16
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Lesson 6 Euler e j90 Imaginary projection Real projection 17
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Lesson 6 Euler e j120 Imaginary projection Real projection 18
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Lesson 6 Euler e j150 Imaginary projection Real projection 19
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Lesson 6 Euler e j180 Imaginary projection (sine) Real projection (cosine) Process Continues 20
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