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Tutorial10_ans - EE 3008 Extra Tutorial Solutions Q1 =100...

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EE 3008. Extra Tutorial Solutions Q1) µ =100, V max =1V log(1 ) ) x y µ + = + (a) When x=0.5V, y=0.85V. (b) When x=0.8V, y=0.95V. Q2) a) N = 2 7 = 128. ) ) x y + = + (1 ) 1 y x +− = For the minimum amplitude of the quantized values, y = 1/ N = 1/128, then 1/128 4 min 256 1 1.74 10 255 x == × . b) For the second minimum amplitude of the quantized values, y = 3/ N = 3/128, then 3/128 256 1 255 x = . c) For the third minimum amplitude of the quantized values, y = 5/ N = 5/128, then 5/128 256 1 255 x = . d) For the n -th minimum amplitude of the quantized values, y = (2 n 1)/ N = (2 n 1)/128, then (2 1)/128 256 1 255 n x = . e) Dynamic range = 20log(|x| max /|x| min )=20log(1/(1.74 × 10 -4 )) = 75.2 (dB)

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Q3) Suppose 0 corresponds to - and 1 corresponds to + (a) The output without lowpass filtering is shown below (b) The output with lowpass filtering is approximately a sin wave. Q4) The DM signal is shown below The digital output is 11110101000010101. .. Q5) () cos 2 t c s tA f t k m t d t π −∞ ⎡⎤ =+ ⎢⎥ ⎣⎦ ⎩⎭ starting point 1v t 8.4ms 0 3.7V 1ms
cos2 cos 2 ( ) sin2 sin 2 ( ) sin 2 2 ( ) tt cc t A f t km t d t A f t t d t Af t

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Tutorial10_ans - EE 3008 Extra Tutorial Solutions Q1 =100...

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