alc_sol_man_ch03 - Solutions Manual Using the product...

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Solutions Manual 12 Introduction to Mechatronics and Meaurement Systems Using the product formula trigonometric identity, Therefore, 2.37 Using the double angle trigonometric identity, Therefore, 2.38 This is a sin wave with half the amplitude of the input with a period of 1s. 2.39 2.40 R L = R i = 100 for maximum power 2.41 The BNC cable is far more effective in shielding the input signals from electromagnetic interference since no loops are formed. ± ²³´ µ · ¸¹ ------------- φ µ φ · º () »¼½ ¸ ω ¾ φ ¿ µ φ · ¿ »¼½ º ¾ À Á ¹  ± µ · ¸ φ µ φ · º »¼½ µ · ¸ θ »¼½  · ö½ Ä ¹ --- · ¸ ω ¾ φ · ¿ ¸ ½ÅÆ ¾ À Á ¹  · · ¸ ¹ ----- Ä ¸ -- ¸ ω ¾ φ · ¿ [] »¼½ º   ¾ À Á ¹  · · ¸ ¹ ¹ ¸ · ¸ Ç ¸È Ç ¸ Ç È Ç ¸ Ç È ¿ ------------------ ÉÊ µ ¼ Ç ¸È Ç Ä Ç ¸È ¿ --------------------- µ Å Ä ¸ ¸ π ¾ ½ÅÆ Ë Ì Ë ½ ------ µ Ì µ ½ ĸÁµ ¸Íµ É Â
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Solutions Manual Introduction to Mechatronics and Measurement Systems 13 3.1 For V i > 0, V o = 0. For V i < 0, V o = V i . The resulting waveform consists only of the negative "humps" of the original cosine wave. Each hump has a duration of 0.5s and there is a 0.5s gap between each hump. 3.2 (a) output passes first (positive) hump only (b) output passes second (negative) hump only (c) output passes first (positive) hump only (d) output passes second (negative) hump only (e) output passes full first (positive) hump and 1/2-scale second hump (f) output passes full wave 3.3 For V i > 0, V o = V i . For V i < 0, V o = -V i . The resulting waveform is a full wave rectified sin wave where there are two positive "humps" for each period of V i (0.5 sec). 3.4 (a) For V i > 0.5V, V o = 0.5V. For V i < 0.5V, V o = V i . The resulting waveform is the original sine wave with the top halves of the positive "humps" (above 0.5 V) clipped off.
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