Applied mathematics for radio and communication engineers by Carl E. Smith

By Carl E. Smith

Arithmetic, consultant, How-to, Communications, Engineering

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This diagram should represent the magnitude of these vectors to a convenient scale and the actual direction with their angle designated. Second, draw perpendiculars from the extreme end of each vector to the x-axis and the y-axis. Mark these components with the vector subscribed with x or y as the case may be. Third, add algebraically the x-components together and the y-components together. Fourth, determine the magnitude of the resultant by IRI = I(ofalgebraic sum )2 + ( algebraic sum )2 x-components of y-components '\I If one component of a vector is known the other component can be easily determined, for if the magnitude IAj = vA;-+ X; then A", = vlAI2 - A~ A.

Find the altitude. Find the altitude. 2. In each of the following examples, find the resultant vector and state in which quadrant it lies. The directions of the vectors are along the x and y axes as shown in Fig. 20. , on a large sheet of paper, will help you to understand the problem clearly and solve it correctly. IBI = 10. el = 14, \D\ = 11. b. IAI = 49, IB = 99. el = 391, IDI = 441. a. IAI c. 9, IDI = d. AI = 900, IBI = = 1,325, IDI e. 4. 3. 750. = 475. 6. 45. c A B x o,J, D FIG. -Direction of vec­ tors in exercise problem 2.

Voltage vector VI ~ 110/0° (110 volts at an angle of 0° with positive x-axis). Voltage vector V 2 ~ 115/75° (115 volts at an angle of 75° with posi­ tive x-axis). 2. Impedance vector ZI = 100/400. Impedance vector Z2 = 50/15° Or 50/3450. 3. Current vector II = 3/2150. Current vector 1 2 ~ 4/3000. 4. Voltage vector VI ~ 150/600. Voltage vector V 2 = 100/1800. Voltage vector V. = 50/3000. 6. Force vector F I = 17/1100. Force vector F 2 = 15/3200. Force vector F. = 10/1900. Work each of the following exercises by the slide-rule method and check for the resultant hypotenuse by the longer method: 6.

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