Problem 7.27 Use phasors to simplify each of the following expressions into a
single term (*hint: see Problem 7.26):
(a) υ1 (t) = 12 cos(6t + 30◦ ) − 6 cos(6t − 45◦ )
(b) υ2 (t) = −3 sin(1000t − 15◦ ) − 6 sin(1000t + 15◦ ) + 12 cos(1000t − 60◦ )
(c) υ3 (t) = 2 cos(377t + 60◦ ) − 2 cos(377t − 60◦ )
(d) υ4 (t) = 10 cos 800t + 10 sin 800t
Solution:
(a)
υ1 (t) = 12 cos(6t + 30◦ ) − 6 sin(6t − 45◦ )
◦
◦
V1 = 12e j30 − 6e− j45
= 12 cos 30◦ + j12 sin 30◦ − 6 cos 45◦ + j6 sin 45◦
= 6.15 + j10.24
◦
= 11.95e j59.0 .
υ (t) = 11.95 cos(6t + 59.0◦ ).
(b)
υ2 (t) = −3 sin(1000t − 15◦ ) − 6 sin(1000t + 15◦ ) + 12 cos(1000t − 60◦ )
= 3 cos(1000t + 75◦ ) + 6 cos(1000t + 105◦ ) + 12 cos(1000t − 60◦ )
◦
◦
◦
V2 = 3e j75 + 6e j105 + 12e− j60
= 3 cos 75◦ + j3 sin 75◦ + 6 cos 105◦ + j6 sin 105◦ + 12 cos 60◦ − j12 sin 60◦
= 5.22 − j1.70
◦
= 5.49e− j18.0
υ2 (t) = 5.49 cos(1000t − 18.0◦ ).
(c)
υ3 (t) = 2 cos(377t + 60◦ ) − 2 cos(377t − 60◦ )
= 2 cos(377t + 60◦ ) + 2 cos(377t + 120◦ )
◦
◦
V3 = 2e j60 + 2e j120
= 2 cos 60◦ + j2 sin 60◦ + 2 cos 120◦ + j2 sin 120◦
= j3.46
◦
= 3.46e j90
υ3 (t) = 3.46 cos(377t + 90◦ )
= −3.46 sin 377t.
(d)
υ4 (t) = 10 cos 800t + 10 sin 800t
= 10 cos 800t + 10 cos(800t − 90◦ )
All rights reserved. Do not reproduce or distribute. © 2016 National Technology and Science Press
◦
V4 = 10 + 10e− j90
= 10 + 10 cos 90◦ − j10 sin 90◦
= 10 + 0 − j10
= 10(1 − j)
√
◦
= 10 2 e− j45
υ4 (t) = 14.14 cos(800t − 45◦ ).
All rights reserved. Do not reproduce or distribute. © 2016 National Technology and Science Press