Question

Determine the spectrum for each of the two signals. Give your answer as two separate plots. (A) x(t) = cos(71Tt -0.17) (B) y(t) = x(t)cos(777t +0.17) Upload a document to show case your work and final answer. Uploads can either be a picture jpg, .png .tiff, etc) or a pdf Upload Choose a File

Answer 1

The spectrum of a signal is represented by (Fra) pair. Here fr is the frequency and az is the complex amplitude. The graphical plot of spectrum has each frequency component represented by a vertical line, the length of which is proportional to the magnitude la and each line is labeled with a The inverse Euler formula gives the expression for the spectrum of a sum of sinusoids is, x(t)= X, + 2 2 Here, X is the complex amplitude and the expression of X is, X = 4,ep . N X + e 12 fut k=1 The spectrum of the set of pair form is, {(0,x),(1 fickx)(-f-13) --(1) From Euler formula, zreje = cos 0 + j sin e The expansion of cos e is, ele +e-18 cos(O) 2 Also, from the trigonometric expression, cos(0, )cos (0,) = [cos(6, +8,) +cos(0, – 0,)] -(2) (a) Draw the spectrum for cos? (Tat -0.17). cosº (77t – 0.17)=[cos(7nt -0.17)][cos(7nt -0.17)] 1 cos[(77t – 0.17) +(7t – 0.17)]+ cosº ( 77t - 0.17)= cos[(77t -0.17)–(7rt -0.17)]

cos' (Trt – 0.17 ) = [cos (147+ – 0.27) + cos(0)] s? (77t – 0.17) = } [cos(14īt – 0.27) +1] 1 1 cos? (7 at -0.17) =- cos (14ft -0.27) 2 2 1 e (1472-0.22) 2 2 2 1 +e-|(14x2-0.28) cos? (7 at -0.17) = + 1 + -e1434e-0.221 1 cos (77t -0.17) +-e e-14 at 0.221 2 4 4 From equation (1), the frequency/complex amplitude pairs are expressed as, 1 [(-3)(14.0-21).(-14, -14.-€0.221 The spectrum of signal cos? (7at -0.17 )is, 1 2 10:27) -0.21) 4 4 -14 0 14 (b) Draw the spectrum for cos? ( 7at -0.17 )cos( 77 At +0.17). In part (a), the equation for cos? (7nt - 0.17)was derived as + -cos(146t – 0.27) 1 2 2

Therefore, 1 2 2 cos? (7rt - 0.17 )cos (777t +0.17) += cos(147t – 0.27) (cos(777t +0.17 ) cos (777t +0.17)+ cos? (7 at -0.17 )cos (77 at +0.17)= | 3[cos(147 -0.27 )cos(777t +0.17)] Further solve form equation (2), cos( 77 7t +0.17)+ cos? (7rt - 0.17 ) cos(777t +0.17 ) = [cos (14ft - 0.21 + 777t +0.17)]+ cos (14mt – 0.21 – 777 -0.17)] cos( 77 7t +0.17 ) + cos? (77t - 0.17 )cos(777t +0.17)= 1 cos(917+ - 0.17)+ -cos(-63nt -0.31) 4 4 cos( 77 7t +0.17)+ cos? (7nt - 0.17 )cos( 77 7t +0.17 ) = 1 4 1 cos(914t -0.17)+cos(634t +0.37 4 e (77 xt +0.13) + e -|(77x1+0.12) + 2 (913t-0.12 ) 2-1(9171-0.10 te cosº (77t -0.17 ) cos(777t +0.17)= + 4 2 e (63x2+0.32) +e -|(63x1+0.31 7 4 2

10.151 CE 2015/p/mat 1 -0.12-j77 at +-e 4 + 1 1 cos? ( 7rt - 0.17 ) cos(777t +0.17 )= + €0.3a1 e 1631 + 1 -0.31 -e 8 le/6321 The above equation is written in the form of, 20.17), jej27 (38.5); -0.121-/28 (38.5) + e + 4 4 cos? (7net - 0.17 )cos (777t +0.17) = 2 -0.15)ę/27 (45.5)* + 2 (0-1) 2=325(45.5) + 8 8 +-e 8 160.36ej27(31.5) 1,-0.321,-12 (31.5) 8 From equation (1), the frequency/complex amplitude pairs are expressed as, 1 1 1 (3.5.f.no) (-3.5.fc2") (15.3. ) 455424) (31.5.fense) (-31.5, 04:1) The spectrum of signal cos? (Tat – 0.17 ) cos( 77 at +0.17 )is, -0.3.11 e 10.31) 120.17) 8 1 8 -0.17) 4 4 1 -0.17 10.11 8 8 -45.5 -38.5-31.5 0 31.5 38.5 45.5