EEN 340 Problems CHAPTER 4, Fall 2011

4.2 (i) x(t) = sin(4t) + cos(8t) + 7 + cos(16¢) (a) Exponential form: wo = 4 a(t) = elt — me + felt + § he —i8t 4 7504 1 desler yb 1 ,—j16¢ = 7+ (-O0.5j)e%ot + (osjye-H0" + (058)e%0t 4 (0.5)e-I0 4 (05)eHH0" 4 (.5)erHHt Co =7 Cy = -0.5j,C_1 = 0.57 Cz = 0.5,C_9 = 0.5 C4 =05,C_4 = 0.5 Cy =0,k 40,1, -1, 2, -2, 4, 4 (b) Combined trigonometric form: D;, = 2\C;|,k > 0 Since sin(4t) = cos(4t — 7/2) a(t) = 7+ cos(wot — 7/2) + cos(2wot) + cos(4wot) Do = Co =7 Dy, = 1,6, = —1/2 Dy = 1,0) =0 Dy = 1,0, =0 Dy =0,k £0,1,2,4 (ii) x(t) = cos?(t) = 4$[1 + cos(2t)] (a) Exponential form: wo = 2 1 1,4 1.- 5 + cos(2t)]} = 5+ fest + te jt C, ro) Il tole Cy C, =0,k £0,1,-1 P ale (b) Combined trigonometric: D, = j,k > 0 (b) (i) wo = 1, a(t) = 0.5e + 0.567! + sel! — be Cy = 0, Cz = C_3 = 0.5, Cs = -0.5 5). C_5 = 0.5). (ii) wo = 2, a(t) =e?! + del 4 be I yer = uy e—J8t Co = 0, Cy = 1, C_1 =0, Cy=C4=, C4 ~s1c4=4 (iii) No Fourier Series (iv) v(t) = sin( Ft) + sin($t), wo = : 1 1 = —j2wi x3(t) _— zee — xe jwot 4 we ei2wot _ ze j2wot J Co = 0, Cl = 3 Cy 4= 5 C2 = 3. C_» = 5, Ch = C_p =0,k >2 4 (a) C42 732 ) B add £1¢,|= due J 647-70" 1 2td= 8, Bp omthnd-1 lb) Cy 52% 5 21Cpl=2e » 4= 90° @) C45 ~ Eb badd, 210.|= PR. 3 Ob 195° oe Aube 2+ 3 Cha (bw,#, 192") = wXo whith - (A Cy = Oe ane 5 ut) = Zi 2 aie (ie) coe buyt £26 48 Se@ L, 1 Yn tol =10 7 TO Xo An WW se o Lf y ; -~ 40 L-2 , ke. ed. . 0) atue= 45. | nt 7, alte\ ai 7 _ : _ | Lo Wo au --40 d) Aue le) . —Jh ws E Ce= Lf alee wt -¥fe we T jie ° (/-<€ > KA? TALS <6 whin KAty K a miltyle of ¢ =H jk — Ske Cry =H (jkwo)C ka = 8 + jkwo 20 wp =— = 7 —j2(10 C,, - 200) Tk C —j20 j2k 20 ‘ku Ttk S84 52k 4n+ jrk 20 C.,| ——_——— ICey V 1672 + k27? k Oey = — tan~'(7) |Co| =0 \C;| =1.5440, 0; = —0.2450rad \Cy| =1.4235, 02 = —0.4636rad |\C3| =1.2732, 63 = —0.6435rad (c) This changes only the value of Co, and therefore only the DC value Co, of the output might change— however, since H (0) = 0 in this case, the DC value of the output does not actually change. Cox = 20 = Co, = 20H(0) = 0 (d)No. The low frequencies get decreased in amplitude—in fact the DC component does not get through at all. (e) This is a lower frequency square wave and so more of its energy will be attenuated by the filter. It will not change part c—the DC output is still 0. 4.28 x(t) | = K(at+b hawctltd a ee] = Zero lok st) jew jywyak KH =O y - [S € Je Taak ab Kx