Question

art 1 sin(5t)) z(t) = (cos(50t) + x(t))?, where x(t) = }. z(t) is passed through a filter with impulse response h(t) in order to pass only the product 2x(t) cos(50t). Which filter below is the correct filter to do that? ST sin(5t) sin(15t) / (a) h(t) = {* tt at (b) h(t) sin(5t) sin(100) L 1. Tt at os(1004) 5 it - Tt J (c) h(t) = {i sinft) sin15)} 2cos(50) (a) h(e) = { i sin tuon )*2 cos (50) (0) ne) = {a sin(10) sin(152) } 2cos(258)

Answer 1

Given ZH) = [Cox Got + 2)2 - Sos) nit) = Singi at ZH) f (1) → 2014) (08 50t Now ZH)= [(op bot + xH)2 x ² (H) + Cor sot + 2xH) (0850+ Elimenate this We need this First hom NH) . Sin 5t) - Helt) 5 Singt 5 Sin 5+ 5 at 52+ 7 / A We know sompling portion Sant Sinal Sa (wt): Sinu Sinuh 25 Se 5+5054 forsier toome form of Gode festiom is Scompling We know the Ponction HA -1/2 A Gr(t) - AT Sa(W.T/2)

A Gy(t) - FT also we know AT Sa (W. 1/2) duality property X (A) X(W) 27 x (W) XH) - this property Using Abr(H) AT SA (W.7/2) Chcete fonction is Even function 6,1-W) = 6,1W) AT Sal to 7/3). 27 AGr (-W) * 1/2 = 5 - To to 5 10 A Sa 5+ - DAXAGO (W) Sa5H) & FT A Crow) FT (S054) TN5 W 45 (H) = 75 (Sinist. Sinist) We know F.T property RH): Y (H) & 13 (X(W) * XW) Roduct Convolution

H(+) = Using this property HA) - 23 Smist. Sonst Sast - A holu) X(w) 3 [oz 610w) *Gors] glu) 2 (196 [ Cho (10) ► now] upper limit - lower limit = +5+5=10 -5-5--10 р, жP) у = AT loxlo – 20x h h = 5 Xlo) To Atte w (tyx 1972 (the IT ET(X? (H) = XW) & XW)

this is a symmetrical groph Ewe see the graph xlw) If we want x²(+) FT Upper limit - lower limit - 10+10= 20 2 bw this ( 8 F.T some 10 to 20 I thing Teledont FF (x² (4) ) FT ( 22 (4) limit of this fournier We don't need the FT(A?(H) We need the and we find that - 20 to 20 Now coming to ZH) Z(+) = 2 (+) + (0x²50t + 2M(+) Corbot B2CH) + COR BOL (op 50f + 294)02504 - We know that F.T of con Wit - & s[w-wo) + d(utwo)] { control - & [ Fw-wo) + Fleuters] Then cossot & [ S[WW-50) + s(lut 50] Cok So. Con 50! - $ $w-lo) + $(w 4100)] 241H) (0850t - [X(W-50) +X (w+50)

Now Z w) fourier trems from of Z(+) ZlW) this band be but lass only Wodon't want Exact Something 2017) FT(24H) a sot) FT/6501.4501) I FT (ZH) 5127 Joo -60 - So -40 -20 - 0795 This is Now from It of 20 40 50 60 100 slube= 10 x FT of Zlo) this group we need such type filter which pass only 2014) A501 or which ten pass only spectrum of 2nHa 50t from the option to pass this band Either cord because there contre Prog. 50 beause of lost form Coe sot Now choch suitable filter from option cold © 757 15 Sin 15t 150+ (5 at 3 7 RESEAB Smst 15517354 ) 1952 (S254. S0154) G+ (4) T Sal W.T/2) T2 = 15 - T= 30 30sa (156) — 22 4 3 6 ) Salist) - TSalt. I/2) 2267(4) 7/2=5 - T = 10 10 Sast 2 Giore) 163016) 15

752 sat. Salt 572 Rus - 25 Sast - Salsi Sa 151 - Sast. Sulist] - 2 hollo) 1 ngglo) how * Ww). 1 Gholu) * Caylu] (1900) * (2016)] IT HRW) = - - A2 = 30 lov3o = kx10 + 20P k= 10 upper limit lower limit = 20 = 20 & 10x1/10 = 1 Slope = 1/10 10 20 Aro ( Salot. Sa lot) (TH) - TSalW.T/2) T Salt. T/2) – 226-16) 7/2=10 =) T=20 20 Sa lot 24 hoollo) Salut - I Caolo)

Roll): 4200 Sa lot. Sotol) VET H1w) = KX1oox to it soto 620w) * Chache] 572 to [Colu) * Graoke) - Area = 20 TO 20x20= upper limit= lower limit= 10 +10=20 -10t0 =-20 ħ xuo h=10 Hw) t 10x1 = 1 I - .05 Slope - 20 20 Slope ofc sul which is greater them slope of Specburn which we want slope in @ option = 105 which is less than slope of spectrum which we want home some bart distorted because Original spectrum . th (f) in option D 40 50 60 option Hence ® is correct