Question

this is signal ....please answer clearly thank you

For each problem below, convolve x[n) and h[n] by hand, and then compare the DTFT magnitude plots of X(12), H(12), and Y(12). In each case, does h[n) represent a low or high pass filter? Does the signal x[n) pass through the filter or get rejected? Why? (8) x[n] - [1 1 1] , h[n] - [1 -1] (9) x[n] - [1 0 -1 0], [n] - [11] (10) x[n] - [1 0 -1 0], h[n] - [1 -1] (11) x[n] = [1 -1 1 -1] , h[n] - [1 11

Answer 1

DTFT stands for Discrete time jourier transform let a [n] be the discrete time signal. Hoen Dter of a th] is x ) Sw) x đa) - 2 3 Top DTFT e xln xle | 30) - P ( x29) ? do © x60] = C, Debb= [] y [n] = a[D] *hin] linear convolution : we can perform Add and the linear convolution using - shift melbod. RTO) hthu y In) = {vo, 0, alo] Longes M = 3 b[n] Lengus N= 2 in y [n] Length is M + Nay = 8+2-1 =H

Ô To) = [ 0 - 0 btn] = [1] (2. A[n] ho] y Eng= {1,1,1,1, og Lengis mtny="ht 2-1 = g © xth] = [104 og broja [-1] - ha an tool ro/ To y cn = {1, -1, l, logo total longos = 5 total Lang15 = 5 © 2Tn)= (1 - 1 ) hong = [1 ] homy and b y yn] = {1,0,0,0,1) Length = 5

q xo -JoD xlan) = - Jw -2jw. 210) + xl) é + e = jus - ajw x(4) = 1+ ē te take mod งu0 - P 1x1) = lite te 15 sub W -D . +1+1+108. [lang] = 3 wao at w zala => 1X1.) I = 1+ (-3). + (-1) 1x103) = 491 = 1 ** Ixikony = lite mit eine W = et e 1x102) = x + (x+ [1] = 1 1x103) = 1 at w = 2T lxue)) = yat yua lit e te 1 + 1x)) = 11+i+'! =D | 1x103) = 8

of FR an 7s りた considor o to X it represents Low pans characteústici Ama hos los loco); c) = i . 146m2) = li-lul sub =0 ) 141= 0 w=)2 = l s21 = 11+i) = Vot1 = V2 W=* 4(3) = 11-24 = 11-[cost-jsina] = 11-[-13) = . 146)), va m of truly - ters hts) тертыеөтab pas #ta No alD] will not be able to pass Abrough the h [o].

juon ah e ga In} = (son o h (9] = [1 0 6 X (2) = -2jw 2 It(1) e 3 x la)= 1es Jx103) a lination substitute warious values of w find maguile de at wzo a 1x121= 1 WEtle =) Ixury li-[cosa-ssin ] || - 11-1-1)) = 2 wat » 1x(2) 1 = 11-1 Tin tuns is low pans signal 1 Hlan) = mus sien it 14.2)) = liteco it'w = 0 Hz)) = 1 at w = 2 = 1H)= litil=v2 at wat » lHerl= 11+11=2 1463 bf 2 tome ñ gw hin) represent Xin) is a 4 High pass filter signal, not able to pass - DO

de p = [1 oy o h (n] = [1 - same as same lXl2)) plot for © also because Plp sean is same o acon) represests Low pass signal b[] = [1 -0 * houly 친웃 x 10) = s am e h m High pass filler 211) not able to pass through hen). 1 ()= [1 / 1-] Bem)= [] already solved Her PD Gun @ Question for hin) hcm represents HPF bigb pass) 13+ ) at com a xls) = 1 anos y 1 xla)) = 11:30 - iwe at w so 1X.) = 11-1-1-11 = 11-31= 2 at w=xl2 => 1x1921 = 11-(-3) - (-1)-(+0 1 = lititx -x] at w=83) 1x) = |-6-3-(1)-(-)) = 114178 +)

хи т, т ак эш hty | чрезми, не раз зяті. фот эх хло) - ). - (0) – (0) - to) - ) - | - > br) pres, Lous poss Stu to an cant pass through hens,