Question

Question2: (40 points]: Consider the system shown in the figure with the input signal xc(t) = 3 cos(100t) + 2 cos(200t), sampling frequency ws = 600 rad/s, and final filter cutoff frequency w1 = 400 rad/s. The filter has an impulse response given byha[n] = 8[n – 1] + 8[n] + 8[n + 1]. a) [10 points] Find and plot the signal Xa(ein) b) [10 points] Find and plot the signals Ya(ejn) and yo (jw) c) [10 points] Find and plot the signal yo(t) d) [10 points] Explain in details how to design the system in order to have yo(t) = xc(t) He (140) p(t) Conversion of xd In] impulse train to sequence Ha len? Ya (n) Conversion of Yp (t) sequence to impulse train Yo (t) 0

Answer 1

sole given Ft) 5 3cos croot) + 2003 (2001) X;(t) = { [56f-100) + 5(8+100)] [SCP-200) + 564+200)] X (H) = [SCP-100)+86f4100] - [SCP-200) A'S EP1200)] TO Yon) get t put n nit 11 costols Tg you Ws = 277 ar = ATT Wg 5 QT at ' given sampling frequeiras Ws = Goorad/s 600 ajan) 3003 (100x 2. n ) + ) +2003 (200x RT.n) 600 600 xn) - 3ơ ( 33 ) ) + 2e8 ( 6G II) aan) 3003 ( 1T) + 2003 (IT)

เฝ (a) x, ceter) Ž [s(2-st) + S69247)] + [s(n-1) + $(2+)] (6) ท] = s[n-) + [n] + s(n+13 Ha (682) er tot el Hacets 1+2.0082 at ii : IM Cedoj 3 at az = TT li cedit j =-11-215 1

at w- FŠ T H Cejš") H 1+ 2009 (3) = 1 + 2x1 1+2 - 2 at WIN TT Hredit) 1+269(3) O, It will pass only one component at w= ग y cedir). - 2x *** [54-4) +(12+ )] y ceday 3 (5(2-şiT) + 5(22+371 2] 3 Lyon) 6 Col ( in)

To get Up (4) put na 600 t Gilt 300+ 2TT 18 (t) – 6x col (thx 300 + fplt - 600 Cloot) Yo (jw) 6 [s(w-100) + s(w4100)] 2 Yplju) - 3 [S(W-100)+ s (W7100)] Yp (ja) 3 3 loo 100 % (pre) 3 - 13 2 →

5 Xcelry $2.37 3:37 allen 12/23 TT - ہمہ ، 0:5 -0.5 (c) Filter تدين 40D of the 11. YcCw) 15 -100 w لی [se (t) 30 Cos (100t) ] (d) To get Yelt X(t) we have to design a filter in Such a way that it posses passes both components.