Question

What are the absolute bandwidths of the signals shown in Problem 4(a) above? (Note: Absolute BW= highest frequency component present in a signal). What is the null-null bandwidth of the waveform in Problem 4(a) - (iii)? What is the 3 dB bandwidth of x(t) = -2e-4tu(t)? Write down an equation to find the 3 dB bandwidth of a filter with transfer function and then solve it to find its 3-dB bandwidth: H(s) = (+2)(+3).

Answer 1

of the signals en Absolute Bandwidths P in - 4t t) - 3+ae ·ult) - Actual Applying founder Transform x() = 3 (29-56] + af htiv) 67.5W + 2 [y-jw) 2. X(w/ stays for wwoo & Band width = 0o. ī) with i 3 sin ( 250 ) - 5 x(w): 37 ( (-250 ) – 5kw 2500)) - 5[20.w] tighest Frequency - Actual BW . 25017 rad/see (D) X(t) = rect (t/4) X(w). 4. Sin(20) - 125 HZ. Hoy 2w = 2 Sin(zw) Stays for w→ → Hlw) Actual Bandwidth = co W) X(t) = 3+ 4 sinc (100 TT) X(w) = 67 56w) +(45 cm ons w Highest Frequency = Actual BW = 100t2 vad/sec. 100 m2

in 4 (a) - (0) b) Null- null bandwidth 460) - it ~ 264 - sect(+/a) FT 12 2 w 2 X = & Sinlaw) . pe; X(w) - 2 Sin (2w) X(W) zo is the positon, where Null will form. x(W)- 0 2 Sin2w), o X(u) a Nul null NO Noll > W=+%, IT,.. le; X(W)=0 = W = IKT Null Bandwidth K=1,2,3...- Null Bandwidth = -(-7) = T rad/ sec.

3) Given a signal a(t) = -26-ttutt) By applying touries transform X(W) = -2 (By properties of . 4 + jw . fourier transform for 3-dB bandwidth Klwel) = (x (80) K(w) ) max = 1X(@))=; mo X ( wal) = 1 2 3 4 1 2 = 4 two =) we = 4=) ( = 2rad/sec 3- dB bandwidth = W (=zrad/sec

Given a transfer function H(s) = -6 (5+2)(5+3) put s = jw, H(SW) = 6 = H (w) z 6 . (jwtz) (w+3) W²+4) (wita) 3-dB bandwidth How dl= (Hiljaa) wa Wan e man pa Clemente (4 (30) = 6:1 ma 4 (3) vast 4) lut:ta) À =) 36x2 = wt 13w136 - wt + 13 w 36 50 wie - 13*17.69 WCF -13 I √169 +144 2 € 2.3459 Wee 1.5316 radisec =) 3 dB Bandwidth . 2