Question

Question #1 [15 points) Consider the following signal, x(t)=rect(t/50) (i) Find X(w) by definition (ii) Sketch the magnitude and phase response of X(w) (iii) Energy of x(t)= Page 1 of 2

Answer 1

yout IT of X(t) = X(W) = act). e at X(t) = A rectly) - 2. reet ( 2 ) A=1, T = 50 (or) 1/2 = 5%= 25

jot XW) S8 A e dt. D er in the unium] su casa ya baris] sults and die sinx. – A Since sin (wys) dealen. +1% -1% 3- = Som. =) Multiply 2 and divide 7/2. - 2x / XA sin(WT) = AT sin (WI)

* AR ૨ (… ' () (x) FT =) AT sin (WI)

1(t) = 1. sect (+/50) 6. A: 1, r = 50, 7% = 25 XW) = AT sin ( I T W/ en substituting above values. - Xw) = (1.50) sin (W (25)) w (25) = 50 sin (250) 25w. Since W = P (based Cou) (24) If w is varied from oo to too. in integral - multiples se k. a k. 1 (2) K-integer = this are points on weaxis XC60) xw) 1 50

Magnitucle X(c) — Amplitude spectrum 1 x) Le this returns onlyzve wall therefoce is ',1x6) 3 (1450)-50- . Ihave just substituted P-50

e plot ! To plot it phase Obtain phase plot we must obtain angle of x w ) Lx(W) = tant (imaginary part of xw] imegteam Real port of X(0) XW)- 50 sin (25w) +( 256 a tot el control) como 8 LxW) = 0 for all values of w. co'angle at every w value. Energy of xet) (BV) 1. crect (t/so) - t -OS exisk 88 but betiseen .Olidt rect fonction -25 to 25 [25-(-25) = +50. 1 -ot. os Energy of *(t) 550

Freegy of acts = fetts) .at 1) act)) = 1 2) limite are from 25 to 25 As fonction Exists during P= 5060) from (-25 1025)