ProbSolv_Chapter17

ProbSolv_Chapter17 - CHAPTER 17 - FOURIER TRANSFORM List of...

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Unformatted text preview: CHAPTER 17 - FOURIER TRANSFORM List of topics for this chapter : Fourier Transform and its Properties Circuit Applications Parseval’s Theorem Applications FOURIER TRANSFORM AND ITS PROPERTIES Problem 17.1 Find the Fourier Transform of the pulse shown in Figure 17.1. f(t) 0 —10 —I Figure 17.1 We begin with the derivative of f(t). 51—29:—6(t+2)+8(t+1)+6(t—1)—5(t—2) Transforming this into the frequency domain yields, jan(03) = —ei2‘” + ej‘” + 6” — e‘jz‘” = 2005(203) — 2005(0)) Therefore, Fan) : 2(cos(2(o.)— cos(o))) J0) Problem 17.2 Find the inverse Fourier transforms of the following, (a) 10/ [003003 + 5)] (b) 5103/ [(-jw + DOG) + 2)l (C) (2 —j®)/(—<02 + 4J0) + 3) . (d) 38(03)/[(ioa + 2)(ico + 3)] 311 Now to find the inVerse transforms. 10 A B . M . . . ' =—+———~, A= lU/5=2andB= 10/—5=—2 s(s+5) _s s+5 (a) F(S) = Therefore, F(co) = (2/jco) — (2/(jco + 5)) Transforming, f(t) = sgnm — 2e“5tugt[ Ss —— 55 A B _._.+._...__ (b) F(s)= m=m=s—1 s+2 A = —5/(1 +2) = —5/3 and B = ~5x(—2)/(—2— 1) = ~10/3 Therefore, ‘ F (0)) = [(—5/3)/(j(0 — 1)] + [(—10/3)/(j(o + 2)] Transforming, f(t) = — §e1u(—t) — gleam“) (2’s) ___A_+i A = 3/2 and B = —5/2 (Q “5’: 5+3, Therefore, F(03) = [Ls/ow + 1)] — 125/00» + 3)] . Transforming, f(t) = 1.5e"u(t)—2.5e‘3‘u(t1 ‘ 1 36 a) ejmt 1 3 1 (d) f(t) = f . ( ), at: _= 2n °°(J0)+2)(_]C0+3) 27:6 11 Problem 17.3 [17. 7] Find the Fourier transform of the "sine-wave pulse" shown in Figure 17.2. sin(7tt) f(t) Figure 17.2 m) = sin(nt) [u(t) — u(t — 2)] . 312 . F(m) = f sin(1tt)e'j"’t dt = f(ej" — e'j" )(e'jm) dt 1 . . 13(0)) : eJ(-m+1r)t +e-J(a)+1:)t dt 1 F 1 . 1 . T F = _. __._____ -J(a)—7r)t 2 +_T_____ -J(m+1:)t 2 (m) 2Ji-J(0)"7T)e ° own)" °J F __ 1(1—e'jz‘” +1—61sz (mu—2 71—0) n+0) F(co) = (2n + 211 e'jz‘”) (2)(7t2“032) F((n)= z” 2(aim—1) CIRCUIT APPLICATIONS . Problem 17.4 Find the transfer function, Vo((o)/Vs(co) for the circuit shown in Figure 17.3. ’ 3 Q 1 H mo i + vs“) 9 V0“) Figure 17.3 First we will solve for I. V V I: ——i—(c—‘))—= _S(°°), and Vo((o) = 51 3+JOJ+5 JCO+8 V0 (03) 5 Vs(co)—jco+8 Therefore, 313 Problem 175 Solve for vc(t) in Figure 17.4, where i(t) = u(t) A. + i(t) 0 1 F vc(t) Figure 17.4 First we transform i(t) into the frequency domain. 10.i I(co) = n5(co)+1/(ico), and vow) = 1(0)) J“) =I(03), 1 JCO+0.1 10+?- JO) 5 co 1 Therefore, Vc(0)) = n .( M‘f—fl = V1 + V2 160 J0)(JO)+0.1) V2 = ———-1-——=é+ B , where A = 1/0.1 = 10 andB = 1/(—0.1)= —10 s(s+0.1) s s+0.1 Therefore, V2(t) = Ssgn(t)—106“‘”°u(t) V1(t = -1~ .n5(w) ejm‘dw=i—n—=5 ' 21: jO)+0.1 211 0.1 ' This leads to vo(t) = 5—Ssgn(t)——IOe"/1°u(t), but sgn(t) = —1+2u(t) Tmmmm, wn)=5~s+mmo—me“%m or V0(t) = 1011 —e‘”1°)um volts Problem 17.6 [17.29] Determine the current i(t) in the circuit of Figure 17.5(b), given the voltage source shown in Figure 17.5(a). v(t) 2 £2 1 i(t) v(t) 1 F (b) Figure 17.5 314 I . 'V(t) = 5(t) — 25a — 1) + 8(t — 2) — (02 V(03)=1— 261‘” + 61°”2 1 — Ze'j‘“ + e‘j‘”2 W») = _m2 1 1+ '20) Now, Z(0)) = 2 + .——- = .J 10) J03 Z(co) 032 1+ J20) I = .—“——.“ 0.5 + 0.5 6'1""2 — e'j‘” (JOJ)(O.5+J0))( ) B t “FL—- 5 B ———> A - 2 B — 2 “ (s)(s+0.5)‘ s s+0.5 ‘ ’ “' 71(0)) = (3)05 + 0.56jmz - e'j“) —[ )(05 + 0.5e'j‘”2 — e'jw) jw 0.5 + jw . i(t) = %sgn(t) + §sgn(t — 2) — sgn(t — 1) — e'o‘s‘u(t) — 6W”) u(t — 2) — 2695‘”) u(t —— 1) PARSEVAL’S THEOREM Problem 17.7 Find the total energy in v(t) where v(t) is the pulse shown below. v(t) 0 —10—{ Inthetimedomain, Wm: l(—10)2dt+ 102dt=100t4+100t2 2 —2 1 II —100 +200 + 200~ 100 = 00 J 315 Problem 17.8 [ I 7. 43] A voltage source VS (t) = e"t sin(2t) u(t) V is applied to a l-Q . resistor. Calculate the energy delivered to the resistor. W152 = Efza) dt= fe'Z‘ sin2(2t) dt But sin2 (A) = E:- [1— cos(2A)] -2t -2t 1 e e __ ~2t __ =—-—°°— wm—fe (0.5)[1 005(401‘“ 2 -2 0 4+16 1 1 WIQ = +(-2-6j(-2) = 0.15 J W0 APPLICATIONS {-2 cos(4t) + 4 sin(4t)|;° Problem 17.9 Given the AM signal, f(t) = 10(1 + 4cos(20001tt))cos(1txlOfit), solve for the: (a) the carrier frequency (b) the lower sideband frequency (0) the upper sideband frequency (om = 20001: = an which leads to f = lkHz (a) (no = 1l:x106 = 21rfc which leads to fc ‘= 500 kHz or 0.5 MHz (b) st = fc—~fm = (500—1)kHz = 499 kHz (0) Usb = fc+fm = (500 + 1) kHz = 501 kHz Problem 17.10 [17.47] A voice signal occupying the frequency band of 0.4 to 3.5 kHz is used to amplitude modulate a 10-MHz carrier. Determine the range of frequencies for the lower and upper sidebands. 316 For the lower sideband, the frequencies range from . 10,000,000 — 3,500 : 9,996,500 Hz to 10,000,000 — 400 = 9,999,600 Hz For the upper sideband, the frequencies range from 10,000,000 + 400 = 10,000,400 Hz to 10,000,000 + 3,500 = 10,003,500 Hz 317 ...
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This note was uploaded on 09/06/2010 for the course EE 50145 taught by Professor Chi during the Spring '10 term at UCSF.

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ProbSolv_Chapter17 - CHAPTER 17 - FOURIER TRANSFORM List of...

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