ECE405HW1Su2010Sol

# ECE405HW1Su2010Sol - ECE 405 HOMEWORK#1 Solutions SU 2010...

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Unformatted text preview: ECE 405 HOMEWORK #1 Solutions SU 2010 ©DR. JAMES S. KANG 1 x(t ) = 12 cos ( 2π ×15, 000t − 60o ) + 20 cos ( 2π × 20, 000t − 150o ) + 16 cos ( 2π × 30, 000t + 110o ) (a) 20 16 12 0 10 20 30 f (kHz) (b) 110 10 –30 –20 0 –10 20 30 f (kHz) –60 –150 (c) 10 8 –30 10 6 6 –20 –10 0 10 20 8 30 f (kHz) (d) 150 110 60 –30 10 –20 –10 0 20 –60 –110 –150 2 (a) 2π 2π −3 1 1 ⎛ n ⎞ − jn −3 ×1×10 ⎛ n ⎞ − jn X n = sinc ⎜ ⎟ e 3×10 = sinc ⎜ ⎟ e 3 3 3 ⎝3⎠ ⎝3⎠ (b) 1 30 f (kHz) 1 ⎛n⎞ sinc ⎜ ⎟ 3 ⎝3⎠ 1 1 1000 fo = = = Hz −3 To 3 × 10 3 Xn = –6fo –5fo –4fo –3fo –2fo –fo (c) n ∠X n 0 0 1 2 -2π/3 2π/3 3 0 0 4 π/3 fo 2fo 5 -π/3 3fo 6 0 4fo 5fo 6fo f 7 8 9 -2π/3 2π/3 0 2π/3 π/3 –6fo –5fo –4fo –3fo –2fo –fo 0 −π/3 fo 2fo 3fo −2π/3 (d) P2 = 2 X 2 2 ⎛ ⎛ 2π ⎜ sin ⎜ 3 2 ⎛2⎞ 2 ⎝ = sinc 2 ⎜ ⎟ = ⎜ 2π 9 ⎝3⎠ 9⎜ ⎜ 3 ⎝ 2 ⎞⎞ ⎟⎟ ⎠ ⎟ = 37.9954 mW ⎟ ⎟ ⎠ (e) 1 ∞2 1000 2π ⎞ ⎛n⎞ ⎛ x(t ) = + ∑ sinc ⎜ ⎟ cos ⎜ 2π × n⎟ × n×t − 3 n =1 3 3 3⎠ ⎝3⎠ ⎝ 1 ∞2 1000 2π ⎞ ⎛ nπ ⎞ ⎛ sin ⎜ n⎟ = +∑ × n×t − ⎟ cos ⎜ 2π × 3 n =1 nπ 3 3⎠ ⎝3⎠ ⎝ (f) r = ωc 2π × 30, 000 = = 90 2π × 1, 000 ωo 3 Yn = 1 1 X n−r + X n+ r 2 2 2 4fo 5fo 6fo f 2π 2π 1 ⎛ n − 90 ⎞ − j ( n −90) 3 1 ⎛ n + 90 ⎞ − j ( n +90) 3 Yn = sinc ⎜ + sinc ⎜ ⎟e ⎟e 6 6 ⎝3⎠ ⎝3⎠ 2π 2π 1 ⎛ n − 90 ⎞ − jn 3 1 ⎛ n + 90 ⎞ − jn 3 = sinc ⎜ e + sinc ⎜ e ⎟ ⎟ 6 6 ⎝3⎠ ⎝3⎠ Yn 1/6 –31 –30 –29 29 30 31 f (kHz) ∠ Yn f (kHz) (g) Let w(t ) be the output of the filter. Then, 1 1 2π ⎞ ⎛1⎞ ⎛ w(t ) = cos ( 2π × 30, 000t ) + sinc ⎜ ⎟ cos ⎜ 2π × 30,333t − ⎟ 3 3 3⎠ ⎝3⎠ ⎝ 1 2π ⎞ 1 π⎞ ⎛2⎞ ⎛ ⎛4⎞ ⎛ + sinc ⎜ ⎟ cos ⎜ 2π × 30, 667t + ⎟ + sinc ⎜ ⎟ cos ⎜ 2π × 31,333t + ⎟ 3 3⎠ 3 3⎠ ⎝3⎠ ⎝ ⎝3⎠ ⎝ 1 π⎞ 1 2π ⎞ ⎛5⎞ ⎛ ⎛7⎞ ⎛ + sinc ⎜ ⎟ cos ⎜ 2π × 31, 667t − ⎟ + sinc ⎜ ⎟ cos ⎜ 2π × 32,333t − ⎟ 3 3⎠ 3 3⎠ ⎝3⎠ ⎝ ⎝3⎠ ⎝ 1 2π ⎞ 1 2π ⎞ ⎛1⎞ ⎛ ⎛2⎞ ⎛ + sinc ⎜ ⎟ cos ⎜ 2π × 29, 667t + ⎟ + sinc ⎜ ⎟ cos ⎜ 2π × 29,333t − ⎟ 3 3⎠ 3 3⎠ ⎝3⎠ ⎝ ⎝3⎠ ⎝ 1 π⎞ 1 π⎞ ⎛4⎞ ⎛ ⎛5⎞ ⎛ + sinc ⎜ ⎟ cos ⎜ 2π × 28, 667t − ⎟ + sinc ⎜ ⎟ cos ⎜ 2π × 28,333t + ⎟ 3 3⎠ 3 3⎠ ⎝3⎠ ⎝ ⎝3⎠ ⎝ 1 2π ⎞ ⎛7⎞ ⎛ + sinc ⎜ ⎟ cos ⎜ 2π × 27, 667t + ⎟ 3 3⎠ ⎝3⎠ ⎝ (h) Z n = Yn e − jn 2π 3×10−3 ×0.25×10−3 = Yn e −j nπ 6 2π 2π nπ ⎡1 ⎛ n − 90 ⎞ − jn 3 1 ⎛ n + 90 ⎞ − jn 3 ⎤ − j 6 = ⎢ sinc ⎜ + sinc ⎜ e e e ⎥ ⎟ ⎟ 6 ⎝3⎠ ⎝3⎠ ⎣6 ⎦ nπ 5 1⎡ ⎛ n − 90 ⎞ ⎛ n + 90 ⎞ ⎤ − j 6 = ⎢sinc ⎜ ⎟ + sinc ⎜ ⎟⎥ e 6⎣ ⎝3⎠ ⎝ 3 ⎠⎦ 3 3 (a) X o = 1 2 ⎧ 0, n even 1 ⎛n⎞ ⎪ = sinc 2 ⎜ ⎟ = ⎨ 2 Xn 2 ⎝ 2 ⎠ ⎪ 2 2 , n odd ⎩n π (b) 1∞ 4 cos [ (2n − 1)2π × 1, 000t ] x(t ) = + ∑ 2 n =1 (2n − 1) 2 π 2 1, 000 (c) H (ω ) = jω + 1, 000 1 1 H ( nωo ) = = ∠ − tan −1 ( 2π n ) 22 1 + j 2π n 1 + 4π n y (t ) = 4 ωo = 1∞ 4 1 cos ⎡ (2n − 1)2π × 1, 000t − tan −1 ( 2π n ) ⎤ +∑ 22 ⎣ ⎦ 2 n =1 (2n − 1) π 1 + 4π 2 n 2 2π = 2π . 1 (a) x(t ) = 10 π ∞ 20 cos ( 2π nt ) 2 n =1 π ( 4n − 1) −∑ (b) y (t ) = 10 π − 2 × 20 10 40 cos ( 2π t − 60o ) = + cos ( 2π t + 120o ) π ( 4 − 1) π 3π 2 × 20 cos (2π t – 60o ) = 10 + 40 cos (2π t + 120o ) y(t) = 10 – π π (4 – 1) π 3π 5 (a) 2e jω 5 X (ω ) . (b) 30 X (3ω ) e − jω 7 . (c) ⎡ ⎛ t + 2 ⎞⎤ jω 2 F ⎢x ⎜ ⎟ ⎥ = 5 X (5ω ) e ⎣ ⎝ 5 ⎠⎦ ⎡ ⎛ t + 2 ⎞⎤ d ⎡5 X (5ω ) e jω 2 ⎤ F ⎢t x ⎜ ⎟⎥ = j ⎦ dω ⎣ ⎣ ⎝ 5 ⎠⎦ ⎡ ⎛ t + 2 ⎞ dx(t ) ⎤ d jω 2 F ⎢t x ⎜ ⎟∗ ⎥ = j d ω ⎡5 X (5ω ) e jω X (ω ) ⎣ ⎦ ⎣ ⎝ 5 ⎠ dt ⎦ (d) 1 ⎛ ω⎞ F ⎡ x* ( 8t ) ⎤ = X * ⎜ − ⎟ ⎣ ⎦8 ⎝ 8⎠ 4 F ⎡t x* ( 8t ) ⎤ = j ⎣ ⎦ d ⎡ 1 * ⎛ ω ⎞⎤ X ⎜ − ⎟⎥ dω ⎢ 8 ⎝ 8 ⎠⎦ ⎣ (e) ⎡ ⎛ t ⎞⎤ F ⎢ x ⎜ − ⎟ ⎥ = 12 X (−12ω ) ⎣ ⎝ 12 ⎠ ⎦ ⎡ ⎛ t + 20 ⎞ ⎤ jω 20 F ⎢− x ⎜ − ⎟ ⎥ = −12 X (−12ω ) e ⎣ ⎝ 12 ⎠ ⎦ ⎡ ⎛ t + 20 ⎞ j1000t ⎤ j (ω −1000 ) 20 F ⎢− x ⎜ − ⎟e ⎥ = −12 X ⎡ −12 (ω − 1000 ) ⎤ e ⎣ ⎦ ⎣ ⎝ 12 ⎠ ⎦ (f) 1 ⎛ ω⎞ F ⎡ x ( −2t ) ⎤ = X ⎜ − ⎟ ⎣ ⎦2 ⎝ 2⎠ 1 ⎛ω ⎞ F ⎡ x* ( −2t ) ⎤ = X * ⎜ ⎟ ⎣ ⎦2 ⎝2⎠ ⎡d ⎤ jω * ⎛ ω ⎞ F ⎢ x* ( −2t ) ⎥ = X⎜ ⎟ ⎣ dt ⎦2 ⎝2⎠ ⎡⎛ d ⎤ j (ω − 500 ) * ⎛ ω − 500 ⎞ j (ω + 500 ) * ⎛ ω + 500 ⎞ ⎞ F ⎢⎜ x* ( −2t ) ⎟ cos(500t ) ⎥ = X⎜ X⎜ ⎟ ⎟+ 4 4 ⎠ ⎝2⎠ ⎝2⎠ ⎣⎝ dt ⎦ (g) 1 F ⎡ x2 (t )⎤ = ⎣ ⎦ 2π X (ω ) ∗ X (ω ) 1 F ⎡ x3 ( t ) ⎤ = X (ω ) ∗ X (ω ) ∗ X (ω ) 2 ⎣ ⎦ ( 2π ) F ⎡ x3 (t ) + 3x 2 (t ) − 9 x(t ) + 15⎤ ⎣ ⎦ = 1 ( 2π ) 2 X (ω ) ∗ X (ω ) ∗ X (ω ) + 3 X (ω ) ∗ X (ω ) − 9 X (ω ) + 15 × 2π × δ (ω ) 2π (h) F ⎡ ( t − 5 ) x ( 3 − t ) ⎤ = F ⎡( t − 5 ) x ( − ( t − 3 ) ) ⎤ = F ⎡t x ( − ( t − 3 ) ) − 5 x ( − ( t − 3 ) ) ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ F [ x(−t ) ] = X (−ω ) F ⎡ x(− ( t − 3)) ⎤ = X (−ω ) e− jω 3 ⎣ ⎦ d ⎡ X (−ω ) e − jω 3 ⎤ ⎦ dω ⎣ d ⎡ X (−ω ) e− jω 3 ⎤ − 5 X (−ω ) e− jω 3 F ⎡( t − 5 ) x ( 3 − t ) ⎤ = j ⎣ ⎦ ⎦ dω ⎣ (i) ⎡ ⎛ t ⎞⎤ F ⎢ x ⎜ ⎟ ⎥ = 2 X (2ω ) ⎣ ⎝ 2 ⎠⎦ F ⎡ t x ( − ( t − 3 )) ⎤ = j ⎣ ⎦ 5 F ⎡ X ( t ) ⎤ = 2π x ( −ω ) ⎣ ⎦ ⎛ω ⎞ x⎜ ⎟ ⎝5⎠ ⎡ ⎛t⎞ ⎤ 2π F ⎢ x ⎜ ⎟ ∗ X ( −5t ) ⎥ = 2 X ( 2ω ) 5 ⎣ ⎝2⎠ ⎦ 6 (a) F ⎡ X ( −5t ) ⎤ = ⎣ ⎦ 2π 5 ⎛ ω ⎞ 4π ⎛ω ⎞ x⎜ ⎟ = X ( 2ω ) x ⎜ ⎟ ⎝5⎠ 5 ⎝5⎠ x'(t) 2 1 0 –1 t 1 –1 –2 ω ω j⎞ ⎛ ω ⎞⎛ − j 2 ⎛ω jω X (ω ) = 2e jω − 2e − jω + sinc ⎜ e − e 2 ⎟ = j 4sin (ω ) − j 2sinc ⎜ ⎟⎜ ⎝ 2π ⎠ ⎝ ⎝ 2π ⎠ Thus ⎛ ω ⎞ ⎛ω ⎞ 4sin (ω ) − 2sinc ⎜ ⎟ sin ⎜ ⎟ ⎝ 2π ⎠ ⎝ 2 ⎠ = 4sin (ω ) − sinc 2 ⎛ ω ⎞ X (ω ) = ⎜ ⎟ ω ω ⎝ 2π ⎠ ⎛ω⎞ 2⎛ ω ⎞ = 4sinc ⎜ ⎟ − sinc ⎜ ⎟ ⎝ 2π ⎠ ⎝ 2π ⎠ (b) x'(t) 2 1 3 1 t 2 –1.5 jω X (ω ) = 2e − jω ⎛ω + sinc ⎜ ⎝ 2π ω ⎞ −j2 ⎛ ω ⎞ − jω 2 ⎟ e − 3sinc ⎜ ⎟ e ⎠ ⎝π ⎠ Thus ⎛ω 2e − jω + sinc ⎜ ⎝ 2π X (ω ) = ω ⎞ −j2 ⎛ ω ⎞ − jω 2 ⎟ e − 3sinc ⎜ ⎟ e ⎠ ⎝π ⎠ jω 6 ⎞ ⎛ω ⎞ ⎟ sin ⎜ ⎟ ⎠ ⎝2⎠ 7 (a) Since 1 , a + jω from the frequency differentiation theorem, d⎡ 1 ⎤ 1 F ⎡t e− at u ( t ) ⎤ = j . ⎢ a + jω ⎥ = 2 ⎣ ⎦ dω ⎣ ⎦ ( a + jω ) F ⎡e− at u ( t ) ⎤ = ⎣ ⎦ (b) Applying modulation property, we obtain ⎤ 1⎡ 1 1 ⎥ F ⎡t e− at u ( t ) cos (ωc t ) ⎤ = ⎢ + 2 2 ⎣ ⎦ 2⎢ ⎥ ⎣ ( a + j (ω − ωc ) ) ( a + j (ω + ωc ) ) ⎦ (c) Since 2a −a t F ⎡e ⎤ = 2 ⎣ ⎦ a + ω2 , ⎡ ⎤ 1 1 ⎡ ⎤ + F ⎣10e− a t cos(200t ) ⎦ = 50 ⎢ ⎥ 2 2 ⎢ 25 + (ω − 200 ) 25 + (ω + 200 ) ⎥ ⎣ ⎦ (d) Since ⎡ ⎛ t ⎞⎤ ⎛ 6ω ⎞ F ⎢ rect ⎜ ⎟ ⎥ = 12 sinc ⎜ ⎟ ⎝ 12 ⎠ ⎦ ⎝π ⎠ ⎣ from frequency shifting property, we get ⎛ 6 (ω − 200 ) ⎞ ⎡ ⎤ ⎛t⎞ F ⎢ rect ⎜ ⎟ e j 200t ⎥ = 12 sinc ⎜ ⎟ π ⎝ 12 ⎠ ⎣ ⎦ ⎝ ⎠ (e) Since 2 , F [sgn(t ) ] = jω we have 1 1 F [sgn(t ) cos(100t ) ] = + j (ω − 100 ) j (ω + 100 ) 8 X(ω) 1 –2.5 0 ω 2.5 5 2π –5 0 5 ω 7 F[x2 (t)] = 1 X(ω) ∗ X(ω) 2π 1 2 F ⎡ x (t ) ⎤ = ⎣ ⎦ 2π X (ω ) ∗ X (ω ) 9 ⎡ ⎛ ω ⎞⎤ 2 ⎛t⎞ F −1 ⎢ 2 rect ⎜ ⎟ ⎥ = sinc ⎜ ⎟ ⎝ 2 ⎠⎦ π ⎝π ⎠ ⎣ ⎡ ⎛ ω ⎞⎤ 4 ⎛t⎞ F −1 ⎢ 4 tri ⎜ ⎟ ⎥ = sinc2 ⎜ ⎟ ⎝ 2 ⎠⎦ π ⎝π ⎠ ⎣ 2 4 ⎛t⎞ ⎛t x(t ) = F −1 ⎣ X (ω ) ⎤ = sinc ⎜ ⎟ cos(4t ) + sinc 2 ⎜ ⎡ ⎦π π ⎝π ⎠ ⎝π 2 ⎛ t ⎞⎡ ⎛ t ⎞⎤ = sinc ⎜ ⎟ ⎢1 + 2sinc ⎜ ⎟ ⎥ cos(4t ) π ⎝π ⎠⎣ ⎝ π ⎠⎦ 10 e−2 = 0.0271 (a) 5 j 2π j π ⎞ ⎟ cos(4t ) ⎠ (b) e 8 = e 4 11 y (t ) = u (t ) + u (t − 1) + u (t − 2) − 5 u (t − 3) + 3 u (t − 4) − u (t − 5) 3 2 1 1 –1 –2 2 3 4 5 t 12 (a) For t ≤ 0 , x(t ) = 0 . For 0 < t ≤ 1 , t 1 x(t ) = ∫ λ d λ = t 2 2 0 For 1 < t , x(t ) = 0.5 . 0.5 0 t 1 (b) 8 t +2 x(t ) = ∫λ 2 dλ = t −2 λ3 3 t +2 t −2 1 16 = ⎡(t + 2)3 − (t − 2)3 ⎤ = 4t 2 + ⎣ ⎦ 3 3 16/3 t 0 (c) For 0 ≤ t < 2 , t 1 x(t ) = ∫ 1× λ d λ = t 2 2 0 For 2 ≤ t < 4 , 2 1 1 2 x(t ) = ∫ 1× λ d λ = − t 2 + 2t = − ( t − 2 ) + 2 2 2 2 −t 2 0 4 2 t (d) 1 0 –1 –2 1 (e) For t < 0 , x(t ) = 0 . For t ≥ 0 , t x(t ) = ∫ e 0 −2 λ 7 ( λ −t ) e dλ = e −7 t t ∫e 0 5λ d λ = e −7 t 1 5t ( e − 1) = 1 ( e−2t − e−7t ) 5 5 9 2 t e–2 λu(λ ) 1 0 λ e7 λu(–λ ) 1 λ 0 e–2 λu(λ ) 1 e7(λ–t)u(t–λ ) 0 λ t x(t) 0 t (f) x(t) 1 –1 0 1 2 3 4 t 13 Since Y (ω ) = 1 ( 2π ) 2 X (ω ) ∗ X (ω ) ∗ X (ω ) 10 the bandwidth of y (t ) is 3B Hz. 14 Y (ω ) = H (ω ) X (ω ) . (a) X (ω ) = F [δ (t ) ] = 1 y (t ) = F −1 ⎡Y (ω ) ⎤ = F −1 ⎡ H (ω ) ⎤ = 20sinc(20t ) ⎣ ⎦ ⎣ ⎦ (b) ⎛ω⎞ X (ω ) = F [10sinc(10t ) ] = rect ⎜ ⎟ ⎝ 20π ⎠ ⎛ω⎞ ⎛ω⎞ ⎛ω⎞ Y (ω ) = H (ω ) X (ω ) = rect ⎜ ⎟ rect ⎜ ⎟ = rect ⎜ ⎟ ⎝ 40π ⎠ ⎝ 20π ⎠ ⎝ 20π ⎠ y (t ) = F −1 ⎡Y (ω ) ⎤ = F −1 ⎡ X (ω ) ⎤ = x(t ) = 20sinc(20t ) ⎣ ⎦ ⎣ ⎦ Η(ω) X(ω) −20π −10π 0 10π 15 From the Parseval's theorem, ∞ 2 1 2 1 2 2 Ex = X (ω ) dω = ∫ ∫ ( −ω + 2 ) dω = π 2π −∞ 2π 0 = 2 ∫ (ω 2 − 4ω + 4 ) d ω 0 2 ⎤ 1 ⎡ω ω 8 ⎢ 3 − 4 2 + 4ω ⎥ = 3π π⎣ ⎦0 3 2 X(ω) 2 −ω + 2 ω+2 −2 2 0 −2 ω X(ω) 2 4 0 2 ω 11 20π ω 16 y (t ) = cos ( 2π × 10t ) + 0.1cos ( 2π ×10 ( t − 0.01) ) α = 0.1 (a) ω0 = 1, ω1 = −α , ω2 = α 2 . Thus, ω1 = −0.1, ω2 = 0.01 . (b) z ( t ) = y ( t ) − 0.1 y ( t − 0.01) + 0.01 y ( t − 0.02 ) = cos ( 2π × 10t ) + 0.1cos ( 2π ×10 ( t − 0.01) ) − 0.1cos ( 2π × 10 ( t − 0.01) ) −0.01cos ( 2π ×10 ( t − 0.02 ) ) + 0.01cos ( 2π ×10 ( t − 0.02 ) ) + 0.001cos ( 2π × 10 ( t − 0.03) ) = cos ( 2π × 10t ) + 0.001cos ( 2π ×10 ( t − 0.03) ) 12 ...
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