561_hmk1_soln

561_hmk1_soln - 2 sinc 2 Tf . From the table: T sinc 2 Tf...

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ECE 561 – Digital Communications Systems Solutions for Homework #1 1. a. x 2 ( t ) = A 2 e 2 at u ( t ) A 2 e 2 at 0 dt < energy signal b. Ε = A 2 e 2 at 0 dt = A 2 2 a c. From the table, X ( f ) = A a + j 2 π f Ψ x ( f ) = X ( f ) 2 = A 2 a 2 + 4 2 f 2 2. a. Power signal since it has infinite energy b. P x = 1 T x 2 ( t ) dt T /2 T /2 = 1 1 cos2 t + 5 cos 4 t [ ] 2 dt 1/2 1/2 = cos 2 2 t + 25 cos 2 4 t + 5 cos2 t cos 4 t dt 1/2 1/2 = ... = 13 W c. The Fourier transform has four components: components at +/- 2 Hz with area 5/2 and components at +/- 1 Hz with area ½ . Thus: G x ( f ) = F n 2 δ f nf o ( ) n = −∞ = 5 2 2 f + 2 ( ) + 5 2 2 f 2 ( ) + 1 2 2 f + 1 ( ) + 1 2 2 f 1 ( ) = 25 4 f + 2 ( ) + 25 4 f 2 ( ) + 1 4 f + 1 ( ) + 1 4 f 1 ( ) d. P x = G x ( f ) −∞ df = 25 4 f + 2 ( ) + 25 4 f 2 ( ) + 1 4 f + 1 ( ) + 1 4 f 1 ( ) −∞ df = 52 4 = 13 W e. Since:
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R x τ ( ) G x ( f ) = 25 4 δ f + 2 ( ) + 25 4 f 2 ( ) + 1 4 f + 1 ( ) + 1 4 f 1 ( ) R x ( ) = 25 2 cos4 πτ + 1 2 cos2 f. R x (0) = 25 2 + 1 2 = 13 3. From the table: a. X ( f ) = T sinc Tf ⇒Ψ x ( f ) = X ( f ) 2 = T 2 sinc 2 Tf b. Ε = 1 dt T /2 T /2 = T c. R x ( ) ↔ Ψ x ( f ) = T
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Unformatted text preview: 2 sinc 2 Tf . From the table: T sinc 2 Tf Thus, the autocorrelation function is: 4. a. = 2 10 6 f 2 df 10000 = ... = 667 KW b. = 2 10 6 f 2 df 5000 10000 = ... = 583 KW 5. a. 10500 9500 = 1KHz b. The total power (area under triangle) is 100 W. Thus, 99% containment is 99W, leaving 0.5 W outside of the bandwidth on either side. The power contained in the band from t-T T 1 !-T T T R x ( ! ) 9KHz to 9KHz + x is: 0.5 x (10 4 x ) = 0.5 x = 100 . So the bandwidth goes from 9100 to 10900 or 1.8 KHz c. 2KHz...
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561_hmk1_soln - 2 sinc 2 Tf . From the table: T sinc 2 Tf...

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