# hw3_soln - ECE 562 Fall 2011 HW 3 SOLUTIONS 1. For a...

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ECE 562 Fall 2011 HW 3 SOLUTIONS 1. For a complex random vector Y , show the following: Σ = (Σ I + Σ Q ) + j QI - Σ IQ ) and ˇ Σ = (Σ I - Σ Q ) + j QI + Σ ) Ans: Σ = E h ( Y - m Y ) ( Y - m Y ) i = E ± ² Y I + j Y Q - ( m Y I + jm Y Q ) ³² Y I + j Y Q - ( m Y I + Y Q ) ³ ´ = E ± ² Y I - m Y I + j ² Y Q - m Y Q ³³² Y I - m Y I + j ² Y Q - m Y Q ³³ ´ = E ± ² Y I - m Y I + j ² Y Q - m Y Q ³³ µ ( Y I - m Y I ) > - j ² Y Q - m Y Q ³ > ¶´ = E ± ( Y I - m Y I )( Y I - m Y I ) > + ² Y Q - m Y Q Y Q - m Y Q ³ > ´ + j E ±µ ² Y Q - m Y Q ³ ( Y I - m Y I ) > - ( Y I - m Y I ) ² Y Q - m Y Q ³ > ¶´ = (Σ I + Σ Q ) + j QI - Σ ) ˘ Σ = E h ( Y - m Y ) ( Y - m Y ) > i = E ± ² Y I + j Y Q - ( m Y I + Y Q ) Y I + j Y Q - ( m Y I + Y Q ) ³ > ´ = E ± ² Y I - m Y I + j ² Y Q - m Y Q Y I - m Y I + j ² Y Q - m Y Q ³³ > ´ = E ± ² Y I - m Y I + j ² Y Q - m Y Q ³³ µ ( Y I - m Y I ) > + j ² Y Q - m Y Q ³ > ¶´ = E ± ( Y I - m Y I Y I - m Y I ) > - ² Y Q - m Y Q Y Q - m Y Q ³ > ´ + j E ±µ ² Y Q - m Y Q ³ ( Y I - m Y I ) > + ( Y I - m Y I ) ² Y Q - m Y Q ³ > ¶´ = (Σ I - Σ Q ) + j QI - Σ ) c ± V.V. Veeravalli, 2011 1

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2. Consider the signal s ( t ) = [sin( πt ) + j cos( πt )]11 0 t 1 . Suppose this signal is corrupted by complex WGN w ( t ) with PSD N 0 = 2 to form the received signal r ( t ) = s ( t ) + w ( t ) . Further suppose we form the random variable Z as: Z = Z 1 0 r ( t ) sin( πt ) dt (a) Find P { Z I > 1 } . (b) Find P { Z I + Z Q > 1 } . (c) Find P ( { Z I > 1 } ∩ { Z Q > 1 } ) Ans: For a complex WGN process w ( t ) and a function ψ ( t ) by deﬁnition h w,ψ i ∼ CN ( 0 ,N o k ψ k 2 ) Then in the case of Z Z = Z 1 0 r ( t ) sin πtdt = Z 1 0 ( s ( t ) + w ( t )) sin = Z 1 0 ( sin 2 πt + j sin πt cos πt ) dt + Z 1 0 w ( t ) sin = 1 2 + h i where ψ ( t ) = sin πt . Since k ψ k 2 = 1 2 , h i ∼ CN (0 , N o 2 ). So Z ∼ CN ( 1 2 , N o 2 ), Z I ∼ N ( 1 2 , N o 4 ), and Z Q ∼ N (0 , N o 4 ) with Z I and Z Q independent. (a) We have P { Z I > 1 } = Q ± 1 - 1 2 p N o / 4 ! = Q ² 1 2 ³ (b) Since Z I and Z Q are independent, Z I + Z Q ∼ N ( 1 2 , N o 2 ). Then P { Z I + Z Q > 1 } = Q ± 1 - 1 2 p N o / 2 ! = Q ² 1 2 ³ (c) Again since Z I and Z Q are independent P { Z I > 1 ,Z Q > 1 } = P { Z I > 1 } P { Z Q > 1 } = Q ± 1 - 1 2 p N o / 4 ! Q ± 1 - 0 p N o / 4 ! = Q ² 1 2 ³ Q ´ 2 µ c ± V.V. Veeravalli, 2011 2
3. Unequal Priors. Describe how the optimal decision regions for BPSK signaling get modiﬁed when the priors on the messages are not the same, say π 1 = 2 / 3 and π 2 = 1 / 3. Ans: For BPSK we have points in the constellation at ± E . The conditional pdfs of r are given by p 1 ( r ) = 1 πN o exp {-| r - E| 2 /N o } p 2 ( r ) = 1 πN o exp {-| r + 2 o } Then we decide on 1 if π 1 p 1 ( r ) > π 2 p 2 ( r ) or equivalently π 1 p 1 ( r ) π 2 p 2 ( r ) > 1 Subsitituting the deﬁnitions gives π 1 π 2 exp ± | r + 2 - | r - 2 N o ² > 1 π 1 π 2 exp ± 4 E r I N o ² > 1 Taking the log of both sides and solving for r I gives r I > - N o 4 E ln ³ π 1 π 2 ´ = - N o 4 E ln2 4. P e for PAM. Compute (the exact) P e as a function of E s for 4-ary PAM. Note that P e ,m is not the same for all m in this case. Figure 1: 4-ary PAM Constellation Ans: The average signal energy is E s = ³ 2 d 2 4 + 2 9 d 2 4 ´ 1 4 = 5 d 2 4 c ² V.V. Veeravalli, 2011 3

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so d = q 4 E s 5 . We need to consider the case of exterior and interior points separately.
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hw3_soln - ECE 562 Fall 2011 HW 3 SOLUTIONS 1. For a...

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