Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 4 docx

The power content is A2 A2 α2 PM + = 200 + 200 · 0.62 · 0.5 = 236 2 PAM = The bandwidth of the signal is WAM = 2W = 20000 Hz 4) If the modulation is FM with kf = 50000, then PFM = A2 = 200 and the effective bandwidth is approximated by Carson’s rule as Bc = 2(β + 1)W = 50000 + W = 120000 Hz W Problem 3.24 f 1) Since F[sinc(400t)] = 400 Π( 400 ), the bandwidth of the message signal is W = 200 and the resulting modulation index βf = kf 10 kf max[|m(t)|] = = =⇒ kf = 120 W W Hence, the modulated signal is t u(t) = A cos(2πfc t + 2πkf −∞ m(τ )dτ ) t = 100 cos(2πfc t + +2π1200 −∞ sinc(400τ )dτ ) 2) The maximum frequency deviation of the modulated signal is ∆fmax = βf W = × 200 = 1200 3) Since the modulated signal is essentially a sinusoidal signal with amplitude A = 100, we have P = A2 = 5000 4) Using Carson’s rule, the effective bandwidth of the modulated signal can be approximated by Bc = 2(βf + 1)W = 2(6 + 1)200 = 2800 Hz Problem 3.25 1) The maximum phase deviation of the PM signal is ∆φmax = kp max[|m(t)|] = kp The phase of the FM modulated signal is t φ(t) = 2πkf      t m(τ )dτ = 2πkf −∞ t 2πkf τ dτ m(τ )dτ = πkf t2 0≤t Y ) = p(Y = 0) [p(X = 1) + p(X = 2) + p(X = 3) + p(X = 4)] + p(Y = 1) [p(X = 2) + p(X = 3) + p(X = 4)] + p(Y = 2) [p(X = 3) + p(X = 4)] + p(Y = 3) [p(X = 4)] 535 = 4096 l 4) In general p(X + Y ≤ 5) = l=0 m=0 p(X = l − m)p(Y = m) However it is easier to find p(X + Y ≤ 5) through p(X + Y ≤ 5) = − p(X + Y > 5) because fewer terms are involved in the calculation of the probability p(X + Y > 5) Note also that p(X + Y > 5|X = 0) = p(X + Y > 5|X = 1) = p(X + Y > 5) = p(X = 2)p(Y = 4) + p(X = 3)[p(Y = 3) + p(Y = 4)] + p(X = 4)[p(Y = 2) + p(Y = 3) + p(Y = 4)] 125 = 4096 Hence, p(X + Y ≤ 5) = − p(X + Y > 5) = − 125 4096 Problem 4.8 1) Since limx→∞ FX (x) = and FX (x) = for all x ≥ we obtain K = 2) The random variable is of the mixed-type since there is a discontinuity at x = lim ) = 1/2 whereas lim →0 FX (1 + ) = 3) 1 P ( < X ≤ 1) = FX (1) − FX ( ) = − = 2 4 73 →0 FX (1 − 4) 1 1 P ( < X < 1) = FX (1− ) − FX ( ) = − = 2 4 5) P (X > 2) = − P (X ≤ 2) = − FX (2) = − = Problem 4.9 1) x < −1 ⇒ FX (x) = x x 1 = x2 + x + 2 −1 x 1 (v + 1)dv + (−v + 1)dv = − x2 + x + ≤ x ≤ ⇒ FX (x) = 2 −1 ≤ x ⇒ FX (x) = −1 ≤ x ≤ ⇒ FX (x) = (v + 1)dv = ( v + v) −1 2) 1 p(X > ) = − FX ( ) = − = 2 8 3) p(X > 0, X < ) FX ( ) − FX (0) 2 = = p(X > 0|X < ) = 1 p(X < ) − p(X > ) 4) We find first the CDF p(X ≤ x, X > ) 1 FX (x|X > ) = p(X ≤ x|X > ) = 2 p(X > ) If x ≤ If x > 2 then p(X ≤ x|X > ) = since the events E1 = {X ≤ } and E1 = {X > } are disjoint 2 then p(X ≤ x|X > ) = FX (x) − FX ( ) so that 2 FX (x) − FX ( ) FX (x|X > ) = − FX ( ) Differentiating this equation with respect to x we obtain fX (x) 1−FX ( ) fX (x|X > ) = x> x≤ 2 5) ∞ E[X|X > 1/2] = = −∞ xfX (x|X > 1/2)dx − FX (1/2) ∞ = = ∞ 1 2 xfX (x)dx 1 x(−x + 1)dx = 8(− x3 + x2 ) 2 74 1 Problem 4.10 1) The random variable X is Gaussian with zero mean and variance σ = 10−8 Thus p(X > x) = x Q( σ ) and p(X > 10−4 ) = Q p(X > × 10−4 ) = Q 10−4 10−4 = Q(1) = 159 × 10−4 10−4 = Q(4) = 3.17 × 10−5 p(−2 × 10−4 < X ≤ 10−4 ) = − Q(1) − Q(2) = 8182 2) p(X > 10−4 |X > 0) = p(X > 10−4 , X > 0) p(X > 10−4 ) 159 = = = 318 p(X > 0) p(X > 0) 3) y = g(x) = xu(x) Clearly fY (y) = and FY (y) = for y < If y > 0, then the equation y = xu(x) has a unique solution x1 = y Hence, FY (y) = FX (y) and fY (y) = fX (y) for y > FY (y) is discontinuous at y = and the jump of the discontinuity equals FX (0) FY (0+ ) − FY (0− ) = FX (0) = In summary the PDF fY (y) equals fY (y) = fX (y)u(y) + δ(y) The general expression for finding fY (y) can not be used because g(x) is constant for some interval so that there is an uncountable number of solutions for x in this interval 4) ∞ E[Y ] = = = −∞ ∞ yfY (y)dy y fX (y)u(y) + δ(y) dy −∞ ∞ y2 σ √ ye− 2σ2 dy = √ 2π 2πσ 5) y = g(x) = |x| For a given y > there are two solutions to the equation y = g(x) = |x|, that is x1,2 = ±y Hence for y > fY (y) = = fX (x2 ) fX (x1 ) + = fX (y) + fX (−y) |sgn(x1 )| |sgn(x2 )| y2 √ e− 2σ2 2πσ For y < there are no solutions to the equation y = |x| and fY (y) = E[Y ] = √ ∞ 2πσ 75 y2 2σ ye− 2σ2 dy = √ 2π Problem 4.11 1) y = g(x) = ax2 Assume without loss of generality that a > Then, if y < the equation y = ax2 has no real solutions and fY (y) = If y > there are two solutions to the system, namely x1,2 = y/a Hence, fY (y) = = = fX (x2 ) fX (x1 ) + |g (x1 )| |g (x2 )| fX ( y/a) fX (− y/a) + 2a y/a 2a y/a y e− 2aσ2 √ √ ay 2πσ 2) The equation y = g(x) has no solutions if y < −b Thus FY (y) and fY (y) are zero for y < −b If −b ≤ y ≤ b, then for a fixed y, g(x) < y if x < y; hence FY (y) = FX (y) If y > b then g(x) ≤ b < y for every x; hence FY (y) = At the points y = ±b, FY (y) is discontinuous and the discontinuities equal to FY (−b+ ) − FY (−b− ) = FX (−b) and FY (b+ ) − FY (b− ) = − FX (b) The PDF of y = g(x) is fY (y) = FX (−b)δ(y + b) + (1 − FX (b))δ(y − b) + fX (y)[u−1 (y + b) − u−1 (y − b)] y2 b = Q (δ(y + b) + δ(y − b)) + √ e− 2σ2 [u−1 (y + b) − u−1 (y − b)] σ 2πσ 3) In the case of the hard limiter p(Y = b) = p(X < 0) = FX (0) = p(Y = a) = p(X > 0) = − FX (0) = Thus FY (y) is a staircase function and fY (y) = FX (0)δ(y − b) + (1 − FX (0))δ(y − a) 4) The random variable y = g(x) takes the values yn = xn with probability p(Y = yn ) = p(an ≤ X ≤ an+1 ) = FX (an+1 ) − FX (an ) Thus, FY (y) is a staircase function with FY (y) = if y < x1 and FY (y) = if y > xN The PDF is a sequence of impulse functions, that is N [FX (ai+1 ) − FX (ai )] δ(y − xi ) fY (y) = i=1 N = Q i=1 σ −Q 76 ai+1 σ δ(y − xi ) Problem 4.12 The equation x = tan φ has a unique solution in [− π , π ], that is 2 φ1 = arctan x Furthermore sin φ cos φ x (φ) = = sin2 φ =1+ = + x2 cos2 φ cos2 φ Thus, fΦ (φ1 ) = |x (φ1 )| π(1 + x2 ) We observe that fX (x) is the Cauchy density Since fX (x) is even we immediately get E[X] = However, the variance is fX (x) = = E[X ] − (E[X])2 ∞ x2 = dx = ∞ π −∞ + x2 σX Problem 4.13 1) ∞ ≥ α ∞ yfY (y)dy ≥ E[Y ] = ∞ α yfY (y)dy α yfY (y)dy = αp(Y ≥ α) Thus p(Y ≥ α) ≤ E[Y ]/α 2) Clearly p(|X − E[X]| > ) = p((X − E[X])2 > question we obtain p(|X − E[X]| > ) = p((X − E[X])2 > ) Thus using the results of the previous )≤ E[(X − E[X])2 ] = σ2 Problem 4.14 The characteristic function of the binomial distribution is n ejvk ψX (v) = k=0 n = k=0 n k n k pk (1 − p)n−k (pejv )k (1 − p)n−k = (pejv + (1 − p))n Thus d = n(pejv + (1 − p))n−1 pjejv (pejv + (1 − p))n j dv j v=0 n−1 = n(p + − p) p = np d (2) E[X ] = mX = (−1) (pejv + (1 − p))n dv v=0 d jv n−1 jv = (−1) n(pe + (1 − p) pje dv v=0 (1) E[X] = mX = = v=0 n(n − 1)(pejv + (1 − p))n−2 p2 e2jv + n(pejv + (1 − p))n−1 pejv v=0 = n(n − 1)(p + − p)p2 + n(p + − p)p = n(n − 1)p2 + np 77 ... = 4 4 = 34 28 72 p(Y = 0) = 4 = 24 1 p(X = 1) = 4 p(X = 2) = p(X = 3) = 4 p(X = 4) = 4 4 3 4 33 28 p(Y = 1) = 1 = 33 28 p(Y = 2) = 2 = 3? ?4 28 p(Y = 3) = 4 = 28 4 p(Y = 4) = = = 24 = 24 = 24 =... = 14 KHz fk3 = 18 KHz fk4 = 22 KHz fk5 = 26 KHz fk6 = 30 KHz fk7 = 34 KHz fk8 = 38 KHz fk9 = 42 KHz fk10 = 46 KHz fl1 = 290 KHz fl2 = 330 KHz fl3 = 370 KHz fl4 = 41 0 KHz fl5 = 45 0 KHz fl6 = 49 0... 74 1 Problem 4. 10 1) The random variable X is Gaussian with zero mean and variance σ = 10−8 Thus p(X > x) = x Q( σ ) and p(X > 10? ?4 ) = Q p(X > × 10? ?4 ) = Q 10? ?4 10? ?4 = Q(1) = 159 × 10? ?4 10−4