Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 3 pptx

Thus,  A sin(2πf 0 t + θ)=−A cos(2πf 0 t + θ) Problem 2.53 Taking the Fourier transform of  e j2πf 0 t we obtain F[  e j2πf 0 t ]=−jsgn(f)δ(f − f 0 )=−jsgn(f 0 )δ(f − f 0 ) Thus,  e j2πf 0 t = F −1 [−jsgn(f 0 )δ(f − f 0 )] = −jsgn(f 0 )e j2πf 0 t Problem 2.54 F   d dt x(t)  = F[  x(t) δ  (t)] = −jsgn(f)F[x(t) δ  (t)] = −jsgn(f)j2πfX(f)=2πfsgn(f)X(f) =2π|f|X(f) Problem 2.55 We need to prove that  x  (t)=(ˆx(t))  . F[  x  (t)] = F[  x(t) δ  (t)] = −jsgn(f)F[x(t) δ  (t)] = −jsgn(f)X(f)j2πf = F[ˆx(t)]j2πf = F[(ˆx(t))  ] Taking the inverse Fourier transform of both sides of the previous relation we obtain,  x  (t)=(ˆx(t))  Problem 2.56 x(t) = sinct cos 2πf 0 t =⇒ X(f)= 1 2 Π(f + f 0 )) + 1 2 Π(f − f 0 )) h(t) = sinc 2 t sin 2πf 0 t =⇒ H(f)=− 1 2j Λ(f + f 0 )) + 1 2j Λ(f − f 0 )) The lowpass equivalents are X l (f)=2u(f + f 0 )X(f + f 0 )=Π(f) H l (f)=2u(f + f 0 )H(f + f 0 )= 1 j Λ(f) Y l (f)= 1 2 X l (f)H l (f)=      1 2j (f +1) − 1 2 <f≤ 0 1 2j (−f +1) 0≤ f< 1 2 0 otherwise Taking the inverse Fourier transform of Y l (f) we can find the lowpass equivalent response of the system. Thus, y l (t)=F −1 [Y l (f)] = 1 2j  0 − 1 2 (f +1)e j2πft df + 1 2j  1 2 0 (−f +1)e j2πft df = 1 2j  1 j2πt fe j2πft + 1 4π 2 t 2 e j2πft      0 − 1 2 + 1 2j 1 j2πt e j2πft     0 − 1 2 − 1 2j  1 j2πt fe j2πft + 1 4π 2 t 2 e j2πft      1 2 0 + 1 2j 1 j2πt e j2πft     1 2 0 = j  − 1 4πt sin πt + 1 4π 2 t 2 (cos πt − 1)  38 The output of the system y(t) can now be found from y(t) = Re[y l (t)e j2πf 0 t ]. Thus y(t)=Re  (j[− 1 4πt sin πt + 1 4π 2 t 2 (cos πt − 1)])(cos 2πf 0 t + j sin 2πf 0 t)  =[ 1 4π 2 t 2 (1 − cos πt)+ 1 4πt sin πt] sin 2πf 0 t Problem 2.57 1) The spectrum of the output signal y(t) is the product of X(f) and H(f). Thus, Y (f)=H(f)X(f)=X(f)A(f 0 )e j(θ(f 0 )+(f−f 0 )θ  (f)| f=f 0 ) y(t) is a narrowband signal centered at frequencies f = ±f 0 . To obtain the lowpass equivalent signal we have to shift the spectrum (positive band) of y(t) to the right by f 0 . Hence, Y l (f)=u(f + f 0 )X(f + f 0 )A(f 0 )e j(θ(f 0 )+fθ  (f)| f=f 0 ) = X l (f)A(f 0 )e j(θ(f 0 )+fθ  (f)| f=f 0 ) 2) Taking the inverse Fourier transform of the previous relation, we obtain y l (t)=F −1  X l (f)A(f 0 )e jθ(f 0 ) e jfθ  (f)| f=f 0  = A(f 0 )x l (t + 1 2π θ  (f)| f=f 0 ) With y(t) = Re[y l (t)e j2πf 0 t ] and x l (t)=V x (t)e jΘ x (t) we get y(t) = Re[y l (t)e j2πf 0 t ] =Re  A(f 0 )x l (t + 1 2π θ  (f)| f=f 0 )e jθ(f 0 ) e j2πf 0 t  =Re  A(f 0 )V x (t + 1 2π θ  (f)| f=f 0 )e j2πf 0 t e jΘ x (t+ 1 2π θ  (f)| f=f 0 )  = A(f 0 )V x (t − t g ) cos(2πf 0 t + θ(f 0 )+Θ x (t + 1 2π θ  (f)| f=f 0 )) = A(f 0 )V x (t − t g ) cos(2πf 0 (t + θ(f 0 ) 2πf 0 )+Θ x (t + 1 2π θ  (f)| f=f 0 )) = A(f 0 )V x (t − t g ) cos(2πf 0 (t − t p )+Θ x (t + 1 2π θ  (f)| f=f 0 )) where t g = − 1 2π θ  (f)| f=f 0 ,t p = − 1 2π θ(f 0 ) f 0 = − 1 2π θ(f) f     f=f 0 3) t g can be considered as a time lag of the envelope of the signal, whereas t p is the time corresponding to a phase delay of 1 2π θ(f 0 ) f 0 . Problem 2.58 1) We can write H θ (f) as follows H θ (f)=      cos θ −j sin θf>0 0 f =0 cos θ + j sin θf<0 = cos θ −jsgn(f) sin θ Thus, h θ (t)=F −1 [H θ (f)] = cos θδ(t)+ 1 πt sin θ 39 2) x θ (t)=x(t) h θ (t)=x(t)  (cos θδ(t)+ 1 πt sin θ) = cos θx(t) δ(t) + sin θ 1 πt x(t) = cos θx(t) + sin θˆx(t) 3)  ∞ −∞ |x θ (t)| 2 dt =  ∞ −∞ |cos θx(t) + sin θˆx(t)| 2 dt = cos 2 θ  ∞ −∞ |x(t)| 2 dt + sin 2 θ  ∞ −∞ |ˆx(t)| 2 dt + cos θ sin θ  ∞ −∞ x(t)ˆx ∗ (t)dt + cos θ sin θ  ∞ −∞ x ∗ (t)ˆx(t)dt But  ∞ −∞ |x(t)| 2 dt =  ∞ −∞ |ˆx(t)| 2 dt = E x and  ∞ −∞ x(t)ˆx ∗ (t)dt = 0 since x(t) and ˆx(t) are orthogonal. Thus, E x θ = E x (cos 2 θ + sin 2 θ)=E x Problem 2.59 1) z(t)=x(t)+j ˆx(t)=m(t) cos(2πf 0 t) − ˆm(t) sin(2πf 0 t) +j[m(t)  cos(2πf 0 t) − ˆm(t)  sin(2πf 0 t) = m(t) cos(2πf 0 t) − ˆm(t) sin(2πf 0 t) +jm(t) sin(2πf 0 t)+j ˆm(t) cos(2πf 0 t) =(m(t)+j ˆm(t))e j2πf 0 t The lowpass equivalent signal is given by x l (t)=z(t)e −j2πf 0 t = m(t)+j ˆm(t) 2) The Fourier transform of m(t)isΛ(f). Thus X(f)= Λ(f + f 0 )+Λ(f −f 0 ) 2 − (−jsgn(f)Λ(f))   − 1 2j δ(f + f 0 )+ 1 2j δ(f − f 0 )  = 1 2 Λ(f + f 0 )[1−sgn(f + f 0 )] + 1 2 Λ(f − f 0 ) [1 + sgn(f − f 0 )] . . . . . . . . . . . . . . ❅ ❅ ❅    1 −f 0 − 1 −f 0 f 0 +1f 0 The bandwidth of x(t)isW =1. 40 3) z(t)=x(t)+j ˆx(t)=m(t) cos(2πf 0 t)+ ˆm(t) sin(2πf 0 t) +j[m(t)  cos(2πf 0 t)+ ˆm(t)  sin(2πf 0 t) = m(t) cos(2πf 0 t)+ ˆm(t) sin(2πf 0 t) +jm(t) sin(2πf 0 t) − j ˆm(t) cos(2πf 0 t) =(m(t) − j ˆm(t))e j2πf 0 t The lowpass equivalent signal is given by x l (t)=z(t)e −j2πf 0 t = m(t) − j ˆm(t) The Fourier transform of x(t)is X(f)= Λ(f + f 0 )+Λ(f −f 0 ) 2 − (jsgn(f)Λ(f))   − 1 2j δ(f + f 0 )+ 1 2j δ(f − f 0 )  = 1 2 Λ(f + f 0 ) [1 + sgn(f + f 0 )] + 1 2 Λ(f − f 0 )[1−sgn(f − f 0 )]    . . . . . . . . . . . . . . . . . . ❅ ❅ ❅ 1 −f 0 +1−f 0 f 0 − 1 f 0 41 Chapter 3 Problem 3.1 The modulated signal is u(t)=m(t)c(t)=Am(t) cos(2π4 × 10 3 t) = A  2 cos(2π 200 π t) + 4 sin(2π 250 π t + π 3 )  cos(2π4 × 10 3 t) = A cos(2π(4 × 10 3 + 200 π )t)+A cos(2π(4 × 10 3 − 200 π )t) +2A sin(2π(4 × 10 3 + 250 π )t + π 3 ) − 2A sin(2π(4 × 10 3 − 250 π )t − π 3 ) Taking the Fourier transform of the previous relation, we obtain U(f)=A  δ(f − 200 π )+δ(f + 200 π )+ 2 j e j π 3 δ(f − 250 π ) − 2 j e −j π 3 δ(f + 250 π )   1 2 [δ(f − 4 × 10 3 )+δ(f +4× 10 3 )] = A 2  δ(f − 4 × 10 3 − 200 π )+δ(f − 4 × 10 3 + 200 π ) +2e −j π 6 δ(f − 4 × 10 3 − 250 π )+2e j π 6 δ(f − 4 × 10 3 + 250 π ) +δ(f +4× 10 3 − 200 π )+δ(f +4× 10 3 + 200 π ) +2e −j π 6 δ(f +4× 10 3 − 250 π )+2e j π 6 δ(f +4× 10 3 + 250 π )  The next figure depicts the magnitude and the phase of the spectrum U(f). s s s s ✻ ✻ ✻ ✻ ✻ ✻✻ ✻ |U(f)|  U(f) −f c − 250 π −f c − 200 π −f c + 200 π −f c + 250 π f c − 250 π f c − 200 π f c + 200 π f c + 250 π − π 6 π 6 A/2 A To find the power content of the modulated signal we write u 2 (t)as u 2 (t)=A 2 cos 2 (2π(4 × 10 3 + 200 π )t)+A 2 cos 2 (2π(4 × 10 3 − 200 π )t) +4A 2 sin 2 (2π(4 × 10 3 + 250 π )t + π 3 )+4A 2 sin 2 (2π(4 × 10 3 − 250 π )t − π 3 ) +terms of cosine and sine functions in the first power Hence, P = lim T →∞  T 2 − T 2 u 2 (t)dt = A 2 2 + A 2 2 + 4A 2 2 + 4A 2 2 =5A 2 42 Problem 3.2 u(t)=m(t)c(t)=A(sinc(t) + sinc 2 (t)) cos(2πf c t) Taking the Fourier transform of both sides, we obtain U(f)= A 2 [Π(f)+Λ(f)]  (δ(f − f c )+δ(f + f c )) = A 2 [Π(f − f c )+Λ(f −f c )+Π(f + f c )+Λ(f + f c )] Π(f − f c ) = 0 for |f − f c | < 1 2 , whereas Λ(f − f c ) = 0 for |f − f c | < 1. Hence, the bandwidth of the bandpass filter is 2. Problem 3.3 The following figure shows the modulated signals for A = 1 and f 0 = 10. As it is observed both signals have the same envelope but there is a phase reversal at t = 1 for the second signal Am 2 (t) cos(2πf 0 t) (right plot). This discontinuity is shown clearly in the next figure where we plotted Am 2 (t) cos(2πf 0 t) with f 0 =3. -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Problem 3.4 y(t)=x(t)+ 1 2 x 2 (t) 43 = m(t) + cos(2πf c t)+ 1 2  m 2 (t) + cos 2 (2πf c t)+2m(t) cos(2πf c t)  = m(t) + cos(2πf c t)+ 1 2 m 2 (t)+ 1 4 + 1 4 cos(2π2f c t)+m(t) cos(2πf c t) Taking the Fourier transform of the previous, we obtain Y (f)=M(f)+ 1 2 M(f) M(f)+ 1 2 (M(f − f c )+M(f + f c )) + 1 4 δ(f)+ 1 2 (δ(f − f c )+δ(f + f c )) + 1 8 (δ(f − 2f c )+δ(f +2f c )) The next figure depicts the spectrum Y (f) 1/4 -2fc -fc -2W 2W fc 2fc 1/8 1/2 Problem 3.5 u(t)=m(t) · c(t) = 100(2 cos(2π2000t) + 5 cos(2π3000t)) cos(2πf c t) Thus, U(f)= 100 2  δ(f − 2000) + δ(f + 2000) + 5 2 (δ(f − 3000) + δ(f + 3000))   [δ(f − 50000) + δ(f + 50000)] =50  δ(f − 52000) + δ(f − 48000) + 5 2 δ(f − 53000) + 5 2 δ(f − 47000) +δ(f + 52000) + δ(f + 48000) + 5 2 δ(f + 53000) + 5 2 δ(f + 47000)  A plot of the spectrum of the modulated signal is given in the next figure ✻ ✻ ✻ ✻ ✻ ✻ ✻ ✻ 125 50 0-53 -52 -48 -47 47 48 52 53 KHz Problem 3.6 The mixed signal y(t) is given by y(t)=u(t) · x L (t)=Am(t) cos(2πf c t) cos(2πf c t + θ) = A 2 m(t) [cos(2π2f c t + θ) + cos(θ)] 44 The lowpass filter will cut-off the frequencies above W , where W is the bandwidth of the message signal m(t). Thus, the output of the lowpass filter is z(t)= A 2 m(t) cos(θ) If the power of m(t)isP M , then the power of the output signal z(t)isP out = P M A 2 4 cos 2 (θ). The power of the modulated signal u(t)=Am(t) cos(2πf c t)isP U = A 2 2 P M . Hence, P out P U = 1 2 cos 2 (θ) A plot of P out P U for 0 ≤ θ ≤ π is given in the next figure. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0 0.5 1 1.5 2 2.5 3 3.5 Theta (rad) Problem 3.7 1) The spectrum of u(t)is U(f)= 20 2 [δ(f − f c )+δ(f + f c )] + 2 4 [δ(f − f c − 1500) + δ(f − f c + 1500) +δ(f + f c − 1500) + δ(f + f c + 1500)] + 10 4 [δ(f − f c − 3000) + δ(f − f c + 3000) +δ(f + f c − 3000) + δ(f + f c + 3000)] The next figure depicts the spectrum of u(t). ✻ ✻ ✻ ✻ ✻ ✻ ✻ ✻ ✻ ✻ -1030-1015-1000 -985 -970 970 985 X 100 Hz 1000 10301015 0 5/2 1/2 10 2) The square of the modulated signal is u 2 (t) = 400 cos 2 (2πf c t) + cos 2 (2π(f c − 1500)t) + cos 2 (2π(f c + 1500)t) +25 cos 2 (2π(f c − 3000)t)+25cos 2 (2π(f c + 3000)t) + terms that are multiples of cosines 45 If we integrate u 2 (t) from − T 2 to T 2 , normalize the integral by 1 T and take the limit as T →∞, then all the terms involving cosines tend to zero, whereas the squares of the cosines give a value of 1 2 . Hence, the power content at the frequency f c =10 5 Hz is P f c = 400 2 = 200, the power content at the frequency P f c +1500 is the same as the power content at the frequency P f c −1500 and equal to 1 2 , whereas P f c +3000 = P f c −3000 = 25 2 . 3) u(t) = (20 + 2 cos(2π1500t) + 10 cos(2π3000t)) cos(2πf c t) = 20(1 + 1 10 cos(2π1500t)+ 1 2 cos(2π3000t)) cos(2πf c t) This is the form of a conventional AM signal with message signal m(t)= 1 10 cos(2π1500t)+ 1 2 cos(2π3000t) = cos 2 (2π1500t)+ 1 10 cos(2π1500t) − 1 2 The minimum of g(z)=z 2 + 1 10 z − 1 2 is achieved for z = − 1 20 and it is min(g(z)) = − 201 400 . Since z = − 1 20 is in the range of cos(2π1500t), we conclude that the minimum value of m(t)is− 201 400 . Hence, the modulation index is α = − 201 400 4) u(t) = 20 cos(2πf c t) + cos(2π(f c − 1500)t) + cos(2π(f c − 1500)t) = 5 cos(2π(f c − 3000)t) + 5 cos(2π(f c + 3000)t) The power in the sidebands is P sidebands = 1 2 + 1 2 + 25 2 + 25 2 =26 The total power is P total = P carrier + P sidebands = 200 + 26 = 226. The ratio of the sidebands power to the total power is P sidebands P total = 26 226 Problem 3.8 1) u(t)=m(t)c(t) = 100(cos(2π1000t) + 2 cos(2π2000t)) cos(2πf c t) = 100 cos(2π1000t) cos(2πf c t) + 200 cos(2π2000t) cos(2πf c t) = 100 2 [cos(2π(f c + 1000)t) + cos(2π(f c − 1000)t)] 200 2 [cos(2π(f c + 2000)t) + cos(2π(f c − 2000)t)] Thus, the upper sideband (USB) signal is u u (t) = 50 cos(2π(f c + 1000)t) + 100 cos(2π(f c + 2000)t) 46 2) Taking the Fourier transform of both sides, we obtain U u (f)=25(δ(f − (f c + 1000)) + δ(f +(f c + 1000))) +50 (δ(f − (f c + 2000)) + δ(f +(f c + 2000))) A plot of U u (f) is given in the next figure. ✻ ✻ ✻✻ -1002 -1001 1001 1002 50 25 0 KHz Problem 3.9 If we let x(t)=−Π  t + T p 4 T p 2  +Π  t − T p 4 T p 2  then using the results of Problem 2.23, we obtain v(t)=m(t)s(t)=m(t) ∞  n=−∞ x(t − nT p ) = m(t) 1 T p ∞  n=−∞ X( n T p )e j2π n T p t where X( n T p )=F  −Π  t + T p 4 T p 2  +Π  t − T p 4 T p 2      f= n T p = T p 2 sinc(f T p 2 )  e −j2πf T p 4 − e j2πf T p 4      f= n T p = T p 2 sinc( n 2 )(−2j)sin(n π 2 ) Hence, the Fourier transform of v(t)is V (f)= 1 2 ∞  n=−∞ sinc( n 2 )(−2j)sin(n π 2 )M(f − n T p ) The bandpass filter will cut-off all the frequencies except the ones centered at 1 T p , that is for n = ±1. Thus, the output spectrum is U(f) = sinc( 1 2 )(−j)M(f − 1 T p ) + sinc( 1 2 )jM(f + 1 T p ) = − 2 π jM(f − 1 T p )+ 2 π jM(f + 1 T p ) = 4 π M(f)   1 2j δ(f − 1 T p ) − 1 2j δ(f + 1 T p )  Taking the inverse Fourier transform of the previous expression, we obtain u(t)= 4 π m(t) sin(2π 1 T p t) 47 [...]... cos(2π(fc + 3 × 1 03 )t) + cos(2π(fc − 3 × 1 03 )t) Taking the Fourier transform of the previous expression, we obtain U (f ) = 50 δ(f − (fc + 1 03 )) + δ(f + fc + 1 03 ) + δ(f − (fc − 1 03 )) + δ(f + fc − 1 03 ) + 25 δ(f − (fc + 3 × 1 03 )) + δ(f + fc + 3 × 1 03 ) + δ(f − (fc − 3 × 1 03 )) + δ(f + fc − 3 × 1 03 ) The spectrum of the signal is depicted in the next figure T 50 T T −10 03 T T... +25 δ(f − 8 × 105 − 1 03 ) − δ(f + 8 × 105 + 1 03 ) j j 1 1 −25 δ(f − 8 × 105 + 1 03 ) − δ(f + 8 × 105 − 1 03 ) j j +125 δ(f − 8 × 105 − 2 × 1 03 ) + δ(f + 8 × 105 + 2 × 1 03 ) +125 δ(f − 8 × 105 − 2 × 1 03 ) + δ(f + 8 × 105 + 2 × 1 03 ) = 50[δ(f − 8 × 105 ) + δ(f + 8 × 105 )] +25 δ(f − 8 × 105 − 1 03 )e−j 2 + δ(f + 8 × 105 + 1 03 )ej 2 π π +25 δ(f − 8 × 105 + 1 03 )ej 2 + δ(f + 8 × 105 − 1 03 )e−j 2 π π +125 δ(f... = 1 2 3) The power of the carrier component is Pcarrier = 400 = 200, whereas the power in the sidebands 2 2 is Psidebands = 400α = 50 Hence, 2 1 50 Psidebands = = Pcarrier 200 4 Problem 3. 15 1) The modulated signal is written as u(t) = 100(2 cos(2π1 03 t) + cos(2 3 × 1 03 t)) cos(2πfc t) = 200 cos(2π1 03 t) cos(2πfc t) + 100 cos(2 3 × 1 03 t) cos(2πfc t) = 100 cos(2π(fc + 1 03 )t) + cos(2π(fc − 1 03 )t)... × 105 − 1 03 )e−j 2 π π +125 δ(f − 8 × 105 − 2 × 1 03 ) + δ(f + 8 × 105 + 2 × 1 03 ) +125 δ(f − 8 × 105 − 2 × 1 03 ) + δ(f + 8 × 105 + 2 × 1 03 ) |U (f )| 125 T T fc −2×1 03 50 T T 25 −fc T T T T fc +2×1 03 fc −2×1 03 U (f ) T T fc fc +2×1 03 π r 2 r fc −1 03 fc +1 03 −π r 2 r 2) The... sidebands is Psidebands = 502 502 2502 2502 + + + = 65000 2 2 2 2 3) The message signal can be written as m(t) = sin(2π1 03 t) + 5 cos(2π2 × 1 03 t) = −10 sin(2π1 03 t) + sin(2π1 03 t) + 5 As it is seen the minimum value of m(t) is −6 and is achieved for sin(2π1 03 t) = −1 or t = 3 1 + 1 03 k, with k ∈ Z Hence, the modulation index is α = 6 4×1 03 4) The power delivered to the load is Pload = 1002 (1 + m(t))2... the absolute value of the message signal Problem 3. 13 1) The modulated signal is u(t) = 100[1 + m(t)] cos(2π8 × 105 t) = 100 cos(2π8 × 105 t) + 100 sin(2π1 03 t) cos(2π8 × 105 t) +500 cos(2π2 × 1 03 t) cos(2π8 × 105 t) = 100 cos(2π8 × 105 t) + 50[sin(2π(1 03 + 8 × 105 )t) − sin(2π(8 × 105 − 1 03 )t)] +250[cos(2π(2 × 1 03 + 8 × 105 )t) + cos(2π(8 × 105 − 2 × 1 03 )t)] 49 Taking the Fourier transform of the previous... −997 997 T 999 2) The average power in the frequencies fc + 1000 and fc − 1000 is Pfc +1000 = Pfc −1000 = 1002 = 5000 2 The average power in the frequencies fc + 30 00 and fc − 30 00 is Pfc +30 00 = Pfc 30 00 = 51 502 = 1250 2 1001 10 03 KHz Problem 3. 16 1) The Hilbert transform of cos(2π1000t) is sin(2π1000t), whereas the Hilbert transform ofsin(2π1000t) is − cos(2π1000t) Thus m(t) = sin(2π1000t) − 2 cos(2π1000t)... m(t))2 cos2 (2πfc t) |u(t)|2 = 50 50 50 The maximum absolute value of 1 + m(t) is 6.025 and is achieved for sin(2π1 03 t) = 1 arcsin( 20 ) 2π1 03 + Since 2 × 1 03 k 1 03 1 20 or t = fc the peak power delivered to the load is approximately equal to max(Pload ) = (100 × 6.025)2 = 72.6012 50 Problem 3. 14 1) u(t) = 5 cos(1800πt) + 20 cos(2000πt) + 5 cos(2200πt) 1 = 20(1 + cos(200πt)) cos(2000πt) 2 The modulating... The bandpass filter will cut-off the low-frequency components M (f ) M (f )+ 1 δ(f ) and the terms with the double frequency components 1 (δ(f −2f0 )+δ(f +2f0 )) 2 4 Thus the spectrum Y2 (f ) is given by Y2 (f ) = M (f − f0 ) + M (f + f0 ) and the bandwidth of y2 (t) is W2 = 2W The signal y3 (t) is y3 (t) = 2m(t) cos2 (2πf0 t) = m(t) + m(t) cos(2πf0 t) with spectrum 1 Y3 (t) = M (f ) + (M (f − f0 )... (f + W ) = 4 +M (f − 2fc − W )u−1 (f − 2fc − W ) + M (f + 2fc + W )u−1 (−f − 2fc − W )) The LPF will cut-off the double frequency components, leaving the spectrum Y (f ) = A2 [M (f − W )u−1 (−f + W ) + M (f + W )u−1 (f + W )] 4 The next figure depicts Y (f ) for M (f ) as shown in Fig P-5.12 Y(f) -W W 2) As it is observed from the spectrum Y (f ), the system shifts the positive frequency components to . f 0 =3. -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0.2 0.4 0.6 0.8 0 0.2. +1)e j2 πft df + 1 2j  1 2 0 (−f +1)e j2 πft df = 1 2j  1 j2 πt fe j2 πft + 1 4π 2 t 2 e j2 πft      0 − 1 2 + 1 2j 1 j2 πt e j2 πft     0 − 1 2 − 1 2j  1 j2 πt fe j2 πft + 1 4π 2 t 2 e j2 πft      1 2 0 + 1 2j 1 j2 πt e j2 πft     1 2 0 =. − 2 j e j π 3 δ(f + 250 π )   1 2 [δ(f − 4 × 10 3 )+δ(f +4× 10 3 )] = A 2  δ(f − 4 × 10 3 − 200 π )+δ(f − 4 × 10 3 + 200 π ) +2e j π 6 δ(f − 4 × 10 3 − 250 π )+2e j π 6 δ(f − 4 × 10 3 + 250 π ) +δ(f