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Principles of Communication - Zeimer, William H Tranter Solutions Manual
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Ziemer & Tranter: Principles of CommunicationsProblem Solutions Chapter 2 Problem 2.2:  x ( t ) = − 8sin(8 πt ) + 4cos(4 πt − π/ 4) . Problem 2.4: (a) Single-sided spectrum Double-sided spectrum f, Hz036|X a (f)|f, Hz03− π  /3  ∠  X a (f) f, Hz−3033|X a (f)|f, Hz−303− π  /3 π  /3 ∠  X a (f) (b) Single-sided spectrum Double-sided spectrum f, Hz05624|X b (f)|f, Hz056− π  /2  ∠  X b (f) f, Hz−6−505612|X b (f)|f, Hz−6−5056− π  /2 π  /2 ∠  X b (f) Problem 2.6:  Area = 2    ∞ 0 1  (1 − t/ ) dt  = 2  (  − / 2) = 1 . Problem 2.8: (a) 65; (b) 380; (c) 0; (d)  12 e − 5 π/ 2 ; (e)  − 4 π 2 + 100 .Problem 2.10: (a)  x ( t ) = 2 Re [ e  j 5 πt ] + 4 Re [ e  j (11 πt − π/ 2) ] =  Re [2 e  j 5 πt + 4 e  j (11 πt − π/ 2) ] . (b)  x ( t ) =  e  j 5 πt + e −  j 5 πt + 2 e  j (11 πt − π/ 2) + 2 e −  j (11 πt − π/ 2) . (c) Single-sided spectrum Double-sided spectrum f, Hz02.55.524|X(f)|f, Hz02.55.5− π  /2  ∠  X b (f) f, Hz−5.5−2.5 0 2.55.512|X b (f)|f, Hz−5.5−2.5 0 2.55.5− π  /2 π  /2 ∠  X b (f) Problem 2.12: (a) Power  ( P   =  A 2 2  ) ; (b) Neither ( E   = ∞  and  P   = 0 );(c) Energy  ( E   =  34 A 2 τ  5 ) ; (d) Energy  ( E   = 5 τ  ) .Problem 2.14: (a)  E   = ∞ ,  P   = ∞ ; (b)  E   = 5  J,  P   = 0 ;(c)  E   = ∞ ,  P   = 32  W; (d)  E   = ∞ ,  P   = 8 . 5  W.Problem 2.16: (a) Break the integral up into contiguous intervals of length  T  0 .Problem 2.18: (a) Expand the integrand, integrate term by term, and simplify, making useof the orthogonality property of the orthonormal functions.(b) Add and subtract the quantity suggested right above (2.31) and simplify.  Problem 2.20: (a)  X  − 2  =  14 , X  0  =  12 , X  2  =  14 ,  all other coeﬃcients zero.(b)  X  − 1  =  12 (1 +  j ) , X  1  =  12 (1 −  j ) ,  all other coeﬃcients zero.(c)  X  4  =  X  − 4  = − 18 , X  2  =  X  − 2  =  14 , X  0  = − 14 ,  all other coeﬃcients zero.(d)  X  − 3  =  18 , X  − 1  =  38 , X  1  =  38 , X  3  =  18 ,  all other coeﬃcients zero.Problem 2.22: Parts (a) through (c) were discussed in the text.For (d), break the integral for  x ( t )  up into a part for  t <  0  and a part for  t >  0 .Then apply the odd half-wave symmetry condition.Problem 2.24: (a)  P  | nf  0 |≤ 1 /τ  /P  total  = 0 . 905 ; (b)  P  | nf  0 |≤ 1 /τ  /P  total  = 0 . 903 .Problem 2.26: (a)  π 2 / 8 ; (b)  π/ 4  (Use Fourier series of pulse train).Problem 2.28: (a) There is no line at DC; otherwise, it looks like a square wave spectrum.(b)  X  An  =  K  (  jnω 0 ) X  Bn  ,  where the superscripts A and B refer to  x a  and  x b .Problem 2.30: By linearity,  F  [ u ( t )] = 1  j 2 πf   + 12 δ  ( f  ) .Problem 2.32: The proof is similar to the proof of the convolution theorem.Problem 2.34: (a)  G 1 ( f  ) =   4 / 31 + [ f/ (3 / 2 π )] 2  2 ; (b)  G 2 ( f  ) =  49 Π( f/ 30); −5 0 500.511.52fG 1 (f) −15 0 1504/9fG 2 (f) (c)  G 3 ( f  ) =  1625 sinc 2 ( f/ 5) ; (d)  G 4 ( f  ) =  0 . 4  sinc  f  − 205  + sinc  f  +205  2 . −15 −10 −5 0 5 10 1500.20.40.60.8fG 3 (f) −20 0 2000.16fG 4 (f) Problem 2.36: (a)  y 1 ( t ) =  0 , t < τ   − 0 . 5 1 α  1 − e − α ( t − τ  +0 . 5)  , τ   − 0 . 5  < t < τ   + 0 . 5 1 α  e − α ( t − τ  − 0 . 5) − e − α ( t − τ  +0 . 5)  , t > τ   + 0 . 5  (b) Plot of   y 2 ( t ) : (c) Normalised plot of   y 3 ( t )  for various values of   α : −1.5−1−0.5 0 0.5 1 1.500.511.52ty 2 (t) −0.5 0 0.50 α =50  α =10 α =2  α =12/  α ty 3 (t) (d)  y 4 ( t ) =    t −∞ x ( λ ) dλ. Problem 2.38: (a)  E  1 ( | f  |≤ W  ) E  total =  2 π  tan − 1 (2 πW/α ) ; (b)  E  1 ( | f  |≤ W  ) E  total = 2   τW  0  sinc 2 ( u ) du .Problem 2.40: Combine the exponents of the two factors in the integrand of the Fouriertransform integral, complete the square, and use the deﬁnite integral given.Problem 2.42: The result is an even triangular wave with zero average value of period  T  0 .It makes no diﬀerence whether the srcinal square wave is even or odd or neither.Problem 2.44: (a)  h ( t ) =  δ  ( t ) − 10 e − 10 t u ( t ) ; (b)  h ( t ) = 0 . 3 e − 0 . 4( t − 3) u ( t − 3) .Problem 2.46: (a) Replace the capacitors with  1 / (  jωC  )  which is their ac-equivalent impedance.Call the junction of the input resistor, feedback resistor and capacitors 1. Callthe junction at the positive input of the op amp 2. Call the junction at thenegative input of the op amp 3. Write down the KCL equations at these three junctions. Use the constraint equation for the op amp  V  2  =  V  3  and thedeﬁnitions for  ω 0 ,  Q , and  K   to get the transfer function expression.(d) Combinations of components giving  RC   = 2 . 3 × 10 − 4 s and  R a R b = 2 . 5757  will work.Problem 2.48: The Paley-Wiener criterion gives the non-convergent integral  I   =    ∞−∞ βf  2 1 + f  2  df  .Problem 2.50: Input ESD  G x ( f  ) =  14+(2 πf  ) 2 ; Output ESD  G y ( f  ) =  400[9+(2 πf  ) 2 ] 2 [4+(2 πf  ) 2 ] .Problem 2.52: (a)  B 90  =  α 2 π  tan(0 . 45 π ) = 1 . 005 α ; (b)  B 90  = 0 . 9 W  ; (c)  B 90  = 0 . 85 /τ  .Problem 2.54: (a) Amplitude distortion only; (b) Phase distortion only;(c) Distortionless; (d) Distortionless.Problem 2.56: The group delay is  T  g ( f  ) =  0 . 11+(0 . 2 πf  ) 2  −  0 . 3331+(0 . 667 πf  ) 2 .The phase delay is  T   p ( f  ) =  12 πf  [tan − 1 (0 . 2 πf  ) − tan − 1 (0 . 667 πf  )] .Problem 2.58: (a)  THD =  (1 . 075) 2 | H  (1000) | 2 (0 . 025) 2 | H  (3000) | 2  = 1849 [4 + 16 × 10 6 Q 2 ][1 + 144 × 10 6 Q 2 ][4 + 144 × 10 6 Q 2 ][(1 + 32 × 10 6 Q 2 ) 2 + 16 × 10 6 Q 2 ] .(b) By trial and error,  Q  = 0 . 55  gives 0.01% THD.Problem 2.60: (a)  | H  BU  ( f  ) | = 1 √  1 + R 2 n ,  where  R  =  f/f  3 , and  R →∞  if   f > f  3  and  R → 0  if   f < f  3 .(b) The ripple in dB is given by  R dB  = 20log 10 √  1 +  2 .  Problem 2.62:  | H  CH  ( ω n ) | 2 = 11 +  2 (2 ω 2 n − 1) 2  =  H  CH  ( s n ) H  CH  ( − s n ) | s n =  jω n ,  where  ω n  =  ω/ω 3 dB .Factor the expression and take the left half plane poles for  H  CH  ( f  ) .Then let  s n  =  jω n . Diﬀerentiate the argument of this expressionwith respect to frequency to get the group delay.Problem 2.64: (a) 0.5 second; Use sketches to show (b) and (c).Problem 2.66: (a) The sampling frequency should be large compared with the signal bandwidth.(b)  Y  ( f  ) = sinc( T  s f  ) ∞  n = −∞ X  ( f   − nf  s ) e −  jπT  s f  ,  where  f  s  = 1 /T  s .For small distortion we want  T  s  much less than the inverse signal bandwidth,which implies that we are sampling much faster than the Nyquist rate.Problem 2.68: The lowpass recovery ﬁlter can cutoﬀ in the range  1 . 9 + kHz to  2 . 1 − kHz.Problem 2.70: Use the form  ˆcos ω 0 t  =    ∞−∞ cos ω 0 ( t − λ ) πλ dλ  and expand the cosine to carry out theintegral. Make use of the fact that the integral of sinc( x ) from  −∞  to  ∞  is unity.Problem 2.72: This is a matter of ﬁnding the integral of the product of the signal and itsHilbert transform, and showing that it is zero.Problem 2.74: The spectra for the various signals are shown below, on normalised axes: −2 −1 0 1 200.51(a)f/W       X                1       (      f      )      /      A 0 f0 f0+W01.5(b)f       X         2       (      f      )      /      A −2 −1 0 1 200.51(c)f/W       X         3       (      f      )      /      A −2 −1 0 1 200.51(d)f/W       X         4       (      f      )      /      A Problem 2.76: For  t < − τ/ 2 . the output is zero. For  | t |≤ τ/ 2 , the result is y ( t ) =  α/ 2   α 2 + (2 π ∆ f  ) 2  cos[2 π ( f  0  + ∆ f  ) t − θ ] − e − α ( t + τ/ 2) cos[2 π ( f  0  + ∆ f  ) t + θ ]  . For  t > τ/ 2 , the result is y ( t ) = ( α/ 2) e − αt   α 2 + (2 π ∆ f  ) 2  e ατ/ 2 cos[2 π ( f  0  + ∆ f  ) t − θ ] − e − ατ/ 2 cos[2 π ( f  0  + ∆ f  ) t + θ ]  . In the above equations,  θ  is given by  θ  = − tan − 1  2 π ∆ f α  .

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