SOLUTIONS - Communication Systems PDF
Preview SOLUTIONS - Communication Systems
Solutions Manual for: Communications Systems, th 5 edition by Karl Wiklund, McMaster University, Hamilton, Canada Michael Moher, Space-Time DSP Ottawa, Canada and Simon Haykin, McMaster University, Hamilton, Canada Published by Wiley, 2009. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Chapter 2 2.1 (a) ⎡−T T⎤ g(t)= Acos(2πf t) t∈ , c ⎢⎣ 2 2⎥⎦ 1 f = c T We can rewrite the half-cosine as: ⎛ t ⎞ Acos(2πf t)⋅rect ⎜ ⎟ c ⎝T ⎠ Using the property of multiplication in the time-domain: G(f)=G (f)∗G (f) 1 2 1 sin(πfT) = [δ(f − f )+δ(f + f )]∗AT 2 c c πfT Writing out the convolution: ∞ AT ⎛sin(πλT)⎞ G(f)= ∫ [δ(λ−(f + f )+δ(λ−(f − f )]dλ ⎜ ⎟ 2 ⎝ πλT ⎠ c c −∞ A ⎛sin(π(f + f )T) sin(π(f − f )T)⎞ 1 = ⎜ c + c ⎟ f = 2π f + f f − f c 2T ⎝ ⎠ c c ⎛ ⎞ A ⎜cos(πfT) cos(πfT)⎟ = ⎜ − ⎟ 2π 1 1 ⎜ f − f + ⎟ ⎝ 2T 2T ⎠ (b)By using the time-shifting property: T g(t−t )(cid:82)exp(−j2πft ) t = 0 0 0 2 ⎛ ⎞ A ⎜cos(πfT) cos(πfT)⎟ G(f)= ⎜ − ⎟⋅exp(−jπfT) 2π 1 1 ⎜ f − f + ⎟ ⎝ 2T 2T ⎠ Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. (c)The half-sine pulse is identical to the half-cosine pulse except for the centre frequency and time-shift. 1 f = c 2Ta A ⎡cos(πfTa) cos(πfTa)⎤ G(f)= − ⋅(cos(πfTa)− jsin(πfTa)) ⎢ ⎥ 2π f − f f + f ⎣ ⎦ c c A ⎡cos(2πfTa) cos(2πfTa) sin(2πfTa) sin(2πfTa)⎤ = − + j − j ⎢ ⎥ 4π f − f f + f f − f f + f ⎣ ⎦ c c c c A ⎡exp(−j2πfTa) exp(−j2πfTa)⎤ = − ⎢ ⎥ 4π f − f f + f ⎣ ⎦ c c (d) The spectrum is the same as for (b) except shifted backwards in time and multiplied by -1. ⎛ ⎞ A ⎜cos(πfT) cos(πfT)⎟ G(f)= ⎜ − ⎟⋅exp(jπfT) 2π 1 1 ⎜ f − f + ⎟ ⎝ 2T 2T ⎠ ⎡ ⎤ A ⎢exp(j2πfT) exp(j2πfT)⎥ = ⎢ − ⎥ 4π 1 1 ⎢ f − f + ⎥ ⎣ 2T 2T ⎦ (e) Because the Fourier transform is a linear operation, this is simply the summation of the results from (b) and (d) ⎡ ⎤ A ⎢exp(j2πfT)+exp(−j2πfT) exp(j2πfT)+(−j2πfT)⎥ G(f)= ⎢ − ⎥ 4π 1 1 ⎢ f − f + ⎥ ⎣ 2T 2T ⎦ ⎡ ⎤ A ⎢cos(2πfT) cos(2πfT)⎥ = ⎢ − ⎥ 2π 1 1 ⎢ f − f + ⎥ ⎣ 2T 2T ⎦ Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2.2 g(t)=exp(−t)sin(2πf t)u(t) c =(exp(−t)u(t))(sin(2πf t)) c 1 ⎡ 1 ⎤ ∴G(f)= ∗ (δ(f − f )−δ(f + f )) ⎢ ⎥ 1+ j2πf ⎣2j c c ⎦ 1 ⎡ 1 1 ⎤ = − ⎢ ⎥ 2j 1+ j2π(f − f ) 1+ j2π(f + f ) ⎣ ⎦ c c 2.3 (a) g(t)= g (t)+g (t) e o 1 g (t)= [g(t)+g(−t)] e 2 ⎛ t ⎞ g (t)= Arect ⎜ ⎟ e ⎝2T ⎠ 1 g (t)= [g(t)−g(−t)] o 2 ⎛ ⎛ 1 ⎞ ⎛ 1 ⎞⎞ t− T t+ T ⎜ ⎜ ⎟ ⎜ ⎟⎟ 2 2 g (t)= A⎜rect⎜ ⎟−rect⎜ ⎟⎟ o T T ⎜ ⎜ ⎟ ⎜ ⎟⎟ ⎜ ⎟ ⎝ ⎝ ⎠ ⎝ ⎠⎠ Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. (b) By the time-scaling property g(-t) (cid:82) G(-f) 1 G (f)= [G(f)+G(−f)] e 2 1 = [sinc(fT)exp(−j2πfT)+sinc(fT)exp(j2πfT)] 2 =sinc(fT)cos(πfT) 1 G (f)= [G(f)−G(−f)] o 2 1 = [sinc(fT)exp(−j2πfT)−sinc(fT)exp(j2πfT)] 2 =−jsinc(fT)sin(πfT) Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2.4. We need to find a function with the stated properties. We can verify that: G(f)=−jsgn(f)+ ju(f −W)− ju(−f −W) meets the stated criteria. By duality g(f)(cid:82)G(-t) 1 ⎛1 1 ⎞ ⎛1 1 ⎞ g(t)= + j δ(t)− exp(−j2πWt)− j δ(t)− exp(j2πWt) ⎜ ⎟ ⎜ ⎟ πt ⎝2 j2πt⎠ ⎝2 j2πt⎠ 1 sin(2πWt) = + j πt 2πt 2.5 1t+T ⎛ πu2 ⎞ g(t)= ∫ exp⎜− ⎟du τ ⎝ τ2 ⎠ t−T 1 0 1t+T = ∫ h(τ)dτ+ ∫ h(τ)dτ τ τ t−T 0 dg(t) 1 1 =− h(t−T)+ h(t+T) dt τ τ By the differentiation property: ⎛dg(t)⎞ F = j2πfG(f) ⎜ ⎟ ⎝ dt ⎠ 1 = [H(f)exp(j2πfτ)−H(f)exp(−j2πfτ)] τ 2j = H(f)sin(2πfτ) τ But H(f)=τexp(−πf 2τ2) 1 ∴G(f)= exp(−πf 2τ2)sin(2πfT) πf sin(2πfT) =exp(−πf 2τ2) πf =2Texp(−πf 2τ2)sinc(2πfT) limG(f)=2Tsinc(2πfT) τ→0 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2.6 (a) 1 If g(t) is even and real then g(t)= [g(t)+g(−t)] 2 and g(t)= g*(t)⇒G(f)=G*(−f) 1 G*(f)= [G*(f)+G*(−f)] 2 1 1 G*(f)= G*(−f) 2 2 G*(f)=G(f) ∴G(f) is all real 1 g(t)= [g(t)−g(−t)] If g(t) is odd and real then 2 and g(t)= g*(t)⇒G(f)=G*(−f) 1 G(f)= [G(f)−G(−f)] 2 1 1 G*(f)= G*(f)− G*(−f) 2 2 G*(f)=−G*(−f) G*(f)=−G(f) ∴G(f) must be all imaginary (b) d (−j2πt)G(t)(cid:82) g(−f) by duality df j d t⋅G(t)(cid:82) g(−f) 2πdf The previous step can be repeated n times so: dn (−j2πft)nG(t)(cid:82) g(−f) df n But each factor (−j2πft) represents another differentiation. n ⎛ j ⎞ tn⋅G(t)(cid:82) g(n)(−f) ⎜ ⎟ ⎝2π⎠ Replacing g with h n ⎛ j ⎞ tnh(t)(cid:82) H(n)(f) ⎜ ⎟ ⎝2π⎠ Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. (c) n ⎛ j ⎞ Let h(t)=tng(t) and H(f)= G(n)(f) ⎜ ⎟ ⎝2π⎠ ∞ ⎛ j ⎞n ∫ h(t)dt = H(0)= G(n)(0) ⎜ ⎟ ⎝2π⎠ −∞ (d) g (t)(cid:82)G (f) 1 1 g*(t)(cid:82)G (−f) 2 2 ∞ g (t)g (t)(cid:82) ∫ G (λ)G (f −λ)dλ 1 2 1 2 −∞ ∞ g (t)g*(t)(cid:82) ∫ G (λ)G (−(f −λ))dλ 1 2 1 2 −∞ ∞ = ∫ G (λ)G (λ− f)dλ 1 2 −∞ (e) ∞ g (t)g*(t)(cid:82) ∫ G (λ)G (λ− f)dλ 1 2 1 2 −∞ ∞ ∫ g (t)g*(t)dt (cid:82)G(0) 1 2 −∞ ∞ ∞ ∫ g (t)g*(t)dt (cid:82)∫ G (λ)G (λ−0)dλ 1 2 1 2 −∞ −∞ ∞ ∞ ∫ g (t)g*(t)dt (cid:82)∫ G (λ)G (λ)dλ 1 2 1 2 −∞ −∞ Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2.7 (a) g(t)(cid:82) ATsinc2(fT) ∞ ∫ g(t) dt = AT −∞ maxG(f)=G(0) = ATsinc2(0) = AT ∴The first bound holds true. (b) ∞ dg(t) ∫ dt =2A dt −∞ j2πfG(f) = 2πfATsinc2(fT) sin(πfT) sin(πfT) = 2πfAT ⋅ πfT πfT sin(πfT) = 2A ⋅sin(πfT) πfT But, sin(πfT) ≤1 ∀f and sinc(πfT) ≤1 ∀f sin(πfT) ∴ 2A ⋅sin(πfT) ≤2A πfT ∴ j2πfG(f) ≤2A Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2.7 c) (j2πf)2G(f) = 4π2f 2G(f) sin2(πfT) = 4π2f 2AT (πfT)2 4A = sin2(πfT) T 4A ≤ T The second derivative of the triangular pulse is plotted as: Integrating the absolute value of the delta functions gives: ∞ d2g(t) 4A ∫ dt = dt2 T −∞ ∞ d2g(t) ∴(j2πf)2G(f) ≤ ∫ dt dt2 −∞ Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
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