# Solution for Communication Systems

**Topics:**Low-pass filter, Frequency modulation, Cutoff frequency

**Pages:**84 (14106 words)

**Published:**November 15, 2011

Communications Systems, 5th 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) g (t ) = A cos(2π f c t ) fc = 1 T ⎡ −T T ⎤ t∈⎢ , ⎥ ⎣ 2 2⎦

We can rewrite the half-cosine as: ⎛t ⎞ A cos(2π f c t ) ⋅ rect ⎜ ⎟ ⎝T ⎠ Using the property of multiplication in the time-domain: G ( f ) = G1 ( f ) ∗ G2 ( f )

1 sin(π fT ) [δ ( f − fc ) + δ ( f + fc )] ∗ AT π fT 2 Writing out the convolution: ∞ AT ⎛ sin(πλT ) ⎞ G( f ) = ∫ ⎜ ⎟ [δ (λ − ( f + f c ) + δ (λ − ( f − f c ) ] d λ 2 ⎝ πλT ⎠ −∞ = A ⎛ sin(π ( f + f c )T ) sin(π ( f − f c )T ) ⎞ + ⎜ ⎟ f + fc f − fc 2π ⎝ ⎠ ⎛ ⎞ A ⎜ cos(π fT ) cos(π fT ) ⎟ = − ⎜ 1 ⎟ 2π ⎜ f − 1 ⎟ f+ 2T 2T ⎠ ⎝ = (b)By using the time-shifting property: T exp(− j 2π ft0 ) g (t − t0 ) t0 = 2 ⎛ ⎞ A ⎜ cos(π fT ) cos(π fT ) ⎟ G( f ) = − ⋅ exp(− jπ fT ) ⎜ 1 ⎟ 2π ⎜ f − 1 ⎟ f+ ⎝ 2T 2T ⎠ fc = 1 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.

fc =

1 2Ta

⎡ cos(π fTa ) cos(π fTa ) ⎤ − ⎢ ⎥ ⋅ (cos(π fTa ) − j sin(π fTa )) f + fc ⎦ ⎣ f − fc A ⎡ cos(2π fTa ) cos(2π fTa ) sin(2π fTa ) sin(2π fTa) ⎤ = − +j −j ⎢ ⎥ 4π ⎣ f − f c f + fc f − fc f + fc ⎦ A 2π A ⎡ exp(− j 2π fTa ) exp(− j 2π fTa ) ⎤ − ⎢ ⎥ 4π ⎣ f − fc f + fc ⎦

G( f ) =

=

(d) The spectrum is the same as for (b) except shifted backwards in time and multiplied by -1. ⎛ ⎞ A ⎜ cos(π fT ) cos(π fT ) ⎟ − ⋅ exp( jπ fT ) G( f ) = ⎜ 1 ⎟ 2π ⎜ f − 1 ⎟ f+ 2T 2T ⎠ ⎝ ⎡ ⎤ A ⎢ exp( j 2π fT ) exp( j 2π fT ) ⎥ = − ⎢ ⎥ 1 4π ⎢ f − 1 ⎥ 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( j 2π fT ) + exp(− j 2π fT ) exp( j 2π fT ) + (− j 2π fT ) ⎥ − G( f ) = ⎢ ⎥ 1 1 4π ⎢ ⎥ f− f+ 2T 2T ⎣ ⎦ ⎡ ⎤ A ⎢ cos(2π fT ) cos(2π fT ) ⎥ = − ⎢ 1 ⎥ 2π ⎢ f − 1 ⎥ f+ 2T 2T ⎦ ⎣

Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.

2.2 g (t ) = exp(−t ) sin(2π f c t )u(t ) = ( exp(−t )u(t ) )( sin(2π f c t ) ) ⎡1 ⎤ 1 ∗ ⎢ (δ ( f − f c ) − δ ( f + f c ) ) ⎥ 1 + j 2π f ⎣ 2 j ⎦ ⎤ 1 ⎡ 1 1 = − ⎢ ⎥ 2 j ⎣1 + j 2π ( f − f c ) 1 + j 2π ( f + f c ) ⎦

∴ G( f ) =

2.3 (a)

g (t ) = g e (t ) + g o (t ) 1 [ g (t ) + g (−t )] 2 ⎛ t ⎞ g e (t ) = Arect ⎜ ⎟ ⎝ 2T ⎠ g e (t ) = 1 [ g (t ) − g (−t )] 2 ⎛ ⎛ 1 ⎞ ⎛ 1 ⎜ ⎜ t − 2T ⎟ ⎜t+ T − rect ⎜ 2 g o (t ) = A ⎜ rect ⎜ ⎟ ⎜ ⎜ T ⎟ ⎜ T ⎜ ⎝ ⎠ ⎝ ⎝ g o (t ) =

⎞⎞ ⎟⎟ ⎟⎟ ⎟⎟ ⎟ ⎠⎠

Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.

(b) By the time-scaling property g(-t)

Ge ( f ) =

G(-f)

1 [G ( f ) + G (− f ) ] 2 1 = [sinc( fT ) exp(− j 2π fT ) + sinc( fT ) exp( j 2π fT ) ] 2 = sinc( fT ) cos(π fT )

Go ( f ) =

1 [G ( f ) − G (− f )] 2 1 = [sinc( fT ) exp(− j 2π fT ) − sinc( fT ) exp( j 2π fT ) ] 2 = − jsinc( fT ) sin(π fT )

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 ) = − j sgn( f ) + ju( f − W ) − ju(− f − W ) meets the stated criteria. By duality g(f) G(-t)

g (t ) =

⎛1 1 ⎞ j ⎜ δ (t ) − ⎟ exp(− j 2π Wt ) − j 2π t ⎠ ⎝2 1 sin(2π Wt ) = +j 2π t πt 1 + πt ⎛ π u2 ⎞ exp ⎜ − 2 ⎟ du ∫ ⎝ τ ⎠ τ t −T 1 1 t +T

⎛1 1 ⎞ j ⎜ δ (t ) − ⎟ exp( j 2π Wt ) j 2π t ⎠ ⎝2

2.5

g (t ) = =

τ

t −T

∫

0

h(τ )dτ +

1

t +T

τ

∫ h(τ )dτ

0

dg (t ) 1 1 = − h(t − T ) + h(t + T ) dt τ τ By the differentiation property: ⎛ dg (t ) ⎞ F⎜ ⎟ = j 2π fG ( f ) ⎝ dt ⎠ 1 = [ H ( f ) exp( j 2π f τ ) − H ( f ) exp(− j 2π f τ )]

τ

=

2j

τ

H ( f ) sin(2π f τ )

But H ( f ) = τ exp(−π f 2τ 2 ) 1 exp(−π f 2τ 2 ) sin(2π fT ) ∴ G( f ) = πf sin(2π fT ) = exp(−π f 2τ 2 ) πf = 2T exp(−π f 2τ 2 )sinc(2π fT ) lim G ( f ) = 2Tsinc(2π fT ) τ →0...

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