Communication Systems Solution Manual 5th Edition

Solutions Manual for:

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 =

⎡ −T T ⎤ t∈⎢ , ⎥ ⎣ 2 2⎦

1 T

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 T πλ ⎝ ⎠ −∞ =

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 ⎠ ⎝ =

fc =

1 2T

(b)By using the time-shifting property: T g (t − t0 ) R exp(− j 2π ft0 ) t0 = 2 ⎛ ⎞ A ⎜ cos(π fT ) cos(π fT ) ⎟ G( f ) = − ⋅ exp(− jπ fT ) ⎜ 1 ⎟ 2π ⎜ f − 1 ⎟ 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.

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 ⎦

G( f ) =

=

A 2π

A ⎡ exp(− j 2π fTa ) exp(− j 2π fTa ) ⎤ − ⎢ ⎥ 4π ⎣ f − fc f + fc ⎦

(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 + 2T − g o (t ) = A ⎜ rect ⎜ rect ⎟ ⎜ ⎜⎜ ⎜ T ⎟ ⎜ T ⎝ ⎠ ⎝ ⎝ g o (t ) =

⎞⎞ ⎟⎟ ⎟⎟ ⎟ ⎟⎟ ⎠⎠

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

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

Ge ( 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 )

Go ( f ) =

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) R G(-t)

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

g (t ) =

2.5

1 + πt

g (t ) = =

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

t +T

⎛ π u2 ⎞ exp ⎜ − 2 ⎟ du τ t −∫T ⎝ τ ⎠ 1

1

τ

0

∫

h(τ )dτ +

t −T

1

τ

t +T

∫ 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

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

2.6 (a)

1 [ g (t ) + g (−t )] 2 and g (t ) = g * (t ) ⇒ G ( f ) = G* (− f ) g (t ) =

If g(t) is even and real then

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 If g(t) is odd and real then

1 [ g (t ) − g (−t )] 2 and g (t ) = g * (t ) ⇒ G ( f ) = G* (− f ) g (t ) =

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)

(− j 2π t )G (t ) R t ⋅ G (t ) R

d g (− f ) by duality df

j d g (− f ) 2π df

The previous step can be repeated n times so: dn (− j 2π ft ) n G (t ) R n g (− f ) df But each factor (− j 2π ft ) represents another differentiation. n

⎛ j ⎞ (n) t n ⋅ G (t ) R ⎜ ⎟ g (− f ) ⎝ 2π ⎠ Replacing g with h n

⎛ j ⎞ (n) t n h(t ) R ⎜ ⎟ H (f) ⎝ 2π ⎠

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(c) n

⎛ j ⎞ (n) Let h(t ) = t g (t ) and H ( f ) = ⎜ ⎟ G (f) ⎝ 2π ⎠ n

∞

n

⎛ j ⎞ (n) ∫−∞ h(t )dt = H (0) = ⎜⎝ 2π ⎟⎠ G (0)

(d) g1 (t ) R G1 ( f ) g 2* (t ) R G2 (− f ) ∞

g1 (t ) g 2 (t ) R

∫ G (λ )G ( f − λ )d λ 1

2

−∞ ∞

g1 (t ) g 2* (t ) R

∫ G (λ )G (−( f − λ ))d λ 1

2

−∞ ∞

=

∫ G (λ )G (λ − f )d λ 1

2

−∞

(e) ∞

g1 (t ) g 2* (t ) R

∫ G (λ )G (λ − f )d λ 1

2

−∞ ∞

∫ g (t ) g (t )dt R G(0) 1

* 2

−∞ ∞

∫

−∞

∞

g1 (t ) g 2* (t )dt R ∫ G1 (λ )G2 (λ − 0)d λ −∞

∞

∞

∫ g (t ) g (t )dt R ∫ G (λ )G (λ )d λ 1

−∞

* 2

1

2

−∞

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2.7 (a) g (t ) R ATsinc 2 ( fT ) ∞

∫

g (t ) dt = AT

−∞

max G ( f ) = G (0) = ATsinc 2 (0) = AT ∴ The first bound holds true.

(b) ∞

∫

−∞

dg (t ) dt = 2 A dt

j 2π fG ( f ) = 2π fATsinc 2 ( fT ) = 2π fAT = 2A

But,

sin(π fT ) sin(π fT ) ⋅ π fT π fT

sin(π fT ) ⋅ sin(π fT ) π fT

sin(π fT ) ≤ 1 ∀f and sinc(π fT ) ≤ 1 ∀f ∴ 2A

sin(π fT ) ⋅ sin(π fT ) ≤ 2 A π fT

∴ j 2π fG ( f ) ≤ 2 A

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

2.7 c) ( j 2π f ) 2 G ( f ) = 4π 2 f 2G ( f ) sin 2 (π fT ) = 4π f AT (π fT ) 2 2

2

4A 2 sin (π fT ) T 4A ≤ T =

The second derivative of the triangular pulse is plotted as:

Integrating the absolute value of the delta functions gives: ∞

∫

−∞

d 2 g (t ) 4A dt = 2 dt T

∴ ( j 2π f ) 2 G ( f ) ≤

∞

∫

−∞

d 2 g (t ) dt dt 2

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

2.8. (a) g1 (t ) ∗ g 2 (t ) R G1 ( f )G2 ( f ) = G2 ( f )G1 ( f ) by the commutative property of multiplication b) g1 ( f ) ∗ [ g 2 ( f ) ∗ g 3 ( f ) ] R G1 ( f ) [G2 ( f )G3 ( f ) ] Because multiplication is commutative, the order of the multiplication doesn't matter. ∴ G1 ( f ) [G2 ( f )G3 ( f ) ] = [G1 ( f )G2 ( f ) ] G3 ( f ) ∴ G1 ( f ) [G2 ( f )G3 ( f ) ] R [ g1 ( f ) ∗ g 2 ( f ) ] ∗ g3 ( f ) c) Taking the Fourier transform gives: G1 ( f ) [G2 ( f ) + G3 ( f )] Multiplication is distributive so: G1 ( f )G2 ( f ) + G2 ( f )G 3 ( f ) R g1 (t ) g 2 (t ) + g1 (t ) g 2 (t )

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

2.9 a) Let h(t ) = g1 (t ) ∗ g 2 (t ) dh(t ) R j 2π fH ( f ) dt = j 2π fG1 ( f )G2 ( f )

= ( j 2π fG1 ( f ) ) G2 ( f )

( j 2π fG1 ( f ) ) G2 ( f ) R ⎡⎢

dg1 (t ) ⎤ ∗ g 2 (t ) ⎣ dt ⎥⎦

∴

d dg (t ) [ g1 (t ) ∗ g 2 (t )] = ⎡⎢ 1 ⎤⎥ ∗ g 2 (t ) dt ⎣ dt ⎦

b)

t

∫ g (t ) ∗ g (t )dt R 1

2

−∞

1 G (0)G2 (0) δ( f ) G1 ( f )G2 ( f ) + 1 j 2π f 2

⎡ 1 ⎤ ⎡ G (0) ⎤ =⎢ G1 ( f ) ⎥ G2 ( f ) + ⎢ 1 δ ( f ) ⎥ G2 ( f ) ⎣ 2 ⎦ ⎣ j 2π f ⎦ ⎡ 1 ⎤ G (0) =⎢ G1 ( f ) + 1 δ ( f ) ⎥ G2 ( f ) 2 ⎣ j 2π f ⎦ t ⎡t ⎤ ∴ ∫ g1 (t ) ∗ g 2 (t )dt = ⎢ ∫ g1 (t ) ⎥ ∗ g 2 (t ) −∞ ⎣ −∞ ⎦ t

2.10.

Y( f ) =

∫ X (ν ) X ( f −ν )dν

−∞

X (ν ) ≠ 0 if ν ≤ W X ( f −ν ) ≠ 0 if f −ν ≤ W

( f −ν ) ≤ W for f ≤ W +ν when ν ≥ 0 and ν ≤ W ( f −ν ) ≥ −W for f ≤ −W +ν when ν ≤ 0 and ν ≥ −W ∴ ( f −ν ) ≤ W for 0 ≤ ν ≤ W when f ≤ 2W ( f −ν ) ≥ −W for -W ≤ ν ≤ 0 when f ≥ −2W ∴ Over the range of integration [ −W ,W ] , the integral is non-zero if

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f ≤ 2W

2.11 a) Given a rectangular function: g (t ) =

1 ⎛t ⎞ rect ⎜ ⎟ , for which the area under g(t) is T ⎝T ⎠

always equal to 1, and the height is 1/T.

1 ⎛t rect ⎜ T ⎝T

⎞ ⎟ R sinc( fT ) ⎠

Taking the limits: 1 ⎛t⎞ lim rect ⎜ ⎟ = δ (t ) T →0 T ⎝T ⎠ 1 lim sinc( fT ) = 1 T →0 T b)

g (t ) = 2Wsinc(2Wt ) ⎛ f ⎞ 2Wsinc(2Wt ) R rect ⎜ ⎟ ⎝ 2W ⎠

lim 2Wsinc(2Wt ) = δ (t )

W →∞

⎛ 2 ⎞ lim rect ⎜ ⎟ =1 ⎝ 2W ⎠

W →∞

2.12. 1 1 + sgn( f ) 2 2 By duality:

G( f ) =

1 1 G ( f ) R δ ( −t ) − j 2π t 2 j 1 ∴ g (t ) = δ (t ) + 2 2π t

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2.13. a) By the differentiation property: 2 ( j 2π f ) G ( f ) = ∑ ki exp(− j 2π fti ) i

∴G( f ) = −

1 4π 2 f 2

∑ k exp(− j 2π ft ) i

i

i

b)the slope of each non-flat segment is: ±

A tb − t a

⎛ 1 ⎞⎛ A ⎞ G( f ) = − ⎜ 2 2 ⎟ ⎜ ⎟ [ exp( j 2π ftb ) − exp( j 2π fta ) − exp( j 2π fta ) + exp( j 2π ftb ) ] ⎝ 4π f ⎠ ⎝ tb − ta ⎠ A =− 2 2 [cos(2π ftb ) − cos(2π fta )] 2π f ( tb − ta ) 1 But: sin(π f (tb − ta )) sin(π f (tb + ta )) = [ cos(2π fta ) − cos(2π ftb ) ] by a trig identity. 2 A ∴ G( f ) = 2 2 [sin(π f (tb − ta )) sin(π f (tb + ta ))] π f (tb − ta )

2.14 a) let g(t) be the half cosine pulse of Fig. P2.1a, and let g(t-t0) be its time-shifted counterpart in Fig.2.1b

ε = G ( f )G* ( f ) = G( f )

2

( G ( f ) exp(− j 2π ft0 ) ) ( G* ( f ) exp( j 2π ft0 ) ) =

G ( f ) exp(− j 2π ft0 ) exp( j 2π ft0 )

( G ( f ) exp(− j 2π ft0 ) ) ( G* ( f ) exp( j 2π ft0 ) ) =

G( f )

2

2

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2.14 b)Given that the two energy densities are equal, we only need to prove the result for one. From before, it was shown that the Fourier transform of the half-cosine pulse was: AT 1 [sinc(( f + fc )T ) + sinc(( f − f c )T )] for fc = 2 2T After squaring, this becomes: sin(π ( f + f c )T ) sin(π ( f − f c )T ) ⎤ A2T 2 ⎡ sin 2 (π ( f + f c )T ) sin 2 (π ( f − f c )T ) + + 2 ⎢ ⎥ 4 ⎣ (π ( f + f c )T ) 2 (π ( f − f c )T ) 2 π 2T 2 ( f + f c )( f − f c ) ⎦ The first term reduces to: π⎞ ⎛ sin 2 ⎜ π fT + ⎟ 2 cos 2 (π fT ) 2 ⎠ cos (π fT ) ⎝ = = 2 2 2 π⎞ π ⎞ π 2T 2 ( f + f c ) ⎛ ⎛ ⎜ π fT + ⎟ ⎜ π fT + ⎟ 2⎠ 2⎠ ⎝ ⎝ The second term reduces to:

π⎞ ⎛ sin 2 ⎜ π fT − ⎟ cos 2 (π fT ) 2⎠ ⎝ = 2 2 2 2 π⎞ π T ( f − fc ) ⎛ − π fT ⎜ ⎟ 2⎠ ⎝ The third term reduces to: sin(π ( f + f c )T ) sin(π ( f − f c )T ) cos(π ) − cos 2 (2π fT ) = 1 ⎞ π 2T 2 ( f + f c )( f − f c ) ⎛ π 2T 2 ⎜ f 2 − 2 ⎟ 4T ⎠ ⎝ −1 − cos(2π fT ) = 1 ⎞ ⎛ π 2T 2 ⎜ f 2 − 2 ⎟ 4T ⎠ ⎝ 2 cos 2 (π fT ) =− 1 ⎞ ⎛ π 2T 2 ⎜ f 2 − 2 ⎟ 4T ⎠ ⎝ Summing these terms gives: ⎡ ⎤ ⎥ 2 2 2 2 ⎢ 2 cos (π fT ) A T ⎢ cos (π fT ) cos (π fT ) ⎥ + −2 2 2 1 1 ⎥ 4π 2T 2 ⎢ ⎛ ⎛ ⎞ ⎛ ⎞ 1 ⎞ 1 ⎛ ⎞ ⎜f+ ⎟⎜ f − ⎟⎥ ⎢⎜ f + ⎟ ⎜f− ⎟ 2T ⎠ ⎝ 2T ⎠ ⎦ ⎝ 2T ⎠ 2T ⎠ ⎝ ⎣⎝ 2

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2.14 b)Cont’d By rearranging the previous expression, and summing over a common denominator, we get: ⎡ ⎤ ⎥ 2 2 ⎢ 2 A T ⎢ cos (π fT ) ⎥ 2 4π 2T 2 ⎢ ⎛ 2 1 ⎞ ⎥ ⎢⎜ f − 2 ⎟ ⎥ 4T ⎠ ⎦ ⎣⎝ ⎡ ⎤ ⎥ A2T 2 ⎢ cos 2 (π fT ) = 2 4⎢ ⎥ 4π T ⎢ 1 1 4T 2 f 2 − 1 2 ⎥ ( ) ⎣ 16 T 4 ⎦ ⎡ ⎤ A2T 2 cos 2 (π fT ) ⎥ = 2 ⎢ π ⎢ ( 4T 2 f 2 − 1)2 ⎥ ⎣ ⎦

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dg (t ) R j 2π fG ( f ) dt

2.15 a)The Fourier transform of Let g '(t ) =

dg (t ) dt ∞

By Rayleigh’s theorem:

∞

∫

g (t ) dt = 2

−∞

∴W T 2

2

∫t = ∫t =

2

2

∫ G( f )

2

df

−∞

g (t ) dt ⋅ ∫ f 2 G ( f ) df 2

(∫

2

2

g (t ) dt

)

2

g (t ) dt ⋅ ∫ g '(t ) g '* (t )dt 2

4π 2

( ∫ g (t ) dt ) 2

2

⎡ t 2 g * (t ) g '(t ) − tg (t )g '* (t )dt ⎤ ∫ ⎦ ≥⎣ 2 2 16π 2 ∫ g (t ) dt

(

)

d ⎡ ⎤ * ⎢⎣ ∫ t ⋅ dt ( g (t ) g (t ) ) dt ⎥⎦ = 2 16π 2 ∫ g (t ) g * (t )dt

(

2

2

)

Using integration by parts, we can show that: ∞ ∞ d 2 2 t ⋅ g ( t ) dt = ∫−∞ dt ∫−∞ g (t ) 1 16π 2 1 ∴WT ≥ 4π ∴W 2T 2 ≥

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2.15 b) For g (t ) = exp(−π t 2 ) g (t ) R exp(−π f 2 ) ∞

∴W 2T 2 =

∞

2 2 2 2 ∫ t exp(−2π t )dt ⋅ ∫ f exp(−2π f )df

−∞

∞

−∞

∫ exp(−2π t

2

)dt

−∞

Using a table of integrals:

∞

∫x

2

exp(−ax 2 )dx =

0

1 π 4a a

∞

∴ ∫ t 2 exp(−2π t 2 )dt = −∞

∞

∫

1 1 4π 2

f 2 exp(−2π t 2 )df =

−∞ ∞

∫ exp(−2π t

2

)=

−∞

⎛ 1 ⎜ 4π 2 2 ∴T W = ⎝

for a > 0

1 4π

1 2

1 2 1⎞ ⎟ 2⎠

2

1 2

⎛ 1 ⎞ =⎜ ⎟ ⎝ 4π ⎠ 1 ∴TW = 4π

2

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2.16. ∞

Given:

∞

∫

x(t ) dt < ∞ and 2

−∞

∫

∞

h(t ) dt < ∞, which implies that

−∞ ∞

However, if

∫

∫

h(t ) dt < ∞ .

−∞ ∞

x(t ) dt < ∞ then 2

−∞

∫

∞

X ( f ) df < ∞ and 2

−∞

∫

X ( f ) df < ∞ . This result also 4

−∞

applies to h(t). Y( f ) = H( f )X ( f ) ∞

∞

∫

Y ( f ) df = 2

−∞

∫ X ( f )H ( f ) ⋅ X

*

( f ) H * ( f )df

−∞ ∞

=

∫

2

2

X ( f ) H ( f ) df

−∞ 2

∞

∫

2

Y ( f ) df

−∞

∞

≤

∫

∞ 4

X ( f ) df

−∞

∫

4

H ( f ) df

−∞

0, the frequencies are shifted inwards by Δf. ∴Vo ( f ) contains {99.98,199.98,399.98} Hz (b) When the lower side-band is transmitted, and Δf>0, then the baseband frequencies are shifted outwards by Δf. ∴Vo ( f ) contains {100.02, 200.02, 400.02} Hz

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f1 = f c − Δf − W

3.22.

f 2 = f c + Δf v1 (t )v2 (t ) = A1 A2 cos(2π f1t + φ1 ) cos(2π f 2t + φ2 ) =

A1 A2 [cos(2π ( f1 − f 2 )t + φ1 − φ2 ) + cos(2π ( f1 + f 2 )t + φ1 + φ2 )] 2

The low-pass filter will only pass the first term. 1 ∴ LFP(v1 (t )v2 (t )) = A1 A2 [cos(−2π (W + 2Δf )t + φ1 − φ2 )] 2 Let v0(t) be the final output, before band-pass filtering. ⎛ W + 2Δf ⎞ 1 φ1 − φ2 A1 A2 [cos(−2π ⎜ ) ⋅ A2 cos(2π f 2t + φ2 )] ⎟t + 2 ⎝ W / Δf + 2 ⎠ W / Δf + 2

vo (t ) =

1 φ −φ φ −φ A1 A22 [cos(−2πΔft + 1 2 − φ2 ) ⋅ cos(2π f 2t + 1 2 + φ2 )] 2 n+2 n+2 φ −φ φ −φ 1 = A1 A22 [cos(−2π ( f c + 2Δf ) + 1 2 − φ2 ) + cos(−2π f c t + 1 2 + φ2 )] 4 n+2 n+2 =

After band-pass filtering, retain only the second term. 1 φ −φ A1 A22 [cos(−2π f c t + 1 2 + φ2 ) 4 n+2

∴ vo (t ) =

φ1

φ2

+ φ2 = 0 n+2 n+2 rearranging and solving for φ2 : −

φ2 = −

φ1 n +1

(b) At the second multiplier, replace v2(t) with v1(t). expression for the phase:

φ1

n+2

φ1 =

−

φ2

n+2

This results in the following

+ φ1 = 0

φ2 n+3

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3.23. Assume that the mixer performs a multiplication of the two signals. y1 (t ) ∈ {1, 2,3, 4,5, 6, 7,8,9} MHz y2 (t ) ∈ {100, 200,300, 400,500, 600, 700,800,900} kHz This system essentially produces a DSB-SC signal centred around the frequency of y1(t). The lowest frequencies that can be produced are: 1 yo (t ) = [cos(2π ( f1 − f 2 )t ) + cos(2π ( f1 + f 2 )t )] 2 f1 = 1 MHz f1 − f 2 = 0.9 MHz f 2 = 100 kHz

f1 + f 2 = 1.1 MHz

The highest frequencies that can be produced are: f1 = 9 MHz

f1 − f 2 = 8.1 MHz

f 2 = 900 kHz

f1 + f 2 = 9.9 MHz

The resolution of the system is the bandwidth of the output signal. Assuming that no branch can be zeroed, the narrowest resolution occurs with a modulation frequency of 100 kHz. The widest bandwidth occurs when there is a modulation frequency of 900 kHz.

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3.24 Given the presence of the filters, only the baseband signals need to be considered. All of the other product components can be discarded. (a) Given the sum of the modulated carrier waves, the individual message signals are extracted by multiplying the signal with the required carrier. For m1(t), this results in the conditions: cos(α1 ) + cos( β1 ) = 0 cos(α 2 ) + cos( β 2 ) = 0 cos(α 3 ) + cos( β 3 ) = 0 ∴α i = β i ± π For the other signals: m2 (t ) : cos(−α1 ) + cos(− β1 ) = 0 cos(α 2 − α1 ) + cos( β 2 − β1 ) = 0 cos(α 3 − α1 ) + cos( β3 − β1 ) = 0

α1 = β1 ± π (α 2 − α1 ) = ( β 2 − β1 ) ± π (α 3 − α1 ) = ( β 3 − β1 ) ± π

Similarly: m3 (t ) : (α1 − α 2 ) = ( β1 − β 2 ) ± π (α 3 − α 2 ) = ( β 3 − β 2 ) ± π m4 (t ) : (α1 − α 3 ) = ( β1 − β 3 ) ± π (α 2 − α 3 ) = ( β 2 − β 3 ) ± π

(b) Given that the maximum bandwidth of mi(t) is W, then the separation between fa and fb must be | fa- fb|>2W in order to account for the modulated components corresponding to fa- fb.

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3.25 b) The charging time constant is (rf + Rs )C = 1μ s

The period of the carrier wave is 1/fc = 50 μs. The period of the modulating wave is 1/fm = 0.025 s. ∴The time constant is much shorter than the modulating wave and therefore should track the message signal very well. The discharge time constant is: Rl C = 100μ s . This is twice the period of the carrier wave, and should provide some smoothing capability. From a maximum voltage of V0, the voltage Vc across the capacitor after time t = Ts is: T Vc = V0 exp(− s ) Rl C Using a Taylor series expansion and retaining only the linear terms, will result in the T linear approximation of VC = V0 (1 − s ) . Using this approximation, the voltage will Rl C decay by a factor of 0.94 from its initial value after a period of Ts seconds. From the code, it can be seen that the voltage decay is close to this figure. However, it is somewhat slower than what was calculated using the linear approximation. In a real circuit, it would also be expected that the decay would be slower, as the voltage does not simply turn off, but rather decreases over time.

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3.25 c)

The output of a high-pass RC circuit can be described according to: V0 (t ) = I (t ) R Qc (t ) = C (Vin (t ) − V0 (t )) dQc dt ⎛ dV (t ) dV (t ) ⎞ V0 (t ) = RC ⎜ in − 0 ⎟ dt ⎠ ⎝ dt Using first order differences to approximate the derivatives results in the following difference equation: RC RC V0 (t ) = V0 (t − 1) + (Vin (t ) − Vin (t − 1)) RC + Ts RC + Ts I (t ) =

The high-pass filter applied to the envelope detector eliminates the DC component.

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Problem 3.25. MATLAB code function [y,t,Vc,Vo]=AM_wave(fc,fm,mi) %Problem 3.25 %Inputs: fc % fm % mi

Carrier Frequency Modulation Frequency modulation index

%Problem 3.25 (a) fs=160000; %sampling rate deltaT=1/fs; %sampling period t=linspace(0,.1,.1/deltaT); %Create the list of time periods y=(1+mi*cos(2*pi*fm*t)).*cos(2*pi*fc*t); %Create the AM wave %Problem 3.25 (b) %%%%Create the envelope detector%%%% Vc=zeros(1,length(y)); Vc(1)=0; %inital voltage for k=2:length(y) if (y(k)>(Vc(k-1))) Vc(k)=y(k); else

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Vc(k)=Vc(k-1)-0.023*Vc(k-1); end end

%Problem 3.25 (c) %%%Implement the high pass filter%%% %%This implements bias removal Vo=zeros(1,length(y)); Vo(1)=0; RC=.001; beta=RC/(RC+deltaT); for k=2:length(y) Vo(k)=beta*Vo(k-1)+beta*(Vc(k)-Vc(k-1)); end

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Chapter 4 Problems

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Problem 4.7. s (t ) = Ac cos(θ (t ))

θ (t ) = 2π f c t + k p m(t ) Let β = 0.3 for m(t) = cos(2πfmt). ∴ s (t ) = A c cos(2π f c t + β m(t )) = Ac [cos(2π f c t ) cos( β cos(2π f mt )) − sin(2π f c t ) sin( β cos(2π f mt ))] for small β : cos( β cos(2π f mt ))  1 sin( β sin(2π f mt ))  β cos(2π f mt ) ∴ s (t ) = Ac cos(2π f ct ) − β Ac sin(2π f c t ) cos(2π fmt ) = Ac cos(2π f c t ) − β

Ac [sin(2π ( f c + f m )t ) + sin(2π ( f c + f m )t ) 2

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Problem 4.14. v2 = av12 s (t ) = Ac cos(2π f c t + β sin(2π f mt )) = Ac cos(2π f c t + β m(t )) v2 = a ⋅ s 2 (t ) = a ⋅ cos 2 (2π f c t + β m(t )) =

a ⋅ cos(4π f c t + 2 β m(t )) 2

The square-law device produces a new FM signal centred at 2fc and with a frequency deviation of 2β. This doubles the frequency deviation.

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4.17. Consider the slope circuit response:

The response of |X1(f)| after the resonant peak is the same as for a single pole low-pass filter. From a table of Bode plots, the following gain response can be obtained: | X 1 ( f ) |=

1 ⎛ f − fB ⎞ 1+ ⎜ ⎟ ⎝ B ⎠

2

Where fB is the frequency of the resonant peak, and B is the bandwidth. For the slope circuit, B is the filter’s bandwidth or cutoff frequency. For convenience, we can shift the filter to the origin (with X 1 ( f ) as the shifted version).

| X 1 ( f ) |=

d | X 1 ( f ) | df

1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝B⎠ =− f = kB

2

k 3 2 2

B(1 + k )

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Because the filters are symmetric about the central frequency, the contribution of the second filter is identical. Adding the filter responses results in the slope at the central frequency being: d | X ( f ) | df

=− f = kB

2k 3 2 2

B(1 + k )

In the original definition of the slope filter, the responses are multiplied by -1, so do this here. This results in a total slope of: 2k 3 2 2

B(1 + k )

As can be seen from the following plot, the linear approximation is very accurate between the two resonant peaks. For this plot B = 500, f1=-750, and f2=750.

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Problem 4. 23

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Problem 4.24 The amplitude spectrum corresponding to the Gaussian pulse p(t ) = c exp ⎡⎣ −π c 2t 2 ⎤⎦ * rect[t / T ]

is given by the magnitude of its Fourier transform.

(

)

P ( f ) = F ⎣⎡ c exp −π c 2t 2 ⎦⎤ F ⎡⎣ rect ( t / T ) ⎤⎦ = c exp ⎡⎣ −π f 2 c 2 ⎤⎦ Tsinc [ fT ] where we have used the convolution theorem

Problem 4.25

The Carson rule bandwidth for GSM is BT = 2 ( Δf + W ) where the peak deviation is given by k c 1 Δf = f = B 2π / log(2) = 0.75 B 2π 4 With BT = 0.3 and T = 3.77 microseconds, the peak deviation is 59.7 kHz From Figure 4.22, the one-sided 3-dB bandwidth of the modulating signal is approximately 50 kHz. Combining these two results, the Carson rule bandwidth is BT = 2 ( 59.7 + 50 ) = 219.4 kHz

The 1-percent FM bandwidth is given by Figure 4.9 with β = vertical axis we find that

Δf 59.7 = = 1.19 . From the W 50

BT = 6 , which implies BT = 6(59.7) = 358.2 kHz. Δf

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Problem 4.26. a)

Beta 1 2 5 10

# of side frequencies 1 2 8 14

b)By experimentation, a modulation index of 2.408, will force the amplitude of the carrier to be about zero. This corresponds to the first root of J0(β), as predicted by the theory.

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Problem 4.27. a)Using the original MATLAB script, the rms phase error is 6.15 % b)Using the plot provided, the rms phase error is 19.83%

Problem 4.28 a)The output of the detected signal is multiplied by -1. This results from the fact that m(t)=cos(t) is integrated twice. Once to form the transmitted signal and once by the envelope detector. In addition, the signal also has a DC offset, which results from the action of the envelope detector. The change in amplitude is the result of the modulation process and filters used in detection.

f ⎞ ⎛ b)If s (t ) = sin(2π f mt ) + 0.5cos ⎜ 2π m t ⎟ , then some form of clipping is observed. 3 ⎠ ⎝

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The above signal has been multiplied by a constant gain factor in order to highlight the differences with the original message signal. c)The earliest signs of distortion start to appear above about fm =4.0 kHz. As the message frequency may no longer lie wholly within the bandwidth of either the differentiator or the low-pass filter. This results in the potential loss of high-frequency message components.

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4.29. By tracing the individual steps of the MATLAB algorithm, it can be seen that the resulting sequence is the same as for the 2nd order PLL. e(t ) is the phase error φe (t ) in the theoretical model. The theoretical model of the VCO is: t

φ2 (t ) = 2π kv ∫ v(t )dt 0

and the discrete-time model is: VCOState = VCOState + 2π kv (t − 1)Ts which approximates the integrator of the theoretical model. The loop filter is a PI-controller, and has the transfer function: a H ( f ) = 1+ jf This is simply a combination of a sum plus an integrator, which is also present in the MATLAB code: Filterstate = Filterstate + e(t ) Integrator v(t ) = Filterstate + e(t )

Integrator +input

b)For smaller kv, the lock-in time is longer, but the output amplitude is greater.

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c)The phase error increases, and tracks the message signal.

d)For a single sinusoid, the track is lost if f m ≥ K 0 where K 0 = k f kv Ac Av For this question, K0=100 kHz, but tracking degrades noticeably around 60-70 kHz. e)No useful signal can be extracted. By multiplying s(t) and r(t), we get: Ac Av ⎡sin(k f φ − VCOState) + sin(4π f c t + k f φ + VCOState) ⎤⎦ 2 ⎣ This is substantially different from the original error signal, and cannot be seen as an adequate approximation. Of particular interest is the fact that this equation is substantially more sensitive to changes in φ than the previous one owing to the presence of the gain factor kv

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Chapter 5 Problems

5.1.

( x − μ x )2 exp(− ) (a) Given f ( x) = 2σ x2 2πσ x2 1

and exp(−π t 2 ) R exp(−π f 2 ) , then by applying the time-shifting and scaling properties: F( f ) =

1 2πσ

2 x

2πσ x2 exp(−π ( 2πσ x2 ) 2 π f 2 ) exp( j 2π f μ x )

= exp(−π 2 2σ x2 f 2 + j μ x 2π f ) 1 = exp( jνμ x − ν 2σ x2 ) 2

and let ν = 2π f

(b)The value of μx does not affect the moment, as its influence is removed. Use the Taylor series approximation of φx(x), given μx = 0. 1 2

φx (ν ) = exp(− ν 2σ x2 ) ∞

x2 exp( x) = ∑ n=0 n ! E[ X n ] =

d nφx (ν ) dν n v =0

2k 2k ∞ ⎛ 1⎞ σ ν ∴ φx (ν ) = ∑ ⎜ − ⎟ x k! 2⎠ k =0 ⎝ k

d nφx (ν ) leaves the lowest non-zero derivative as ν2k-n. dν n When this derivative is evaluated for v=0, then E[ X n ] =0.

For any odd value of n, taking

For even values of n, only the terms in the resulting derivative that correspond to ν2k-n = ν0 are non-zero. In other words, only the even terms in the sum that correspond to k = n/2 are retained. ∴ E[ X n ] =

n! σ x2 (n / 2)!

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5.2. (a) All the inputs for x ≤0 are mapped to y = 0. However, the probability that x > 0 is unchanged. Therefore the probability density of x ≤0 must be concentrated at y=0. ∞

(b) Recall that

∫

f x x)dx = 1 where f x ( x) is an even function.

Because fy(y) is a

−∞

probability distribution, its integral must also equal 1. ∞

∴

∫ 0

∞

f x ( x)dx = 0.5 and

∫

0

f y ( y )dy = 0.5

+

Therefore, the integral over the delta function must be 0.5. This means that the factor k must also be 0.5.

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5.3 (a)

p y ( y ) = p y ( y | x0 ) P( x0 ) + p y ( y | x1 ) P( x1 ) Assume: P( x0 ) = P( x1 ) = 0.5 1 ∴ p y ( y ) = [ p y ( y | x0 ) + p y ( y | y1 ) 2 1 ( y + 1) 2 ( y − 1) 2 py ( y) = [exp(− ) exp( )] + − 2σ 2 2σ 2 2 2πσ 2 ∞

(b) P ( y ≥ α ) = ∫ p y ( y )dy α

Use the cumulative Gaussian distribution, y

Φ μ ,σ 2 ( y ) =

∫

−∞

( y − μ )2 exp(− )dy 2σ 2 2πσ 2 1

1 ∴ P( y ≥ α ) = [Φ −1,σ 2 (−α ) + Φ1,σ 2 (−α )] 2 1 y−μ )] But, Φ μ ,σ 2 ( y ) = [1 + erf ( 2 σ 2 1 ⎛ −α + 1 ⎞ ⎛ −α − 1 ⎞ ∴ P( y ≥ α ) = [2 + erf ⎜ + erf ⎜ ⎟ ⎟] 2 ⎝σ 2 ⎠ ⎝σ 2 ⎠

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Problem 5.4

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Problem 5.5

If, for a complex random process Z(t) RZ (τ ) = E [ Z *(t ) Z (t + τ ) ] then (i) The mean square of a complex process is given by RZ (0) = E [ Z *(t ) Z (t ) ] 2 = E ⎡ Z (t ) ⎤ ⎣ ⎦

(ii) We show RZ (τ ) has conjugate symmetry by the following RZ (−τ ) = E [ Z *(t ) Z (t − τ ) ]

= E [ Z *( s + τ ) Z ( s ) ] = E [ Z ( s ) Z ( s + τ )] * = RZ* (τ ) where we have used the change of variable s = t - τ. (iii) Taking an approach similar to that of Eq. (5.67) 2 0 ≤ E ⎡⎢ ( Z (t ) ± Z (t + τ ) ) ⎤⎥ ⎣ ⎦

= E ⎡⎣( Z (t ) ± Z (t + τ ) )( Z *(t ) ± Z *(t + τ ) ) ⎤⎦

= E [ Z (t ) Z *(t ) ± Z (t ) Z *(t + τ ) ± Z *(t ) Z (t + τ ) + Z (t + τ ) Z *(t + τ ) ] 2 2 = E ⎡ Z (t ) ⎤ ± E [ Z (t ) Z *(t + τ )] ± E [ Z *(t ) Z (t + τ ) ] + E ⎡ Z (t + τ ) ⎤ ⎣ ⎦ ⎣ ⎦ 2 = 2E ⎡ Z (t ) ⎤ ± 2 Re {E [ Z *(t ) Z (t + τ ) ]} ⎣ ⎦ = 2 RZ (0) ± 2 Re { RZ (τ )}

Thus Re { RZ (τ )} ≤ RZ (0) .

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Problem 5.6 (a)

E[ Z (t1 ) Z * (t2 )] = E[( A cos(2π f1t1 + θ1 ) + jA cos(2π f 2t1 + θ 2 )) ⋅ ( A cos(2π f1t2 + θ1 ) + jA cos(2π f 2t2 + θ 2 ))] Let ω1=2πf1 ω2=2πf2 After distributing the terms, consider the first term: A2 E[cos(ω1t1 + θ1 ) cos(ω1t2 + θ1 )] =

A2 E[cos(ω1 (t1 − t2 )) + cos(ω1 (t1 + t2 ) + 2θ1 )] 2

The expectation over θ1 goes to zero, because θ1 is distributed uniformly over [-π,π]. This result also applies to the term A2 [cos(ω2t1 + θ 2 ) cos(ω2t2 + θ 2 )] . Both cross-terms go to zero. ∴ R(t1 , t2 ) =

A2 [cos(ω1 (t 1 −t2 )) + cos(ω2 (t1 − t2 ))] 2

(b) If f1 = f2, only the cross terms may be different: E[ jA2 (cos(ω1t1 + θ 2 ) cos(ω1t2 + θ1 ) + cos(ω1t1 + θ 2 ) cos(ω1t2 + θ1 )] But, unless θ1=θ2, the cross-terms will also go to zero. ∴ R(t1 , t2 ) = A2 cos(ω1 (t1 − t2 )) (c) If θ1=θ2, then the cross-terms become: − jA2 E[cos((ω1t1 − ω2t2 )) + cos((ω1t1 + ω2t2 ) + 2θ1 ) + jA2 E[cos((ω2t1 − ω1t2 )) + cos((ω1t1 + ω2t2 ) + 2θ1 )] After computing the expectations, the cross-terms simplify to: jA2 [cos(ω2t1 − ω1t2 ) − cos(ω1t1 − ω2t2 )] 2 ∴ RZ (t1 , t2 ) =

A2 [cos(ω1 (t1 − t2 )) + cos(ω2 (t1 − t2 )) + j cos(ω2t1 − ω1t2 ) − j cos(ω1t1 − ω2t2 )] 2

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Problem 5.7

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Problem 5.8

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Problem 5.9

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Problem 5.10

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Problem 5.11

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Problem 5.12

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Problem 5.13

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Problem 5.14

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Problem 5.15

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Problem 5.16

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Problem 5.17

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Problem 5.18

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Problem 5.19

Problem 5.20

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Problem 5.21

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Problem 5.22

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Problem 5.23

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Problem 5.24

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Problem 5.25

Problem 5.26

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Problem 5.27

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Problem 5.28

c)For a given filter, H ( f ) , let α = ln H ( f ) ∞

and the Paley-Wiener criterion for causality is:

α( f )

∫ 1 + (2π f ) df < ∞ 2

−∞

For the filter of part (b) 1 α ( f ) = [ ln(2) + ln( S x ( f ) − ln( N 0 )] 2 The first and the last terms have no impact on the absolute integrability of the previous expression, and so do not matter as far as evaluating the above criterion. This leaves the only condition: ∞ ln S x ( f ) ∫−∞ 1 + (2π f )2 df < ∞

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Problem 5.29

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Problem 5.30

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Problem 5.31

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Problem 5.32

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Problem 5.33 (a) The receiver position is given by x(t) = x0+vt Thus the signal observed by the receiver is ⎡ ⎛ x ⎞⎤ r (t , x) = A( x) cos ⎢ 2π f c ⎜ t − ⎟ ⎥ ⎝ c ⎠⎦ ⎣ ⎡ ⎛ x + vt ⎞ ⎤ = A( x) cos t ⎢ 2π f c ⎜ t − 0 c ⎟⎠ ⎥⎦ ⎝ ⎣ ⎡ ⎛ f v⎞ x ⎤ = A( x) cos ⎢ 2π ⎜ f c − c ⎟ t − f c 0 ⎥ c ⎠ c⎦ ⎣ ⎝ The Doppler shift of the frequency observed at the receiver is f D =

fc v . c

(b) The expectation is given by E ⎡⎣exp ( j 2π f nτ ) ⎤⎦ =

1 2π

=

1 2π

π

∫ exp ( j 2π f τ cosψ ) dψ D

n

n

−π

π

∫π exp ( j 2π f τ sinψ ) dψ D

n

n

−

= J 0 ( 2π f Dτ ) where the second line comes from the symmetry of cos and sin under a -π/2 translation. Eq. (5.174) follows directly from this upon noting that, since the expectation result is real-valued, the right-hand side of Eq.(5.173) is equal to its conjugate.

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Problem 5.34 The histogram has been plotted for 100 bins. Larger numbers of bins result in larger errors, as the effects of averaging are reduced. Distance 0σ 1σ 2σ 3σ 4σ

Relative Error 0.94% 2.6 % 4.8 % 47.4% 60.7%

The error increases further out from the centre. It is also important to note that the random numbers generated by this MATLAB procedure can never be greater than 5. This is very different from the Gaussian distribution, for which there is a non-zero probability for any real number.

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5.34 Code Listing %Problem 5.34 %Set the number of samples to be 20,000 N=20000 M=100; Z=zeros(1,20000); for i=1:N for j=1:5 Z(i)=Z(i)+2*(rand(1)-0.5); end end sigma=sqrt(var(Z-mean(Z))); %Calculate a histogram of Z [X,C]=hist(Z,M); l=linspace(C(1),C(M),M); %Create a gaussian function with the same variance as Z G=1/(sqrt(2*pi*sigma^2))*exp(-(l.^2)/(2*sigma^2)); delta2=abs(l(1)-l(2)); X=X/(20000*delta2);

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5.35 (a) For the generated sequence:

μˆ y = −0.0343 + j 0.0493 σˆ y2 = 5.597 The theoretical values are: μy = 0 (by inspection). The theoretical value of σ y2 =5.56. See 5.35 (c) for the calculation. 5.35 (b) From the plots, it can be seen that both the real and imaginary components are approximately Gaussian. In addition, from statistics, the sum of tow zero-mean Gaussian signals is also Gaussian distributed. As a result, the filter output must also be Gaussian.

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5.35 (c)

y (n) = ay (n − 1) + w(n) Y ( z ) = aY ( z ) z −1 ∴ H ( z) =

1 n R h( n) = a u ( n) −1 1 − az

1 (1 − az )(1 − az )

-1

Rh(z) = H(z)H(z ) =

−1

=

a z −1 1 1 + 2 −1 2 1 − a 1 − az 1 − a 1 − az

But, Ry(z) = Rh(z)Rw(z) Taking the inverse z-transform: ry (n) =

σ w2 1 − a2

an

−∞ < n < ∞

From the plots, the measured and observed autocorrelations are almost identical.

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Chapter 6 Solutions

Problem 6.3

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Problem 6.4

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Problem 6.5

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Problem 6.6 Problem 6.7

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Problem 6.8

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Problem 6.9 Problem 6.10

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Problem 6.11

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Problem 6.12

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Problem 6.13

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Problem 6.24

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Problem 6.15

Problem 6.16 Problem 6.17

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Chapter 7 Problems Problem 7.1

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Problem 7.2

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Problem 7.3

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Problem 7.4

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Problem 7.5

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Problem 7.6

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Problem 7.7

Problem 7.8

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Problem 7.9

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Problem 7.10 Problem 7.11 Problem 7.12

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Problem 7.13

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Problem 7.14

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Problem 7.15

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Problem 7.16

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Problem 7.17

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Problem 7.18

Problem 7.19

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Problem 7.20

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Problem 7.21

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Problem 7.22 The maximum slope of the signal s (t ) = A sin ( 2π ft ) is 2πfA. Consequently, the

maximum change during a sample period is approximately 2πAfTs. To prevent slope overload, we require 100mV > 2π AfTs = 2π A(1kHz ) /(68kHz ) = 0.092 A or A < 1.08 V.

Problem 7.23 (a) Theoretically, the sampled spectrum is given by

Ss ( f ) =

∞

∑ H ( f − nf )

n =−∞

s

s

where Hs(f) is the spectrum of the signal H(f) limited to f ≤ f s / 2 . For this example, the sample spectrum should look as below.

0

-5 kHz

5 kHz

f

(b) The sampled spectrum is given by 2.5

x 10

5

Amplitude Spectrum

2

1.5

1

0.5

0 -5

-4

-3

-2

-1 0 1 Frequency (kHz)

2

3

4

5

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There are several features to comment on: (i) The component at +4 kHz is due to aliasing of the -6 kHz sinusoid; and the component at -4kHz is due to aliasing of the +6 kHz sinusoid. The lower frequency is at 2 kHz is six times larger than the one at 4 kHz. One would expect the power ratio to be 4:1, not 6:1. The difference is due to relationship between the FFTsize (period) and the sampling rate. (Try a sampling rate of 10.24 kHz and compare.)

(ii)

(b) The spectrum with a 11 kHz sampling rate is shown below. 2.5

x 10

5

Amplitude Spectrum

2

1.5

1

0.5

0 -6

-4

-2

0 Frequency (kHz)

2

4

6

As expected the 2kHz component is unchanged in frequency, while the aliased component is shifted to reflect the new sampling rate.

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Problem 7.23 (a) The expanding portion of the μ-law compander is given by exp ⎡⎣log(1 + μ ) υ ⎤⎦ − 1 m=

μ

=

(1 + μ ) exp ⎡⎣ υ ⎤⎦ − 1 μ

(b) (i) For the non-companded case, the rms quantization error is determined by step size. The step size is given by the maximum range over the number of quantization steps 2A Δ= Q 2 For this signal the range is from +10 to -1, so A = 10 and with Q = 8, we have Δ = 0.078. From Eq. ( ) , the rms quantization error is then given by 1 2 −2 R 2 σ Q2 = mmax 3 1 = (10) 2 2−16 3 = 0.0005086 and the rms error is σQ – 0.02255. (ii) For a fair comparison, the signal must have similar amplitudes. The rms error with companding is 0.0037 which is significantly less. The plot is shown below. Note that the error is always positive. 0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 -0.005

0

50

100

150

200

250

300

350

400

450

Rest TBD.

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Problem 7.24 Problem 7.25

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Chapter 8

Problem 8.1

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Problem 8.2 Problem 8.3

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Problem 8.4

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Problem 8.5

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Problem 8.6

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Problem 8.7

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Problem 8.8

Problem 8.9

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Problem 8.10

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Problem 8.11

Problem 8.12

. Problem 8.13

Problem 8.14

Problem 8.15

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Problem 8.16

Problem 8.17 Problem 8.18 Problem 8.19

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Problem 8.20

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Problem 8.21

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Problem 8.22

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Problem 8.23

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Problem 8.24

Problem 8.25 Problem 8.26 Problem 8.29

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Chapter 9

Problem 9.1 The three waveforms are shown below for the sequence 0011011001. (b) is ASK, (c) is PSK; and (d) is FSK.

Problem 9.2 The bandpass signal is given by s (t ) = g (t ) cos ( 2π f c t )

The corresponding amplitude spectrum, using the multiplication theorem for Fourier transforms, is given by S ( f ) = G ( f ) * [δ ( f − f c ) + δ ( f + f c ) ] = G( f − fc ) + G( f + fc )

For a triangular spectrum G(f), the corresponding sketch is shown below. Problem 9.3

To be done

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Problem 9.4

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Problem 9.5

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Problem 9.6

**The problem here is solved as “erfc” here and in the old edition, but listed in the textbook question as “Q(x)”.

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Problem 9.7

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Problem 9.8

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Problem 9.9

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Problem 9.10

Problem 9.11

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Problem 9.12

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Problem 9.13 Problem 9.14

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Problem 9.15

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Problem 9.16

Problem 9.17

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Problem 9.18

Problem 9.19

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Problem 9.20

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Problem 9.21

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Problem 9.22

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Problem 9.23

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Chapter 10 Problems Problem 10.1

Problem 10.2

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Problem 10.3

Problem 10.4

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Problem 10.5

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Problem 10.6

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Problem 10.7

a)

b)To be done

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Problem 10.8

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Problem 10.9

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Problem 10.10

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Problem 10.11

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Problem 10.12

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Problem 10.13

Problem 10.14

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Problem 10.15

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Problem 10.16

Problem 10.17

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Problem 10.18

Problem 10.19

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Problem 10.20

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Problem 10.21

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Problem 10.22

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Problem 10.23

Problem 10.24

Problem 10.25

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Problem 10.26

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Problem 10.27

Problem 10.28

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Problem 10.29

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Problem 10.30

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