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)
⎡ −T T ⎤
g (t ) = A cos(2π f c t ) t∈⎢ , ⎥
⎣ 2 2⎦
1
fc =
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
=
2 π fT
Writing out the convolution:
∞
AT ⎛ sin(πλT ) ⎞
G( f ) = ∫ ⎜ ⎟ [δ (λ − ( f + f c ) + δ (λ − ( f − f c ) ] d λ
−∞
2 ⎝ πλ T ⎠
A ⎛ sin(π ( f + f c )T ) sin(π ( f − f c )T ) ⎞ 1
= ⎜ + ⎟ fc =
2π ⎝ f + fc f − fc ⎠ 2T
⎛ ⎞
A ⎜ cos(π fT ) cos(π fT ) ⎟
= ⎜ −
2π ⎜ f − 1 1 ⎟
f+ ⎟
⎝ 2T 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 )
2π ⎜ f − 1 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.
1
fc =
2Ta
⎡ cos(π fTa ) cos(π fTa ) ⎤
A
G( f ) = ⎢ − ⎥ ⋅ (cos(π fTa ) − j sin(π fTa ))
2π
⎣ 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 ⎡ 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 ) ⎟
G( f ) = ⎜ − ⋅ exp( jπ fT )
2π ⎜ f − 1 1 ⎟
f+ ⎟
⎝ 2T 2T ⎠
⎡ ⎤
A ⎢ exp( j 2π fT ) exp( j 2π fT ) ⎥
= ⎢ − ⎥
4π ⎢ f − 1 f+
1
⎥
⎣ 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 ) = ⎢ − ⎥
4π ⎢ f−
1
f+
1
⎥
⎣ 2T 2T ⎦
⎡ ⎤
A ⎢ cos(2π fT ) cos(2π fT ) ⎥
= ⎢ −
2π ⎢ f − 1 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 ⎤
∴ G( f ) = ∗ ⎢ (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 ) ⎦
2.3 (a)
g (t ) = g e (t ) + g o (t )
1
g e (t ) = [ g (t ) + g (−t )]
2
⎛ t ⎞
g e (t ) = Arect ⎜ ⎟
⎝ 2T ⎠
1
g o (t ) = [ g (t ) − g (−t )]
2
⎛ ⎛ 1 ⎞ ⎛ 1 ⎞⎞
⎜ ⎜ t − 2T ⎟ ⎜ t + 2T ⎟⎟
g o (t ) = A ⎜ rect ⎜ ⎟ − rect ⎜ ⎟⎟
⎜⎜ ⎜ T ⎟ ⎜ T ⎟ ⎟⎟
⎝ ⎝ ⎠ ⎝ ⎠⎠
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
,(b)
By the time-scaling property g(-t) R G(-f)
1
Ge ( f ) = [G ( f ) + G (− f ) ]
2
1
= [sinc( fT ) exp(− j 2π fT ) + sinc( fT ) exp( j 2π fT ) ]
2
= sinc( fT ) cos(π fT )
1
Go ( f ) = [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) R G(-t)
1 ⎛1 1 ⎞ ⎛1 1 ⎞
g (t ) = j ⎜ δ (t ) −
+ ⎟ exp(− j 2π Wt ) − j ⎜ δ (t ) − ⎟ exp( j 2π Wt )
πt⎝2 j 2π t ⎠ ⎝2 j 2π t ⎠
1 sin(2π Wt )
= +j
πt 2π t
t +T
2.5 1 ⎛ π u2 ⎞
τ t −∫T
g (t ) = exp ⎜ − 2 ⎟ du
⎝ τ ⎠
0 t +T
1 1
=
τ ∫
t −T
h(τ )dτ +
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
= H ( f ) sin(2π f τ )
τ
But H ( f ) = τ exp(−π f 2τ 2 )
1
∴ G( f ) = exp(−π f 2τ 2 ) sin(2π fT )
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
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
If g(t) is odd 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 ) − G* (− f )
2 2
G ( f ) = −G (− f )
* *
G* ( f ) = −G ( f )
∴ G ( f ) must be all imaginary
(b)
d
(− j 2π t )G (t ) R g (− f ) by duality
df
j d
t ⋅ G (t ) R 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 ⎠
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
,(c)
n
⎛ j ⎞ (n)
Let h(t ) = t g (t ) and H ( f ) = ⎜
n
⎟ G (f)
⎝ 2π ⎠
∞ 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
−∞ −∞
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
,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
dt = 2 A
j 2π fG ( f ) = 2π fATsinc 2 ( fT )
sin(π fT ) sin(π fT )
= 2π fAT ⋅
π fT π fT
sin(π fT )
= 2A ⋅ sin(π fT )
π fT
But, sin(π fT ) ≤ 1 ∀f and sinc(π fT ) ≤ 1 ∀f
sin(π fT )
∴ 2A ⋅ 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 2 2
(π fT ) 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
∞
d 2 g (t )
∴ ( j 2π f ) 2 G ( f ) ≤ ∫
−∞
dt 2
dt
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
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)
⎡ −T T ⎤
g (t ) = A cos(2π f c t ) t∈⎢ , ⎥
⎣ 2 2⎦
1
fc =
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
=
2 π fT
Writing out the convolution:
∞
AT ⎛ sin(πλT ) ⎞
G( f ) = ∫ ⎜ ⎟ [δ (λ − ( f + f c ) + δ (λ − ( f − f c ) ] d λ
−∞
2 ⎝ πλ T ⎠
A ⎛ sin(π ( f + f c )T ) sin(π ( f − f c )T ) ⎞ 1
= ⎜ + ⎟ fc =
2π ⎝ f + fc f − fc ⎠ 2T
⎛ ⎞
A ⎜ cos(π fT ) cos(π fT ) ⎟
= ⎜ −
2π ⎜ f − 1 1 ⎟
f+ ⎟
⎝ 2T 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 )
2π ⎜ f − 1 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.
1
fc =
2Ta
⎡ cos(π fTa ) cos(π fTa ) ⎤
A
G( f ) = ⎢ − ⎥ ⋅ (cos(π fTa ) − j sin(π fTa ))
2π
⎣ 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 ⎡ 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 ) ⎟
G( f ) = ⎜ − ⋅ exp( jπ fT )
2π ⎜ f − 1 1 ⎟
f+ ⎟
⎝ 2T 2T ⎠
⎡ ⎤
A ⎢ exp( j 2π fT ) exp( j 2π fT ) ⎥
= ⎢ − ⎥
4π ⎢ f − 1 f+
1
⎥
⎣ 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 ) = ⎢ − ⎥
4π ⎢ f−
1
f+
1
⎥
⎣ 2T 2T ⎦
⎡ ⎤
A ⎢ cos(2π fT ) cos(2π fT ) ⎥
= ⎢ −
2π ⎢ f − 1 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 ⎤
∴ G( f ) = ∗ ⎢ (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 ) ⎦
2.3 (a)
g (t ) = g e (t ) + g o (t )
1
g e (t ) = [ g (t ) + g (−t )]
2
⎛ t ⎞
g e (t ) = Arect ⎜ ⎟
⎝ 2T ⎠
1
g o (t ) = [ g (t ) − g (−t )]
2
⎛ ⎛ 1 ⎞ ⎛ 1 ⎞⎞
⎜ ⎜ t − 2T ⎟ ⎜ t + 2T ⎟⎟
g o (t ) = A ⎜ rect ⎜ ⎟ − rect ⎜ ⎟⎟
⎜⎜ ⎜ T ⎟ ⎜ T ⎟ ⎟⎟
⎝ ⎝ ⎠ ⎝ ⎠⎠
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
,(b)
By the time-scaling property g(-t) R G(-f)
1
Ge ( f ) = [G ( f ) + G (− f ) ]
2
1
= [sinc( fT ) exp(− j 2π fT ) + sinc( fT ) exp( j 2π fT ) ]
2
= sinc( fT ) cos(π fT )
1
Go ( f ) = [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) R G(-t)
1 ⎛1 1 ⎞ ⎛1 1 ⎞
g (t ) = j ⎜ δ (t ) −
+ ⎟ exp(− j 2π Wt ) − j ⎜ δ (t ) − ⎟ exp( j 2π Wt )
πt⎝2 j 2π t ⎠ ⎝2 j 2π t ⎠
1 sin(2π Wt )
= +j
πt 2π t
t +T
2.5 1 ⎛ π u2 ⎞
τ t −∫T
g (t ) = exp ⎜ − 2 ⎟ du
⎝ τ ⎠
0 t +T
1 1
=
τ ∫
t −T
h(τ )dτ +
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
= H ( f ) sin(2π f τ )
τ
But H ( f ) = τ exp(−π f 2τ 2 )
1
∴ G( f ) = exp(−π f 2τ 2 ) sin(2π fT )
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
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
If g(t) is odd 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 ) − G* (− f )
2 2
G ( f ) = −G (− f )
* *
G* ( f ) = −G ( f )
∴ G ( f ) must be all imaginary
(b)
d
(− j 2π t )G (t ) R g (− f ) by duality
df
j d
t ⋅ G (t ) R 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 ⎠
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
,(c)
n
⎛ j ⎞ (n)
Let h(t ) = t g (t ) and H ( f ) = ⎜
n
⎟ G (f)
⎝ 2π ⎠
∞ 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
−∞ −∞
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
,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
dt = 2 A
j 2π fG ( f ) = 2π fATsinc 2 ( fT )
sin(π fT ) sin(π fT )
= 2π fAT ⋅
π fT π fT
sin(π fT )
= 2A ⋅ sin(π fT )
π fT
But, sin(π fT ) ≤ 1 ∀f and sinc(π fT ) ≤ 1 ∀f
sin(π fT )
∴ 2A ⋅ 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 2 2
(π fT ) 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
∞
d 2 g (t )
∴ ( j 2π f ) 2 G ( f ) ≤ ∫
−∞
dt 2
dt
Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.
