M. J. Roberts - 8/17/04
Chapter 8 - Correlation, Energy Spectral Density
and Power Spectral Density
Selected Solutions
(In this solution manual, the symbol, ⊗, is used for periodic convolution because the
preferred symbol which appears in the text is not in the font selection of the word processor
used to create this manual.)
1. Plot correlograms of the following pairs of CT and DT signals.
(a)
x1 ( t) = cos(2πt)
and
x 2 ( t) = 2 cos( 4πt)
(b)
 2πn 
x1[ n ] = sin

 16 
and
 2πn 
x 2 [ n ] = 2 cos

 8 
(c)
x1 ( t) = e − t u( t)
and
x 2 ( t) = e −2 t u( t)
x2
1
-1
x
1
1
-1
(d)
2.
x1[ n ] = e
−
n
10
 2πn 
cos
 u[ n ]
 10 
and
x1[ n ] = e
−
n
10
 2πn 
sin
 u[ n ]
 10 
Plot correlograms of the following pairs of CT and DT signals.
(a)
x1 ( t) = cos(2πt)
and
x 2 ( t) = cos2 (2πt)
(b)
x1[ n ] = n
and
x2[n] = n 3
(c)
x1 ( t) = t
and
x 2 ( t) = 2 − t 2 ,
,
−10 < n < 10
−4 < t < 4
3. In MATLAB generate two vectors, x1 and x2, representing DT signals using the
following code fragment,
x1 = randn(100,1) ; x2 = randn(100,1) ; x3 = randn(100,1) ;
Then plot correlograms of the following pairs of DT signals.
(a)
x1
and
x2
(b)
x1
and
x1+x2
Solutions 8-1
M. J. Roberts - 8/17/04
x1 + x2
3
-3
3
x1
-3
(c)
x1+x2 and
(d)
x1+x3
x1+x2/10
and
-x1+x3/10
4. Plot the correlation function for each of the following pairs of energy signals.
(a)
x1 ( t) = 4 rect ( t)
Using
x 2 ( t) = −3 rect (2 t)
and
R xy (τ ) = x(−τ ) ∗ y(τ ) ,
R12 (τ ) = 4 rect (−τ ) ∗ (−3 rect (2τ )) = −12 rect (τ ) ∗ rect (2τ )
Express the wider rectangle as the sum of two narrower rectangles, each of
which is the same width as the other rectangle, then do two easy convolutions
instead of one hard one.

1 
1
 
 
R12 (τ ) = −12 rect  2τ +   + rect  2τ −    ∗ rect (2τ )
 
 
4 
4

πf
πf
−j


1
1 
 
 
 f  j
F
−12 rect  2  τ +   + rect  2  τ −    ∗ rect ( 2τ ) ←
→ −3 sinc 2    e 2 + e 2 
 2
 
 
4
4 


  
1
 
R12 (τ ) = −6 tri 2τ +   + tri 2τ −
 
4
  
1 
 
4 
R (τ)
12
-2
2
τ
-6
(b)
x1[ n ] = 2 rect 3 [ n ]
and
x 2 [ n ] = 5 rect 8 [ n ]
(c)
x1 ( t) = 4 e − t u( t)
and
x 2 ( t) = 4 e − t u( t)
 2 
= 8 e− τ
R12 (τ ) = 8F −1 
 1 + ω 2 
Solutions 8-2
M. J. Roberts - 8/17/04
x1[ n ] = 2e
(d)
−
n
16
For m ≥ 0,
R12 [ m] = −3e
−
 2πn 
sin
 u[ n ]
 8 
m
16
0
∑e
n
8
n =−∞
and
x 2 [ n ] = −3e
−
n
16
 2πn π 
sin
−  u[ n ]
 8
4
  2πm π 
 4πn 2πm π  
+  − cos −
+
− 
 cos −
 8

8
4
8
4 
For m < 0,
R12 [ m] = −3e
−
m
16
n
  2πm π 
 4πn 2πm π  
8
e
+  − cos −
+
− 
 cos −
∑
 8

8
4
8
4 
n =−∞
m
5. Plot the correlation function for each of the following pairs of power signals.
x1 ( t) = 6 sin(12πt)
(a)
LCM of the two periods is
CTFS representation.
and
x 2 ( t) = 5 cos(12πt)
1
. Therefore both signals are at the fundamental of the
6
FS
R12 (τ ) ←
→ X1* [ k ] X 2 [ k ]
X1[ k ] = j 3(δ [ k + 1] − δ [ k − 1])
FS
R12 (τ ) ←
→− j
and
5
(δ[k − 1] + δ[k + 1])
2
X 2[k ] =
15
(δ [ k + 1] − δ [ k − 1])(δ [ k − 1] + δ [ k + 1])
2
FS
R12 (τ ) ←
→− j
15
(δ [ k + 1] − δ [ k − 1])
2
R12 (τ ) = −15 sin(12πτ )
Alternate Solution:
1
30
6 sin(12πt)5 cos(12π ( t + τ )) dt = lim ∫ sin(12πt) cos(12π ( t + τ )) dt
∫
T
T →∞ T
T →∞ T T
R12 (τ ) = lim
15
15
sin(−12πτ ) + sin(24πt + 12πτ )]dt = lim
[
∫
T
T →∞ T
T →∞ T
R12 (τ ) = lim
T
2
∫ [sin(−12πτ ) + sin(24πt + 12πτ )]dt
−
T
2
T
cos(24πt + 12πτ )  2
15 
R12 (τ ) = lim t sin(−12πτ ) −
 T
T →∞ T
24π
−

2
Solutions 8-3
M. J. Roberts - 8/17/04

 24πT

 24πT

cos
+ 12πτ 
cos −
+ 12πτ  





15 T
T
2
2

R12 (τ ) = lim  sin(−12πτ ) −
+ sin(−12πτ ) +
T →∞ T
24π
24π
2

2


R12 (τ ) = 15 sin(−12πτ ) = −15 sin(12πτ )
R12(τ)
15
-0.5
0.5
τ
-15
(b)
 2πn 
x1[ n ] = 6 sin

 12 
and
 2πn 
x 2 [ n ] = 5 sin

 12 
(c)
x1 ( t) = 6 sin(12πt)
and
π

x 2 ( t) = 5 sin12πt − 

4
6. Find the autocorrelations of the following CT and DT energy and power signals and
show that, at zero shift, the value of the autocorrelation is the signal energy or power and
that all the properties of autocorrelation functions are satisfied.
(a)
x( t) = e −3 t u( t)
(b)
x[ n ] = rect 5 [ n − 5]
R x [ m ] = rect 5 [ − m − 5 ] ∗ rect 5 [ m − 5 ] = rect 5 [ m + 5 ] ∗ rect 5 [ m − 5 ]
F
R x [ m ] ←
→ 11 drcl ( F,11) e j10 π F 11 drcl ( F,11) e− j10 π F = 121 drcl2 ( F,11)
 m
R x [ m] = rect 5 [ m] ∗ rect 5 [ m] = 11tri 
 11
Even function, maximum at m = 0. Energy is
Ex =
(c)
∞
∑ rect [n − 5]
n =−∞
5
2
= 11 = R x [0] . Check.
  1
  3 
x( t) = rect  2 t −   − rect  2 t −  
  4
  4
7. Find the autocorrelation functions of the following power signals.
(a)
x( t) = 5 sin(24πt) − 2 cos(18πt)
Solutions 8-4
M. J. Roberts - 8/17/04
The period of this signal is
X[ k ] = j
1
second. Therefore f 0 = 3 and its CTFS is
3
5
(δ[k + 4] − δ[k − 4]) − (δ[k − 3] + δ[k + 3])
2
The autocorrelation can be found from
5
R x (τ ) ←→ X [ k ] X [ k ] = j (δ [ k + 4 ] − δ [ k − 4 ]) − (δ [ k − 3] + δ [ k + 3])
2
FS
FS
R x (τ ) ←
→
25
(δ [ k + 4 ] + δ [ k − 4 ]) + (δ [ k − 3] + δ [ k + 3])
4
R x (τ ) =
(b)
2
*
25
cos(24πt) + 2 cos(18πt)
2
 2πn 
 2πn 
x[ n ] = −4 sin
 − 2 cos

 36 
 40 
8. A signal is sent from a transmitter to a receiver and is corrupted by noise along the way.
The signal shape is of the functional form,
f 
1 
x( t) = A sin(2πf 0 t) rect  0  t −

 4  2 f0 
What is the transfer function of a matched filter for this signal?
f 
1 
h( t) = K sin(2πf 0 ( t0 − t)) rect  0  t0 − t −

2 f 0  
4
The sine is odd and the rect is even. Therefore
f 
1 
h( t) = −K sin(2πf 0 ( t − t0 )) rect  0  t − t0 +

2 f 0  
4

1 
 4( f + f 0 )
 4( f − f 0 ) 
2K − j 2πf  t 0 − 2 f 0  
e
− sinc
H( f ) = j
sinc


f0
f0
f0



 

9. Find the signal power of the following sums or differences of signals and compare it to
the power in the individual signals. How does the comparison relate to the correlation
between the two signals that are summed or differenced?
(a)
x( t) = sin(2πt) + cos(2πt)
Solutions 8-5
M. J. Roberts - 8/17/04
Px =

1
1  2
2
2
sin
(
π
t
)
+
cos
(
π
t
)
dt
=
sin
(
π
t
)
+
cos
(
π
t
)
+
sin
(
π
t
)
cos
(
π
t
)
2
2
2
2
2
2
2

 dt
T0 ∫T0
T0 ∫T0  144424443

=1
Px = ∫ dt + 2 ∫ sin(2πt) cos(2πt) dt = 1
1
1 4442444
1
3
=0
1
1
and the power of the cosine is . The power of the sum
2
2
equals the sum of the powers because these two signals are uncorrelated at zero shift.
The power of the sin is
(b)
π

x( t) = sin(2πt) + cos 2πt − 

4
(c)
 n
x[ n ] = rect 2 [ n ] ∗ comb10 [ n ] − tri  ∗ comb10 [ n ]
 2
The signal power of the sum is less than sum of the signal powers because the two
signals are negatively correlated at zero shift.
(d)
 n − 5
x[ n ] = rect 2 [ n ] ∗ comb10 [ n ] + tri
 ∗ comb10 [ n ]
 2 
The signal power of the sum equals the sum of the signal powers because the two
signals are uncorrelated at zero shift.
10. Find the crosscorrelation functions of the following pairs of periodic signals.
(a)
 t
 t 
x1 ( t) = rect   ∗ comb  and
 6
 24 
 t − 3
 t 
x 2 ( t) = rect 
 ∗ comb 
 6 
 24 
∞
 τ − 3
 τ − 3
τ 
R12 (τ ) = 6 tri
δ (τ − 24 n )
∗
 ∗ comb  = 144 tri
 6  n∑
 6 
 24 
=−∞
(b)
 2πn 
x1[ n ] = sin 2 

 8 
and
 2πn 
x 2 [ n ] = sin 2 

 10 
(c)
x1 ( t) = e − j10πt and
x 2 ( t) = cos(10πt)
11. Find the ESD’s of the following energy signals.
(a)
x[ n ] = Aδ [ n − n 0 ]
(b)
x( t) = e −100 t u( t)
Solutions 8-6
M. J. Roberts - 8/17/04
X( jω ) =
1
1
⇒ Ψx ( jω ) = 4
100 + jω
10 + ω 2
 7
 2πn 
x[ n ] = 10  sin
 u[ n ]
 8
 12 
 t − t0 
x( t) = A tri

 w 
n
(c)
(d)
12. Find the ESD of the response, y( t) or y[ n ] , of each system with impulse response, h( t)
or h[ n ], to the excitation, x( t) or x[ n ] .
n
(a)
 9
x[ n ] = δ [ n ] , h[ n ] =  −  u[ n ]
 10 
X( jΩ) = 1 ⇒ Ψx ( jΩ) = 1
Ψy ( jΩ) = Ψx ( jΩ) H( jΩ) =
and
2
H( jΩ) =
10
1
=
9 − jΩ 10 + 9e − jΩ
1+ e
10
100
(10 + 9 cos(Ω))
2
+ 81sin (Ω)
2
=
100
181 + 180 cos(Ω)
(b) x( t) = e −100 t u( t) , h( t) = e −100 t u( t)
(c) x[ n ] = rect 3 [ n ] , h[ n ] = rect 2 [ n − 2]
 1
(d) x( t) = 4 e − t cos(2πt) u( t) , h( t) = rect  t − 
 2
13. Find the PSD’s of these signals.
(a)
x( t) = A cos(2πf 0 t + θ )
R x (τ ) =
[
]
A2
A2
cos(2πf 0τ ) ⇒ Gx ( f ) =
δ ( f − f0 ) + δ ( f + f0 )
2
4
In finding PSD’s, it is usually easier to find the autocorrelation first and then
Fourier transform it to find the PSD rather than taking the direct route using
the definition.
(b)
x( t) = 3 rect (100 t) ∗ comb(25 t)
 f 
 f 
G x ( f ) = 3.6 × 10 −5 sinc 2 
 comb 
 100 
 25 
(c)
 2πn 
x[ n ] = 8 sin

 12 
(d)
x[ n ] = 3 rect 4 [ n ] ∗ comb 20 [ n ]
Solutions 8-7
M. J. Roberts - 8/17/04
9 sin 2 (9πF )
comb(20 F )
G x (F ) =
20 sin 2 (πF )
14. Find the PSD of the response, y( t) or y[ n ] , of each system with impulse response, h( t)
or h[ n ], to the excitation, x( t) or x[ n ] .
−
π

x( t) = 4 cos 32πt −  , h( t) = e 10 u( t)

4
t
(a)
[
]
G x ( f ) = 4 δ ( f − 16) + δ ( f + 16)
and
H( f ) =
1
10
=
1
+ j 2πf 1 + j 20πf
10
G y ( f ) = 4[δ ( f − 16) + δ ( f + 16)]
G y ( f ) = 400
100
1 + (20πf )
2
 δ ( f − 16)
δ ( f + 16) 
= 400 
2 +
2
1 + (20πf ) 1 + (20πf ) 
δ ( f − 16) + δ ( f + 16)
2
1 + ( 320π )
(b)
x( t) = 4 comb(2 t) , h( t) = rect ( t − 1)
(c)
 11
x[ n ] = 2 comb 8 [ n ] , h[ n ] =   u[ n − 1]
 12 
n
G y (F ) =
(d)
0.42 comb(8 F )
1 − 1.8333 cos(2πF ) + 0.8403
x[ n ] = (−0.9) u[ n ] , h[ n ] = (0.5) u[ n ]
n
n
15. Plot correlograms of the following pairs of CT and DT signals.
(a)
(b)
   1
  3  
x1 ( t) = tri 4  t −   − tri 4  t −    ∗ comb( t)
  4 
   4
  n − 8
 n − 24  
x1[ n ] = tri
 ∗ comb 32 [ n ]
 − tri
 8  
  8 
 2πn 
x 2 [ n ] = cos

 32 
Solutions 8-8
and
and
x 2 ( t) = sin(2πt)
M. J. Roberts - 8/17/04
(c)
(d)
   1
  3  
x1 ( t) = tri 4  t −   − tri 4  t −    ∗ comb( t)
  4 
   4

  1  
x 2 ( t) = tri( 4 t) − tri 4  t −    ∗ comb( t)
  2  


 n − 24  
 n − 8
x1[ n ] = rect 
 ∗ comb 32 [ n ]
 − rect 
 16  
 16 

 2πn 
x 2 [ n ] = sin

 32 
and
and
16. Plot a correlogram for the following sets of samples from two signals, x and y. In each
case, from the nature of the correlogram what relationship, if any, exists between the two
sets of data?
(a)
x = {6,5,8,-2,3,-10,9,-2,-4,3,-2,6,0,-5,-7,1,9,9,4,-6}
y = {-1,-10,-4,4,5,-2,-3,-5,-9,2,6,-5,-1,-10,-9,0,4,-10,9,-1}
(b)
x = {4,6,0,0,5,-6,8,-9,0,8,7,2,-5,-3,-4,-4,8,0,4,7}
y = {-11,-13,3,-1,-8,10,-16,16,1,-17,-14,-3,9,7,12,9,-17,1,-8,-17}
(c)
x = {0,6,11,16,19,20,19,16,11,6,-0,-7,-12,-17,-20,-20,-20,-17,-12,-7}
y = {19,15,10,8,3,-9,-12,-19,-19,-25,-19,-17,-12,-5,-1,5,8,12,17,20}
17. Plot the correlation function for each of the following pairs of energy signals.
(a)
x1 ( t) = rect ( t) sin(10πt)
and
x 2 ( t) = rect ( t) cos(10πt)
(b)
x1[ n ] = δ [ n − 1] − δ [ n + 1]
and
x 2 [ n ] = −δ [ n − 1] + δ [ n + 1]
(c)
x1 ( t) = e − t
2
and
x 2 ( t) = e −2 t
2
π − 23 τ 2
R12 (τ ) =
e
3
18. Plot the correlation function for each of the following pairs of power signals.
(a)
 2πn 
x1[ n ] = −3 sin

 20 
(b)
x1 ( t) = rect ( 4 t) ∗ comb( t)
 2πn 
x 2 [ n ] = 8 sin

 10 
and
and
x 2 ( t) = rect ( 4 t) ∗ comb( t)
(c)
1
 t
x1 ( t) = 4 rect ( t) ∗ comb  − 2
 2
2
and
1
 t
x 2 ( t) = 4 rect ( t − 1) ∗ comb  − 2 .
 2
2
Solutions 8-9
M. J. Roberts - 8/17/04
Graphical Solution:
1
 t
x1 ( t) = 4 rect ( t) ∗ comb  − 2
 2
2
1
 t
x 2 ( t) = 4 rect ( t − 1) ∗ comb  − 2 .
 2
2
and
The easiest way to find the correlation function for these two signals is to first graph them,
then find the correlation graphically.
x1(t)
x2(t)
2
2
-2
2
t
2
t
-2
For periodic functions, the correlation function is the average value of the product of the two
functions over one period of the product of the two functions. In this case the periods of the
two functions are the same, 2, and the period of the product is also 2. With zero shift of
x 2 ( t) the product is -4 everywhere and therefore the average of the product is also -4. If
x 2 ( t) is shifted by exactly half a period, 1, the product is +4 everywhere and the average
value is +4. For any shift between these two shifts, the average value of the product varies
linearly with the shift amount between these two extreme values of the correlation. Also the
correlation function is periodic with the same period as these two functions. Therefore the
correlation function is all illustrated below.
R12(τ)
4
2
-4
τ
∞
R12 (τ ) = −4 + 8 ∑ tri( t − 1 − 2 n )
n =−∞
19. Find the autocorrelations of the following CT and DT energy and power signals and
show that, at zero shift, the value of the autocorrelation is the signal energy or power and
that all the properties of autocorrelation functions are satisfied.
(a)
x[ n ] = δ [ n ] + δ [ n − 1] + δ [ n − 2] + δ [ n − 3]
δ [− m] + δ [− m − 1]
 δ [ m] + δ [ m − 1]

R x [ m] = x[− m] ∗ x[ m] = 
 ∗

 +δ [− m − 2] + δ [− m − 3]  +δ [ m − 2] + δ [ m − 3]
Solutions 8-10
M. J. Roberts - 8/17/04
δ [− m] + δ [− m + 1] + δ [− m + 2] + δ [− m + 3] 


+δ [− m − 1] + δ [− m] + δ [− m + 1] + δ [− m + 2] 
R x [ m] = 

+δ [− m − 2] + δ [− m − 1] + δ [− m] + δ [− m + 1] 
+δ [− m − 3] + δ [− m − 2] + δ [− m − 1] + δ [− m]


4δ [ m] + 3δ [− m + 1] + 2δ [− m + 2] + δ [− m + 3]
R x [ m] = 

+3δ [− m − 1] + 2δ [− m − 2] + δ [− m − 3]

R x [ m] = δ [ m + 3] + 2δ [ m + 2] + 3δ [ m + 1] + 4δ [ m] + 3δ [ m − 1] + 2δ [ m − 2] + δ [ m − 3]
Ex =
∞
∑
x[ n ] =
2
n =−∞
∞
∑ δ[n] + δ[n − 1] + δ[n − 2] + δ[n − 3]
n =−∞
2
= 4 = R x [0] . Check.
x( t) = A cos(2πf 0 t + θ )
(b)
FS
R x (τ ) ←
→ X* [ k ] X [ k ]
A2
R x (τ ) =
cos(2πf 0τ )
2
Alternate Solution:
R x (τ ) = lim
T →∞
1
T
T
2
∫ x(t) x(t + τ )dt = lim
T →∞
T
−
2
R x (τ ) = lim
T →∞
2
A
2T
T
2 2
A
T
∫ cos(2πf t + θ ) cos(2πf (t + τ ) + θ )dt
0
0
T
−
2
T
2
∫ [cos(−2πf τ ) + cos(4πf t + 2πf τ + 2θ )]dt
0
−
0
o
T
2
T
sin( 4πf 0 t + 2πf oτ + 2θ )  2
A 
t cos(−2πf 0τ ) +
R x (τ ) = lim


T →∞ 2T
4πf 0

− T
2
2
A2
R x (τ ) =
cos(2πf 0τ )
2
R x (0) =
(c)
A2
which is the average signal power of any sinusoid of amplitude A
2
x[ n ] = comb12 [ n ]
FS
R x [ m ] ←
→ X* [ k ] X [ k ]
Solutions 8-11
M. J. Roberts - 8/17/04
FS
Using comb N0 [ n ] ←
→
1
N0
*
 1  1
1
R x [ m ] ←→  
= 2
 N0  N0 N0
FS
R x [ m] =
R x [0] =
Alternate Solution:
1
N →∞ N
R x [ m] = lim
comb12 [ m]
12
1
. Check.
12
1
∑ x[n] x[n + m] = lim N ∑ comb [n]comb [n + m]
N →∞
n= N
n= N
12
12
The product, comb12 [ n ] comb12 [ n + m], is non-zero only if m is a integer multiple of
12. The summation is over a period of length N. There are two cases.
Case I. N even
Let the summation be from −
N
N
to
− 1. Then
2
2
1
R x [ m] = lim
N →∞ N
N
−1
2
∑ comb [n]comb [n + m] .
12
n =−
12
N
2
In that range, for m an integer multiple of 12, the number of impulses in the product
N
is the greatest integer in . Therefore
12
N


1 Greatest Integer in  , m an integer multiple of 12
12 
R x [ m] = lim 

N →∞ N
0 , otherwise




1
, m an integer multiple of 12 comb12 [ m]

R x [ m] = 12
=
12

0 , otherwise
Case II.
N odd
Let the summation be from −
N −1
N −1
to
. Then
2
2
Solutions 8-12
M. J. Roberts - 8/17/04
1
N →∞ N
R x [ m] = lim
N −1
2
∑
N −1
n =−
2
comb12 [ n ] comb12 [ n + m] .
The rest of the analysis is the same and so is the answer (in the limit).

1
, m an integer multiple of 12 comb12 [ m]

R x [ m] = 12
.
=
12

0 , otherwise
20. Find and sketch the autocorrelation function of
1
 t
x( t) = 10 rect (2 t) ∗ comb  .
 4
4
FS
R12 (τ ) ←
→ X1* [ k ] X 2 [ k ]
R12 (τ ) =
1
25
 t
tri(2 t) ∗ comb 
 4
4
2
R12 (0) =
25
. Check.
2
Alternate Solution:
Check to be sure that its value at a shift of zero is the same as its average signal power.
The signal is graphed below.
x(t)
10
1
4
4
t
This is a periodic signal. Therefore its autocorrelation function is also periodic with the
same period. The maximum autocorrelation value occurs at zero shift and that value is the
average of the square of the signal (its average power), 12.5. The autocorrelation varies
linearly with shift up to a shift of one-half at which time it is zero. This is true for positive or
1
negative shifts of up to one-half. The autocorrelation for shifts, < τ < 2 , is zero. This
2
pattern repeats periodically. Therefore the autocorrelation function is as illustrated below.
Solutions 8-13
M. J. Roberts - 8/17/04
Rx(τ)
12.5
1
2
R12 (τ ) =
τ
4
1
25
 t
tri(2 t) ∗ comb 
 4
4
2
21. Find all cross correlation and autocorrelation functions for these three signals:
x1 ( t) = cos(2πt) , x 2 ( t) = sin(2πt) , x 3 ( t) = cos( 4πt)
Check your autocorrelation answers by finding the average power of each signal.
22. Find and sketch the crosscorrelation between a unit-amplitude, one Hz cosine and a 50%
duty-cycle square wave which has a peak-to-peak amplitude of two, a period of one, an
average value of zero and is an even function.
FS
R12 (τ ) ←
→ X1* [ k ] X 2 [ k ]
x1 ( t) = cos(2πt)
and
x 2 ( t) = 2 rect (2 t) ∗ comb( t) − 1
 1
R12 (τ ) = sinc  cos(2πτ )
 2
Old Solution:
T0
4
T0
4
4  sin(2π ( kf 0 ) t)  4
2
 kπ 
 k
X c [ k ] = ∫ cos(2π ( kf 0 ) t) dt = 
sin  = sinc 
 =
 2
T0 0
T0  2π ( kf 0 ) 
kπ  2 
0
X s[ k ] = 0 , because the function is even
All the cosines in the trigonometric expression for the square wave, except the one at the
fundamental frequency of one Hz, are orthogonal to the one-Hz cosine over one period.
Therefore the crosscorrelation is simply
R(τ ) =
1
2
1
2
1
 1
 1
cos(2πt) sinc  cos(2π ( t + τ )) dt = 2 sinc  ∫ [cos(−2πτ ) + cos( 4πt + 2πτ )]dt
∫


 2 0
2
1 1
−
2
1
sin( 4πt + 2πτ )  2
 1
 1 
= sinc  cos(2πτ )
R(τ ) = 2 sinc  t cos(2πτ ) +

 2
 2 
4π
0
Solutions 8-14
M. J. Roberts - 8/17/04
23. Find and sketch the ESD of each of these signals:
 t
x( t) = A rect  
 w
(a)
Ψx ( f ) = X( f ) = Aw sinc( wf ) = A 2 w 2 sinc 2 ( wf )
2
(b)
 t + 1
x( t) = A rect 

 w 
(c)
 t
x( t) = A sinc 
 w
2
Ψx ( f ) = X( f ) = Aw rect ( wf ) = A 2 w 2 rect ( wf )
2
2
2
x( t) =
(d)
1 − t2
e
2π
Ψ x ( f ) = e− ( 2 π f )
2
24. Find the PSD's of
(a)
x( t ) = A
R x (τ ) = lim
T →∞
1
T
T
2
∫ x (t ) x (t + τ ) dt = lim
−
T →∞
T
2
2
A
T
G x ( f ) = F  R x (τ )  = A 2δ ( f )
(b)
x(t ) = A cos(2πf0 t )
(c)
T
2
∫ dt = A
−
2
T
2
x(t ) = A sin(2πf0 t )
25. Which of the following functions could not be the autocorrelation function of a real
signal and why?
An autocorrelation function has the properties,
and
(a)
(b)
1.
It is an even function,
2.
Its maximum value occurs at zero shift,
3.
Its Fourier transform is strictly non-negative for all f (or ω ).
R (τ ) = tri (τ )
R (τ ) = A sin ( 2π f0τ )
Solutions 8-15
M. J. Roberts - 8/17/04
(c)
(d)
R (τ ) = rect (τ )
R (τ ) = A sinc ( Bτ )
Solutions 8-16