M. J. Roberts - 8/28/04
Chapter 4 - The Fourier Series
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. Using MATLAB plot each sum of complex sinusoids over the time period indicated.
(a)
x( t) =
1 30
 k
sinc  e j 200πkt , −15ms < t < 15ms
∑
 10 
10 k =−30
The MATLAB program can exactly follow the mathematical process in the
equation.
For example:
f0 = 100 ; T0 = 1/f0 ; fmax = 30*f0 ;
Tmin = 1/fmax ; dt = Tmin/16 ;
t = -0.015:dt:0.015 ; x = t*0 ;
for k = -30:30,
x = x + sinc(k/10)*exp(j*200*pi*k*t) ;
end
x = real(x/10) ;
plot(t,x,'k') ;
(Although the vector, x, should be real, it may have some very small, but non-zero
imaginary parts because of round-off in the calculations. That is why the ‘real’ command
is used in the next-to-last line.)
x(t)
1
-15
15
Solutions 4-1
t (ms)
M. J. Roberts - 8/28/04
(b)
j 9 
 k − 2   j10πkt
 k + 2
x( t) = ∑ sinc
 e
 − sinc
 2  
 2 
4 k =−9 
,
−200ms < t < 200ms
2. Show by direct analytical integration that the integral of the function,
g( t) = A sin(2πt) B sin( 4πt)
1
1
is zero over the interval, − < t < .
2
2
1
2
1
2
∫ g(t)dt = AB ∫ sin(2πt) sin(4πt)dt
−
Using sin( x ) sin( y ) =
1
2
∫ g(t)dt =
1
−
2
1
2
AB
∫1 g(t)dt = 2
2
1
2
∫
−
1
2
1
[cos( x − y ) − cos( x + y )] ,
2
1
2
−
−
∫ [cos(2πt − 4πt) − cos(2πt + 4πt)]dt
−
1
2
1
2
1
AB  sin(−2πt) sin(6πt)  2
∫1 [cos(−2πt) − cos(6πt)]dt = 2  −2π − 6π  1
−
−
g( t) dt =
1
2
AB
2
1
2
2
2
AB  sin(−π ) sin( 3π )  sin(π ) sin(−3π )  
−
−
−
 = 0

 −2π
2  −2π
6π
6π  
3. Convert the function, g( t) = (1 + j )e j 4 πt + (1 − j )e − j 4 πt , to an equivalent form in which
“j” does not appear.
Two very important trigonometric identities to remember are
e jx + e− jx
cos ( x ) =
2
,
e jx − e− jx
sin ( x ) =
j2
Solutions 4-2
M. J. Roberts - 8/28/04
4. Using MATLAB plot these products over the time range indicated and observe in
each case that the net area under the product is zero.
1
4
(a)
x( t) = −3 sin(16πt) × 2 cos(24πt) , 0 < t <
(b)
x( t) = −3 sin(16πt) × 2 cos(24πt) , 0 < t < 1
(c)
x( t) = −3 sin(16πt) × 2 cos(24πt) , −
(d)
x( t) = x1 ( t) x 2 ( t)
where
1
3
<t<
16
16
x1 ( t) is an even, 50%-duty-cycle square wave with a fundamental period
of
4 seconds and an average value of zero
and
x 2 ( t) is an odd, 50%-duty-cycle square wave with a fundamental period of
4 seconds and an average value of zero
x(t)
6
4
t
-6
(e)
x( t) = x1 ( t) x 2 ( t)

  1  1
 t 1
where x1 ( t) = rect (2 t) ∗ comb( t) and x 2 ( t) = rect  4  t −   ∗ comb   −
 2  8
  8  2

5. A sine function can be written as
sin(2πf 0 t) =
e j 2πf 0 t − e − j 2πf 0 t
.
j2
Solutions 4-3
M. J. Roberts - 8/28/04
This is a very simple complex CTFS in which the harmonic function is only non-zero at
two harmonic numbers, +1 and –1. Verify that we can write the harmonic function
directly as
X[ k ] =
j
(δ[k + 1] − δ[k − 1]) .
2
Write the equivalent expressions for sin(2π (− f 0 ) t) and show that the harmonic function
is the complex conjugate of the previous one for sin(2πf 0 t) .
If the harmonic function is correct this equality should hold:
sin(2πf 0 t) =
∞
∞
j
e j 2πf 0 t − e − j 2πf 0 t
= ∑ X[ k ]e j 2π ( kf 0 ) t = ∑ (δ [ k + 1] − δ [ k − 1])e j 2π ( kf 0 ) t
j2
k =−∞
k =−∞ 2
Finish simplifying and see whether it does.
6. For each signal, find a complex CTFS which is valid for all time, plot the magnitude
and phase of the harmonic function versus harmonic number, k, then convert the
answers to the trigonometric form of the harmonic function.
(a)
x( t) = 4 rect ( 4 t) ∗ comb( t)
This form is in Appendix E. But let’s do one directly from the definition.
X[ k ] =
1
2
1
8
1
x( t)e − j 2π ( kf 0 ) t dt = 4 ∫ rect ( 4 t)e − j 2πkt dt = 4 ∫ e − j 2πkt dt
∫
T
T0 0
1
1
−
2
−
8
Finish and check your answer against Appendix E.
(b)
1
 t
x( t) = 4 rect ( 4 t) ∗ comb 
 4
4
(c)
A periodic signal which is described over one fundamental period by
sgn( t) , t < 1
x( t) = 
, 1< t < 2
0
Solutions 4-4
M. J. Roberts - 8/28/04
7 . Using the CTFS table of transforms and the CTFS properties, find the CTFS
harmonic function of each of these periodic signals using the time interval, TF ,
indicated.
1
10
(a)
x( t) = 10 sin(20πt) , TF =
(b)
x( t) = 2 cos(100π ( t − 0.005)) , TF =
1
50
In this case, the representation time is the same as the fundamental period. From
Appendix E,
1
FS
cos ( 2π f0t ) ←
→ (δ [ k − 1] + δ [ k + 1]) , TF = T0
2
Linearity Property:
1
FS
2 cos (100π t ) ←
→ δ [ k − 1] + δ [ k + 1] , TF = T0 =
50
Time Shifting Property:
1
FS
2 cos (100π ( t − 0.005 )) ←
→ (δ [ k − 1] + δ [ k + 1]) e− j 0.5 π k , TF = T0 =
50
1
1
f0t 0 = 50 ×
=
200 4
πk
−j
1
FS
2 cos (100π ( t − 0.005 )) ←
→ (δ [ k − 1] + δ [ k + 1]) e 2 , TF = T0 =
50
Using the equivalence property of the impulse,
g [ n ] Aδ [ n − n0 ] = g [ n0 ] Aδ [ n − n0 ] ,
we can write
e− j 0.5 π kδ [ k − 1] + e− j 0.5 π kδ [ k + 1] = e
−j
−j
π
2
π
δ [ k − 1] + e 2 δ [ k + 1]
j
j
FS
2 cos (100π ( t − 0.005 )) ←
→ j (δ [ k + 1] − δ [ k − 1]) , TF = T0 =
From Appendix E,
FS
2 sin (100π t ) ←
→ j (δ [ k + 1] − δ [ k − 1]) , TF = T0 =
(c)
x( t) = −4 cos(500πt) , TF =
1
50
Solutions 4-5
1
50
1
50
M. J. Roberts - 8/28/04
f 0 = 250 ⇒ T0 =
Use the table entry,
(d)
x( t) =
1
⇒ TF = 5T0
250
TF = mT0
1
FS
cos ( 2π f0t ) ←
→ (δ [ k − m ] + δ [ k + m ])
2
( m an integer )
d − j10πt
1
, TF =
e
(
)
dt
5
TF = T0 =
1
5
Start with
FS
e j 2 π f0 t ←
→ δ [ k − 1]
,
TF = T0 =
1
5
from Appendix E. Then apply the derivative property of the CTFS.
(e)
 t
x( t) = rect ( t) ∗ comb 
 4
(f)
x( t) = rect ( t) ∗ comb( t) , TF = 1
, TF = 4
Use the fact that sinc ( k ) = δ [ k ] .
(g)
x( t) = tri( t) ∗ comb( t) , TF = 1
8. If a periodic signal, x( t) , has a fundamental period of 10 seconds and its harmonic
function is
 k
X[ k ] = 4 sinc  ,
 20 
with a representation period of 10 seconds, what is the harmonic function of z( t) = x( 4 t)
using the same representation period, 10 seconds?
Solutions 4-6
M. J. Roberts - 8/28/04
 k  k
, an integer
X
Using the time-scaling property, Z[ k ] =   a  a
,
0
, otherwise

 k  k
k
 

 k
an integer  4 sinc  ,
an integer 
X   ,
 80 
4
Z[ k ] =   4  4
= 

0



, otherwise  0
, otherwise


This can be written more compactly as
 k
Z[ k ] = 4 sinc  comb 4 [ k ] .
 80 
9. A periodic signal, x( t) , has a fundamental period of 4 ms and its harmonic function is
X[ k ] = 15(δ [ k − 1] + δ [ k + 1]) ,
with a representation period of 4 ms. Find the integral of x( t) .
Start with
FS
cos ( 2π f0t ) ←
→
1
(δ [ k − 1] + δ [ k + 1])
2
,
TF = T0 = 4 ms
Linearity
FS
30 cos ( 500π t ) ←
→ 15 (δ [ k − 1] + δ [ k + 1])
,
TF = T0 = 4 ms
Then use the integration property.
10. If X[ k ] is the harmonic function over one fundamental period of a unit-amplitude
50%-duty-cycle square wave with an average value of zero and a fundamental period
of 1 µs , find an expression consisting of only real-valued functions for the signal
whose harmonic function is X[ k − 10] + X[ k + 10] .
Find the time-domain function which corresponds to the CTFS harmonic function,
X[ k ], then apply the frequency-shifting (harmonic-number-shifting) property.
Solutions 4-7
M. J. Roberts - 8/28/04
11. Find the harmonic function for a sine wave of the general form, A sin(2πf 0 t) . Then,
using Parseval’s theorem, find its signal power and verify that it is the same as the
signal power found directly from the function itself.
A
(δ [ k + 1] − δ [ k − 1])
2
FS
A sin ( 2π f0t ) ←
→j
Px =
1
T0
T0
2
∫
−
A sin(2πf 0 t) dt =
2
T0
2
2
A
T0
T0
2
, TF = T0
T0
2
∫ sin (2πf t)dt = 2T ∫ [1 − cos(4πf t)]dt
A
2
2
0
0
−
0
T0
2
Px =
−
T0
2
A2
2
By Parseval’s theorem,
Px =
∞
∑
k =−∞
2
j
A
A2
A2
δ [ k + 1] − δ [ k − 1]) =
(1 + 1) =
. Check.
(
2
4
2
12. Show for a cosine and a sine that the CTFS harmonic functions have the property,
X[ k ] = X* [− k ] .
For a cosine,
X[ k ] =
X[− k ] =
1
(δ[k − 1] + δ[k + 1])
2
1
1
δ [− k − 1] + δ [− k + 1]) = (δ [ k + 1] + δ [ k − 1]) = X[ k ] = X* [ k ] . Check.
(
2
2
1 3 . Find and sketch the time functions associated with these harmonic functions
assuming TF = 1.
(a)
X[ k ] = δ [ k − 2] + δ [ k ] + δ [ k + 2]
(b)
 k
X[ k ] = 10 sinc 
 10 
14. Find the even and odd parts, x e ( t) and x o ( t) , of
Solutions 4-8
M. J. Roberts - 8/28/04
π

x( t) = 20 cos 40πt +  .

6
Then find the harmonic functions, X e [ k ] and X o [ k ], corresponding to them. Then, using
the time shifting property find the harmonic function, X[ k ] and compare it to the sum of
the two harmonic functions, X e [ k ] and X o [ k ].
Use cos( x + y ) = cos( x ) cos( y ) − sin( x ) sin( y ) to find the even and odd parts.
15. Using the direct summation formula find and sketch the DTFS harmonic function of
comb N 0 [ n ] with N F = N 0 .
1
X[ k ] =
∑ comb N0 [ n ] e− j 2π (kF0 )n
N 0 n = N0
One can choose any convenient period for the summation. In the period chosen below
there is exactly one non-zero impulse in the comb function at n = 0 .
1
X[ k ] =
N0
N0
−1
2
∑
n=−
comb N0 [ n ] e− j 2 π ( kF0 ) n =
N0
2
1
N0
16. Using the DTFS table of transforms and the DTFS properties, find the DTFS
harmonic function of each of these periodic signals using the representation period,
N F , indicated.
(a)
 2πn 
x[ n ] = 6 cos
 , N F = 32
 32 
(b)
The solution is X[ k ] = j 5(comb12 [ k + 1] − comb12 [ k − 1])e
−j
π
k
3
We can demonstrate that this solution is correct by reconstituting the signal using
x[ n ] =
∑ X[k ]e
k = N0
j 2π
kn
N0
=
∑ j 5(comb [k + 1] − comb [k − 1])e
12
12
−j
π
k
3
j 2π
e
kn
N0
k = N0
Since the summation extends only over one period, N 0 = 12 , choose the simplest period,
−6 ≤ k < 6 . In that period the two comb functions are simply two impulses at k = ±1.
Solutions 4-9
M. J. Roberts - 8/28/04
x[ n ] =
5
∑ j 5(δ[k + 1] − δ[k − 1])e
 n 1
j 2πk  − 
 12 6 
k =−6
 n 1
 n 1
 j 2π  12n − 16 
 − j 2π  12n − 16 
j 2π  −  
− j 2π  −  
 12 6 
 12 6 
−e
x[ n ] = j 5 e
−e

 = − j 5 e





  n 1  
  n − 2 
x[ n ] = − j 5  j 2 sin 2π  −    = 10 sin 2π 




  12  
12 6  

We could have chosen a different period, for example 4 ≤ k < 16 . Then
15
x[ n ] = ∑ j 5(δ [ k − 11] − δ [ k − 13])e
 n 1
j 2πk  − 
 12 6 
k =4
 n 1
 13 n 13  
 j 2π  1112n −116 
 j 22π  12n − 16 
j 26π  −  
j 2π 
− 
 12 6 
− e  12 6  
x[ n ] = j 5 e
−e
 = j 5 e




x[ n ] = j 5e
 12 n 12 
j 2π 
− 
 12 6 
 n 1
 n 1
 − j 2π  12n − 16 
 j 2π  −12n + 16 
j 2π  −  
j 2π  −  
 12 6 
j 2πn − j 4 π
− e  12 6  
−e
 = j 5e{ e{ e
e
=1
=1 





  n 1  
  n − 2 
x[ n ] = j 5 − j 2 sin 2π  −    = 10 sin 2π 

  12 6   
  12  

which is the same answer as before (with somewhat more effort).
(c)
 n  n
an integer
,
x
, N F = 48
x[ n ] =  1  8  8
0
, otherwise

 2πn 
where x1[ n ] = sin

 6 
Use the time-scaling property of the DTFS.
(d)
x[ n ] = e j 2πn , N F = 6
Solutions 4-10
M. J. Roberts - 8/28/04
Notice that this function, although written as x[ n ] = e j 2πn could be more
simply written as x [ n ] = 1 because for any integer value of discrete time, n, the result is
always the same, 1. Therefore
FS
e j 2 π n = 1←
→ comb 6 [ k ]
|X[k]|
1
-40
40
k
Phase of X[k]
π
-40
40
k
-π
If we take a more straightforward approach to finding the DTFS harmonic function using
N F = mN 0
e
we get
n
j 2π
N0
FS
←
→ comb N F [ k − m ]
FS
e j 2 π n ←
→ comb 6 [ k − 6 ]
Observe that this answer and the previous answer are identical. That is,
comb 6 [ k ] = comb 6 [ k − 6 ] ,
as must be true.
(e)
(f)
 2π ( n − 1) 
 2πn 
x[ n ] = cos
 − cos
 , N F = 16
 16 
 16 
 33 × 2π n 
 33π n 
x [ n ] = − sin 
= − sin 
 , N F = 64


 32 
64 
The period of this signal is 64.
Solutions 4-11
M. J. Roberts - 8/28/04
This form does not appear directly in Appendix E. Therefore we must either use
what is in Appendix E with some properties to get to this form or simply apply the
definition of the DTFS harmonic function directly.
From Appendix E,
j 2π
e
n
N0
FS
←
→ comb N0 [ k − 1]
Using the frequency-shifting property,
j 2π
e
j 2π
e
mn
N0
j 2π
e
n
( m +1)
N0
n
N0
FS
←
→ comb N0 [ k − m − 1]
FS
←
→ comb N0  k − ( m + 1) 
or
j 2π
e
mn
N0
FS
←
→ comb N0 [ k − m ]
Then
j 2π
e
mn
N0
+e
2
− j 2π
mn
N0
− j 2π
mn
N0
 2π mn  FS 1
←→ comb N0 [ k − m ] + comb N0 [ k + m ]
= cos 
2
 N 0 
(
)
and
j 2π
e
mn
N0
−e
j2
 2π mn  FS j
←→ comb N0 [ k + m ] − comb N0 [ k − m ]
= sin 
2
 N 0 
(
X[ k ] = −
)
j
(comb64 [k + 33] − comb64 [k − 33])
2
We can demonstrate that this solution is correct by reconstituting the signal using
x[ n ] =
∑ X[k ]e
j 2π
k = N0
kn
N0
=
∑
k = N0
kn
j 2π
j
− (comb 64 [ k + 33] − comb 64 [ k − 33])e N 0
2
Since the summation extends only over one period, N 0 = 64 , choose the simplest period,
−32 ≤ k < 32 .
kn
j 2π
j
x[ n ] = ∑ − (comb 64 [ k + 33] − comb 64 [ k − 33])e N 0
k =−32 2
31
Solutions 4-12
M. J. Roberts - 8/28/04
We must now determine for which values of k, −32 ≤ k < 32 , the comb functions are not
zero. Take the first comb function,
comb 64 [ k + 33] .
Its impulses occur whenever k + 33 is an integer multiple of 64. The only value of k in
the range, −32 ≤ k < 32 , for which that is true is k = 31. Similarly, for
comb 64 [ k − 33]
the only value for which it is non-zero is k = −31 . Therefore we can write the summation
as
−31n
j 2π
j 2π

j
j  j 2π 31n
64
x[ n ] = ∑ − (δ [ k − 31] − δ [ k + 31])e N 0 = −  e 64 − e

2

k =−32 2
kn
31
31n
j 2π
 j
j  − j 2π 3164n
 31n  
 31πn 
 31n 
x[ n ] =  e
− e 64  =  − j 2 sin 2π
  = sin 2π
 = sin



 32 
64 
64  
2
 2
But this can also be written as
64 n
33 n
33 n
 j  j 2π 3364n
j 2π
− j 2π
− j 2π

j  − j 2π 6464n j 2π 3364n
64
64
64
x[ n ] =  e12
e
−
e
e
=
e
−
e



4
4
3
123
2  =1

 2
=1
j
 33n  
 33πn 
x[ n ] =  j 2 sin 2π
  = − sin


 32 
64  
2
If these two results are to both be correct
 31πn 
 33πn 
sin
 = − sin

 32 
 32 
for any integer value of n. We can write
 33πn 
 33πn 
 31πn − 64πn 
 31πn 
 31πn

− 2πn = sin
sin
 = sin −
 = − sin

 = sin


 32 
 32 
 32 
 32

32
Solutions 4-13
M. J. Roberts - 8/28/04
proving that the two expressions are indeed equivalent for integer values of n.
(g)
x[ n ] = rect 5 [ n ] ∗ comb11[ n ]
 k

2W + 1
drcl 
, 2W + 1
N0
 N0

 n

, 2 m + 1 = comb 2 m +1 [ n ] and the fact that
And, from Appendix A, drcl
 2m + 1

2W + 1 = N 0 ,
X[ k ] = comb11[ k ]
FS
Using rectW [ n ] ∗ comb N0 [ n ] ←
→
FS
→ comb N0 [ k ] because x[ n ] = rect 5 [ n ] ∗ comb11[ n ] = 1
Agrees with 1←
FS
→ comb N0 [ k ] , N 0 can be arbitrarily chosen. The meaning of
and in 1←
the result is the same regardless of the choice of N 0 .
This result was obtained for a period of N 0 = 11. But when the function is a
constant, we can choose any period we like and get an equivalent result (because a
constant repeats exactly in any “period” you choose). For example, if we let N 0 = 4 the
FS
→ comb N0 [ k ] , yields X[ k ] = comb 4 [ k ] . Then, reconstituting the
transform pair, 1←
signal from its DTFS,
x[ n ] =
∑ X[k ]e
j 2π
kn
N0
=
k = N0
∑
k = N0
comb 4 [ k ]e
j 2π
kn
N0
Summing over the period, −2 ≤ k < 2 , yields
x[ n ] =
1
∑ comb [k ]e
k =−2
j 2π
kn
4
4
=e
j 2π
( 0) n
4
=1 .
If we chose the period, 2 ≤ k < 6 we would get
5
x[ n ] = ∑ comb 4 [ k ]e
j 2π
kn
4
=e
j 2π
( 4) n
4
=1 ,
k =2
which is exactly the same result. In general, for any choice of period and any range of k
covering one period,
x[ n ] =
k0 + N 0 −1
∑
k = k0
comb N 0 [ k ]e
j 2π
kn
N0
Solutions 4-14
=e
j 2π
qN 0 n
N0
=1
M. J. Roberts - 8/28/04
where the integer, q, lies in the range, k0 ≤ q < k0 + N 0 .
(h)
x[ n ] = rect 2 [ n ] ∗ comb 21[ n − 3] , N F = 21
17. Find the DTFS harmonic function of
x[ n ] =
n
∑ comb [m] − comb [m − 1]
3
m =−∞
3
with N F = N 0 = 3 .
18. Find the average signal power of
x[ n ] = rect 4 [ n ] ∗ comb 20 [ n ]
directly in the DT domain and then find its harmonic function, X[ k ], and find the signal
power in the “k” domain and show that they are the same.
In the DT domain:
Px =
9
20
In the “k” domain:
Do the summation numerically in MATLAB to get Px =
9
.
20
19. Using the frequency shifting property of the DTFS find the DT-domain signal, x[ n ] ,
corresponding to the harmonic function,
X[ k ] =
7
 k − 16 
drcl
, 7 .
 32

32
This frequency shifting causes the sign of the DT-domain function to alternate.
20. Find the DTFS harmonic function for
x[ n ] = rect 3 [ n ] ∗ comb 8 [ n ]
Solutions 4-15
M. J. Roberts - 8/28/04
with the representation period, N F = 8 .
representation,
Then, using MATLAB, plot the DTFS
7
x F [ n ] = ∑ X[ k ]e
j 2π
kn
8
k =0
over the DT range, −8 ≤ n < 8 . For comparison, plot the function,
xF 2[n] =
20
∑ X[k ]e
j 2π
kn
8
k =13
over the same range. The plots should be identical.
21. A periodic signal, x(t), with a period of 4 seconds is described over one period by
x( t) = 3 − t , 0 < t < 4 .
Plot the signal and find its trigonometric CTFS description. Then plot on the
same scale approximations to the signal, x N ( t) , given by
x N ( t) = X c [0] + ∑ X c [ k ] cos(2π ( kf F ) t) + X s[ k ] sin(2π ( kf F ) t)
N
k =1
for N = 1, 2 and 3 . (In each case the time scale of the plot should cover at least two
periods of the original signal.)
Find the trigonometric harmonics series by direct integration using
1
Xc [ 0 ] =
T0
n
,
∫ x (t ) cos ( 2π ( kf ) t ) dt
0
t0
2
Xs [ k ] =
T0
∫x
∫ x (t ) dt
t0
t 0 + T0
2
Xc [ k ] =
T0
Use
t 0 + T0
t 0 + T0
∫ x (t ) sin ( 2π ( kf ) t ) dt
0
t0
cos( x ) dx = x n sin( x ) − n ∫ x n −1 sin( x ) dx ,
or
Solutions 4-16
M. J. Roberts - 8/28/04
∫x
n
sin ( x ) dx = − x n cos ( x ) + n ∫ x n −1 cos ( x ) dx
from Appendix A, and make the change of variable
λ=
πk
t
2
to finish the integrals.
22. A periodic signal, x(t), with a period of 2 seconds is described over one period by
1

sin(2πt) , t < 2
.
x( t) = 
1
0
, < t <1

2
Plot the signal and find its complex CTFS description. Then plot on the same scale
approximations to the signal, x N ( t) , given by
x N ( t) =
N
∑ X[k ]e
j 2π ( kf F ) t
k =− N
for N = 1, 2 and 3 . (In each case the time scale of the plot should cover at least two
periods of the original signal.)
23. Find and plot two periods of the complex CTFS description of cos(2πt)
and
(a)
Over the interval, 0 < t < 1,
(b)
Over the interval, 0 < t < 1.5 .
(a)
The period of cos(2πt) is 1, therefore the complex Fourier series description is
simply the exponential description of the cosine function
e j 2πt + e − j 2πt
.
cos(2πt) =
2
That is,
1
X[1] = X[−1] = and X[ k ] = 0 , k ≠ 1 .
2
Solutions 4-17
M. J. Roberts - 8/28/04
The plot is simply the plot of two periods of the cosine because the complex Fourier
series description is exact everywhere.
(b)
TF =
1
X[ k ] =
TF
3
2
and f F =
2
3
t 0 + TF
∫ x (t ) e
− j 2 π ( kfF ) t
dt
t0

   2k  
  2k  

 sin 3π  − 1  sin 3π  + 1 
3
3

+ 

 
 2k 
 2k 


2π  + 1
2π  − 1






3
3
1
X[ k ] = 

3
  2k   
  2k  
cos3π  − 1  − 1
cos3π  + 1  − 1

 
 3

  3

+j

+ j
 2k 
 2k 


2π  − 1
2π  + 1


 3

 3

24. Using MATLAB, plot the following signals over the time range, −3 < t < 3 .
(a)
x 0 ( t) = 1
(b)
x1 ( t) = x 0 ( t) + 2 cos(2πt)
(c)
x 2 ( t) = x1 ( t) + 2 cos( 4πt)
(d)
x 20 ( t) = x19 ( t) + 2 cos( 40πt)
For each part, (a) through (d), numerically evaluate the area of the signal over the time
1
1
range, − < t < .
2
2
The “Nth” member of this sequence of functions can be represented by the general form
x N ( t) = x N −1 ( t) + 2 cos(2πNt) .
Based on what you observed in parts (a) through (d) what is the limit as “N” approaches
infinity? Find the trigonometric Fourier series expression for the unit comb function and
compare it to this result.
Solutions 4-18
M. J. Roberts - 8/28/04
Successive Cumulative Sums of Higher Harmonic Cosines
so(t), N=1
5
0
so(t), N=2
-5
-3
5
so(t), N=3
-1
0
1
2
3
-2
-1
0
1
2
3
-2
-1
0
1
2
3
-2
-1
0
Time, t (s)
1
2
3
0
-5
-3
10
0
-10
-3
50
so(t), N=20
-2
0
-50
-3
25. Using the CTFS table of transforms and the CTFS properties, find the CTFS
harmonic function of each of these periodic signals using the time interval, TF ,
indicated.
(a)
 
x( t) = 3 rect  2 t −
 
1
  ∗ comb( t) , TF = 1
4
Remember:
If
Then
And
g( t) = g 0 ( t) ∗ δ ( t)
g( t − t0 ) = g 0 ( t − t0 ) ∗ δ ( t) = g 0 ( t) ∗ δ ( t − t0 )
g( t − t0 ) ≠ g 0 ( t − t0 ) ∗ δ ( t − t0 ) = g( t − 2 t0 )
(b)
1
 t
x( t) = 5[tri( t − 1) − tri( t + 1)] ∗ comb  , TF = 4
 4
4
 t  FS w
w
t 1
Using tri   ∗ comb   ←
→ sinc 2 
 w  T0
T0
 T0 
 T0

k

, with w = 1 and T
0
1
 t  FS 1
 k
tri ( t ) ∗ comb   ←
→ sinc 2  
 4
 4
4
4
FS
→e
Then using the time-shifting property, x ( t − t 0 ) ←
πk
−j
1
 t  FS 1
2 k
tri ( t − 1) ∗ comb   ←→ sinc   e 2
 4
 4
4
4
Solutions 4-19
− j 2 π ( kf0 ) t 0
X[ k ] ,
= 4,
M. J. Roberts - 8/28/04
and
πk
1
 t  FS 1
 k j
tri ( t + 1) ∗ comb   ←
→ sinc 2   e 2
 4
 4
4
4
Then, using linearity,
πk
j
1
 − j π2k
1
 k
 t  FS
5  tri ( t − 1) −tri ( t + 1)  ∗ comb   ←→ 5  e
− e 2  sinc 2  
 4


4
4
4

or
1
5  πk
 k
 t  FS
5  tri ( t − 1) −tri ( t + 1)  ∗ comb   ←
→ − j sin   sinc 2  
 4
 4
4
2  2 
(c)
x( t) = 3 sin(6πt) + 4 cos(8πt) , TF = 1
(d)
x( t) = 2 cos(24πt) − 8 cos( 30πt) + 6 sin( 36πt) , TF = 2
(e)
x( t) =
t


1 
∫ comb(λ ) − comb λ − 2 dλ
, TF = 1
−∞
1
50
(f)
x( t) = 4 cos(100πt) sin(1000πt) , TF =
(g)

 t
 t  
 t
 t
x( t) = 14 rect   ∗ comb   ⊗ 7 rect   ∗ comb   , TF = 24

 8 





8
12  
5

Let
 t
 t
x1 ( t) = rect   ∗ comb 
 8
 12 
and
 t
 t
x 2 ( t) = rect   ∗ comb 
 5
 8
Then, using
 t  FS w
w
t 1
→ sinc 
rect   ∗ comb   ←
 w  T0
T0
 T0 
 T0
x( t) = 98 x1 ( t) ⊗ x 2 ( t)
 2k 
X1[ k ] = 8 sinc 
 3
, TF = 12
Solutions 4-20

k

,
M. J. Roberts - 8/28/04
k

 k
8 sinc  , an integer
3
2
X1[ k ] = 
0
, otherwise

 5k 
X 2 [ k ] = 5 sinc 
 8
, TF = 24
, TF = 8
k

 5k 
5 sinc  , an integer
24
3
X 2[k ] = 
0
, otherwise

, TF = 24
k

 k
 5k 
40 sinc  sinc  , an integer
3
24
6
X[ k ] = 98 × 24 
0
, otherwise

, TF = 24
All the values of this function are zero except when k = 0 . Then
X[ k ] = 94080δ [ k ]
, TF = 24 .
(This result indicates that the periodic convolution of the two signals is a constant, 94080.
That can be confirmed by graphically periodically convolving the two signals. That is by
finding the area under the product over one cycle and observing that, as one signal is
shifted, the area does not change. Signal one is a periodic sequence of pulses of height
168, width 8 and period 12. Signal two is a periodic sequence of pulses of height 56,
width 5 and period 8. The overall overlap width is a constant 10. That product is
168 × 56 × 10 = 94080 .)
(h)

 t
 t 
 t
 t 
x( t) = 8 rect   ∗ comb   ⊗ −2 rect   ∗ comb   , TF = 20
 2
 5  
 6
 20  

k

 k
 3k 
an integer
sinc  sinc  ,
10
20
4
X[ k ] = −192
0
, otherwise

26. A signal, x( t) , is described over one period by
Solutions 4-21
, TF = 20
M. J. Roberts - 8/28/04
T0

−
A
,
−
<t<0

2
.
x( t) = 
T
0
 A, 0< t<

2
Find its complex CTFS and then, using the integration property find the CTFS of its
integral and plot the resulting CTFS representation of the integral.
Find its harmonic function using the integral definition. You should get
X [ k ] = jA
cos ( kπ ) − 1
kπ
Then apply the integration property. Verify that the harmonic is correct at k = 0 . (You
will need to apply L’Hôpital’s rule.)
27. In some types of communication systems binary data are transmitted using a
technique called binary phase-shift keying (BPSK) in which a “1” is represented by a
“burst” of a CT sine wave and a “0” is represented by a burst which is the exact
negative of the burst that represents a “1”. Let the sine frequency be 1 MHz and let
the burst width be 10 periods of the sine wave. Find and plot the CTFS harmonic
function for a periodic binary signal consisting of alternating “1’s” and “0’s” using its
fundamental period as the representation period.
The signal can be represented by
[
]
x( t) = sin(2 × 10 6 πt) 2 rect (10 5 t) ∗ 5 × 10 4 comb(5 × 10 4 t) − 1
Using mulitplication-convolution duality,
X[ k ] =
j


 k
comb[ k + 20] − comb[ k − 20]) ⊗ 10 5 × 10 −5 sinc  − δ [ k ]
(
 2


2
Remember that a periodic convolution is the same as the aperiodic convolution of one
period of one of the periodic signals with the entire other signal. Choose one period of
the difference of two combs.
Solutions 4-22
M. J. Roberts - 8/28/04
28. Using the DTFS table of transforms and the DTFS properties, find the harmonic
function of each of these periodic signals using the representation period, N F ,
indicated.
(a)
x[ n ] = e
−j
2πn
16
⊗ comb 24 [ n ] , N F = 48
From the table,
e
j
2π n
16
FS
←
→ comb16 [ k − 1] , N F = 16
Therefore, using the change of period property,
e
−j
2π n
16
k

k

an integer
comb16  − 1 ,
←→ 
, N F = 48
3
3 
0
, otherwise

FS
Also, from the table,
N F = mN 0
j
e
2π n
N0
FS
←
→ comb N F [ k − m ]
48 = 3 × 16
e
j
2π n
16
FS
←
→ comb 48 [ k − 3]
k 
The function, comb16  − 1 , is a sequence of impulses. The impulses occur wherever
3 
k
− 1 = 16 m , m any integer. Rearranging, the impulses occur where k − 3 = 48 m . This
3
describes a comb of the form, comb 48 [ k − 3] . Therefore the two answers agree.
From the table,
FS
comb N0 [ n ] ←
→
1
, N F = 24
N0
Therefore, using the time scaling property,
Solutions 4-23
M. J. Roberts - 8/28/04
k

1
an integer 
,

comb N0 [ n ] ←→  24
2
 , N F = 48
0 , otherwise 
FS
Then, since the two DTFS’s are both done with reference to the same representation
period, using the multiplication-convolution duality property,
k
k
 1


an integer  
,
an integer 
comb 48 [ k − 3] ,
X[ k ] = 2 
2
3
  24

0



, otherwise
0
,
otherwise

The non-zero impulses in the first harmonic function occur at values of k for which k + 3
is an integer multiple of 48. Therefore all these k’s must be odd. The values of k for
which the second harmonic function is non-zero are all even. Therefore
X[ k ] = 0 , for all k.
−j
This result implies that the original DT function, x[ n ] = e
Show that to be true by actually doing the convolution directly.
(b)
(c)
2πn
16
⊗ comb 24 [ n ] , is zero.
 2πn 
x[ n ] = (rect 5 [ n ] ∗ comb 24 [ n ]) sin
 , N F = 24
 6 
 n
x[ n ] = x1[ n ] − x1[ n − 1] where x1[ n ] = tri  ∗ comb 20 [ n ] , N F = 20
 8
29. Find the signal power of
 14πn 
 26πn 
x[ n ] = 5 sin
 − 8 cos
 .
 15 
 30 
Use Parseval’s theorem to find the signal power from the harmonic function.
3 0F i .n d
the
DTFS
harmonic
x[ n ] = (rect1[ n − 1] − rect1[ n − 4 ]) ∗ comb 6 [ n ].
xN [n] =
N
∑
k =− N
X[ k ]e
j
πnk
3
X[ k ],
partial
function,
Then plot the
5
, for N = 0,1, 2 and then the total sum, x[ n ] = ∑ X[ k ]e
k =0
x[ n ] = rect1[ n − 1] ∗ comb 6 [ n ] − rect1[ n − 4 ] ∗ comb 6 [ n ]
Solutions 4-24
j
πnk
3
of
sum,
.
M. J. Roberts - 8/28/04
Using rect NW

kπ 
sin  ( 2 NW + 1) 
N0 

1
FS
→
[ n ] ∗ comb N0 [ n ] ←
N0
 kπ 
sin  
 N0 
 kπ 
 kπ 
 kπ 
sin 
sin 
sin 
4 πk
4 πk
πk
−
j
−
j
−j

1  2
1  2
1  2   − j π3k
3
3
X[ k ] =
−
=
−e 3 
e
e
e

6  kπ 
6  kπ 
6  kπ  

sin 
sin 
sin 
 6
 6
 6
X[ k ] =
1
e
6
5π k
−j
6
 kπ 
2  kπ 
sin  
πk
πk
5 π k sin 
−j
 j −j 6
 2 
 2  j2
2
e −e
 = 3 e
 kπ  
 kπ 
sin  
sin  
 6 
 6 
 kπ 
2  kπ 
sin 2  
 πk 5
5πk
πnk
N sin 
 2  −j 6 j 3
 2  j 3  n − 2 
j
j
xN [n] = ∑
e
e
= ∑
e
3 k =− N
3 k =− N
 kπ 
 kπ 
sin 
sin 
 6
 6
N
N=0
x[n]
2
-12
12
n
12
n
12
n
12
n
-2
N=1
x[n]
2
-12
-2
N=2
x[n]
2
-12
-2
Total Sum from 0 to 5
x[n]
2
-12
-2
31. Find and sketch the magnitude and phase of the DTFS harmonic function of
Solutions 4-25
M. J. Roberts - 8/28/04
 2π 
 2π 
x[ n ] = 4 cos n + 3 sin n
 7 
 3 
which is valid for all discrete-time.
The least common period of these two signals is N 0 = 21. Use the tables and the
change-of-period property of the DTFS, to find the DTFS harmonic function.
X[ k ] = 2(comb 21[ k − 3] + comb 21[ k + 3]) + j
3
(comb21[k + 7] − comb21[k − 7])
2
|X[k]|
2
-21
21
k
Phase of X[k]
π
-21
21
k
-π
32. The sun shining on the earth is a system in which the radiant power from the sun is
the excitation and the atmospheric temperature (among many other things) is a
response. A simplified model of the radiant power falling on a typical mid-latitude
location in North America is that it is periodic with a fundamental period of one year
and that every day the radiant power of the sunlight rises linearly from the time the
sun rises until the sun is at its zenith then falls linearly until the sun sets. The earth
absorbs and stores the radiant energy and re-radiates some of the energy into space
every night. To keep the model of the excitation as simple as possible assume that
the energy loss every night can be modeled as a continuation of the daily linear
radiant power pattern except negative at night. There is also a variation with the
seasons caused by the tilt of the earth’s axis of rotation. This causes this linear riseand-fall pattern to rise and fall sinusoidally on a much longer time scale as illustrated
below.
Solutions 4-26
M. J. Roberts - 8/28/04
Radiant Power
June 21
2
1.75
Day
-1
(a)
365
-0.75
2
December 21
Write a mathematical description of the radiant power from the sun.
1  1  2πt 

p( t) = 2 tri(2 t) ∗ comb( t) −  + sin

2  4  365 

(b)
Assume that the earth is a first-order system with a time constant of 0.16 years.
What day of the year should be the hottest according to this simplified model?
The differential equation for the rate of heat flow into the earth is
K T( t)
d
K T( t)) = p( t) −
(
dt
58.4
where K is a proportionality constant relating the temperature of the earth to its stored
energy (t in days). Since the radiant power striking the earth is periodic the temperature
is also periodic and both can be expressed in a CTFS with a representation period of 365
days. Find the harmonic function for the radiant power, P[k]. The temperature must also
be periodic with the same fundamental period as the radiant power. Solve for the
harmonic function of the temperature, T[k], by substituting P[k] into the differential
equation. Then simplify it by realizing that the higher order harmonic content in T[k] is
negligible compared to the fundamental. Then find the time-domain temperature
expression, T(t), and find the time at which it reaches a maximum.

1
k

2 k
an integer  j
sinc   − δ [ k ] ,
2
2
365

 + (δ [ k + 1] − δ [ k − 1])

 8
, otherwise
1 0

Τ[ k ] =
2
π
k
K
+ 0.01712
j
365
Solutions 4-27
M. J. Roberts - 8/28/04
 2π t 
 2π t 
+ 0.03424 sin 
0.0344 cos 

 365 
 365 
1
1
Τ (t ) ≅
−
2
0.03424 K
8K
2
 2π 

 + ( 0.01712 )
365
The maximum temperature occurs at about
t = 45.625 .
So the hottest day should occur about 46 days after the summer solstice or about August
6.
33. The speed and timing of computer computations are controlled by a clock. The clock
is a periodic sequence of rectangular pulses, typically with 50% duty cycle. One
problem in the design of computer circuit boards is that the clock signal can interfere
with other signals on the board by being coupled into adjacent circuits through stray
capacitance. Let the computer clock be modeled by a square-wave voltage source
alternating between 0.4 and 1.6 V at a frequency of 2 GHz and let the coupling into
an adjacent circuit be modeled by a series combination of a 0.1 pF capacitor and a 50
Ω resistance. Find and plot over two fundamental periods the voltage across the 50
Ω resistance.
Describe the excitation as a rectangle convolved with a comb, plus a constant. Find its
harmonic function. Put that into the differential equation for the response voltage,
RC[ v′i ( t) − v′o ( t)] = v o ( t)
 k
j 2.4 × 10 9 kπ sinc 
 2
and solve for the harmonic function of the response, Vo [ k ] =
.
1
+ j 4 × 10 9 kπ
RC
Then plot the response voltage by computing its CTFS. It should look like the graph
below.
Solutions 4-28
M. J. Roberts - 8/28/04
v (t)
o
1
-6.25e-10
6.25e-10
t
-1
34. Find and plot versus F the magnitude of the response, y[ n ] , to the periodic excitation,
x[ n ] = cos(2πFn ) ,
in the system below.
x[n]
y[n]
0.9
D
y[ n ] = x[ n ] + 0.9 y[ n − 1]
or
y[ n ] − 0.9 y[ n − 1] = x[ n ] .
The final graph should look like the graph below.
Amplitude
12
F
1
1
This is a lowpass DT filter.
Solutions 4-29