Homework4 Solutions
Problem 1: Given the following signal:
𝑥[𝑛] = 1 + sin (
2𝜋
2𝜋
4𝜋
𝜋
) 𝑛 + 3 sin ( ) 𝑛 + cos ( 𝑛 + )
𝑁
𝑁
𝑁
2
This signal is periodic with fundamental period N and fundamental frequency 𝜔0 =
2𝜋
𝑁
Define Fourier series of this signal, Fourier series coefficients of this signal and sketch the
plots for Real and Imaginary Parts of Fourier Series Coefficients and for Magnitude and
Phase of the same coefficients.
Solution:
a) Fourier Series:
1
1
Using sin 𝜔0 𝑛 = 2𝑗 (𝑒 𝑗𝜔0 𝑛 − 𝑒 −𝑗𝜔0 𝑛 ) and cos 𝜔0 𝑛 = 2 (𝑒 𝑗𝜔0 𝑛 + 𝑒 −𝑗𝜔0 𝑛 )
𝑥[𝑛] = 1 +
2𝜋
2𝜋
2𝜋
1 𝑗(2𝜋)𝑛
3
(𝑒 𝑁 − 𝑒 −𝑗( 𝑁 )𝑛 ) + (𝑒 𝑗( 𝑁 )𝑛 − 𝑒 −𝑗( 𝑁 )𝑛 )
2𝑗
2𝑗
4𝜋𝑛 𝜋
1 𝑗(4𝜋𝑛+𝜋)
−𝑗(
+ )
𝑁 2 )
+ (𝑒 𝑁 2 + 𝑒
2
3 1 𝑗(2𝜋)𝑛
3 1 −𝑗(2𝜋)𝑛
1 𝑗𝜋 𝑗2(2𝜋)𝑛
𝑁
𝑁
𝑥[𝑛] = 1 + ( + ) 𝑒
+ ( − )𝑒
+ ( 𝑒 2) 𝑒 𝑁
2 2𝑗
2 2𝑗
2
2𝜋
𝜋
1
−𝑗2( )𝑛
𝑁
+ ( 𝑒 −𝑗 2 ) 𝑒
2
b) Fourier series coefficients:
𝑎0 = 1
𝑎1 =
3 1 3 1
+ = − 𝑗
2 2𝑗 2 2
𝑎2 =
1
𝑗
2
𝑎𝑘 = 0
𝑎−1 =
1
𝑎−2 = − 𝑗
2
𝑓𝑜𝑟
1
𝑎3𝑁+2 = 𝑗
2
𝑘 ≠ −2, −1,0,1,2
1
𝑎−2−𝑁 = − 𝑗
2
3 1 3 1
− = + 𝑗
2 2𝑗 2 2
c) |𝒂𝒌 | Magnitude:
|𝑎0 | = 0
3
1
√10
|𝑎1 | = √( )2 + ( )2 =
2
2
2
3
1
√10
|𝑎−1 | = √( )2 + ( )2 =
2
2
2
1
1
|𝑎2 | = √( )2 =
2
2
|𝑎−2 | =
|𝑎𝑘 | = 0
𝑓𝑜𝑟
𝑘 ≠ −2, −1, 0, 1, 2
d) ≮ 𝒂𝒌 – Phase
≮ 𝑎0 = 0
1
−
1
≮ 𝑎1 = arctan ( 2 ) = −𝑎𝑟𝑐𝑡𝑎𝑛 ( )
3
3
2
≮ 𝑎−1
1
1
= arctan ( 2 ) = 𝑎𝑟𝑐𝑡𝑎𝑛 ( )
3
3
2
1
𝜋
≮ 𝑎2 = arctan ( 2 ) = 𝑎𝑟𝑐𝑡𝑎𝑛∞ =
0
2
≮ 𝑎−2 = −𝑎𝑟𝑐𝑡𝑎𝑛∞ = −
≮ 𝑎𝑘 = 0
𝑓𝑜𝑟
1
2
𝑘 ≠ −2, −1, 0, 1, 2
1
2
Problem 2: Given the following signal
𝜋
𝑥(𝑡) = 1 + 𝑠𝑖𝑛𝜔0 𝑡 + 2𝑐𝑜𝑠𝜔0 𝑡 + cos(2𝜔0 𝑡 + )
4
This signal is periodic with fundamental frequency 𝜔0 . Determine the Fourier series of this
signal, Fourier series coefficients of this signal and sketch the plots for Real and Imaginary
Parts of Fourier series coefficients and for magnitude and phase of the same coefficients.
Solution:
a) Fourier Series:
1
1
Using sin 𝜔0 𝑡 = 2𝑗 (𝑒 𝑗𝜔0 𝑡 − 𝑒 −𝑗𝜔0 𝑡 ) and cos 𝜔0 𝑡 = 2 (𝑒 𝑗𝜔0 𝑡 + 𝑒 −𝑗𝜔0 𝑡 )
𝜋
𝜋
1 𝑗𝜔 𝑡
1
[𝑒 0 − 𝑒 −𝑗𝜔0 𝑡 ] + [𝑒 𝑗𝜔0 𝑡 + 𝑒 −𝑗𝜔0 𝑡 ] + [𝑒 𝑗(2𝜔0 𝑡+ 2 ) + 𝑒 −𝑗(2𝜔0 𝑡+ 2 ) ]
2𝑗
2
1
1
1 𝜋
𝑥(𝑡) = 1 + (1 + ) 𝑒 𝑗𝜔0 𝑡 + (1 − ) 𝑒 −𝑗𝜔0 𝑡 + ( 𝑒 𝑗( 4 ) ) 𝑒 𝑗2𝜔0 𝑡
2𝑗
2𝑗
2
𝜋
1
+ ( 𝑒 −𝑗( 4 ) ) 𝑒 −𝑗2𝜔0 𝑡
2
𝑥(𝑡) = 1 +
b) Fourier series coefficients:
𝑎0 = 1
𝑎1 = 1 +
𝑎2 =
1
1
=1− 𝑗
2𝑗
2
𝑎−1 = 1 −
1 𝑗(𝜋) √2 √2
𝑒 4 =
+
𝑗
2
4
4
𝑎𝑘 = 0
𝑓𝑜𝑟
1
1
=1+ 𝑗
2𝑗
2
𝜋
1
√2 √2
𝑎−2 = − 𝑒 −𝑗( 4 ) =
−
𝑗
2
4
4
|𝑘| > 2
c) |𝒂𝒌 | Magnitude:
|𝑎0 | = 1
1
√5
|𝑎1 | = √(1)2 + (− )2 =
2
2
1
√2
√2
|𝑎2 | = √( )2 + ( )2 =
4
4
2
1
=
2
|𝑎𝑘 | = 0
d)
𝑓𝑜𝑟
≮ 𝒂𝒌 – Phase
≮ 𝑎0 = 0
|𝑘| > 2
1
√5
|𝑎−1 | = √(1)2 + ( )2 =
2
2
|𝑎−2 | = √(
√2 2
√2
) + (− )2
4
4
1
1
≮ 𝑎1 = arctan ( 2) = −𝑎𝑟𝑐𝑡𝑎𝑛 ( )
1
2
−
≮ 𝑎−1
1
1
= arctan ( 2 ) = 𝑎𝑟𝑐𝑡𝑎𝑛 ( )
1
2
√2
≮ 𝑎2 = arctan ( 4 ) = 𝑎𝑟𝑐𝑡𝑎𝑛1
√2
4
≮ 𝑎−2
√2
= arctan ( 4 ) = −𝑎𝑟𝑐𝑡𝑎𝑛1
√2
4
≮ 𝑎𝑘 = 0 ,
−
𝑓𝑜𝑟
|𝑘| > 2
Problem 3: Given the following signal
2𝜋
5𝜋
𝑥(𝑡) = 2 + cos ( 𝑡) + 4 sin ( 𝑡)
3
3
Determine the fundamental frequency 𝜔0 and Fourier series coefficients 𝑎𝑘 such that
∞
𝑥(𝑡) = ∑ 𝑎𝑘 𝑒 𝑗𝑘𝜔0 𝑡
𝑘=−∞
Solution:
a) Fourier Series:
1
1
Using sin 𝜔0 𝑡 = 2𝑗 (𝑒 𝑗𝜔0 𝑡 − 𝑒 −𝑗𝜔0 𝑡 ) and cos 𝜔0 𝑡 = 2 (𝑒 𝑗𝜔0 𝑡 + 𝑒 −𝑗𝜔0 𝑡 )
2𝜋
2𝜋
5𝜋
5𝜋
1
4
𝑥(𝑡) = 2 + (𝑒 𝑗( 3 )𝑡 + 𝑒 −𝑗( 3 )𝑡 ) + (𝑒 𝑗( 3 )𝑡 − 𝑒 −𝑗( 3 )𝑡 )
2
2𝑗
𝑥(𝑡) = 2 +
2𝜋
2𝜋
1 𝑗(2𝜋)𝑡 1 −𝑗(2𝜋)𝑡
3
𝑒 3 + 𝑒
− 2𝑗𝑒 𝑗5( 6 )𝑡 + 2𝑗𝑒 −𝑗5( 6 )𝑡
2
2
b) Fundamental Frequency
𝜔0 =
2𝜋 𝜋
=
6
6
c) Fourier coefficients
The non-zero Fourier coefficients of 𝑥(𝑡) are:
1
∗
𝑎0 = 2; 𝑎2 = 𝑎−2 = ; 𝑎5 = 𝑎−5
= −2𝑗
2
Problem 4:
𝑗𝜋
𝑗𝜋
∗
∗
Given 𝑎0 = 2; 𝑎2 = 𝑎−2
= 2𝑒 6 ; 𝑎4 = 𝑎−4
= 𝑒 3 of a discrete-time periodic signal 𝑥[𝑛]
which is real valued and has a fundamental period 𝑁 = 5.
Express 𝑥[𝑛]in the form
∞
𝑥[𝑛] = 𝐴0 + ∑ 𝐴𝑘 sin(𝜔𝑘 𝑛 + 𝜑𝑘 )
𝑘=1
Solution:
Using Fourier Series Synthesis Equation
2𝜋
2𝜋
2𝜋
2𝜋
𝑥[𝑛] = 𝑎0 + 𝑎2 𝑒 𝑗2( 𝑁 )𝑛 + 𝑎−2 𝑒 −𝑗2( 𝑁 )𝑛 + 𝑎4 𝑒 𝑗4( 𝑁 )𝑛 + 𝑎−4 𝑒 −𝑗4( 𝑁 )𝑛
𝜋
4𝜋
𝜋
4𝜋
𝜋
8𝜋
𝜋
8𝜋
𝑥[𝑛] = 2 + 2𝑒 𝑗( 6 ) 𝑒 𝑗( 5 )𝑛 + 2𝑒 −𝑗(6 ) 𝑒 −𝑗( 5 )𝑛 + 𝑒 𝑗 3 𝑒 𝑗( 5 )𝑛 + 𝑒 −𝑗 3 𝑒 −𝑗( 5 )𝑛
4𝜋
𝜋
8𝜋
𝜋
𝑥[𝑛] = 2 + 4 cos [( ) 𝑛 + ] + 2 cos [( ) 𝑛 + ]
5
6
5
3
𝑥[𝑛] = 2 + 4 sin [(
4𝜋
2𝜋
8𝜋
5𝜋
) 𝑛 + ( )] + 2 sin [( ) 𝑛 + ( )]
5
3
5
6
Problem 5:
Each of the two sequences 𝑥1 [𝑛] and 𝑥2 [𝑛] has a period 𝑁 = 4 and the corresponding
Fourier coefficients are specified as
𝑥1 [𝑛] ⇒ 𝑎𝑘 ,
𝑥2 [𝑛] ⇒ 𝑏𝑘
Where
1
1
𝑎0 = 𝑎3 = 𝑎1 = 𝑎2 = 1 𝑎𝑛𝑑 𝑏0 = 𝑏1 = 𝑏2 = 𝑏3 = 1
2
2
Using multiplication property, determine the Fourier Series coefficients 𝑐𝑘 for the signal
𝑔[𝑛] = 𝑥1 [𝑛]𝑥2 [𝑛].
Solution:
Using multiplication property we have
3
𝑥1 [𝑛]𝑥2 [𝑛] ⟺ 𝐹𝑆 ⟺
∑ 𝑎𝜄 𝑏𝑘−𝑙 = ∑ 𝑎𝜄 𝑏𝑘−𝑙
𝑙= <𝑁>
𝑙=0
= 𝑎0 𝑏𝑘 + 𝑎1 𝑏𝑘−1 + 𝑎2 𝑏𝑘−2 + 𝑎3 𝑏𝑘−3 = 𝑏𝑘 + 2𝑏𝑘−1 + 2𝑏𝑘−2 + 𝑏𝑘−3 = 6
So, 𝑥1 [𝑛]𝑥2 [𝑛] ⟺ 𝐹𝑆 ⟺ 𝑐𝑘 = 6 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘.
Problem 6:
Given
𝑥(𝑡) = cos(4𝑡) + cos(7𝑡)
and
Represent 𝑥(𝑡) and 𝑦(𝑡) using complex exponentials
Solution:
For given 𝑥(𝑡) = cos(4𝑡) + cos(7𝑡) we can write
𝑦(𝑡) = cos(4(𝑡 − 3)) + cos(7(𝑡 − 3))
𝑦(𝑡) = 𝑥(𝑡 − 3)
Using Euler’s relationship:
1
1
1
1
𝑥(𝑡) = 𝑒 𝑗4𝑡 + 𝑒 −𝑗4𝑡 + 𝑒 𝑗7𝑡 + 𝑒 −𝑗7𝑡
2
2
2
2
𝑦(𝑡) =
1 −𝑗12 𝑗4𝑡 1 𝑗12 −𝑗4𝑡 1 −𝑗21 𝑗7𝑡 1 𝑗21 −𝑗7𝑡
𝑒
𝑒 + 𝑒 𝑒
+ 𝑒
𝑒 + 𝑒 𝑒
2
2
2
2
𝑦(𝑡) =
1 𝑗4(𝑡−3) 1 −𝑗4(𝑡−3) 1 𝑗7(𝑡−3) 1 −𝑗7(𝑡−3)
𝑒
+ 𝑒
+ 𝑒
+ 𝑒
2
2
2
2
𝑦(𝑡) = cos(4(𝑡 − 3)) + cos(7(𝑡 − 3))
Problem7:
Given
𝑥[𝑛] = cos [
2𝜋𝑛
𝑁
]
ℎ[𝑛] = 𝛼 𝑛 𝑈[𝑛] ,
and
−1<𝛼 <1
Determine 𝑦[𝑛]using DT Fourier Series.
Solution:
Let us write 𝑥[𝑛] in complex exponentials form:
𝑥[𝑛] =
1 𝑗(2𝜋)𝑛 1 −𝑗(2𝜋)𝑛
𝑒 𝑁 + 𝑒 𝑁
2
2
We know that
∞
𝐻(𝑒
𝑗𝜔
∞
) = ∑ ℎ[𝑛]𝑒
−𝑗𝜔𝑛
∞
𝑛 −𝑗𝜔𝑛
= ∑𝛼 𝑒
𝑛=0
= ∑(𝑎𝑒 −𝑗𝜔 )𝑛 =
𝑛=0
𝑛=0
Again we know that
∞
2𝜋
2𝜋
𝑦[𝑛] = ∑ 𝑎𝑘 𝐻 (𝑒 𝑗𝑘 𝑁 ) 𝑒 𝑗𝑘 𝑁 𝑛
𝑘=<𝑁>
1
1
From 𝑥[𝑛]we found that 𝑎1 = 2 , 𝑎−1 = 2
2𝜋
1
2𝜋
1
2𝜋
2𝜋
So , 𝑦[𝑛] = 2 𝐻 (𝑒 𝑗 𝑁 ) 𝑒 𝑗( 𝑁 )𝑛 + 2 𝐻 (𝑒 −𝑗 𝑁 ) 𝑒 −𝑗( 𝑁 )𝑛
1
𝑦[𝑛] = (
2
For,
1
1 − 𝛼𝑒
1
1−𝛼𝑒
−𝑗
2𝜋
𝑁
2𝜋
1
𝑗( )𝑛
𝑁
)𝑒
+ (
2𝜋
2
−𝑗
𝑁
= 𝑟𝑒 𝑗𝜃
1
2𝜋
−𝑗( )𝑛
𝑁
2𝜋 )𝑒
1 − 𝛼𝑒 𝑗 𝑁
1
1 − 𝛼𝑒 −𝑗𝜔
We get 𝑦[𝑛] = 𝑟𝑐𝑜𝑠(
2𝜋
𝑁
𝑛 + 𝜃)
Problem8:
Given
𝑥(𝑡) = cos(4𝜋𝑡) , 𝑦(𝑡) = sin(4𝜋𝑡) , 𝑧(𝑡) = 𝑥(𝑡)𝑦(𝑡)
a) Determine the Fourier series coefficients of 𝑥(𝑡).
b) Determine the Fourier series coefficients of 𝑦(𝑡).
c) Use results of parts (a) and (b) along with the multiplication property of the CT
Fourier series, to determine the Fourier coefficients of 𝑧(𝑡) = 𝑥(𝑡)𝑦(𝑡).
d) Determine the Fourier series coefficients of 𝑧(𝑡) in trigonometric form, and compare
your results with that of part (c).
Solution:
a) Using Euler Relation
𝑥(𝑡) =
1 𝑗4𝜋𝑡 1 −𝑗4𝜋𝑡
𝑒
+ 𝑒
2
2
1
The non-zero Fourier series coefficients of 𝑥(𝑡) are 𝑎1 = 𝑎−1 = 2
b) Using Euler Relation
𝑦(𝑡) =
1 𝑗4𝜋𝑡 1 −𝑗4𝜋𝑡
𝑒
− 𝑒
2𝑗
2𝑗
1
∗
The non-zero Fourier series coefficients of 𝑥(𝑡) are 𝑏1 = 𝑏−1
= 2𝑗
c) Using the multiplication property we know that
𝑧(𝑡) = 𝑥(𝑡)𝑦(𝑡) ↔ 𝐹𝑆 ↔ 𝑐𝑘 = ∑ 𝑎𝜄 𝑏𝑘−𝑒
𝑙= −∞
Therefore,
𝑐𝑘 = 𝑎𝑘 ∗ 𝑏𝑘 =
1
1
𝛿[𝑘 − 2] − 𝛿[𝑘 + 2]
4𝑗
4𝑗
1
d) We have 𝑧(𝑡) = 𝑥(𝑡)𝑦(𝑡) = cos(4𝜋𝑡) sin(4𝜋𝑡) = 2 sin(8𝜋𝑡)
Using Euler Relation:
1 𝑗8𝑡 1 −𝑗8𝑡
𝑒 − 𝑒
4𝑗
4𝑗
Therefore, the non-zero Fourier series coefficients of 𝑧(𝑡) are:
1
∗
𝑐2 = 𝑐−2
=
4𝑗
𝑧(𝑡) =