Coding and transformation
EE040-4-3-CT
(Individual Assignment)
Student Name: Maxime Frédéric FABRE
Student Number/ID: TP029408
Submission Date: 08/11/2012
Maxime FABRE
TP029408
Coding and transformation
Individual Assignment
EE020-4-3-NP
UCFEFREI1209
Exercice 1 :
The Linear convolution of two sequences x(n) & h(n) is given by,
y(n) = x(n) * h(n)

=
 x(k ).h(n  k )
k  
The circular convolution of two sequences x(n) & h(n) is defined by,
N 1
y(m) =
 x (k ).h ((m  k ))
k 0
N
Where x(n) – input sequence; h(n) – impulse response; y(n), y(m) – output sequence.
Given: x(n) = [1,1,1,2,1] and h(n) = [1,1,2,1]
So, as we learn during the lectures, the linear convolution output function is defined as :
𝒀(𝒏) = 𝒙(𝒏) × 𝒉(𝒏)
Where x(n) and h(n) are the linear input sequences. The linear convolution can be done using
two different methods:

Tabulation Method

Matrix Method
We are going to solve the problem with one of these two methods, more precisely Tabulation
method and after we will verify the result using a MATLAB program.
Page 2
Maxime FABRE
TP029408
Coding and transformation
Individual Assignment
EE020-4-3-NP
UCFEFREI1209
1) Linear convolution
We’re going to resolve the linear convolution with the tabulation method. However we could have
been done the matrix method.
𝑥(𝑛) = [1,1,1,2,1]
ℎ(𝑛) = [1,1,2,1]
h(n)
1
1
2
1
1
1
1
2
1
1
1
1
2
1
2
2
2
4
2
1
1
1
2
1
x(n)
1
1
1
2
1
Then, we add the diagonals in order to have the final answer:
First Diagonal:
y(1) = 1
Second diagonal:
y(2) = 1+1 = 2
Third Diagonal:
y(3) = 1+1+2 = 4
Fourth Diagonal:
y(4) = 2+1+2+1 = 6
Fifth diagonal:
y(5) = 1+2+2+1 = 6
Sixth Diagonal:
y(6) = 1+4+1 = 6
Seventh Diagonal:
y(7) = 2+2 = 4
Eight Diagonal:
y(8) = 1
So, the output sequence y(n) = {1,2,4,6,6,6,4,1}
Let’s check the results by Matlab:
x=[1,1,1,2,1];
h=[1,1,2,1];
conv(x,h)
Result
Page 3
Maxime FABRE
TP029408
Coding and transformation
Individual Assignment
2) Circular convolution
𝟏
𝟏
(
𝟐
𝟏
𝟏
𝟏
𝟏
𝟐
𝟐
𝟏
𝟏
𝟏
𝟏
𝟐
𝟏
𝟏
𝟏
𝟏
) *
𝟐
𝟏
𝟏
𝟕
𝟏
𝟖
𝟏 = ( )
𝟖
𝟐
𝟕
(𝟏)
The output sequence y(n) = {7, 8, 8, 7}
Program Matlab:
x=[1,1,1,2,1];
h=[1,1,2,1];
cconv(x,h,4)
Results obtained by Matlab:
The execution time of linear convolution is faster than
the circular convolution.
Page 4
EE020-4-3-NP
UCFEFREI1209
Maxime FABRE
TP029408
Coding and transformation
Individual Assignment
EE020-4-3-NP
UCFEFREI1209
Exercice 2
To resolve the problem we have to put the sequence in bit reversed order.
The sequence X(n) in bit reversed order :








X(0) = 2
X(4) = 1
X(2) = 2
X(6) = 1
X(1) = 2
X(5) = 1
X(3) = 2
X(7) = 1
In order to continue the problem, we have to identify how many stages we need. As we have 8
points, we require 3 stages of computation. In the first stage we have only 𝑊20
First stage Computation:
The input sequence for the first stage is {2,1,2,1,2,1,2,1}.
A = a + b 𝑊𝑁𝑛𝑘
B = a - b 𝑊𝑁𝑛𝑘
X(0) = 2
2+1=3
X(1) = 2
2+1=3
X(4) = 1
2-1=1
X(5) = 1
2-1=1
X(2) = 2
2+1=3
X(3) = 1
2+1=3
X(6) = 1
2-1 =1
X(7) = 2
2-1=1
So, now we have this input sequence after the first stage {3,1,3,1,3,1,3,1}. According to the lecture, in
the second stage computation, we have 𝑊40 and 𝑊41 .
Page 5
Maxime FABRE
TP029408
Coding and transformation
Individual Assignment
EE020-4-3-NP
UCFEFREI1209
Second stage Computation
3
3+3=6
1
6
1
1+1(-j) =1-j
3
1-j
3
3-3 = 0
1
0
1
1 – 1(-j) = 1 + j
3
1+j
After the second stage computation, the output DFT sequence is:
{6,1-j,0,1+j,6,1-j,0,1+j}
Now, we have to take this sequence and put it as the input sequence in the third stage
computation. Moreover, the phase factors involved in third stage computation are𝑊80 , 𝑊81 , 𝑊82
and 𝑊83 .
𝑊80 = 1, 𝑊81 = 0.707 − 0.707𝑗, 𝑊82 = −𝑗, 𝑊83 = −0.707 − 0.707𝑗
Third stage Computation
6
6+6=12
1-j
(1-j) + (1-j) * (0.707-0.707j) = 1 - 2.414j
0
0+0 (-j) = 0
1+j
(1+j) + (1+j) * (-0.707-0.707j) =1 - 0.414j
6
6-6=0
1-j
(1-j) - (1-j) * (0.707-0.707j) = 1 + 0.414j
1
0-0(-j) = 0
1+j
(1+j) - (1+j) * (-0.707-0.707j) = 1 + 2.414j
So, we obtain finally the output sequence :{12, 1-2.414j, 0, 1-0.414j, 0, 1+0.414j, 0, 1+2.414j} this is
the final solution :
Page 6
Maxime FABRE
TP029408
Coding and transformation
Individual Assignment
X(n) = {12, 1-2.414j, 0, 1-0.414j, 0, 1+0.414j, 0, 1+2.414j }
Now, we have to verify these results with MATLAB program:
disp('FFT of x(n)=[2,2,2,2,1,1,1,1');
x=[2,2,2,2,1,1,1,1];
y=fft(x,8);
disp(y);
Let’s look the result obtained by Matlab.
We obtain the same results, so there is no problem
Exercice 3:
Figure Q3
Page 7
EE020-4-3-NP
UCFEFREI1209
Maxime FABRE
TP029408
Coding and transformation
Individual Assignment
EE020-4-3-NP
UCFEFREI1209
2 𝑖𝑓 0 < 𝑥 < 10
0 𝑖𝑓 10 < 𝑥 < 20
x(t) = {
∞
𝑋(𝑗𝑤) = ∫ 𝑥(𝑡)𝑒 −𝑗𝑤𝑡 𝑑𝑡
−∞
10
20
𝑋(𝑗𝑤) = ∫0 2𝑒 −𝑗𝑤𝑡 dt + ∫10 𝑥(𝑡)𝑒 −𝑗𝑤𝑡 𝑑𝑡
10
𝑋(𝑗𝑤) = ∫0 2𝑒 −𝑗𝑤𝑡 dt
𝑋(𝑗𝑤) = −
𝑋(𝑗𝑤) = −
𝑋(𝑗𝑤) = −
2
× (𝑒 −10𝑗𝑤 − 1)
𝑗𝑤
2
× (𝑒 −5𝑗𝑤 ∗ 𝑒 5𝑗𝑤 − 1)
𝑗𝑤
2
𝑒 −5𝑗𝑤 − 𝑒 5𝑗𝑤
×(
)
𝑗𝑤
𝑒 5𝑗𝑤
𝑋(𝑗𝑤) = −
𝑋(𝑗𝑤) =
2
10
× [𝑒 −𝑗𝑤 ]0
𝑗𝑤
2
× (− 𝑒 −5𝑗𝑤 + 𝑒 5𝑗𝑤 )
𝑗𝑤𝑒 5𝑗𝑤
4
𝑒 5𝑗𝑤 − 𝑒 −5𝑗𝑤
𝑋(𝑗𝑤) =
×(
𝑤𝑒 5𝑗𝑤
𝑋(𝑗𝑤) =
4
× sin(5𝑤)
𝑤𝑒 5𝑗𝑤
𝑋(𝑗𝑤) =
2𝑗
4 −5𝑗𝑤
𝑒
× sin(5𝑤)
𝑤
Program Matlab:
syms t;
syms w;
f = int((2*exp(-i*w*t)),t,0,10)
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