Discrete Mathematics Assignment: Detailed Task Solutions

This document presents a comprehensive solution to a Discrete Mathematics assignment, covering various tasks related to signal processing, Discrete Fourier Transform (DFT), and L-norms. The solutions include detailed calculations and explanations for each task, such as calculating DFTs, analy...Read More

DISCRETE MATHEMATICS

Table of Contents
Task - 4.1.........................................................................................................................................2
Task – 4.2.........................................................................................................................................3
Task - 4.3........................................................................................................................................3
Task - 4.4.........................................................................................................................................4
Task - 4.5.........................................................................................................................................4
Task - 4.6.........................................................................................................................................5
Task – 4.7.........................................................................................................................................6
Task - 4.8.........................................................................................................................................7
Task - 4.9.........................................................................................................................................8
Task - 4.10.......................................................................................................................................8
Task - 4.11.......................................................................................................................................9
2

Task - 4.1
x(n) =cos (2πkn/N+π/4), n=0, to n=N-1
X(0) =cos(2π(0)(0)/1 + π/4)
=cos (2π(0) + π/4)
=Cos(0 + π/4)
=cos (π/4)
x(0) =0.70710
x(1) = cos(2πkn/N + π/4)
=COS (2π(1)(1)/2 + π/4)
=COS (2π/2 +π/4)
=-0.70710
x(2) = cos(2πkn/N + π/4)
=cos(2π(2)(2)/3 + π/4)
= cos(8π/3 + π/4)
=-0.96590
x(3) =cos (2πkn/N +π/4)
=cos (2π(3)(3/4) +π/4)
=cos (18π/4 +π/4)
=-70670
x(N-1) = cos(2π(n)(N-1)/n + π/4
=cos (2πnN-2πnN + π/4)
(a) x(n)={0.70710,-0.70710,-0.96590,-0.70670…….cos (2π(n)(N-1)/n +π/4}
(b) ||x||=0.70710
3

Task – 4.2
X=[1,-1,1,-1]
(a) L-2 norm
‖ X ‖2=√x12+……..+x02
|x|=∑
k =1
n
|xk|2
‖ X ‖2=√112+122+….(1)n2
(b) orthogonal
1
(-1,1,-1,1)
-1
1 (1,-1,1,-1)
-1
1 (-1,1,-1,1)
-1
(c) ||x||=|1,-1,1,-1|
4

Task - 4.3
‖x‖1=∑
n =0
N−1
|x [ n]|
L1-norm ‖x‖1=|x|
L2-norm ‖x‖2=√x12+……..xn2
(a)
‖x‖1= ∑
n =0
N−1
|x |0+|x|1+|x|2+….|x|n
‖x‖2=√x02+x12+x22+……..xn2
(b)
Φk=e2 πj∈¿ N and (k=0,…..N-1),(n=1,2,N-1),(N=1,2,..N-1)
Φk=e2 πj (0 ) (1)/ (1)+e2 πj (1 )(2 )/(2)+e2 πj ( 2 ) (3 )/(3)+……+e2 πj ( N−1 ) (N−1)/(N −1)
Euclidean norm of a complex number is also called as modulus.
Φk=e0+e2 πj+e4 πj+……+ e2 πj ( N−1 )
Task - 4.4
X(n)= Ãφk[n]+ B ̃ φk[n]
K≠l,and complex Ã& B ̃
The absolute value norm
(a)
⟦ X ⟧2=√(x 1+ x 2+… xn)2
‖ x ‖=√ | A|2φk+|B|2φk+…..|Z|2φN-1
(b)
‖x‖=√ |A |2φk+|B|2φk+|c|2φk+……. |Z|2φN-1
(c)
‖x n‖= ∑
K=0
N −1
Dkφk[n]
5

=∑
k =0
N−1
Dkφk[0]+Dkφk[1]+Dkφk[2]…..
‖x n‖=(|D+ D kφ +D k φ 2+ ….. D k φ N-1|)
Task - 4.5
x(n)=Asin(2 πf 0nTs), n=0,1,….(N-1)
Fs=1/Ts
(a)
Power series
∑
n=0
∞
an(x-c)n=a0+a1(x-c)1+a2(x-c)2+….
DFT
x(k)=∑
n =0
N−1
x (n) e− j 2 πkn /w
x(k)= ∑
n =0
N−1
Asin¿ ¿0nTs)e− j2 πkn/ w
=A0+…..A1(sin2 πf 0(N-1)Ts)e− j2 π ( N −1 )(N −1)/( N−1)
=A0+A1(sin2 πf 0(N-1)Ts-e− j2 π (N−1)+…
(b)
X(k)=A0+A1(sin2 πf 0(N-1)Ts- e− j2 π (N−1)+….
=Asin(2πf 0nTs)[e jAN /2.e− jAN / 2]
Task - 4.6
6

X(n)= {
0 :n=0,1, .. ( N −2 )
−1
2 : n=(N −1) }
Solution
(a)
DFT
X(k)=∑
n =0
N−1
x ( n ) . e−2 πjkn/ N
X(k)=∑
n =0
N−1
x (n)[cos2 πkn
N − jsin( 2 πkn
N )]
X(0)=∑
n =0
N−2
x ( 0 ) [ cos 0− jsino ] + ∑
n=1
N −2
x ( 1 ) [cos 2 π − jsin 2 π ]
X(k)={0,1,….(N-2)cos2 π}
(b)
Magnitude
0 1 2 3 N-1
(C)
7

3
0
3.5 2π
-1 phase diagram
Task – 4.7
(a)
DFT
x(k)=∑
N=0
N−1
x ( n ) e− j 2 πkn / N
(a)
X(k)= ∑
k =0
N−1
x (n) e− j 2 πkn /N N(0,1,2,…..N-1) a(0,1,2,….N-1) n(0,1,2,,N-1)
X(0) = -1/2e− j2 π ( 0 ) (0 )/ 0 +0 + 1/2 e− j2 πan
=1/2¿ ¿)
=1/2[1+cos2πan/N – jsin 2πan/N]
=1/2 [1+cos2π(0)/0 – jsin 2π(0)/0] =1/2[1+1]
=1
8

X(1)= 1/2[1+cos2π(1)/1 – jsin 2π(1)/1] =1/2[1+1+0]
= 1
X(K)={1,1,….N-1}
(b)
X(k)= ½{[1+cos2π(0)/0 – jsin 2π(0)/0] [1+cos2π(1)/1 – jsin 2π(1)/1]
[1+cos2π(2(2))/2 – jsin 2π(2)()2/2] ,,,…….. [1+cos2π(N-1)/2 – jsin 2π(N-1)(N-1)/(N-
1)]
=1/2[(1+1)(1+1)(-1),..,….(1+N-1)]
=1/2[4.,..(N-1)]
=[2+N-1]
(C)
0 1 2 (N-1)
Frequency (HZ)
9
Magnitude

Task - 4.8
Given:
Xk =
{ 1 , k=0 , N
2 −1
0 , k= N
2 , N −1 }
Solution:
N
√1 n = nth
X0 = 1 & X0 = N/2 as per the given data
X[n] = 1/N ( 1−(−1)n/−N
√ 1 n )
N
√ 1n indicate nth of N roots of unity ( N
√ 1n = 1)
x(n) = 1/N ( 1−(−1) n
1−1 ) & X0 = 1+….+N/2-1
XN/2 = 0+….+N-1
If n=0,1,2………N-1
X(0) =1/N
X(1) = 1/N (1+1/1-1)
X(1) = 1/N (0)
X(1) = 0
Task - 4.9
Given data:
F(x) =∑
n =0
N−1
x [n]e−2 πjkn/ N (k=0,1,2,……N-1)
Solution:
Gn x[n] =δn
10

If k=0
f(x)=∑
n =0
N−1
x [n]e0
f(x)=∑
n =0
N−1
x [n]
f(x)=∑
n=0
N −1
δn
f(x)=δ0 +δ1+δ2+…….fN-1
F(f(x))= F(f(x))
Result:-
F(f(x)) = F(δ0+ δ1+ δ2+……….+fN-1)
Task - 4.10
Given data:
X[n]=1/N ∑
k =0
N−1
X [ K ]e2 πjnK / N (n=0,1,…..N-1)
Solution:-
If n=0
X[0]=1/N ∑
k =0
N−1
X [ K ]e0
X[0]=1/N ∑
k =0
N−1
X [ K ]
11

If n=1
X[1]=1/N∑
k =0
N−1
X [ K ]e2 πjnK / N
X[1]=1/N
X[1]=1/N x(0) e0+1/Nx(1) e2 πj/ N +1/Nx(2) e4 πj/ N +……
If n=N-1
X[N-1]=1/N ∑
k =0
N−1
X [ K ]e2 πjk ( N −1 ) / N
X[N-1]=1/N[1/Nx[0]+1/Nx[1]e2 πj( N−1)/ N+1/N[1/N2x[0]+1/N2x[1]+1/N2x[2]+….]e2 πj( N−1)/ N
+1/Nx[2] e4 πj(N −1 )/ N +….
X[N-1]=[1/N 2x[0]+1/N 2x[1]e2 πj( N−1)/ N+[1/N3x[0]+1/N3x[1]+1/N3x[2]+….]e2 πj( N−1)/ N+1/N
x[2]e4 πj(N −1 )/ N +….
Task - 4.11
formula:-
DFT
Xk = ∑
n =0
N−1
x ( n ) . e−2 πjkn/ N
=∑
n =0
N−1
x n . [cos ( 2 πkn
N )− jsin ( 2 πkn
N )]
IDFT
Xn=1/N ∑
k =0
N−1
Xk . ei 2 πkn / N
W=e−2 πj / N
12

We know that
Xk = ∑
n =0
N−1
x ( n ) . e−2 πjkn/ N and WN=e− j2 π / N
X(k)=∑
n =0
N−1
x ( n ) W Nkn
DFT Result:-
[X]=[WNkN][x]
X(n)=1/N∑
n =0
N−1
X ( k ) W N
IDFT Result:-
[X]=1/N[WNkN][x]
13