Answer:
Applying Kirchhoff's current law at inverting terminal
Explanation:
Detailed explanation is given in the attached document.
Answer:
a) it is periodic
N = (20/3)k = 20 { for K =3}
b) it is Non-Periodic.
N = ∞
c) x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.
then the period of the signal is given as
N = ( 2π/w₀)K
k is least integer for which N is also integer
Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2
now
a) cos(2π(0.15)n)
w₀ = 2π(0.15)
Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3
so, it is periodic
N = (20/3)k = 20 { for K =3}
b) cos(2n);
w₀ = 2
Now, 2π/w₀ = 2π/2) = π
so, it is Non-Periodic.
N = ∞
c) cos(π0.3n) + cos(π0.4n)
x(n) = x1(n) + x2(n)
x1(n) = cos(π0.3n)
x2(n) = cos(π0.4n)
so
w₀ = π0.3
2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3
∴ N1 = 20
AND
w₀ = π0.4
2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5
∴ N² = 5
so, x(n) is periodic
N = LCM ( 5, 20 )
Answer:
Java program explained below
Explanation:
FindSpecialNumber.java
import java.util.Scanner;
public class FindSpecialNumber {
public static void main(String[] args) {
//Declaring variable
int number;
/*
* Creating an Scanner class object which is used to get the inputs
* entered by the user
*/
Scanner sc = new Scanner(System.in);
//getting the input entered by the user
System.out.print("Enter a number :");
number = sc.nextInt();
/* Based on user entered number
* check whether it is special number or not
*/
if (number == -99 || number == 0 || number == 44) {
System.out.println("Special Number");
} else {
System.out.println("Not Special Number");
}
}
}
_______________
Output#1:
Enter a number :-99
Special Number
Output#2:
Enter a number :49
Not Special Number
Answer:
11 toes on one foot? and 5 one the other or just 11 toes?
your NO HINT threw me off
Explanation:
