Answer:
import java.util.Scanner;
class Main {
public static int calcSeries(int n) {
int sum = 0;
for(int i=10; i>=n; i--) {
sum += i;
}
return sum;
}
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
int n = 0;
do {
System.out.print("Enter n: ");
n = reader.nextInt();
if (n >= 10) {
System.out.println("Please enter a value lower than 10.");
}
} while (n >= 10);
reader.close();
System.out.printf("sum: %d\n", calcSeries(n));
}
}
Answer:
1) 10cos ( 2πfct + 10sin 2πfmt )
2) 50 watts
3) 1000 Hz
Explanation:
m(t) = 10cos(1000πt)
c(t) = 10cos(2πfct )
modulation index ( = 10
1) expression for modulated signal ( u(t) )
u(t) = Ac(cos 2πfct ) + A (cos 2πfct m(t) ) Am cos2πf mt
= 10cos 2πfct + 100 cos 2πfct cos2π500t
= 10cos ( 2πfct + 10sin 2πfmt )
2) power of modulated signal
power of modulated signal μ^2 / 2 ]
= 10^2/2 ]
= 50 watts
3) Bandwidth
B = 2fm = 2 * 500 = 1000 Hz
Resolution is the correct option
Have a great day
In the C language, letters can be printed as their ASCII value (which is an integer). So there, something like this would print 65 for A.