Electronics and Communication Engineering - Exam Questions Papers

For u(n), a right handled sequence,

|z| > , |z| =

∴ |3z| > 1; |2z| > 1

∴ < 1; < 1.

1 + G(s) H(s) = 0

s² + 2s + 2 = 0

ξ < 1, under damped.

The given digits are 2, 2, 3, 3, 3, 4, 4, 4, 4 we have to find the numbers that are greater than 300

∴ The first digit can be 3 or 4 but not 2.

Now, let us fix the first, second and third digits as 3, 2, 2, then the fourth place can be filled in 3 ways.

∴ The number of ways is 3 similarly, we fix first third and fourth place as 3, 2 and 2 respectively (4) so the second place can be filled in 3 ways again,

The number of ways is 3

Now, we fix first, second and fourth, previous cases and we obtain the same result.

∴ The number of ways is 3 so, the total number of ways is 9 similarly this can done by fixing the numbers as 3 and 4 (instead of 2) and thereby we obtain the a ways each

The number of numbers starting with 3 is 27

Similarly by taking 4 as the first digit we get 27 numbers

∴ The number of numbers that are greater than 3000 is 27 + 27 = 54

But, 3222, 4222, is not possible as there are only two 2's, 3333 is not possible as there are only three 3's

∴ The total number of numbers that are greater than 3000 is 54 - 3 = 51.

f₁ = y₂ + y₆ + y₁₁ + y₁₂

and f₂ = y₁ + y_l0 + y_l3

f₁(m, n, o, p) = (3, 4, 9, 13)

f₂(m, n, o, p) = (2, 5, 14)

∴ = Π(2, 3, 4, 5, 9, 13, 14).

Compare this with

where s = 3.