Electronics and Communication Engineering - Exam Questions Papers

(s³ + 2s² + 5s + 6)y(s) = (s² + 4) x (s)

s³y(s) + 2s²y(s) + 5sy(s) + 6y(s) = s² x (s) + 4 x (s)

Replacing s by and y(s) by y(t) and x(s) by x(t) we get

y(t) + 2y(t) + 5y(t) + 6y(t)

= x(t) + 4x(t)

+ 2 + 5 + 6 = + 4x

This is required differential equation.

Since y is at logic '1' initially, y' must be at logic '1'.

Also, one input of the EXOR gate is tied to zero therefore 'f' will be directly transferred to the output.

The logic gates will introduce gate delays which will lead to the answer .

I_c = 8 A

m = 0.8

= 8 x 1.149 = 9.19 A.

Maximum data rate for noiseless channel is given by Nyquist Theorem

= 2H log₂ V bits/sec

V - discrete levels (Here Binary i.e. 2)

H - Bandwidth

∴ Maximum data rate = 2H log₂ V

= 2(3 KHz) log₂ 2 = 2 x 3 x 10³ x 1

= 6 x 10³ = 6000 bps.

Apply Mesh analysis

Apply KVL to mesh 2

-2 + V_y + 1 = 0

∴ V_y = 1

Apply KVL to mesh 1

∴ -8 + V_x + 2 = 0

∴ V_x = 6 V

Apply KVL to mesh 3

-1 + V_z -2 = 0

∴ V_z = 3 V.