Electronics and Communication Engineering - Exam Questions Papers

Using Mesh analysis

Loop 1 : 6 = 9i₁ - 3i₄ ...(i)

Loop 2: 6 + 6 = - 6i₂

i₂ = - 2A ...(ii)

Loop 3: 6 = 9i₃ - 3i₄ ...(iii)

Loop 4: 6 = 9i₄ - 3i₃ - 3i₁ ...(iv)

Solving the above equations simultaneously

i₁ = i₃ = 1.1428 A

i₄ = 1.432 A

∴ i_x = i₄ - i₃ = 0.2892 A.

Picture elements (pixel) = 512 x 512 = 262144

Information per picture element N = log₂ M

M = 2^N

8 = 2^N

N = 3

Information for 262144 pixels = 262144 x 3 = 786432 bits

Entropy is same as average information per message.

Since there is no independent source in the network

V_TH = O V

To find R_Th apply : A current source of 1 A across A and B.

Apply KCL at node n₁

Apply KCL at node Vn₂

Vn₁ = 2V_x put in (i)

∴ Vn₂ = 4v_x

put in (ii)

100 + V_x = 2V_x

100 = V_x

∴ Vn₂ = 4V_x = 400 V

∴ .

Consider x(t) = e^j2pft

y(t) = h(t) * x (t)

⇒ y(t) = h(t). e^{j2pf(t - t)} . dt

= e^i2pfth(t) . e^{- j2pft} . dt

= e^j2pft x H(f) = x(t) x H(f)

⇒ .

Dual of R → G