ECN 4th Chapter PDF

# Unit-y Network Parameters ## Two port network parameters ### Network Equation - V₁ and V₂ depend on I₁ and I₂ (independent) - Convert from terms of V₁ , V₂ to terms of I₁ , I₂ - Z = |[Z₁₁ Z₁₂] [Z₂₁ Z₂₂]| - Y = |[Y₁₁ Y₁₂] [Y₂₁ Y₂₂]| - T = |[A B] [C D]| - h = |[h₁₁ h₁...

# Unit-y Network Parameters ## Two port network parameters ### Network Equation - V₁ and V₂ depend on I₁ and I₂ (independent) - Convert from terms of V₁ , V₂ to terms of I₁ , I₂ - Z = |[Z₁₁ Z₁₂] [Z₂₁ Z₂₂]| - Y = |[Y₁₁ Y₁₂] [Y₂₁ Y₂₂]| - T = |[A B] [C D]| - h = |[h₁₁ h₁₂] [h₂₁ h₂₂]| ### Impedence parameters - V₁ = z₁₁I₁ + z₁₂I₂ - V₂ = z₂₁I₁ + z₂₂I₂ ### Admittance Parameters - I₁ = Y₁₁V₁ + Y₁₂V₂ - I₂ = Y₂₁V₁ + Y₂₂V₂ ### Transmission parameters (A, B..CO) parameters - V₁ = AV₂ - BI₂ - I₁ = CV₂ - DI₂ ### Inverse transmission parameters (A', B', C', D') - V₁ = A'V₂ - B'I₂ - I₁ = C'V₂ - D'I₂ ### Hybridetes parameters - V₁ = h₁₁I₁ + h₁₂V₂ - I₂ = h₂₁I₁ + h₂₂V₂ ### Inverse hybrid parametrs - I₁ = g₁₁V₁ + g₁₂I₂ - I₂ = g₂₁V₁ + g₂₂I₂ ## Z - parameters - |[V₁] [V₂]| = |[Z₁₁ Z₁₂] [Z₂₁ Z₂₂]| * |[I₁] [I₂]| - V₁ = z₁₁I₁ + z₁₂I₂ - V₂ = z₂₁I₁ + z₂₂I₂ ### Step 1 - Port 2-2' is opened. - Z₁₁ = V₁/I₁ - Z₂₁ = V₂/I₁ - Z₁₂ = V₁/I₂ - Z₂₂ = V₂/I₂ ### Step 2 - Find Z parameters of the network given. - **Step 0** - I₂=0 - 2-2' open circuited - Apply KVL: - V₁ = 3I₁ + 2I₂ → 0 - V₂ = 2I₁ (V.D across 2Ω is V₂) - Z₁₁ = V₁/I₁ = 5 - Z₂₁ = V₂/I₁ = 2 - Z₂₂ = V₂/I₁ = 2 - **Step 1** - I₁ = 0 - 1-1' open circuit - Apply KVL: - V₂ = 4I₂ + 2I₁ - V₁ = 2I₂ (V.D across 2Ω is V₁) - Z₁₂ = V₁/I₂ = 2 - Z₂₂ = V₂/I₂ = 6 - Z = |[5 2] [2 6]| ### Step 3 - Find the Z parameters - **Step 0** - I₂ = 0 (opencircuit) - V₁ = 3I₁ + 2I₁ + 2I₁ - V₂ = 2I₁ + 2I₁ - V₂ = 4I₁ - **Step 1** - I₁ = 0 (opencircuit) - V₁ = 2I₁ - V₂ = 4I₁ + 2I₁ = 6I₁ - Z₁₂ = V₁/I₂ = 2 - Z₂₂ = V₂/I₂ = 6 - Z = |[4 2] [2 6]| ## Y → parameters - |[I₁] [I₂]| = |[Y₁₁ Y₁₂] [Y₂₁ Y₂₂]| * |[V₁] [V₂]| - I₁ = Y₁₁V₁ + Y₁₂V₂ - I₂ = Y₂₁V₁ + Y₂₂V₂ ### Step 1 - Port 2-2' is shorted (input admittance) - Y₁₁ = I₁/V₁ - Y₂₁ = I₂/V₁ (S.C transfer admittance) ### Step 2 - Port 1-1' is shorted - Y₁₂ = I₁/V₂ (S.C Reverse transfer admittance) - Y₂₂ = I₂/V₂ (S.C output admittance) ### Step 3 - Find Y parameters - **Step 0** - V₂ = 0 (short circuited) - V₁ = 3I₁ + 2(I₁ + I₂) + 2I₁ - V₁ = 7I₁ + 2I₂ - Apply KVL: - 2(I₁ + I₂) + 2I₁ = 0 - I₂ = -6I₁ - I₁ = -2/3 I₂ - I₂ = -3/2 I₁ - V₁ = 7I₁ + 2[-3/2 I₁] - V₁ = [1/3] I₁ = -[1/2] I₂ - V₁ = -[1/2] I₂ - **Step 1** - V₁ = 0 - 1-1' shorted - 3I₁ + 2(I₁ + I₂) + 2I₁ = 0 - 3I₁ = (-2/3)I₂ = 0 - V₂ = 4I₂ + 2(I₁ + I₂) + 2I₁ - V₂ = 4I₁ + 6 I₂ - V₂ = 4I₁ + 6 (-2/3)I₁ - V₂ = -17I₁ - V₂ = -17(1/2)I₂ - V₂ = -34/2 I₂ - Y₁₂ = I₁/V₂ = -1/17 - Y₂₂ = I₂/V₂ - Y = |[1/3 -1/17] [-1/2 1/34]| ## 3. Transmission parameters - Transmission parameters, ABCD parameters. - |[V₁] [I₁]| = |[A B] [C D]| * |[V₂] [I₂]| - V₁ = AV₂ - BI₂ - I₁ = CV₂ - DI₂ ### Step 1 - I₂=0 port 2-2' opened - A = V₁/V₂ (open circuit Reverse voltage gain) - C = I₁/V₂ (open circuit reverse transfer admittance) ### Step 2 - V₂ = 0 port 2-2' is shorted. - B = -V₁/I₂ (short circuit Reverse transfer impedance) - D = -I₁/I₂ (short circuit Reverse current gain) ### Step 3 - Find A.BCD parameters - **Step 1** - I₂ =0 (opencircuit) - V₁ = I₁ + 2(I₁ - I₃ ) - V₂ = 0.5I₃ - I₃ + 0.5I₃ - 2( I₁ - I₃) = 0 - I₃ + 0.5I₃ - 2I₁ + 2I₃ = 0 - I₃ = (2/3.5)I₁ - I₃ = (4/7)I₁ - V₁ = I₁ + 2I₁ - 2(4/7)I₁ - V₁ = I₁ + 2I₁ - (8/7)I₁ - V₁ = 13/7 I₁ - V₂ = 0.5I₃ = 0.5(4/7)I₁ - V₂ = 2/7I₁ - A = V₁/V₂ = 13/7 * 7/2 = 13/2 - C = I₁/V₂ = 7/2/2/7 = 49/4 ### Step 2 - V₂ = 0 2-2' shorted - V₁ = I₁ + 2(I₁ + I₂) - V₂ = I₁ + 2(I₁ + I₂) - I₁ + 2(I₁ + I₂) = 0 - I₂ + 2I₁ + 2I₂ = 0 - I₁ = -2/3 I₂ - I₂ = -1/2 I₁ - V₁ = I₁ + 2(I₁ ) + 2*[2(-1/2)I₁] - V₁ = [2/3] I₂ + 4[-1/3]I₂ + 2. I₂. 3I₁-4I₂ - V₁ = -[2/3]I₂ - 3I₁ + 2I₂ - V₁ = -[2/3]I₂ + 2 I₂ - V₁ = -[2/3] I₂ - I₂ - V₁ = -[5/3] I₂ - V₁ = -5/3 I₂ - B = -V₁/I₂ = 5/3 = 5/2 - D = -I₁/I₂ = (2/3) * 3 = 3/2 ## Hybrid parameters [h] - |[V₁] [I₂]| = |[h₁₁ h₁₂] [h₂₁ h₂₂]| * |[I₁] [V₂]| - V₁ = h₁₁I₁ + h₁₂V₂ - I₂ = h₂₁I₁ + h₂₂V₂ ### Step 1 - V₂=0 port 2-2' is shorted - h₁₁ = V₁/I₁ (short circuit input impedance) - h₂₁ = I₂/I₁ (short circuit current gain) ### Step 2 - I₁ = 0 port 1-1' open circuit - h₁₂ = V₂/I₁ (open circuit reverse voltage gain - h₂₂ = I₂/V₂ (open circuit output admittance. ### Step 3 - Write the Expression for I₂ and determine all h and q parameters for figure. - **Step 1** - V₂ = 0 shorted - Apply KVL at mesh ① - V₁ = I₁ + I₂ + 2V₁ - -V₁ = I₁ + I₂ - Apply KVL at mesh ② - 2V₁ + (I₂) +(I₁ + I₂ + 2V₁) = 0 - 4V₁ + 2I₂ + I₁ = 0 - 4V₁ = -(I₁ + 2I₂) - Sub in ② - 4(-I₁ + I₂) = -(I₁ + 2I₂) - 4I₁ + 4I₂ = -(I₁ + 2I₂) - 5I₁ + 6I₂ = 0 - I₁ = -6/5 I₂ - I₂ = -5/6 I₁ - -V₁ = I₁ + I₂ - V₁ = -0.5I₁ - V₁ = 0.5I₁ - h₁₁ = V₁/I₁ = 0.5 - h₂₁ = I₂/I₁ = -5/6 = -1.5 - **Step 2** - I₁ =0 (open circuited) - Apply KVL at mesh ① - V₁ = (I₂ - I₃ + 2V₁) - -V₁ = I₂ - I₃ + 2V₁ - Apply KVL at mesh ② - 2V₁ + (I₂) + (I₂ - I₃ + 2V₁) - 2I₃ = 0 - Apply KVL at mesh ③ - V₂ = 2I₃ - Substitute (①) in (₂) - 2(-I₂ + I₃) + (I₂ - I₃ ) + 2I₃ = 0 - 2(-I₂ + I₃) + (I₂ - I₃) + I₂ - I₃+ 2(-I₂ + I₃) = 0 - -2I₂+2I3 + I₂- I₃ + I₂- I₃ - 2I₂+2I₃ = 0 - -2I₂ + 2I₂ = 0 - I₂ = 0 - -V₁ = I₂ - I₃ - V₁ = -I₃ - V₁ = I₃ - V₂ = 2I₃ - h12 = V₂/I₁ = 2I₃/I₁ = 2 - h₂₂ = I₂/V₂ = 0/2 = 0 ## T Network (Y parameters) - **Step 1** - Z₁₁ = Za + Zb + Zc - Z₂₂ = Zb + Zc - Z₁₂ = Zc - Z₂₁ = Zc - Za = Z₁₁ - Z₁₂ - Zb = Z₂₂ - Z₂₁ - Zc = Z₁₂ = Z₂₁ - **Step 2** - Y = |[1/(Za+Zb+Zc) -Zc/(Za+Zb+Zc)] [-Zc/(Za+Zb+Zc) 1/(Zb+Zc)]| - **Step 3** - Z₁₁ = Y₁₁ + Y₁₂ - Z₂₂ = Y₂₂ + Y₂₁ - Z₂₁ = -Y₁₂ - Z₁₂ = -Y₂₁ ## Example - Y = |[6 2] [2 1]| - Z = Y⁻¹ = |[1 -2] [-2 6]| - Y = |[6 -2] [-2 5]| - Z = Y⁻¹ = |[5/18 -1/9] [-1/9 2/9]| ## Two port network ### Reciprocal - z₁₁ = z₁₂ - y₂₁ = y₁₂ - h₂₁ = -h₁₂ - g₂₁ = -g₁₂ ### Symmetric - z₂₂ = z₁₁ - y₂₂ = y₁₁ - h₁₁h₂₂ - h₁₂h₂₁ = 1 - g₁₁h₂₂ - g₁₂g₂₁ = 1 - AD - BC = 1 - A = D ### Inverse transmission parameters ### Reciprocal - A'D'B'C' = 1 ### Symmetric - A'= D' ## Note: If the circuit consists of dependent sources we cannot say whether the network is symmetrical (or) reciprocal. ## Two port parameters conversion - Z ↔ Y - Z ↔ T - T ↔ Y - T ↔ H - H ↔ Z - A = Z₁₁ - B = Δ/Z₂₁ - C = 1/Z₂₁ - D = Z₂₂/Z₂₁ ## Given: - Z = |[3 -2] [2 -1]| - V₁ = 3I₁ + 2I₂ - V₂ = 2I₁ + 3I₂ - **Step 1** - I₂ = 0 - V₁ = AV₂ + BI₂ - I₁ = CV₂ + DI₂ - A= V₁/V₂ = 1.5 - C= I₁/V₂ = 0.5 - From equ's ① V₂ = 2I₁, - V₂ = 3I₁ - **Step 2** - V₂ = 0 - B = -V₁/I₂ = 2.5 - D = -I₁/I₂ = 1.5 - **Step 3** - T = |[1.5 -2.5] [0.5 1.5]| - AD - BC = 1 - (1.5 x 1.5) - (2.5)(0.5) - 2.25 - 1.25 - 1 ## Example - T = |[1.5 -2.5] [0.5 1.5]| - V₁ = 1.5V₂ - 2.5I₂ - I₁ = 0.5V₂ - 1.5I₂ - **Step 1** - h₁₁ = V₁/I₁ = 1.667 - h₂₁ = I₂/I₁ = -0.667 - **Step 2** - h₁₂ = V₂/I₁ = 1 - h₂₂ = I₂/V₂ = 0.33 - **Step 3** - V₁ = 1.5V₂ - 2.5I₂ - V₂ = 0.5V₂ - 1.5I₂ - V₂ = (2/3)I₂ - V₁ = 1.5V₂ - 2.5I₂ - V₁ = (2/3)V₂ = 0.667 - V₂ = 1/3 V₁ - h = |[1.667 0.667] [-0.667 0.33]| - h₁₂ = -h₂₁ = 2/3 (Reciprocal) - Symmetrical - AH = |[1.667 0.667] [0.33 0.33]| - |[2/2 2/2] [2/3 2/3]| = |[3/3 3/3] [3/3 3/3]| = 1 ## Series connection of two port Network - **Step 1** - Z = Z₁ + Z₂ - |[Z]| = |[Z₁]| + |[Z₂]| - |[Z]| = |[5 2] [2 6]| + |[3 1] [1 1]| - |[Z]| = |[8 3] [3 10]| ## Parallel connection - **Step 1** - Z = Z₁ + Z₂ - [Z] = [YA] + [YB] ## Example Bridged T' Network - **Step 1** - The Y21 parameters of the network shown in given figure → Bridged T - Y₁ = |[1/3 -1/6] [-1/6 1/3]| + |[1/3 -1/6] [-1/6 1/3]| - Y₁ = |[2/3 -1/3] [-1/3 2/3]| ## Twin T Network - Determine the Z-parameters for the two-port network in the figure given below. - Voltages are Same so parallel connection - **Network A** - Z = |[2/5 + 1/(1/2+1/2)] [1/(1/2+1/2)]| |[1/(1/2+1/2)] [1/(1/2+1/2)]| - Z = |[2.4 1] [1 1]| - [Y] = Z⁻¹ - **Network B** - Z = |[25+1/(1/2+1/2)] 1/(1/2+1/2)]| |[1/(1/2+1/2)] [1/(1/2+1/2)]| - Z = |[1 + 25] 1| |[1 1]| - Y = Z⁻¹ - **Network 2** - Z = |[25+1/1] 1/1| |[1/1 1/1]| = |[26 1] [1 1]| - Y = Z⁻¹ - **Network 3** - Z = |[25+1] 1| |[1 1]| - Z = |[1+25 1| |[1 1]]| = |[26 1| |[1 1]| - Y = Z⁻¹ ## Example for cascade connection of two networks - Find transmission parameters. - **Step 1** - I₂=0 - V₁ = 2I₁ - I₂ - V₂ = I₁ - A = V₁/V₂ = 3 - C = I₁/V₂ = 1 - **Step 2** - V₂ = 0 - V₁ = 2I₁ - B = -V₁/I₂ = -2 - D = -I₁/I₂ = 1 - **Step 3** - T₁ = T₂ = |[3 2] [1 1]| - T = [T₁] * [T₂] = |[3 2] [1 1]| * |[3 2] [1 1]| - T = |[9+2 6+2] [3+1 2+1]| = |[11 8] [4 3]| ## Note: - For Series-parallel connection of 2 Nlw [h] = [ha] + [hb] - For parallel - series connection of 2 Nlw [G] = [ga] + [gb] ## J - Notation - w = 5 rad/sec - JS(1) = JUL = SL - ω = 2πf - C = 1/(ωC) - Z = |[2-J0.2 -J0.2] [-J0.2 J14.8]| - Y = Z⁻¹ - Z = -0.04