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Q: G1 phase s pase and g2 phase is called?
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Draw XOR using NAND gate?

Given two inputs, A and B. Use one NAND as an inverter (one source to both inputs) for A. Do the same for B. Use one NAND gate with inputs \A (not A, the output from the first gate) and B. Use one NAND gate with inputs A and \B. Use one last NAND gate with inputs coming from the two previous gates. Its output will behave like an XOR gate. Alternatively: Gates G1, G2, G3, G4, arrange in a diamond with G1 and G4 at the left and right vertices, A input along top edge, B input along bottom edge. G2 at top center, G3 at bottom center. A goes to G1 and G2, B goes to G1 and G3, G2 and G3 go to G4. Net list: A - G1i1 A - G2I1 B - G1i2 B - G3I2 G1o - G2i2 G1o - G3i1 G2o - G4i1 G3o - G4i2 G4o - A^B


The vector representation of a convolutional code is g1=[1 1]and g2=[1 0]. If the received sequence is 11 11 01 10, please use Viterbi Algorithm to decode it. Please show the full trellis diagram, including the updated trellis state metrics?

The vector representation of a convolutional code is g1=[1 1] and g2=[1 0]. If the received sequence is 11 11 01 10, we can use Viterbi Algorithm to decode it. The full trellis diagram, including the updated trellis state metrics, is shown below: Time t1: State 0 (00): Metric = 0 State 1 (10): Metric = 0 Time t2: State 0 (00): Metric = 0 + (1-1)^2 + (1-1)^2 = 0 State 1 (10): Metric = 0 + (1-1)^2 + (1-0)^2 = 1 Time t3: State 0 (00): Metric = 0 + (1-1)^2 + (1-1)^2 = 0 State 1 (10): Metric = 1 + (1-1)^2 + (1-0)^2 = 2 State 0 (01): Metric = 0 + (1-1)^2 + (0-1)^2 = 1 State 1 (11): Metric = 0 + (1-1)^2 + (0-0)^2 = 0 Time t4: State 0 (00): Metric = 0 + (1-1)^2 + (1-1)^2 = 0 State 1 (10): Metric = 2 + (1-1)^2 + (1-0)^2 = 3 State 0 (01): Metric = 1 + (1-1)^2 + (0-1)^2 = 2 State 1 (11): Metric = 0 + (1-1)^2 + (0-0)^2 = 0 Therefore, the decoded sequence is 11 11 01 10.