Circuit Analysis
Assignment 1 Answers
Year 2025
0027 63 985 5033
,
, QUESTION 1
1.1.1.
𝐹𝑜𝑟 𝑡 < 2: 𝑏𝑜𝑡ℎ 𝑢(𝑡 − 2)𝑎𝑛𝑑 𝑢(𝑡 − 4)𝑎𝑟𝑒 𝑧𝑒𝑟𝑜 𝑠𝑜 𝑓(𝑡) = 0
𝐹𝑜𝑟 2 ≤ 𝑡 < 4: 𝑢(𝑡 − 2) = 1(𝑠𝑡𝑒𝑝 𝑡𝑢𝑟𝑛𝑠 𝑜𝑛 𝑎𝑡 𝑡 = 2); 𝑢(𝑡 − 4) = 0(𝑠𝑡𝑒𝑝 𝑖𝑠 𝑠𝑡𝑖𝑙𝑙 𝑜𝑓𝑓 𝑢𝑛𝑡𝑖𝑙 𝑡 = 4)
𝑠𝑜 𝑓(𝑡) = 5(1) − 3(0) = 5
𝐹𝑜𝑟 𝑡 ≥ 4: 𝐵𝑜𝑡ℎ 𝑠𝑡𝑒𝑝 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝑂𝑁: 𝑢(𝑡 − 2) = 1, 𝑢(𝑡 − 4) = 1; 𝑠𝑜 𝑓(𝑡) = 5(1) − 3(1) = 2
Assignment 1 Answers
Year 2025
0027 63 985 5033
,
, QUESTION 1
1.1.1.
𝐹𝑜𝑟 𝑡 < 2: 𝑏𝑜𝑡ℎ 𝑢(𝑡 − 2)𝑎𝑛𝑑 𝑢(𝑡 − 4)𝑎𝑟𝑒 𝑧𝑒𝑟𝑜 𝑠𝑜 𝑓(𝑡) = 0
𝐹𝑜𝑟 2 ≤ 𝑡 < 4: 𝑢(𝑡 − 2) = 1(𝑠𝑡𝑒𝑝 𝑡𝑢𝑟𝑛𝑠 𝑜𝑛 𝑎𝑡 𝑡 = 2); 𝑢(𝑡 − 4) = 0(𝑠𝑡𝑒𝑝 𝑖𝑠 𝑠𝑡𝑖𝑙𝑙 𝑜𝑓𝑓 𝑢𝑛𝑡𝑖𝑙 𝑡 = 4)
𝑠𝑜 𝑓(𝑡) = 5(1) − 3(0) = 5
𝐹𝑜𝑟 𝑡 ≥ 4: 𝐵𝑜𝑡ℎ 𝑠𝑡𝑒𝑝 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝑂𝑁: 𝑢(𝑡 − 2) = 1, 𝑢(𝑡 − 4) = 1; 𝑠𝑜 𝑓(𝑡) = 5(1) − 3(1) = 2
