Week 23

Week 23 Physics, Circuits:

  • Physics reading: Ch 35.
    • Focus on 35.1-35.3 (but definitely read the rest, especially 35.5). The main point of 35.4 is that there are things called inductors that have reactance ZL = iωL (Knight’s “XL” is Horowitz & Hill’s “ZL” – they use different letters for reactance). You can think of inductors as a sort of “inverse capacitor” in terms of their circuit behavior (though not why they work).
    • Note that Knight doesn’t use imaginary numbers… hopefully you will start to see why they are useful. Otherwise, how can you get results 35.24 and 35.30? (definitely know equation 35.30 – that shows up a lot in circuits).
    • Due online Friday, 11:59 pm:
      • Ch 31, problems 39, 43, 73, 74
      • Ch 35, problems 7, 8, 13, 14, 16, 19, 28, 42, 43
        • Notes:
          • You may find it easier to do all of these problems using the Horowitz approach (using complex impedances). I would!
          • 13, 14: “crossover frequency” is the same as f3db
          • 42: watch out; Knight is asking for the delta-V (voltage difference) across the “top” resistor… the Vout you are used to would be ΔV across the bottom two.
  • Circuits Reading:
    • AOE 2nd ed, pp. 20-42.
    • Horowitz and Hayes, Lab Manual, pp. 32 – 50.
  • Due Wednesday, 5pm, box outside my office: AOE Ch 1: 1.13, 1.14 (read the section immediately before the question :), 1.15, 1.16, 1.17, 1.21, 1.22, 1.24
    • Notes:
      • 1.14: You need to make the assumption that the Vin = 0 until you connect it at t = 0, so at t = 0 Vout (V(t) in Fig 1.34) = 0 too. Then You connect Vin(t) , which may vary in time… what is Vout(t)? Hint: consider the Thevinin equivalent looking into the two resistors, then compare to the ‘Time Constant’ example on p. 34.
      • 1.21: They are looking for the magnitude of the response here, like they found for the high-pass filter on pp. 35-6. Show this by following that example, adapting it to the low-pass filter geometry (complex voltage divider in Fig 1.58), using the complex impedance for a capacitor.
      • 1.22 (show by calculating the phase shift (in degrees – you’ll have to convert from radians) at 0.1 f3DB (for a high-pass filter) and 10 f3DB (for a low-pass filter).

Calculus

  • Reading: 14.1 – 14.3
  • Problems:
    • Monday Spring Drills 3, due 10 am Mon. April 18 via WileyPlus:
      • 14.1: 6, 12, 24, 27
      • 14.2: 2, 3, 5, 14, 16, 26, 34, 38
      • 14.3: 1, 2,4, 6, 8, 15, 16, 20
      • Solutions
    • Wednesday Spring Problem Set #3, due noon Wed. April 20 outside Lab 2 3255:
      • 14.1: 8, 18
      • 14.2: 40, 44, 48, 49
      • 14.3: 22, 26, 31, 35
      • Solutions
  • Quiz 3 on Week 3 Material