1. ELE1110C Tutorial 3 27/09/2006 27/09/2006 Cathy, KAI Caihong

2. Outline Lecture Reviews Examples

3. How to represent a sine wave (sinusoid)? Complex number notation (L2.137) –a+jb Represent a vector with a magnitude and phase –Magnitude = –Phase = –Unit vector = a jb  1 

4. Rotation: multiplied by j (rotate anti-clockwise) => Period = 2π/w, Frequency = 1/Period = w/2π => w : Angular Frequency (i.e. w = 1000 x 2π for a 1000Hz sine wave)

5. Impedance Calculation Capacitor:

6. Impedance Calculation Impedance of capacitor: Magnitude of impedance : Low frequency => Z C is large => open circuit High frequency => Z C is small => short circuit

7. Circuit Analysis If V 1 is an varying voltage, what is V 2 ? R CV2V2 V1V1

8. Capacitor – Transfer Function Definition –Impedance of capacitor : By KVL So that Hence

9. Capacitor – Transfer Function Transfer function T(w)= Output / Input = Note that the output response depends on the frequency of the input sinusoid High frequency ( ), –High frequency signals are cut Low frequency ( ), –Low frequency signals are passed –The circuit is a low pass filter

10. Usually we focus more on power ration in filter design: Therefore, 3dB point (half-power, or cut-off point):

11. At half power point, =>  C CR = 1 =>  C = 1/RC since  C = 2f, therefore f C = 1/2RC 3dB point of a LPF

12. Magnitude and phase response of a low pass filter (LPF) –Magnitude response = –Phase response =

13. dB is the logarithm of a ratio between powers –3dB => P OUT is 2 times P IN –10dB => P OUT is 10 times P IN –20dB => P OUT is 100 times P IN –-3dB => P OUT is half of P IN, and so on

14. Examples from Problem Sheet 3 Example 1: Find the transfer function of the following circuit

15. Let Y = (1/ jwC) As no current will flow into R 2, V 1 = V out, I 2 = 0 I 0 = I 1 = V in /(R 1 + Y) V out = V 1 = I 0 * Y =YV in /(R 1 +Y) Then you can find V out /V in

16. Example 2: Find the transfer function of the circuit below.

17. Let Y 1 = (1/ jwC 1 ), Y 2 = (1/ jwC 2 ) V 1 = V in *(Y 1 //(Y 2 +R 2 )) / ( (Y 1 //(Y 2 +R 2 )) + R 1 ) V out = V 1 *(R 2 /(R 2 +Y 2 )) Therefore, V out = [V in *(Y 1 //(Y 2 +R 2 )) / ( (Y 1 //(Y 2 +R 2 )) + R 1 ) ] *(R 2 /(R 2 +Y 2 )) (V out /V in )= (Y 1 //(Y 2 +R 2 )) *(R 2 /(R 2 +Y 2 )) / ( (Y 1 //(Y 2 +R 2 )) + R 1 ) *(R 2 /(R 2 +Y 2 ))

18. Example 3: The following diagram shows an elevator call button - an example of proximity switch. The button has two sides separated by insulator. One side is a plate conductor and the other side is a ring conductor. The button behaves as a capacitor (C1). When a finger is placed somewhere near the middle of the ring, the finger acts like an earth as the body can absorb a fair amount of charges quickly. The capacitance between the ring and the finger is C2 and the capacitance between the plate and the finger is C3. Also shown are the equivalent circuits for the two states.

20. The circuit for the elevator call button is as follows. Suppose C1=30pF, and C2=C3=10pF, find V with i) no finger present, and ii) finger present.

21. As Q = CV, we can deduce that the ratio of voltage across 2 capacitors in series is: 1)

22. 2) As no current will flow into the earth point (finger), the earthing has no effect to this circuit.