EE 105 Spring 2005 Discussion Notes written by Amin Monday March 28 2005 EE 105 Discussion Section 101 Announcements Response of Linear Circuits to Sinusoidal Input Signals A linear circuit is a circuit that consists of resistors capacitors inductors voltage sources independent and or dependent and or current sources independent and or dependent Now suppose that a linear circuit is excited by a sinusoidal input signal either current or voltage In other words suppose that the input to a linear circuit is x t A cos t where A amplitude of input sin usoidal 1 frequency of input sin usoidal in rad s phase offset of input sin usoidal in rad Voltage and current signals will be generated in the linear circuit in response to the aforementioned sinusoidal input These voltage and current signals are also sinusoidal in nature In fact the frequencies of the generated voltage and current signals is identical to the frequency of the input signal Hence one can mathematically characterize each voltage and current signal generated by the input sinusoid as follows y t B cos t where B amplitude of generated voltage current 2 frequency of generated voltage current in rad s phase offset of generated voltage current in rad There are various methods that one can employ in order to calculate B and Here we will focus on using the Laplace Transform Using the Laplace Transform to Find the Sinusoidal Response of a Linear Circuit For the theory behind using the Laplace Transform as it pertains to circuit system analysis please see EE 20 120 Here we will simply use the results by following certain steps To also demonstrate the process we will carry these steps out using the circuit in Figure 1 as an example 1 EE 105 Spring 2005 Discussion Notes written by Amin v X t i S t i1 t R1 i2 t C1 i3 t L1 Figure 1 Parallel RLC circuit 1 Represent every current and voltage in the circuit using Laplace notation Carrying this out on the circuit in Figure 1 results in the circuit depicted in Figure 2 V X s I S s I 1 s R1 I 2 s C1 I 3 s L1 Figure 2 Representing signals using Laplace notation 2 In the Laplace domain the concept of impedance i e Z is used rather than resistance i e R to describe Ohm s Law More specifically the following transformation takes place in the Laplace domain v t R i t becomes V s Z s I s 3 Like resistance impedance has units of ohms The impedance formulas for the various possible circuit elements are as follows Impedance of a resistor with resistance R Z R s R 1 s C Impedance of an inductor with inductance L Z L s s L Impedance of a capacitor with capacitance C Z C s 4 5 6 2 EE 105 Spring 2005 Discussion Notes written by Amin Hence in step 2 replace each resistor capacitor and inductor in the circuit with its corresponding impedance Carrying this out on the circuit in Figure 2 results in the circuit depicted in Figure 3 V X s I S s I 1 s R1 I 2 s 1 s C1 I 3 s s L1 Figure 3 Complete Laplace representation of original circuit 3 Use the circuit calculation techniques learned in EE 40 node voltage mesh current voltage divider current divider Th venin and Norton equivalents etc to obtain sdomain expressions for all of the voltages and currents of interest Upon doing so find the s domain transfer function that relates each voltage and current of interest to the applied input More specifically output transfer function input Y s H s X s 7 Let us then try to find appropriate expressions for V X s I 1 s I 2 s I 3 s and their respective transfer functions The equivalent impedance to the right of the input I S s is as follows 1 Z eq s R1 s L1 s C1 8 s L1 R1 Z eq s 2 s C1 L1 R1 s L1 R1 Thus s L1 R1 I S s V X s Z eq s I S s 2 s C1 L1 R1 s L1 R1 9 V X s s L1 R1 H 1 s 2 I S s s C1 L1 R1 s L1 R1 I 1 s V X s s L1 R1 1 I S s 2 Z R1 s C1 L1 R1 s L1 R1 R1 I s s L1 H 2 s 2 1 I S s s C1 L1 R1 s L1 R1 10 3 EE 105 Spring 2005 Discussion Notes written by Amin I 2 s V X s s L1 R1 2 s C1 I S s Z C1 s C1 L1 R1 s L1 R1 I s s 2 C1 L1 R1 H 3 s 2 2 I S s s C1 L1 R1 s L1 R1 I 3 s V X s s L1 R1 1 2 I S s Z L1 s C1 L1 R1 s L1 R1 s L1 I s R1 H 4 s 2 3 I S s s C1 L1 R1 s L1 R1 11 12 4 Using the transfer function one can compute the B and see 2 for each voltage current of interest as follows B H s j evaluated at the frequency of H s j evaluated at the frequency of the input signal A the input signal 13 14 Example Let the frequency of the input signal i S t be 2 1kHz 6 28e3 rad its s amplitude A 1 and its phase 0 Let s also R1 1k C1 1 pF L1 25mH Then using 13 and 14 yields assume that H 2 s j 6 28e3 0 1551 H 2 s j 6 28e3 81 i1 t 0 1551 sin 6 28e3 rad t 81 s H 4 s j 6 28e3 0 9879 H 4 s j 6 28e3 8 9 i3 t 0 9879 sin 6 28e3 rad t 8 9 s More on Transfer Functions Every transfer function can be represented as a ratio of two polynomials specifically H s P s Q s More 15 where 4 EE 105 Spring 2005 Discussion Notes written by Amin P s Q s are each polynomial functions of s The roots of P s and Q s have special names More specifically the roots of P s are referred to as the zeros of the transfer function H s The roots of Q s on the other hand are referred to as the poles of the transfer function H s Common Transfer Functions The circuits depicted in Figure 4 occur frequently in circuit analysis design Hence it is useful to have their associated transfer functions memorized It can be shown that V1 s I 1 s R1 R2 C1 V2 s V3 s C2 Figure 4 Common circuits V1 s R1 I 1 s 1 s R1 C1 16 V3 s 1 V2 s 1 s R2 C 2 17 5
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