EE 05 Dr. A. Zidouri Electric Circuits Two-Port Circuits Two-Port Parameters Lecture #4-1 -
EE 05 Dr. A. Zidouri The material to be covered in this lecture is as follows: o ntroduction to two-port circuits o The Terminal Equations o The Two-Port -parameters o The Two-Port y-parameters - -
EE 05 Dr. A. Zidouri After finishing this lecture you should be able to: Understand the mportance of Two-Port Circuits Relate the Current and oltage at One Port to the Current and oltage at the Other Port. Determine the Two-Port -parameters Determine the Two-Port y-parameters - 3 -
EE 05 Dr. A. Zidouri ntroduction to Two-Port Circuits n analying some electrical systems, focusing on two pairs of terminal is convenient. Often, a signal is fed into one pair of terminals and then after being processed, is extracted at a second pair of terminals. The terminal pairs represent the points where signals are either fed in or extracted. They are referred to as ports of the system. Fig. 4-1 illustrates the basic two-port building block. Use of this building block is subject to several restrictions: o There can be no energy stored within the circuit o There can be no independent sources within the circuit o The current into the port must equal the current out of the port o All external connections must be made to either the input port or output port, no connections are allowed between the ports. The fundamental principle underlying two-port modeling of a system is that only the terminal variables (i 1, v 1, i, and v ) are of interest. - 4 -
EE 05 Dr. A. Zidouri The Terminal Equations n two-port network we are interested in relating the current and voltage at one port to the current and voltage at the other port. Fig. 4-1 shows the reference polarities of the terminal voltages and the reference directions of the terminal currents. Most general description is carried out in the s domain. We write all equations in the s domain, resistive networks and sinusoidal steady state solutions become special cases. Fig. 4- shows the basic building block in terms of the s-domain variables 1, 1,, and. 1 1 Fig. 4- The s-domain Two-Port Basic Building Block Out these four terminal variables, only two are independent. Thus we can describe a two-port network with just two simultaneous equations. However there are six ways in which to combine the four variables: mpedance Parameters (-parameters): = + 1 11 1 1 = + 1 1 (4-1) - 5 -
EE 05 Dr. A. Zidouri Admittance Parameters (y-parameters): Hybrid Parameters (h-parameters): nverse Hybrid Parameters (g-parameters): Transmission Parameters (a-parameters): nverse Transmission Parameters (b-parameters): = y + y 1 11 1 1 = y + y 1 1 = h + h 1 11 1 1 = h+ h = g + g 1 1 1 11 1 1 = g + g 1 1 = a a 1 11 1 = a a = b b 1 1 11 1 1 1 = b b 1 1 1 (4-) (4-3) (4-4) (4-5) (4-6) These six sets of equations may also be considered as three pairs of mutually inverse relations. The coefficients of the variables are called the parameters of the two-port circuit. We refer to the -parameters, y-parameters, a-parameters, b-parameters, h-parameters and g-parameters of the network. - 6 -
EE 05 Dr. A. Zidouri The Two-Port Parameters -parameters: or in matrix form: 1 11 1 1 = 1 = + 1 11 1 1 = + 1 1 [ ] 1 = (4-7) The values of the parameters can be evaluated by setting 1 =0 (input port open-circuited) or =0 (output port open-circuited). Thus, =, = = 0 = 0 1 1 11 1 1 1 =, = = 0 = 0 1 1 1 (4-8) The -parameters are also called the open-circuit impedance parameters: 11 = Open-circuit input impedance 1 = Open-circuit transfer impedance from port 1 to port 1 = Open-circuit transfer impedance from port to port 1 = Open-circuit output impedance Example 4-1 illustrates the determination of the -parameters for a resistive circuit. - 7 -
EE 05 Dr. A. Zidouri Example 4-1 Find the -parameters for a resistive circuit shown in Fig. 4-3 Solution: To obtain 11 and 1 we connect a voltage 1 (or a current source 1 ) to port 1 with port open circuited as in Fig. 4-4a. i 1 v 1 v i Fig. 4-3 The Circuit for Example 4-1 11 0 0 1 = = = 10 Ω 40 1 = 0 When is ero, 15 0.75 15 + 5 1 = = 1, therefore 1 1 = = = 7.5Ω = 0 1 1 0.75 0 10 To obtain 1 and we connect a voltage (or a current source ) to port with port 1 open circuited as in Fig. 4-4b. - 8 -
EE 05 Dr. A. Zidouri 1 =0 v 1 Fig. 4-4b Circuit for finding 1 an = 0 = = = 9.375Ω 40 1 15 5 = 0 = 0.8 5+ 0 When 1 is ero, ( ) 1, and 9.375 = hence = 1 7.5 1 = 9.375 = Ω = 0 1 0.8 Note that each of these parameters is the ratio of a voltage to a current and therefore is an impedance with the dimension of ohms; this is why they are called -parameters. When 11 =, the two-port network is said to be symmetrical. When the two-port network is linear and has no dependent sources, the transfer impedances are equal ( 1 = 1 ), and the two-port network is said to be reciprocal. - 9 -
EE 05 Dr. A. Zidouri y-parameters: or in matrix form: 1 y y 11 1 1 = y y 1 = y + y 1 11 1 1 = y + y 1 1 [ y] 1 = (4-9) The values of the parameters can be evaluated by setting 1 =0 (input port short-circuited) or =0 (output port short-circuited). Thus, y y =, y = = 0 = 0 1 1 11 1 1 1 =, y = = 0 = 0 1 1 1 (4-10) The y-parameters are also called the short-circuit admittance parameters: y 11 = Short-circuit input admittance y 1 = Short -circuit transfer admittance from port to port 1 y 1 = Short -circuit transfer admittance from port 1 to port y = Short -circuit output admittance Example 4- illustrates the determination of the y-parameters for a resistive circuit. - 10 -
EE 05 Dr. A. Zidouri Example 4- Obtain the y-parameters for the resistive circuit shown in Fig. 4-5 Solution: To obtain y 11 and y 1 we connect a current 1 (or a voltage source 1 ) to input port 1 with output port short circuited as in Fig. 4-6a. y 1 = 0 1 ( 4) 4 = 0 1 3 = 0 = 1 = 1 = 1 = 11 4 4+ 3 0.75S When is ero, = = and = hence 1 1 1 1 4 3, y 1 ( ) ( 4 ) 1 = = 3 = 0.5 1 = 0 1 3 0 = S - 11 -
EE 05 Dr. A. Zidouri To obtain y 1 and y we connect a current source (or a voltage source ) to port with port 1 short circuited as in Fig. 4-6b. y = = ( 8) = 8 = = 0 = 0 5 = 0 1 1 = 8 = 0.8 8+ When 1 is ero, ( ) 1 and 0.65S 8 5 = hence y 0.8 1 = = = 1 1.6 = 0 = 0 1 1 Note that each of these parameters is the ratio of a current to a voltage and therefore is an admittance with the dimension of siemens; this is why they are called y-parameters. 0.5S - 1 -
EE 05 Dr. A. Zidouri Self Test 4: a) Determine the -parameters for the circuit in Fig. 4-7 b) Determine the y-parameters for the circuit in Fig. 4-8 Fig. 4-7 Circuit for self test 4a Fig. 4-8 Circuit for self test 4b Answer: a) 60 11 = Ω = 40Ω = 70Ω = 40Ω 1 1 b) y = 0.73S y = y = 0.0909S y = 0.1364S 11 1 1-13 -