Advanced Analysis Techniques for Circuits in EMLAB

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Explore additional analysis techniques like superposition, Thevenin's and Norton's theorems, maximum power transfer, and application examples in EMLAB. Learn about linearity in systems, equivalent circuits, and how to use superposition to simplify linear circuits. Dive into examples showcasing the application of these techniques for circuit analysis.

  • Analysis Techniques
  • EMLAB Circuits
  • Superposition
  • Thevenins Theorem
  • Linearity

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  1. 1 Chapter 5. Additional analysis techniques EMLAB

  2. Contents 2 1. Introduction 2. Superposition 3. Thevenin s and Norton s theorems 4. Maximum power transfer 5. Application Examples EMLAB

  3. 1. Introduction 3 Examples of equivalent circuits EMLAB

  4. Linearity 4 Linear system L ?1(t) ?1(t) Linear system L ?2(t) ?2(t) Linear system L ??1+ ??2(t) ??1(t) + ??2(t) A system satisfying the above statements is called as a linear system. Resistors, Capacitors, Inductors are all linear systems. An independent source is not a linear system. All the circuits in the circuit theory class are linear systems! EMLAB

  5. Example of linear system 5 R1 output Resistor +?1(t) 1k R1 ?1(t) ?1(t) = i1(?) ?1 1k input Capacitor +?2(t) C1 C1 1n 1n ?2(t) t ?2(t) =1 ? i2(?)?? 0 EMLAB

  6. Example 5.1 6 For the circuit shown in the figure, determine the output voltage ????using linearity. First, arbitrarily assume that the output voltage is ????= 1 [?]. ?2=?2 ????= ?2= 1 [?] 2?= 0.5 [??] ?1= 4? ?2+ ?2= 3 [?] ?1=?1 ??= 2? ??+ ?1= 6 [?] 3?= 1 [??] ?0= ?1+ ?2= 1.5 [??] For the arbitrary assumption that Vout= 1 [V], the source voltage Voshould be 6 V. Then from linearity, the actual output voltage should satisfy the following relation. ??????= 6:1 ??????= 2 [?] ??:????= 12:???? ???? EMLAB

  7. 2. Superposition 7 Source superposition Circuit with voltage source set to zero (Short circuited) ?? ?? ??,2 ??,1 ?? = + + ??,2 + ?? + ??,1 Circuit Circuit Circuit Circuit with current source set to zero(Open) ?? ?? ??= ??,1+ ??,2 ??= ??,1+ ??,2 Superposition is utilized to simplify the original linear circuits. If a voltage source is eliminated, it is replaced by a short circuit connected to the original terminals. If a current source is eliminated, it is replaced by an open circuit. EMLAB

  8. Example 5.2 8 To provide motivation for this subject, let us examine the simple circuit below, in which two sources contribute to the current in the network. The actual values of the sources are left unspecified so that we can examine the concept of superposition. = + 3? ?1+ 3? (?1 ?2) = 0 3? (?2 ?1) + 6? ?2+ ?2= 0 ?1+ 3? ?1+ 3? (?1 ?2) = 0 3? (?2 ?1) + 6? ?2= 0 ?1+ 3? ?1+ 3? (?1 ?2) = 0 3? (?2 ?1) + 6? ?2+ ?2= 0 ?1+ (3? + 3?) ?1 3? ?2= 0 3? ?1+ (3? + 6?) ?2+ ?2= 0 ?1 ?2 ?1 ?2 6? 3? 3? 9? = ?1 ?2 ?1 ?2 6? 3? 3? 9? = 1 1 ?1 ?2 ?1 ?2 3?1 ?2 ?1 2?2 3 1 1 2 = = 15? 15? EMLAB

  9. Example 5.3 9 Let us use superposition to find ??in the circuit in the following figure. 1? + 2? 1? + 2? + 6? =2 ??= (2 10 3) 3[??] ?? = ??(6?) = 4[?] 6? ?? = 3 = 2[?] 1? + 2? + 6? ??= ?? + ?? = 6[?] EMLAB

  10. Example 5.4 10 Consider now the network in Fig. 5.4a. Let us use superposition to find Vo. EMLAB

  11. 11 Superposition : 8? 3 =24 6? =18 ?? = ?1 ?1= 6 [?] 7[?] 8? 3+ 2? 7 6? + 2? 10 3?||6? =30 ??= ?? + ?? =48 ?? = (2 10 3) 7[?] 7[?] Node analysis : 2 10 3+?? 6 ?1 +?? ?1 4? +?? 6?= 0 2? ?1 ?? 4? +?1 (?? 6) 2? +?1 2?= 0 EMLAB

  12. Thevenins and Nortons theorems 12 ? a + ? ?? ?? + ???? ?? Single port network ? = ?(?) ? = ?? + ? = ?? ? + ?? ? = ?(?) ? = ?? + ? = ????? + ???? 1. To find D, measure current with =0. (short circuit current) 2. To find C, measure the variation of i with changing. (admittance) 1. To find B, measure voltage with i =0. (open circuit voltage) 2. To find A, measure the variation of with i changing. (impedance) EMLAB

  13. How to construct Thevenins equivalent circuit 13 (1) Measure open circuit voltage (VOC) with a volt meter. Input resistance of a voltmeter is infinite. ?? ? = 0 + ??? Circuit V ??? ?? = + ??? (2) Measure resistance (RTH) with sources suppressed. Voltage sources short circuited and current sources open circuited. short Circuit Ohm meter open EMLAB

  14. How to construct Nortons equivalent circuit 14 (1) Measure short circuited current (ISH). ?? Input resistance of an ammeter is zero. ??? Circuit A ?? = ??? ??? (2) Measure resistance (RTH) with sources suppressed. Voltage sources short circuited and current sources open circuited. short Circuit Ohm meter open EMLAB

  15. Example 5.6 15 Let us use Th venin s and Norton s theorems to find Voin the network below. 6? ??= 3? + 6? 9 = 6 [?] 1 ??= 3 [??] = 6 [?] 3?+1 1 6? EMLAB

  16. Example 5.7 16 Let us use Th venin s theorem to find in the network in Fig. 5.9a. 6? ???1= 12 = 8[?] 6? + 3? ?? 1= 2? +6? 3? 6? + 3?= 4?[ ] ???1= (2 10 3)(4?) + 8 = 16[?] EMLAB

  17. Circuits containing only dependent sources 17 ?1+ ??= 1 ??=3 ?1 1?+?1 2?? +?1 1 2? 7[?] = 0 2? ?1=?? 1?=3 ?2=1 2?? =1 1 7[??], 7[??], ?3= 2?= 0.5[??] 1? ??= ?1+ ?2+ ?3=15 14[??] ?? =1 =14 15[? ] ?? EMLAB

  18. Example 5.10 18 Let us determine at the terminals A-B for the network in Fig. 5.12a. ?1 2000?? 2? +?1 1?+?1 ?2 = 0 3? ?2 ?1 3? +?2 2?= 1 10 3 ??=?1 1? ?2=10 7[?] 1 10 3=10 ?2 ?? = 7[? ] EMLAB

  19. Circuits containing both independent and dependent sources (Example 5.11) In these types of circuits we must calculate both the open-circuit voltage and short-circuit current to calculate the Th venin equivalent resistance. 19 KCL for the supernode around the 12-V source is ) (???+ 12) ( 2000?? 1? +???+ 12 2? +??? 2?= 0 =??? ???= 12 ?? 2?, ???= 6[?], = 18[??] 2? 3 ?? =??? =1 1? = 18 3[? ], ??= 6 7[?] 1? + 1? +1 ??? 3? EMLAB

  20. Example 5.12 20 Let us find in the network in Fig. 5.14a using Th venin s theorem. ?1=?? 2?, ?2= 2[??] ?? 2? 2 10 3 ?? = 4? ?1= 4[??] ???= 2??1+ 3 = 11[?] ?? 4?=?? 2? 2 10 3 2?=3 3 ?? = 8[?], ?3= 2[??] ???=?? 2?+ ?3=11 2[??] ?? =??? 6? =33 = 2[? ], ??= 11 4[?] ??? 2? + 6? EMLAB

  21. Example 5.14 21 We will now demonstrate how to find Vo in the circuit in Fig. 5.16a using the repeated application of source transformation. 4? ??= (4 10 3) = 1[??] 4? + 4? + 8? ??= (1 10 3)(8?) = 8[?] EMLAB

  22. 4. Maximum Power Transfer 22 By employing Th venin s theorem, we can determine the maximum power that a circuit can supply and the manner in which to adjust the load to effect maximum power transfer. 2 ? ?????= ?2??= ?? ? + ?? =(? + ??)2?2 2?2??(? + ??) (? + ??)4 ?????? ??? = 0 ??= ? In other words, maximum power transfer takes place when the load resistance RL=R. EMLAB

  23. Example 5.16 23 Let us find the value of RL for maximum power transfer in the network in Fig. 5.20a and the maximum power that can be transferred to this load. ?1= 2 10 3, 3?(?2 ?1) + 6??2+ 3 = 0 ?2=1 3[??] ???= 4??1+ 6??2= 10[?] ?? = 6[? ] ??= ?? = 6[? ] 2 10 12? (6?) =25 ??= 6[??] EMLAB

  24. Example 5.17 24 Let us find RLfor maximum power transfer and the maximum power transferred to this load in the circuit in Fig. 5.21a. ??? 2000?? 1? + 3? + ( 4 10 3) +??? 2?= 0 =??? ?? 2? ???= 8[?], ???= 4[??] ?? =??? = 2[? ] ??? 2 8 (6?) =8 ??= 3[??] 12? EMLAB

  25. Example 5.21 25 Let us design the pad so that it has an equivalent resistance of and divides (i.e., attenuates) the input voltage by a factor of 10. Coaxial cable is often used in very-high-frequency systems. For example, it is commonly used for signal transmission with cable television. In these systems resistance matching, the kind we use for maximum power transfer, is critical. In the laboratory, a common apparatus used in high-frequency research and development is the attenuator pad. The attenuator pad is basically a voltage divider, but the equivalent resistance at both its input ports is carefully designed for resistance matching. Given the network in Fig. 5.26 in which a source, modeled by VSand RS(50 ), drives an attenuator pad, which is connected to an equivalent load. EMLAB

  26. Measurement example 26 EMLAB

  27. 27 ?? ??= ?2+ [?1||(?2+ 50)] = 50 ?? ???= ?2+ [?1||(?2+ 50)] = 50 ?1 ?? = ?? ?1+ ?2+ 50 50 =?? ????= ?? ?? + 50 2 ???? ?? ???? ?? ?? ?? =1 ?1 1 10 = = 2 ?1+ ?2+ 50 ?1= 20.83[ ], ?2= 33.33[ ] EMLAB

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