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ADDITIONAL ANALYSIS TECHNIQUES LEARNING GOALS REVIEW LINEARITY The property has two equivalent definitions. We show and application of homogeneity APPLY SUPERPOSITION We discuss some implications of the superposition property in linear circuits DEVELOP THEVENIN’S AND NORTON’S THEOREMS These are two very powerful analysis tools that allow us to focus on parts of a circuit and hide away unnecessary complexities MAXIMUM POWER TRANSFER This is a very useful application of Thevenin’s and Norton’s theorems THE METHODS OF NODE AND LOOP ANALYSIS PROVIDE POWERFUL TOOLS TO DETERMINE THE BEHAVIOR OF EVERY COMPONENT IN A CIRCUIT The techniques developed in chapter 2; i.e., combination series/parallel, voltage divider and current divider are special techniques that are more efficient than the general methods, but have a limited applicability. It is to our advantage to keep them in our repertoire and use them when they are more efficient. In this section we develop additional techniques that simplify the analysis of some circuits. In fact these techniques expand on concepts that we have already introduced: linearity and circuit equivalence SOME EQUIVALENT CIRCUITS ALREADY USED LINEARITY THE MODELS USED ARE ALL LINEAR. MATHEMATICALLY THIS IMPLIES THAT THEY SATISFY THE PRINCIPLE OF SUPERPOSITION THE MODEL y Tu IS LINEAR IFF T (1u1 2 u2 ) 1Tu1 2Tu2 for all possible input pairs u1 , u2 and all possible scalars 1 , 2 v IS A VECTOR CONTAINING ALL THE NODE VOLTAGES AND f IS A VECTOR DEPENDING ONLY ON THE INDEPENDENT SOURCES. IN FACT THE MODEL CAN BE MADE MORE DETAILED AS FOLLOWS Av Bs AN ALTERNATIVE, AND EQUIVALENT, DEFINITION OF LINEARITY SPLITS THE SUPERPOSITION PRINCIPLE IN TWO. THE MODEL y Tu IS LINEAR IFF 1. T (u1 u2 ) Tu1 Tu2 , u1, u2 additivity 2. T (u) Tu, , u USING NODE ANALYSIS FOR RESISTIVE CIRCUITS ONE OBTAINS MODELS OF THE FORM Av f HERE, A,B, ARE MATRICES AND s IS A VECTOR OF ALL INDEPENDENT SOURCES FOR CIRCUIT ANALYSIS WE CAN USE THE LINEARITY ASSUMPTION TO DEVELOP SPECIAL ANALYSIS TECHNIQUES homogeneit y NOTICE THAT, TECHNICALLY, LINEARITY CAN NEVER BE VERIFIED EMPIRICALLY ON A SYSTEM. BUT IT COULD BE DISPROVED BY A SINGLE COUNTER EXAMPLE. IT CAN BE VERIFIED MATHEMATICALLY FOR THE MODELS USED. FIRST WE REVIEW THE TECHNIQUES CURRENTLY AVAILABLE A CASE STUDY TO REVIEW PAST TECHNIQUES DETERMINE VO SOLUTION TECHNIQUES AVAILABLE?? Redrawing the circuit may help us in recognizing special cases The procedure can be made entirely algorithmic USING HOMOGENEITY 1. Give to Vo any arbitrary value (e.g., V’o =1 ) 2. Compute the resulting source value and call it V’_s 3. Use linearity. REQ V1 4. The given value of the source (V_s) corresponds to k Assume that the answer is known. Can we Compute the input in a very easy way ?!! R1 R2 V0 R2 … And Vs using a second voltage divider VS R4 REQ REQ R4 REQ R1 R2 V1 V0 REQ R2 Solve now for the variable Vo VS VS' Hence the desired output value is If Vo is given then V1 can be computed using an inverse voltage divider. V1 VS' V0' kVS' kV0' , k V0 kV0' VS ' V ' 0 VS This is a nice little tool for special problems. Normally when there is only one source and when in our judgement solving the problem backwards is actually easier SOLVE USING HOMOGENEITY ASSUME Vout V2 1[V ] I1 VO NOW USE HOMOGENEITY VO 6[V ] Vout 1[V ] VO 12[V ] Vout 2[V ] LEARNING EXTENSION COMPUTE IO USING HOMOGENEITY. USE I 6mA VS 1.5[mA] 2k V1 6[V ] VS 0.5[mA ] 2mA 1.5[mA] V1 3[V ] 0.5[mA ] USE HOMOGENEITY I 2mA I O 1mA I 6mA I O ____ ASSUME IO 1mA Source Superposition This technique is a direct application of linearity. It is normally useful when the circuit has only a few sources. VS FOR CLARITY WE SHOW A CIRCUIT WITH ONLY TWO SOURCES + - IL Due to Linearity circuit V 1 V 2 IS L Can be computed by setting the current source to zero and solving the circuit L VL _ VL a1VS a2 I S CONTRIBUTION BY VS CONTRIBUTION BY I S 1 VL V L2 + Can be computed by setting the voltage source to zero and solving the circuit Circuit with voltage source set to zero (SHORT CIRCUITED) SOURCE SUPERPOSITION I I L2 1 L = V 1 L + Circuit with current source set to zero(OPEN) Due to the linearity of the models we must have I L I L1 I L2 VL VL1 VL2 Principle of Source Superposition The approach will be useful if solving the two circuits is simpler, or more convenient, than solving a circuit with two sources We can have any combination of sources. And we can partition any way we find convenient VL2 LEARNING EXAMPLE WE WISH TO COMPUTE THE CURRENT = i 1 + Req 3 3 || 6 [k ] R 6 (3 || 3) [k ] eq i2" Loop equations v2 Req Contribution of v1 Once we know the “partial circuits” we need to be able to solve them in an efficient manner Contribution of v2 LEARNING EXAMPLE Compute V0 using source superposition We set to zero the voltage source Current division Ohm’s law Now we set to zero the current source Voltage Divider 2[V ] 6k 3V V0" V0 V0' V0" 6[V ] + - 3k LEARNING EXAMPLE Compute V0 using source superposition We must be able to solve each circuit in a very efficient manner!!! If V1 is known then V’o is obtained using a voltage divider V1 can be obtained by series parallel reduction and divider Set to zero current source + - V1 V1 _ Set to zero voltage source 6k 4k||8k V1 2k V1 _ 8/3 (6) 2 8/3 V'0 _ 2k VO' 6k 18 V1 [V ] 6k 2k 7 The current I2 can be obtained using a current divider and V”o using Ohm’s law I2 2k||4k 2mA + I2 6k V"0 I2 2k (2k || 4k ) (2)mA 2k 6k (2k || 4k ) VO" 6kI 2 VO VO' VO" 2k _ WHEN IN DOUBT… REDRAW! + + + Sample Problem COMPUTE I0 USING SOURCE SUPERPOSITION 1. Consider only the voltage source I 01 1.5mA 2. Consider only the 3mA source 3. Consider only the 4mA source Current divider I 02 1.5mA I 03 0 Using source superposition I 0 I 01 I 02 I 03 3mA SUPERPOSITION APPLIED TO OP-AMP CIRCUITS TWO SOURCES. WE ANALYZE ONE AT A TIME CONTRIBUTION OF V1. Basic inverter circuit R2 VO 1 V1 R1 Principle of superposition VO VO 1 VO 2 R2 R V1 1 2 V2 R1 R1 CONTRIBUTION OF V2 Basic non-inverting amplifier V O2 Notice redrawing for added clarity R2 1 V2 R1 I1 I O1 I1 2 2 2 1 3 1 3 Linearity