Linearity. If kx is applied to the element, the output must be ky. kx ky. 2. additivity property. x 1 y 1, x 2 y 2

Transcription

1 Lnearty An element s sad to be lnear f t satsfes homogenety (scalng) property and addte (superposton) property. 1. homogenety property Let x be the nput and y be the output of an element. x y If kx s appled to the element, the output must be ky. 2. addtty property kx ky x 1 y 1, x 2 y 2 If (x 1 x 2 ) s appled to the element, the output must be y 1 y 2. (x 1 x 2 ) y 1 y 2

2 Lnear Elements Resstor It s a lnear element because ts oltage-current relaton satsfes both homogenety and addtty property. = R Inductor It s a lnear element.. Its oltage-current relatonshp 1 s = L d dt Capactor It s also a lnear element 2. It relatonshp s = C d dt 1 (dfferental operator s lnear) 2 Inductor and capactor should not hae any ntal energy

3 Dependent Sources Dependent oltage and current sources are lnear as long as ther nput-output relatonshp s lnear. o = k, o = r, o = k, o = g A lnear crcut conssts of only lnear elements and lnear dependent sources. In lnear crcut, the output and nput are related by a lnear relatonshp. I V V Lnear Network I R

4 What about Independent Sources? They are nonlnear as ther relatonshp does not satsfy the lnearty. I V V I Fgure: Independent oltage source Fgure: Independent current source = V = I

5 Test yourself Fnd whether the element s lnear or not. (A) 2 (V) -2 = = 1 2, 2 2 = 2 2 ( 1 2 ) 3 = It s nonlnear. For an element to be lnear, ts characterstcs must not only be a straght lne but also pass through orgn.

6 Superposton It states that n a lnear network wth a number of ndependent sources, the response can be found by summng the responses to each ndependent source actng alone, wth all other ndependent sources set to zero. 1. Consder one ndependent source at a tme whle all other ndependent sources are turned off. It means that oltage source s replaced by short crcut and current source by open crcut. 2. Leae dependent sources n the crcut as they are controlled by other arables.

7 R 1 e V R 2 I Let us fnd e usng nodal analyss. By KCL, e V e = I R 1 R 2 e = R 2 R 1 R 2 V R 1R 2 R 1 R 2 I e = αv βi where α and β are constants. Notce that e has two components as there are two ndependent sources.

8 Let us analyze the same crcut usng superposton. 1. Voltage source actng alone: R 1 e V V R 2 e V = R 2 R 1 R 2 V 2. Current source actng alone: R 1 e I R 2 I e I = R 1R 2 R 1 R 2 I By superposton, e = e V e I = R 2 R 1 R 2 V R 1R 2 R 1 R 2 I

9 Test yourself Use superposton to fnd the laue of current x. 6 Ω x 3 Ω 8 V 2 A 3 x Voltage source actng alone: 6 Ω x 3 Ω 8 V 3 x 8 6 x 3 x 3 x = 0 x = 2 3 A

10 Current source actng alone: 6 Ω x e 3 Ω By KCL, e x e 3 = 0 2 A 3 x x = e 6 By superposton, By solng ths, x = 1 2 A. x = xv xi x = = 1 6 A

11 Theenn s Theorem Any lnear two-termnal crcut can be replaced by an equalent crcut consstng of a oltage source V Th n seres wth a resstor R Th, where V Th s the open-crcut oltage at the termnals and R Th s the nput or equalent resstance at the termnals when the ndependent sources are turned off. R R Th V m I n V Th

12 Proof Let us apply a current source and fnd the oltage. R V m I n Snce the network s lnear, ts termnal oltage can be found by superposton. The oltage wll hae as many components as the number of ndependent sources. = m α m V m n β n I n R

13 = V Th R Th Ths equaton can be represented n a network form as shown below. R Th V Th Ths makes us replace any lnear network wth ts Theenn equalent. R R Th V m I n V Th

14 The Thenn equalent crcut for any lnear network at a gen par of termnals conssts of a oltage source V Th n seres wth a resstor R Th. The oltage V Th and R Th can be obtaned as follows. 1. V Th can be found by calculatng the open crcut oltage at the desgnated termnal par on the orgnal network. 2. R Th can be found by calculatng the resstance of the open-crcut network seen from the desgnated termnal par wth all ndependent sources nternal to the network set to zero.

15 If the network has dependent sources, R Th can be found as follows. 1. Turn off all ndependent sources. Do not turn off dependent sources. 2. Apply a test oltage o at the termnals and calculate the current o. o Network wth all ndependent sources set to zero o R Th = o o

16 Example R 1 e V R 2 I V Let us fnd e usng Theenn equalent. To fnd V Th : R 1 V oc e I V Th = V IR 1

17 To fnd R Th : R 1 R Th R Th = R 1 Theenn equalent s R Th e V Th R 2 By oltage dson, R 2 R 2 R 2 R 1 R 2 e = V Th = (V IR 1 ) = V I R 2 R Th R 2 R 1 R 1 R 2 R 1 R 2

18 Norton s Theorem It states that a lnear two-termnal crcut can be replaced by an equalent crcut consstng of a current source I N n parallel wth a resstor R N, where I N s the short-crcut current through the termnals and R N s the nput or equalent resstance at the termnals when the ndependent sources are turned off. R V m I n I N R N R N = R Th

19 The Norton equalent crcut for any lnear network at a gen par of termnals conssts of a current source I N n parallel wth a resstor R N. The current I N and resstance R N can be obtaned as follows: 1. I N can be found by applyng a short at the desgnated termnal par on the orgnal network and calculatng or measurng the current through the short crcut. R V m I n I sc I N = I sc 2. R N can be found n the same manner as R Th.

20 Proof Let us apply a oltage source and fnd the current. R V m I n Snce the network s lnear, the current can be found by superposton. The current wll hae as many components as the number of ndependent sources. ( = α m V m ) β n I n R m n

21 = I N R N Ths equaton can be represented n a network form as shown below. I N R N Ths makes us replace any lnear network wth ts Norton equalent. R V m I n I N R N

22 Source Transformaton a a R I R V (a) b (b) b These two crcuts are equal as long as they hae same characterstcs at ther termnals. V = IR I = V R

23 Theenn and Norton Transformaton a a R Th I N R Th V Th (c) b (d) b V Th = I N R Th I N = V Th R Th

24 Maxmum Power Transfer R Th V Th R L The power delered to the load s p p max p = 2 R L = V 2 Th (R Th R L ) 2 R L R Th R L

25 V 2 Th dp = 0 dr L ( (RTh R L ) 2 ) 2R L (R Th R L ) (R Th R L ) 4 = 0 R L = R Th Maxmum power s transferred to the load when the load resstance equals the Theenn resstance as seen from the load (R L = R Th ). The maxmum power transferred to load s p max = V 2 Th 4R Th