Linear Algebra and Complex Variables
- 1. East West University PRESENTATION GROUP 3 COURSE CODE: MAT205 COURSE TITLE: LINEAR ALGEBRA AND COMPLEX VARIABLES SECTION: 3 1
- 2. Group 3 Members NAME: MD. RIAZUL ISLAM ID: 2016-1-80-028 NAME: UTHPOL KISOR MITHU ID: 2016-2-55-009 NAME: PRANTO SAHA UTTSOB ID: 2016-2-80-019 NAME: ABU MD.MAHIR HASSAN ID: 2016-1-55-012 NAME: KM. RAKIBUR RAHMAN ID: 2015-2-53-032 2
- 3. Introduction This presentation is mainly about to let us all know that the application of linear algebra in the field of electrical circuits. 3
- 4. What is Linear Algebra? Linear algebra is the branch of mathematics concerning vector spaces and linear mappings between such spaces. 4
- 5. Electrical circuits Electrical Circuit is nothing but just a combination resistor, capacitor, inductor, diodes, etc. including some logic gates. Each component has its own specification. And through which we get to know what currents and voltages are. An electrical circuit is a path in which electrons from a voltage or current flows. 5
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- 9. Linear Algebra in Linear Circuits Linear algebra most apparently uses by electrical engineers. When ever there is system of linear equation arises the concept of linear algebra. Various electrical circuits solution like Kirchhoff's law, Ohms law are conceptually arise linear algebra. 9
- 10. Internal Linear Circuit In Mobile phone 10
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- 12. After Using Nodal analysis We get the following 4 equations Again After simplifying the following equations we get, 380V1-150V2= 555 55V1-118V2+55V3= 0 15V2-380V3+15V4= 0 5V1+10V3-24V4= 0 12
- 14. 66 . 3 16 . 12 36 . 4 0 0 0 1 93 . 24 0 0 0 0 3936 . 0 1 0 4473 . 1 0 0 3947 . 0 1 83 . 0 78 . 2 1 0 0 0 04 . 0 1 0 0 0 0 3936 . 0 1 0 4473 . 1 0 0 3947 . 0 1 R2’=R2/-139.71 ; R3’=R3/15 & R4’=R4/-0.3947 R3’=R3-R2 & R4’=R4-R2 R3’=R3/-24.93 & R4’=R4/4.36 66 . 3 16 . 12 06 . 5 1 0 0 1 33 . 25 1 0 0 0 3936 . 0 1 0 4473 . 1 0 0 3947 . 0 1 14
- 15. 83 . 0 74 . 2 0 0 0 0 04 . 0 1 0 0 0 0 3936 . 0 1 0 4473 . 1 0 0 3947 . 0 1 R4’= R4+R3 Now Backward Substituting the Equation from Augment matrix we get .. .. . . .. . . . V4= 0.3029 Volt V3=0.04967 Volt V2=0.8626 Volt V1=1.8010 Volt 15
- 16. Now we have found the lower voltage as we need for our external device(Flash light; Camera; Speaker or any other else) of mobile phone .For doing these we had some linear equations and we already have solved them using Gaussian Elimination method. That is the applicable relation with Linear Algebra in Circuit of mobile phone. 16
- 18. This is a Drone 18
- 19. Internal Linear Circuit In Drones 19
- 21. FINDING NODE EQUATIONS NODE EQUATIONS: For Node V1: (V1-8)/2 + V1/3 + V1/4 + (V1-V2)/2 = 0 For Node V2: (V2-V1)/2 + V2/3 + (V2-V3)/2 = 0 For Node V3: (V3-V2)/2 + V3/3 + (V3-V4)/2 = 0 For Node V4: (V4-V3)/2 + V4/3 +V4/5 + (V4-8)/2=0 21
- 22. So, the equation becomes, 19V1-6V2 = 48 -3V1+8V2-3V3 = 0 -3V2 + 8V3-3V4 = 0 -15V3+46V4 = 120 22
- 23. Finding Node Voltages Using Cramer’s Rule The equation becomes, 19V1-6V2 = 48 -3V1+8V2-3V3 = 0 -3V2 + 8V3-3V4 = 0 -15V3+46V4 = 120 Let, A = 46 15 0 0 3 8 3 0 0 3 8 3 0 0 6 19 A1= 46 15 0 120 3 8 3 0 0 3 8 0 0 0 6 48 A2 = 46 15 120 0 3 8 0 0 0 3 0 3 0 0 48 19 A3= 46 120 0 0 3 0 3 0 0 0 8 3 0 48 6 19 A4= 120 15 0 0 0 8 3 0 0 3 8 3 48 0 6 19 23
- 24. NOW, V1= |𝑨𝟏| |𝑨| = 𝟏𝟏𝟎𝟔𝟒𝟎 𝟑𝟓𝟒𝟏𝟔 =3.1240 V V2= |𝑨𝟐| |𝑨| = 𝟔𝟕𝟎𝟑𝟐 𝟑𝟓𝟒𝟏𝟔 =1.8927 V V3= |𝑨𝟑| |𝑨| = 𝟔𝟖𝟏𝟏𝟐 𝟑𝟓𝟒𝟏𝟔 =1.8949 V V4= |𝑨𝟒| |𝑨| = 𝟏𝟏𝟒𝟔𝟎𝟎 𝟑𝟓𝟒𝟏𝟔 =3.2358 V 24
- 25. CRAMER’S RULE An explicit formula for the solution of a system of linear equations with as many equations as unknowns, valid whenever the system has a unique solution. Expresses the solution in terms of the determinants of the square matrix and of matrices obtained from it by replacing one column by the vector of right-hand-sides of the equations. 25
- 26. APPLICATION If the network is complex, the number of equation (unknowns) increase. In such case, the solution of simultaneous equations can be obtained by Cramer's Rule for determinants. We can apply this application to solve Kirchhoff's voltage law and determine the value of current flowing through different branches. 26
- 27. EXAMPLE Apply Kirchhoff's current law and voltage law to the circuit and Indicate the various branch currents (I1 & I2). 27
- 28. Applying KVL in loop 1 : A-B-E-F-A., - 15 i1 + 20 i2 + 50 = 0 ...........(1) And for loop 2 : B-C-D-E-B., -30 (i1 - i2 ) - 100 + 20 i2 = 0 ............(2) Rewriting all the equations, taking constants on one side. 15 i1 + 20 i2 = 50 …......(3) - 30 i1 + 50 i2 = 100 ........(4) 28
- 29. I1=D1/D= 500/1350 = 0.37 A I2= D2/D = 3000/1350 = 2.22 A ∴ I1 = 0.37 A and I2 = 2.22 A 29
- 30. BENEFIT TO USE LINEAR ALGEBRA IN LINEAR CIRCUIT In the linear circuit there are lots of node voltages , which we need to know .If we express them by linear equation then the finding result become much easier. It can also be used to find the result of mesh current by using linear equation. 30
- 32. Limitations The equations become for each passive component i(c)= 𝒊(𝒕) 𝒅𝒕 +i(0) v(C)=C 𝒅𝒗 𝒅𝒕 these are the differentiate and integral equation so they are not linear as we see … If we also try to convert them in phasor domain then they become v(L)=i(L)jѠL v(C)=i(L) 𝟏 𝒋Ѡ𝑪 here we also see that there are some complex entry. Because j= −𝟏 That’s why these complicated capacitor and inductive Circuit cannot be solved by using of the method of Linear Algebra.So this is the simple Limitation of relation between Linear algebra and non-Linear Circuit. 32
- 33. Thank you all………. Are there any question? 33