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a. Compute the transfer matrix of the network in the figure.b. Let $A=\left[\begin{array}{rr}{4 / 3} & {-12} \\ {-1 / 4} & {3}\end{array}\right] .$ Design a ladder networkwhose transfer matrix is $\overline{A}$ by finding a suitable matrix factorization of $A .$

$A=\left[\begin{array}{cc}{1} & {0} \\ {-1 / R_{3}} & {1}\end{array}\right]\left[\begin{array}{cc}{1} & {-R_{2}} \\ {0} & {1}\end{array}\right]\left[\begin{array}{cc}{1} & {0} \\ {-1 / R_{1}} & {1}\end{array}\right]$$=\left[\begin{array}{cc}{1} & {0} \\ {-1 / 6} & {1}\end{array}\right]\left[\begin{array}{cc}{1} & {-12} \\ {0} & {1}\end{array}\right]\left[\begin{array}{cc}{1} & {0} \\ {-1 / 36} & {1}\end{array}\right]$

Algebra

Chapter 2

Matrix Algebra

Section 5

Matrix Factorizations

Introduction to Matrices

Missouri State University

Campbell University

Lectures

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In mathematics, the absolu…

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in part, they were asked to compute the transfer matrix. Uh, the network given in the figure for problem 29. So looking at figure, we see that it is a network with we have a shunt circuit, followed by a serious circuit, followed by another shun circuit. So it follows the transfer matrix of the network is the product of matrices for each circuit and rightto left order. So we're starting with the shunts circuit. Resistance are three. This is 10 negative. One of our our three one times. And this is going to be the serious circuit with resistance. R two, this is one in the Siri's circuit has transfer matrix one negative are too 01 and followed by the transfer matrix for the Sean circuit with resistance R one just 10 negative one for our one one. Then we see that multiplying these matrices together we get if I start with the one here on the left in the middle, we have one negative are too negative. One over r three and are two of our three plus one. Multiply this by 10 negative one over our 11 and we get one plus are two over our one negative are too negative. One over R three minus are to over. Our one are three minus one over R one and finally are two of our our three plus one. And this could be rewritten if we'd like as r one plus R two over our one negative are too the opposite of here. We have least common denominator is R one R three speak it R one plus r two plus our three over are one are three and then we have our two plus r three over our three. So this is the transfer matrix for the whole network now in part B, whereas to design a ladder network with a transfer matrix of A by finding a suitable matrix factory ization of the matrix A now a is the matrix four thirds negative, 12 negative. 1/4 three. So we want to find resistances R one R two and r three such that a equals this matrix up here. So we see that immediately are two must be 12 homes and plugging that in. We see that four thirds must equal r one plus 12 over our one so one plus 12 over our one or thirds minus one is one third So 1 30 equals 12 over our one. So our one must be equal to 36 homes and we see that are two of our our three plus one has to equal three. So are to ever are three as equal to or are three as two equal are to over to which is 12 over to or six owns. And so the matrix a could be factored since this, it's this transfer matrix. They can be written as 10 negative one of her are three. Where are three with six homes? One followed by one negative are too so negative. 12 01 followed by 10 negative one over R one, which is negative. One of her 36 one. So we have given a factor ization for a

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