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# homeworkProblems - EE263 Autumn 2007-08 Prof S Boyd EE263...

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EE263 Autumn 2007-08 Prof. S. Boyd EE263 homework problems Lecture 2 – Linear functions and examples 2.1 A simple power control algorithm for a wireless network. First some background. We consider a network of n transmitter/receiver pairs. Transmitter i transmits at power level p i (which is positive). The path gain from transmitter j to receiver i is G ij (which are all nonnegative, and G ii are positive). The signal power at receiver i is given by s i = G ii p i . The noise plus interference power at receiver i is given by q i = σ + summationdisplay j negationslash = i G ij p j where σ> 0 is the self-noise power of the receivers (assumed to be the same for all receivers). The signal to interference plus noise ratio (SINR) at receiver i is defined as S i = s i /q i . For signal reception to occur, the SINR must exceed some threshold value γ (which is often in the range 3 – 10). Various power control algorithms are used to adjust the powers p i to ensure that S i γ (so that each receiver can receive the signal transmitted by its associated transmitter). In this problem, we consider a simple power control update algorithm. The powers are all updated synchronously at a fixed time interval, denoted by t = 0 , 1 , 2 ,... . Thus the quantities p , q , and S are discrete-time signals, so for example p 3 (5) denotes the transmit power of transmitter 3 at time epoch t = 5. What we’d like is S i ( t ) = s i ( t ) /q i ( t ) = αγ where α> 1 is an SINR safety margin (of, for example, one or two dB). Note that increasing p i ( t ) (power of the i th transmitter) increases S i but decreases all other S j . A very simple power update algorithm is given by p i ( t + 1) = p i ( t )( αγ/S i ( t )) . (1) This scales the power at the next time step to be the power that would achieve S i = αγ , if the interference plus noise term were to stay the same. But unfortunately, changing the transmit powers also changes the interference powers, so it’s not that simple! Finally, we get to the problem. (a) Show that the power control algorithm (1) can be expressed as a linear dynamical system with constant input, i.e. , in the form p ( t + 1) = Ap ( t ) + b, where A R n × n and b R n are constant. Describe A and b explicitly in terms of σ,γ,α and the components of G . (b) Matlab simulation. Use matlab to simulate the power control algorithm (1), starting from various initial (positive) power levels. Use the problem data G = 1 . 2 . 1 . 1 2 . 1 . 3 . 1 3 , γ = 3 , α = 1 . 2 , σ = 0 . 01 . 1

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Plot S i and p as a function of t , and compare it to the target value αγ . Repeat for γ = 5. Comment briefly on what you observe. Comment: You’ll soon understand what you see. 2.2 State equations for a linear mechanical system. The equations of motion of a lumped me- chanical system undergoing small motions can be expressed as M ¨ q + D ˙ q + Kq = f where q ( t ) R k is the vector of deflections, M , D , and K are the mass , damping , and stiffness matrices, respectively, and f ( t ) R k is the vector of externally applied forces. Assuming M is invertible, write linear system equations for the mechanical system, with state x = [ q T ˙ q T ] T , input u = f
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