Prelim 1 Solutions - ECE220 Exam 1 Solutions(CLOSED BOOK...

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ECE220 Exam 1 Solutions (CLOSED BOOK) Spring 2008 February 21, 2008 PRINT name, netid, lab section: Kevin Problem 1 (10 points total) Using Euler’s identity show that 2 cos(ˆ ω 0 ) cos(ˆ ω 0 n + φ ) = cos(ˆ ω 0 ( n + 1) + φ ) + cos(ˆ ω 0 ( n 1) + φ ) (1) ......................................................................... ......................................................................... Recall the inverse Euler formula for cosines, cos( ˆ w 0 ) = ( e j ˆ w 0 + e - j ˆ w 0 ) / 2 Consider the left hand side of the equation (1), LHS = 2( e j ˆ w 0 + e - j ˆ w 0 ) / 2 · (( e j w 0 n + φ ) + e - j ( ˆ w 0 n + φ ) / 2) = 1 2 ( e j ( ˆ w 0 n + φ ω 0 ) + e j ( ˆ w 0 n + φ - ˆ ω 0 ) + e - j ( ˆ w 0 n + φ - ˆ ω 0 ) + e - j ( ˆ w 0 n + φ ω 0 ) ) = 1 2 ( e j ( ˆ w 0 ( n +1)+ φ ) + e j ( ˆ w 0 ( n - 1)+ φ ) + e - j w 0 ( n - 1)+ φ ) + e - j ( ˆ w 0 ( n +1)+ φ ) ) = cos(ˆ ω 0 ( n + 1) + φ ) + cos(ˆ ω 0 ( n 1) + φ ) = RHS 1
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PRINT name, netid, lab section: Dr. Hutchins Problem 2 (20 points total) Measurements of a discrete-time signal x [ n ] known to be a sinusoid of the following form x [ n ] = a cos(ˆ ω 0 n + φ ) are listed in Table 1 below for 0 n 5 n 0 1 2 3 4 5 x [ n ] 1.8478 0.6141 -1.0521 -1.9773 -1.5099 0.0209 Table 1. Use the identity established in Problem 1 to find values of a , φ and ˆ ω 0 . ......................................................................... ......................................................................... Rewriting the equation of Problem 1 we see that it is a difference equation of the form: x [ n + 1] = 2 cos(ˆ ω 0 ) x [ n ] x [ n 1] where x [ n ] = cos(ˆ ω 0 n + φ ). Thus it is capable (given the result of problem 1) of computing the next term of a sinusoidal sequence from the current sample and one past sample. It is an oscillator. Note that the unknown constant in the difference equation is 2 cos(ˆ ω 0 ). Any three consecutive samples of x [ n ] are sufficient to find this constant, and from it, ˆ ω 0 . For example: cos(ˆ ω 0 ) = x [ n + 1] + x [ n 1] 2 = 1 . 0521 + 1 . 8478 2 · 0 . 6441 = 0 . 6497 giving ˆ ω = sqrt 3 2 . This frequency defines the essential nature of the system - its poles in fact - it’s IIR. We now know the frequency at which the system wants to oscillate, and the exact amplitude a and phase φ are a matter of establishing ”initial conditions”. Since there are two numbers to determine, two values of the sequence are required. It is most convenient to use the smallest values of n : n = 0 and n = 1. For n = 0 we have a cos( φ ) = 1 . 8478 (2) and for n = 1 we have a cos(ˆ ω 0 + φ ) = 0 . 6141 (3) 2
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These are two equations in two unknowns, but are not linear equations. A famous procedure (called Prony’s method) will always solve them, but in some simple cases, the use of trig identities will work. You may recall (and if not, you certainly are capable to derive it from Euler’s formula) the identity: cos(ˆ ω 0 + φ ) = cos(ˆ ω ) cos( φ ) sin(ˆ ω ) sin( φ ) which by (3) leads to a (cos(ˆ ω ) cos( φ ) sin(ˆ ω ) = 0 . 6141 This can be divide, side by side, by (2) to give 0 . 6479 0 . 7618 tan( φ ) = 0 . 3323 or φ = π 8 and
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