This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ECE220 Exam 1 Solutions (CLOSED BOOK) Spring 2008 February 21, 2008 PRINT name, netid, lab section: Kevin Problem 1 (10 points total) Using Eulers identity show that 2 cos( ) cos( n + ) = cos( ( n + 1) + ) + cos( ( n 1) + ) (1) ......................................................................... ......................................................................... Recall the inverse Euler formula for cosines, cos( w ) = ( e j w + e j w ) / 2 Consider the left hand side of the equation (1), LHS = 2( e j w + e j w ) / 2 (( e j ( w n + ) + e j ( w n + ) / 2) = 1 2 ( e j ( w n + + ) + e j ( w n +  ) + e j ( w n +  ) + e j ( w n + + ) ) = 1 2 ( e j ( w ( n +1)+ ) + e j ( w ( n 1)+ ) + e j ( w ( n 1)+ ) + e j ( w ( n +1)+ ) ) = cos( ( n + 1) + ) + cos( ( n 1) + ) = RHS 1 PRINT name, netid, lab section: Dr. Hutchins Problem 2 (20 points total) Measurements of a discretetime signal x [ n ] known to be a sinusoid of the following form x [ n ] = a cos( n + ) are listed in Table 1 below for 0 n 5 n 1 2 3 4 5 x [ n ] 1.8478 0.61411.05211.97731.5099 0.0209 Table 1. Use the identity established in Problem 1 to find values of a , and . ......................................................................... ......................................................................... Rewriting the equation of Problem 1 we see that it is a difference equation of the form: x [ n + 1] = 2 cos( ) x [ n ] x [ n 1] where x [ n ] = cos( n + ). 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( ). Any three consecutive samples of x [ n ] are sufficient to find this constant, and from it, . For example: cos( ) = x [ n + 1] + x [ n 1] 2 = 1 . 0521 + 1 . 8478 2 . 6441 = 0 . 6497 giving = sqrt 3 2 . This frequency defines the essential nature of the system  its poles in fact  its IIR. We now know the frequency at which the system wants to oscillate, and the exact amplitude a and phase 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....
View
Full
Document
This note was uploaded on 03/13/2008 for the course ECE 2200 taught by Professor Johnson during the Spring '05 term at Cornell University (Engineering School).
 Spring '05
 JOHNSON

Click to edit the document details