220Final

# 220Final - ECE 220, Section 001 Final Exam Solutions 13...

This preview shows pages 1–5. Sign up to view the full content.

ECE 220, Section 001 Final Exam Solutions 13 December 2005 Problem 1. (10 points) Consider the pulse, p ( t ), in Equation 1: p ( t )= ± 1 , 0 t< 10 , 0 , otherwise. (1) Express the signal s ( t ) in Equation 2 in terms of p ( t ). s ( t 0 , t < 0 , - 2 , 0 5 , 1 , 5 15 , 0 ,t 15. (2) Solution The signal contains two pulses, one of width 5-0 = 5, and another of width 15-5 =10. The scaled signal s 1 ( t p (2 t ± 1 , 0 2 10 , 0 , otherwise. = ± 1 , 0 5 , 0 , otherwise, has a width of Fve. The signal s 2 ( t p ( t - 5) = ± 1 , 0 t - 5 < 10 , 0 , otherwise. = ± 1 , 5 15 , 0 , otherwise, has a width of 10. With these two signals we can write s ( t - 2 s 1 ( t )+ s 2 ( t ), so Fnally, s ( t - 2 p (2 t p ( t - 5) 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 2. (10 points) Consider the complex-valued function z ( t ) in Equation 3; t is a real number: z ( t )= ± 1+ j + e - j π 16 ² e - j 6 . (3) Determine the period, T , of this function. Clearly explain your answer. Guesses won’t count. Solution Let T denote the period of this function; T satisFes the property z ( t + T z ( t ) , t ( -∞ , ) . (4) We have z ( t + T ± j + e - j π 16 ² e - j ( t + T ) π 6 = ± j + e - j π 16 ² e - j 6 e - j 6 = z ( t ) e - j 6 (5) ±rom equations 4 and 5, we see that T must be the smallest positive number that satisFes the property: e - j 6 =1 . (6) Since 1 = e - j 2 π , we can write e - j 6 = e - j 2 π and thus 6 =2 π So, Fnally, T = 12 2
Problem 3. (10 points) Find all of the solutions to the equation z 3 - 27 j = 0 (7) where z is a complex number. Express the solutions in exponential format. Solution Express j in exponential format as j = e jπ/ 2 Using the fact that e j 2 πk = 1 for any integer k , rewrite equation 7 as z 3 = 27 j = 27 e 2 = 27 e 2 e j 2 From this last equation, we can write (after taking the third root): z = 27 1 / 3 ± e 2 ² 1 / 3 ± e j 2 ² 1 / 3 =3 e 6 e j 2 πk/ 3 (8) Let now k =0 , 1 , 2 in equation 8: z 1 e 6 e j 2 π 0 / 3 e 6 z 2 e 6 e j 2 π 1 / 3 e j 5 π/ 6 z 3 e 6 e j 2 π 2 / 3 e j 9 6 So, ±nally, z 1 e 6 ,z 2 e j 5 6 3 e j 9 6 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 4. (10 points) Consider the following set of equations: bx 1 + x 2 =0 x 2 - x 3 =1 x 1 + bx 3 b is a real-valued constant. Determine, if possible, all values for b that will make the system of equations have no solutions. Justify your answer.
This is the end of the preview. Sign up to access the rest of the document.

## 220Final - ECE 220, Section 001 Final Exam Solutions 13...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online