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Problem 1 | Problem 1 | ||
− | a1. E = 6 so P = 0 | + | <br> |
− | a2. P = 6 / N so P is | + | a1. by using integral over function square E = 6 so P = 0 |
− | b. N must less than L + M - 1 | + | <br> |
+ | a2. P = 6 / N so P is infinite | ||
+ | <br> | ||
+ | b. N must less than L + M - 1 | ||
+ | <br> | ||
c. 2*PI/T greater than 20000PI | c. 2*PI/T greater than 20000PI | ||
− | so that T <= 1/10000 | + | <br> |
− | d. We did not cover the material | + | so that T <= 1/10000 |
− | Problem 2 | + | <br> |
− | a. | + | d. We did not cover the material |
− | b. | + | <br> |
− | c. | + | Problem 2 |
− | d. | + | <br> |
− | e. F | + | a. F(unit circle must be in ROCK) |
− | Problem 3: | + | <br> |
− | a.m = K/A | + | b. T(summable) |
+ | <br> | ||
+ | c. T F(cause of the square root) T F | ||
+ | <br> | ||
+ | d. F(ak=a*-k) T(a-k = ak) | ||
+ | <br> | ||
+ | e. F | ||
+ | <br> | ||
+ | Problem 3: | ||
+ | <br> | ||
+ | a.m = K/A | ||
+ | <br> | ||
b&c y(t) can be rewritten into 2 parts and has an energy of A^2/2 +k^2/32\t | b&c y(t) can be rewritten into 2 parts and has an energy of A^2/2 +k^2/32\t | ||
+ | <br> | ||
− | + | [[Back to Final Exam Sp 2005 solutions, ECE301 Spring 2013]] | |
− | Back to Final Exam Sp 2005 solutions, ECE301 Spring 2013 | + |
Latest revision as of 06:35, 3 May 2013
Problem 1
a1. by using integral over function square E = 6 so P = 0
a2. P = 6 / N so P is infinite
b. N must less than L + M - 1
c. 2*PI/T greater than 20000PI
so that T <= 1/10000
d. We did not cover the material
Problem 2
a. F(unit circle must be in ROCK)
b. T(summable)
c. T F(cause of the square root) T F
d. F(ak=a*-k) T(a-k = ak)
e. F
Problem 3:
a.m = K/A
b&c y(t) can be rewritten into 2 parts and has an energy of A^2/2 +k^2/32\t