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ECE 437 Digital Signal Processing I FALL 2002 ECE 436 Digital Signal Processing I with Laboratory FALL 2002
Exam # November 25, 2002
Name:
There are six questions on the exam, weighted as shown below in determining the final score. Do all you work on these pages and indicate your final answer clearly. There are two extra blank pages at the back if you need the space, and you can use the backs of the sheets too if necessary. Neatness and clarity are important and can influence your grade! The exam is closed book, closed notes.
Grades
- (10 pts.)
- (10 pts.)
- (20 pts.)
- (20 pts.)
- (20 pts.)
- (20 pts.)
Total (100 pts.)
- The DFT of x = { 3 , 0 , 3 , 0 , − 1 , 0 , − 1 , 0 } is X = { 4 , 4 − j, 0 , 4 + j, 4 , 4 − j, 0 , 4 + j}. Let the signal x(n) be defined by the following plot.
−10 −5 0 5 10 15
−
−
0
1
2
3
4
x(n)
With X(ω) the DTFT of x(n), what is the value of X(π/4)?
- For the filter
H(z) =
2 + 4z−^1 + 2z−^2 1 − 34 z−^1 + 18 z−^2 draw the direct form II implementation and a cascade implementation.
- For the system given by
???
∂≥
∂≥
∂≥
x(n)
y(n)
write down a state space description. Mark on the diagram the location of the signals corresponding to the system states.
- The impulse response h(n) is nonzero only for n = 0 and n = 1 and the input signal x(n) is zero for n < 0. We are given that
{H(k)} is the DFT of {h(0), h(1), 0 , 0 } {X 1 (k)} is the DFT of { 0 , x(0), x(1), x(2)} {X 2 (k)} is the DFT of {x(2), x(3), x(4), x(5)}
With Y 1 (k) = H(k)X 1 (k) and Y 2 (k) = H(k)X 2 (k) we have
Y 1 = {− 1 , − 2 − j, 9 , −2 + j} Y 2 = { 0 , 0 , − 12 , 0 }
For y(n) = h(n) ∗ x(n) (linear convolution), find y(n) for n = 0, 1 , 2 , 3 , 4 , 5.
Extra worksheet.
