| Line 3: | Line 3: | ||
Grading format: | Grading format: | ||
<br>Similar to the grading format of HW1, HW2 is graded for completeness as well as theoretical understanding of course material. | <br>Similar to the grading format of HW1, HW2 is graded for completeness as well as theoretical understanding of course material. | ||
| + | |||
| + | Correction to original solution Q3: | ||
| + | <br> | ||
| + | The graph for x[n-k] should be 0 at n, and 1's from n+1 to n+10 | ||
| + | <br>The correct solution (as Kim described) is as follows: | ||
| + | <br>y[n]=0 for n<-10 | ||
| + | <br>y[n]=n+11 for -10<=n<=-1 | ||
| + | <br>y[n]=9-n for 0<=n<=8 | ||
| + | <br>y[n]=0 for n>8 | ||
<br>Comments: | <br>Comments: | ||
Latest revision as of 07:36, 10 February 2009
Grading format:
Similar to the grading format of HW1, HW2 is graded for completeness as well as theoretical understanding of course material.
Correction to original solution Q3:
The graph for x[n-k] should be 0 at n, and 1's from n+1 to n+10
The correct solution (as Kim described) is as follows:
y[n]=0 for n<-10
y[n]=n+11 for -10<=n<=-1
y[n]=9-n for 0<=n<=8
y[n]=0 for n>8
Comments:
- In Q2, $ (-1)^n = e^{j\pi n} $.
- When computing the DTFT using the summation formula, note that some expressions are in the form of geometric series.
- In Q3, convolution must be separated into various cases. The analytical expression will vary depending on the case. Also, drawing the signals is very helpful.
