Revision as of 11:49, 30 September 2013

Homework 6, ECE438, Fall 2013, Prof. Boutin

Please leave all mistakes and instructor's comments as is, unless otherwise noted by Prof. Mimi.

Question 1

The sampling rate should be at least twice of the highest frequency of the signal to avoid aliasing. Let's make it just twice of the highest frequency for illustration.

$ \begin{align} f_s=2 \cdot 2500=5000Hz \end{align} $

• Instructor's comment: Actually, strictly speaking, it should be greater than 5000HZ. -pm

We'd like a high pass filter that filters our everything below 60 Hz.

$ \begin{align} \omega_c=\frac{2\pi \cdot f_c }{f_s}=\frac{2\pi \cdot 60 }{5000} \end{align} $

Question 2

We can think of the long term trend as low frequency component and annual cycle as high frequency component. In order to remove the annual cycle, we need a low pass filter. The sampling rate $ f_s=12 $samples/year. The periodic component has frequency of $ f_c=1 $cycle/year.

\\ So the low pass filter has cutoff frequency of $ \omega_s=\frac{2\pi \cdot 1}{12}=\frac{\pi}{6} $

Question 3

Question 4

a) For $ k=0,1,...,N-1 $

$ \begin{align} X_N(k) &= \sum_{k=0}^{N-1}x[n]e^{-\frac{j2\pi nk}{N}} \\ &= x[0]e^{-\frac{j2\pi 0\cdot k}{N}} \\ &= 1 \end{align} $

• Instructor's comments: How about the other values of k? -pm

b) Using Euler Formula, we have

$ \begin{align} x[n] &= e^{\frac{j\pi n}{3}}(\frac{ e^{\frac{j\pi n}{6}} + e^{-\frac{j\pi n}{6}} }{2}) \\ &= \frac{1}{2}e^{\frac{j\pi n}{2}} + \frac{1}{2}e^{\frac{j\pi n}{6}} \end{align} $

Observing that $ x[n] $ has fundamental period $ N=12 $. Using IDFT, we have

$ \begin{align} x[n] &= \frac{1}{N}\sum_{n=0}^{N-1}e^{\frac{j2\pi nk}{N}} \\ \frac{1}{2}e^{\frac{j\pi n}{2}} + \frac{1}{2}e^{\frac{j\pi n}{6}} &= \frac{1}{12}\sum_{n=0}^{11}e^{\frac{j2\pi nk}{12}} \end{align} $

By comparison, we know for $ k=0,1,...,11 $

$ X_{12}[k] = \left\{ \begin{array}{ll} 6, & k=1,3 \\ 0, & otherwise. \end{array} \right. $

• Instructor's comments: How about k=12, k=13, and all the other values of k? -pm

c)

$ x[n]=(\frac{1}{\sqrt 2} + j\frac{1}{\sqrt 2})^n = (e^{\frac{j\pi}{4}})^n $

Then $ x[n] $ has fundamental period $ N=8 $. Using IDFT, we have

$ \begin{align} x[n] &= \frac{1}{N}\sum_{n=0}^{N-1}e^{\frac{j2\pi nk}{N}} \\ e^{\frac{j\pi n}{4}} &= \frac{1}{8}\sum_{n=0}^{7}e^{\frac{j2\pi nk}{8}} \end{align} $

By comparison, we know for $ k=0,1,...,7 $

$ X_{8}[k] = \left\{ \begin{array}{ll} 8, & k=1 \\ 0, & otherwise. \end{array} \right. $

• Instructor's comments: Don't forget to say that K[k} repeats periodically with period 8. THat way, all values of k are covered. -pm

Question 5

Observing that $ X(k) $ has a fundamental period $ N=4 $

$ \begin{align} x[n] &= \frac{1}{N}\sum_{k=0}^{N-1}(e^{j \pi k }+e^{-j \frac{\pi}{2} k})e^{\frac{j2\pi nk}{N}} \\ &= \frac{1}{4}\sum_{k=0}^{3}(e^{\frac{j2\pi (n+2)k}{4}} + e^{\frac{j2\pi (n-1)k}{4}}) \\ &= \frac{1}{4}\sum_{k=0}^{3}(e^{\frac{j2\pi (n+2)k}{4}-j2\pi k} + e^{\frac{j2\pi (n-1)k}{4}}) \\ &= \frac{1}{4}\sum_{k=0}^{3}(e^{\frac{j2\pi (n-2)k}{4}} + e^{\frac{j2\pi (n-1)k}{4}}) \\ \end{align} $

when $ n\neq 1 \text{ or } 2 $, using geometric series summation formula we have

$ x[n]=\frac{1}{4}( \frac{1-e^{j2\pi (n-2)}}{1-e^{\frac{j2\pi (n-2)}{4}}} + \frac{1-e^{j2\pi (n-1)}}{1-e^{\frac{j2\pi (n-1)}{4}}} ) = 0 $

when $ n=1 \text{ or } 2 $

$ x[n]=\sum_{k=0}^{3}1=4 $

$ x[n] $ will be periodic with 4.

NOTE: In general, $ X(k) $ does not need to have a length equal to the fundamental period. Suppose N is an arbitrary number, we can still derive the IDFT using argument that is similar to the one described above.

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