Solution to HW7
Q1.
Recall, the Discrete Fourier Transform is defined as follows -
Definition: let x[n] be a DT signal with Period N. Then,
$ X [k] = \sum_{k=0}^{N-1} x[n].e^{-j2\pi kn/N} $
$ x [n] = (1/N) \sum_{k=0}^{N-1} X[k].e^{j2\pi kn/N} $
$ x_1[n]= e^{j \frac{2}{3} \pi n}; $
Function's period N = 3,
Using IDFT,
$ \begin{align} x[n] &= \frac{1}{3} \sum_{k=0}^{2} X[k].e^{j2\pi kn/3} \\ e^{j \frac{2}{3} \pi n} &= \frac{1}{3} \sum_{k=0}^{2} X[k].e^{j2\pi kn/3} \\ &= \frac{1}{3} \left[ X[0].e^{j0} + X[1].e^{j2\pi n/3} + X[2].e^{j2\pi (2/3)} \right] \\ &= \frac{1}{3} \left[ X[0] + X[1].e^{j2\pi n/3} + X[2].e^{j4\pi /3} \right] \end{align} $
For the two sides to be equal,
X[0] = 0
X[1] = 3
X[2] = 0
Plugging in we can verify,
$ \begin{align} e^{j \frac{2}{3} \pi n} &= \frac{1}{3} \left[ 0 + 3.e^{j2\pi n/3} + 0 \right]\\ e^{j \frac{2}{3} \pi n} &= \frac{1}{3} 3.e^{j2\pi n/3} \\ e^{j \frac{2}{3} \pi n} &= e^{j \frac{2}{3} \pi n} \end{align} $
So our three selected values for X[k] are correct. Thus
$ X[k] = \begin{cases} 3, & k = 1 \\ 0, & \mbox{else} \end{cases} $
$ x_2[n]= e^{j \frac{2}{\sqrt{3}} \pi n}; $
Function x2[n] is aperiodic. Let's see why -
Assume x2[n] is periodic, then
$ e^{j \frac{2}{\sqrt{3}} \pi n} = e^{j \frac{2}{\sqrt{3}} \pi (n + N)} $ for function to be periodic, where N is an integer
$ e^{j \frac{2}{\sqrt{3}} \pi n} = e^{j \frac{2}{\sqrt{3}} \pi n}e^{j \frac{2}{\sqrt{3}} \pi N} $
$ e^{j \frac{2}{\sqrt{3}} \pi n} = e^{j \frac{2}{\sqrt{3}} \pi n}.(1) $
$ e^{j \frac{2}{\sqrt{3}} \pi N} = 1 $
For this to be true -
$ j \frac{2}{\sqrt{3}} \pi N = j 2\pi n, $ where n is an integer
$ N = n\sqrt{3} $
N is not an integer and this contradicts our assumption proving that it cannot be true.
Thus, the x_2[n] is aperiodic and we cannot apply the DFT.
$ x_3[n]= e^{j \frac{4}{3} \pi n}; $
Function's period N = 3,
Using IDFT,
$ \begin{align} x[n] &= \frac{1}{3} \sum_{k=0}^{2} X[k].e^{j2\pi kn/3} \\ e^{j \frac{4}{3} \pi n} &= \frac{1}{3} \sum_{k=0}^{2} X[k].e^{j2\pi kn/3} \\ &= \frac{1}{3} \left[ X[0].e^{j0} + X[1].e^{j2\pi n/3} + X[2].e^{j2\pi n(2/3)} \right] \\ &= \frac{1}{3} \left[ X[0] + X[1].e^{j2\pi n/3} + X[2].e^{j4\pi n/3} \right] \end{align} $
For the two sides to be equal,
X[0] = 0
X[1] = 0
X[2] = 3
$ X[k] = \begin{cases} 3, & k = 2 \\ 0, & \mbox{else} \end{cases} $
$ x_4[n]= e^{j \frac{2}{1000} \pi n}; $
Function's period N = 1000,
Using IDFT,
$ \begin{align} x[n] &= \frac{1}{1000} \sum_{k=0}^{999} X[k].e^{j2\pi kn/1000} \\ e^{j \frac{2}{1000} \pi n} &= \frac{1}{1000} \sum_{k=0}^{999} X[k].e^{j2\pi kn/1000} \\ &= \frac{1}{1000} \left[ X[0].e^{j0} + X[1].e^{j2\pi n/1000} + X[2].e^{j2\pi n(2/1000)} + ... \right] \\ &= \frac{1}{1000} \left[ X[0] + X[1].e^{j2\pi n/1000} + X[2].e^{j4\pi n/1000} + ... \right] \end{align} $
For the two sides to be equal,
X[0] = 0
X[1] = 1000
X[2] = 0
$ X[k] = \begin{cases} 1000, & k = 1 \\ 0, & \mbox{else} \end{cases} $
$ x_5[n]= e^{-j \frac{2}{1000} \pi n}; $
Function's period N = 1000,
$ \begin{align} x_5[n]&= e^{-j \frac{2}{1000} \pi n}.1 \\ &= e^{-j \frac{2}{1000} \pi n}.e^{-j 2\pi n} \\ &= e^{j 2\pi n(1 - (1/1000))} \\ &= e^{j 2\pi n\frac{999}{1000} } \\ \end{align} $
Using IDFT,
$ \begin{align} x[n] &= \frac{1}{1000} \sum_{k=0}^{999} X[k].e^{j2\pi kn/3} \\ e^{j 2\pi n\frac{999}{1000}} &= \frac{1}{1000} \sum_{k=0}^{999} X[k].e^{j2\pi kn/1000} \\ &= \frac{1}{1000} \left[ X[0].e^{j0} + X[1].e^{j2\pi n/1000} + X[2].e^{j2\pi n(2/1000)}+ ... + X[999].e^{j2\pi (999/1000)} \right] \\ &= \frac{1}{1000} \left[ X[0] + X[1].e^{j2\pi n/1000} + X[2].e^{j2\pi n (2/1000)} + ... + X[999].e^{j2\pi (999/1000)} \right] \\ \end{align} $
For the two sides to be equal,
X[0] = 0
X[1] = 0
X[2] = 0
X[999] = 1000
$ X[k] = \begin{cases} 1000, & k = 999 \\ 0, & \mbox{else} \end{cases} $
$ x_6[n]= \cos\left( \frac{2}{1000} \pi n\right) ; $
- $ x_7[n]= \cos^2\left( \frac{2}{1000} \pi n\right) ; $.
- $ x_8[n]= (-j)^n . $
Back to HW7
Back to [[2010_Fall_ECE_438_Boutin|ECE 438 Fall 2010]
