Homework 2, ECE438, Fall 2011, Prof. Boutin
Question 1
Pick a note frequency f0 = 392Hz
| x(t) = 'cos'(2πf0t) = 'cos'(2π * 392t) |
| $ a.\ Assign\ sampling\ period\ T_1=\frac{1}{1000} $ |
| $ 2f_0<\frac{1}{T_1}, \ No\ aliasing\ occurs. $ |
$ \begin{align} x_1(n) &=x(nT_1)=cos(2\pi *392nT_1)=cos(2\pi *\frac{392}{1000}n) \\ &=\frac{1}{2}\left( e^{-j2\pi *\frac{392}{1000}n} + e^{j2\pi *\frac{392}{1000}n} \right) \\ \end{align} $
| $ 0<2\pi *\frac{392}{1000}<\pi $ |
| $ -\pi<-2\pi *\frac{392}{1000}<0 $ |
$ \begin{align} \mathcal{X}_1(\omega) &=2\pi *\frac{1}{2} \left[\delta (\omega -2\pi *\frac{392}{1000}) + \delta (\omega + 2\pi *\frac{392}{1000})\right] \\ &=\pi \left[\delta (\omega -2\pi *\frac{392}{1000}) + \delta (\omega + 2\pi *\frac{392}{1000})\right] \\ \end{align} $
| $ for\ all\ \omega $ |
| $ \mathcal{X}_1(\omega)=\pi* rep_{2\pi} \left[\delta (\omega -2\pi *\frac{392}{1000}) + \delta (\omega + 2\pi *\frac{392}{1000})\right] $ |
| In this situation, no aliasing occurs. In the interval of [ − π,π], which represents one period, the frequcy spectrum remains the same as Fig a-1. |
| $ b.\ Assign\ sampling\ period\ T_2=\frac{1}{500} $ |
| $ 2f_0>\frac{1}{T_2}, \ Aliasing\ occurs. $ |
$ \begin{align} x_2(n) &=x(nT_2)=cos(2\pi *392nT_2)=cos(2\pi *\frac{392}{500}n) \\ &=\frac{1}{2}\left( e^{-j2\pi *\frac{392}{500}n} + e^{j2\pi *\frac{392}{500}n} \right) \\ \end{align} $
| $ \pi<2\pi *\frac{392}{500}<2\pi $ |
| $ -2\pi<-2\pi *\frac{392}{500}<\pi $ |
| $ \mathcal{X}_2(\omega)=\pi \left[\delta (\omega -2\pi *\frac{392}{500}) + \delta (\omega + 2\pi *\frac{392}{500})\right] $ |
| $ X_2(f)=\frac{1}{2}\left[\delta (f -\frac{392}{500}) + \delta (f + \frac{392}{500})\right] $ |
| $ for\ all\ \omega $ |
| $ \mathcal{X}_2(\omega)=\pi* rep_{2\pi} \left[\delta (\omega -2\pi *\frac{392}{500}) + \delta (\omega + 2\pi *\frac{392}{500})\right] $ |
| $ X_2(f)=\frac{1}{2}rep_2\left[\delta (f -\frac{392}{500}) + \delta (f + \frac{392}{500})\right] $ |
| In this situation, aliasing DO occurs. In the interval of [ − π,π], which represents one period, the frequcy spectrum is different from Fig b-1. |
Question 2
$ (1)\ x[n]=a^{n+1}u[n-1],\ a>0 $
Compute Z transform
$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty} x[n]z^{-n} \\ &= \sum_{n=-\infty}^{\infty} a^{n+1} u[n-1]z^{-n} \\ &= a\sum_{n=1}^{\infty} a^{n}z^{-n} \\ &= \frac{a^2z^{-1}}{1-az^{-1}} \end{align} $
with ROC: $ |z|>a $
Compute Inverse Z transform
The power series expansion of the given function is
$ \begin{align} X(z) &= a^2 z^{-1}\sum_{n=0}^{\infty} a^n z^{-n},\ |z|>a \\ &= a\sum_{n=0}^{\infty} a^{n+1}z^{-n-1} \end{align} $
Substitute n=m-1
$ \begin{align} X(z) &= a\sum_{m=1}^{\infty} a^{m}z^{-m} \\ &= \sum_{m=-\infty}^{\infty} a^{m+1}u[m-1]z^{-m},\ \text{and by comparison with } X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} \end{align} $
$ \begin{align} x[n] &= a^{n+1} u[n-1] \end{align} $
$ (2)\ x[n]=-a^{n}u[-n-1],\ a>0 $
Compute Z transform
$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty} x[n]z^{-n} \\ &= -\sum_{n=-\infty}^{\infty} a^{n} u[-n-1]z^{-n} \\ &= -\sum_{n=-\infty}^{-1} a^{n}z^{-n} \\ \end{align} $
Substitute m=-n
$ \begin{align} X(z) &= -\sum_{n=1}^{\infty} a^{-n}z^{n} \\ &= -\frac{a^{-1}z}{1-a^{-1}z} \\ &= \frac{1}{1-az^{-1}} \end{align} $
with ROC: $ |z|<a $
Compute Inverse Z transform
$ \begin{align} X(z) &= \frac{1}{1-az^{-1}} \\ &= \frac{a^{-1}z}{a^{-1}z-1} \\ &= -a^{-1}z\frac{1}{1-a^{-1}z} \end{align} $
The power series expansion of the given function is
$ \begin{align} X(z) &= -a^{-1}z\sum_{n=0}^{\infty} a^{-n}z^{n} \\ &= -\sum_{n=0}^{\infty} a^{-n-1}z^{n+1} \\ \end{align} $
Substitute n+1=-m
$ \begin{align} X(z) &= -\sum_{m=-1}^{-\infty} a^{m}z^{-m} \\ &= -\sum_{m=-\infty}^{\infty} a^{m}u[-m-1]z^{-m},\ \text{and by comparison with } X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} \end{align} $
$ \ x[n]=-a^{n}u[-n-1],\ a>0 $
