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Exam 3 Material Summary

Previous Yet Pertinent Material

CT Fourier Transform Pair $\mathcal{X}(\omega) = \int_{-\infty}^{+\infty}x(t)e^{-j\omega t} \,dt$

$x(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}\mathcal{X}(\omega)e^{j\omega t} \,dt$

DT Fourier Transform Pair

$X(e^{j\omega}) = \sum_{n=-\infty}^{+\infty} x[n]e^{-j\omega n}$

$x[n] = \frac{1}{2\pi}\int_{2\pi} X(e^{j\omega})e^{j\omega n}$

An infinite geometric series converges iff |r| < 1

$\int_{T_1}^{T_2} = -\int_{T_2}^{T_1}$

$\int_{-\infty}^{+\infty}x(t)u(t)\,dt = \int_{0}^{+\infty}x(t)\,dt$ $\sum_{n = -\infty}^{+\infty} x[n]u[n] = \sum_{n = 0}^{+\infty} x[n]$ Something with discrete signals vs CT signals results in the changing of the limits to include one additional value.

All DT Fourier transforms are periodic with period $2\pi$

Chapter 7

Sampling
1. Impulse Train Sampling
2. The Sampling Theorem and the Nyquist
Signal Reconstruction Using Interpolation: the fitting of a continuous signal to a set of sample values
1. Sampling with a Zero-Order Hold (Horizontal Plateaus)
2. Linear Interpolation (Connect the Samples)
Undersampling: Aliasing
Processing CT Signals Using DT Systems (Vinyl to CD)
1. Analog vs. Digital: The Show-down (A to D conversion -> Discrete-Time Processing System -> D to A conversion
Sampling DT Signals (CD to MP3 albeit a complicated sampling algorithm, MP3 is less dense signal)

Sampling Theory

Let x(t) be a band-limited signal with

X(j\omega) = 0

for

|\omega| > \omega_M

. Then x(t) is uniquely determined by its samples

x(nT), n = 0, \pm 1, \pm 2,..., \mbox{ if}

\,\!\omega_s > 2\omega_M

where

\omega_s = \frac{2\pi}{T}

.

Given these samples, we can reconstruct x(t) by generating a periodic impulse train in which successive impulses have amplitudes that are successive sample values. This impulse train is then processed through an ideal lowpass filter with gain T and cutoff frequency greater than \omega_M and les than \omega_s - \omega_M. The resulting output signal will exactly equal x(t)

Observations

To determine if x(t) is band-limited, one must exam

X(\omega)

the Fourier transform of x(t).

\,\!p(t) = \sum_{n = -\infty}^{+\infty}\delta (t - nT)

|

v

- - ->X- - ->|H(j\omega)|- - -> x_r(t)

Recommended Exercises: 7.1, 7.2, 7.3, 7.4, 7.5, 7.7, 7.10, 7.22, 7.29, 7.31, 7.33

Chapter 8

Complex Exponential and Sinusoidal Amplitude Modulation (You Can Hear the Music on the Amplitude Modulation Radio -Everclear) Systems with the general form $ y(t) = x(t)c(t) $ where $ c(t) $ is the carrier signal and $ x(t) $ is the modulating signal. The carrier signal has its amplitude multiplied (modulated) by the information-bearing modulating signal.
1. Complex exponential carrier signal: $c(t) = e^{\omega_c t + \theta_c}$
2. Sinusoidal carrier signal: $c(t) = cos(\omega_c t + \theta_c )$
Recovering the Information Signal $ x(t) $ Through Demodulation
1. Synchronous
2. Asynchronous
Frequency-Division Multiplexing (Use the Entire Width of that Frequency Band!)
Single-Sideband Sinusoidal Amplitude Modulation (Save the Bandwidth, Save the World!)
AM with a Pulse-Train Carrier Digital Airwaves
1. $c(t) = \sum_{k=-\infty}^{+\infty}\frac{sin(k\omega_c \Delta /2)}{\pi k}e^{jk\omega_c t}$
2. Time-Division Multiplexing "Dost thou love life? Then do not squander time; for that's the stuff life is made of." -Benjamin Franklin)

Recommended Exercises: 8.1, 8.2, 8.3, 8.5, 8.8, 8.10, 8.11, 8.12, 8.21, 8.23

Chapter 9

1. The Laplace Transform "Here I come to save the day!"

$X(s) = \int_{-\infty}^{+\infty}x(t)e^{-st}\, dt$

s is a complex number of the form $\sigma + j\omega$ and if $\sigma = 0$ then this equation reduces to the Fourier Transform of $$ x(t) $$ . Indeed, the LT can be viewed as the FT of the signal $x(t)e^{-\sigma t}$ as follows:

$\mathcal{F}\lbrace x(t)e^{-\sigma t} \rbrace = \mathcal{X}(\omega) = \int_{-\infty}^{+\infty}x(t)e^{-\sigma t}e^{-j\omega t}\, dt$

2. The Region of Convergence for Laplace Transforms (To Infinity or Converge!)

Definitions

A signal x(t) is:

right sided if there exists a t_0 such that x(t) = 0 for t < t_0
left sided if there exists a t_0 such that x(t) = 0 for t > t_0
two sided if it extends infinitely for both t > 0 and t < 0
of finite duration if there exist two values of t, T_1 and T_2 such that x(t) = 0 for t < T_1 and t > T_2

From 4: A two sided signal can be represented as the sum of a right sided signal and a left sided signal if the signal is divided at any arbitrary T_0. The two sided signal conver

A Laplace transform is rational if it is of the form X(s) = \frac{N(s)}{D(s)} Property

: The ROC of X(s) consists of strips parallel to the $j\omega$ -axis in the s-plane.
: For rational Laplace transforms, the ROC does not contain any poles.
: If x(t) is of finite duration and is absolutely integrable, then the ROC is the entire s-plane.
: If x(t) is right sided, and if the line Re{s} = $\sigma_0$ is in the ROC, then all values of s for which Re{s} > $\sigma_0$ will also be in the ROC.
: If x(t) is left sided, and if the line Re{s} = \sigma_0 is in the ROC, then all values of s for which Re{s} < \sigma_0 will also be in the ROC.
: If x(t) is two sided, and if the line Re{s} = \sigma_0 is in the ROC, then the ROC will consist of a strip in the s-plane that includes the line Re{s} = \sigma_0.
: If the Laplace transform X(s) of x(t) is rational, then its ROC is bounded by poles or extends to infinity. In addition, no poles of X(s) are contained in the ROC.
: If the Laplace Transform X(s) of x(t) is rational, then if x(t) is right sided, the ROC is the region in the s-plane to the right of the rightmost pole. If x(t) is left sided, the ROC is the region in the s-plane to the left of the leftmost pole.

Notes:

If giving a Laplace Transform for an answer to a question, the definition is incomplete without providing a ROC.
In order to determine the inverse Laplace transform of a LT X(s), one must consider its ROC. The ROC coupled with properties 1-8 will be used to distinguish between the signals that produce the same LT X(s)

Partial Fraction Expansion

Any rational function $X(s) = \frac{N(s)}{D(s)}$ can be expressed as a linear combination of LOWER ORDER terms.

Example

X(s) = \frac{(s - z_1)(s-z_2)}{(s-p_1)(s-p_2)^2} = \frac{A}{s-p_1} + \frac{B}{s-p_2} + \frac{C}{(s-p_2)^2}

z_1 and z_2 are referred to as the zeroes of the function because X(z_1) = 0;

p_1 and p_2 are referred to the poles of the function because X(p_1) is infinity creating a large "pole" on the graph

p_2 is a second order pole because it occurs twice

To obtain the coefficients you can use the relationship:

R = (s-p_R)X(s) \Bigg|_{s=p_{r}}\,\,\,

for this example the 2nd order pole creates a special case

The coefficient B cannot be computed directly because (s-p_2)X(s) still leaves a pole at p_2 and therefore cannot be

evaluated at s=p_2. A and C can be computed easily however, and once those are computed there is only one unknown left

in the equation and can clearly be obtained through direct algebraic manipulation.

If that method proves fruitless or too hard to compute, then a system of equations can be obtained by acquiring a

common denominator for the RHS of the equation resulting in

$ A(s-p_2)^2 + B(s-p_2)(s-p_1) + C(s-p_1) $

. This equation is

precisely equal to the numerator of the LHS of the equation therefore, after algebraically expanding all the terms. You

can obtain a system of 3 equations and 3 unknowns which may be solved using a variety of methods, including those

learned in linear algebra, like Kramer's(need to double check that name) rule. Methods learned in high school algebra

also apply.

3. The Inverse Laplace Transform

$x(t) = \frac{1}{2\pi}\int_{\sigma - j\infty}^{\sigma + j\infty} X(s)e^{st}\,ds$

for values of $s = \sigma + j\omega$ in the ROC. The formal evaluation of the integral requires contour integration in the complex plane which is beyond the scope of this course.

3.1 The Laplace Transforms we will consider will fall into several categories that can be inverted using tables.

X(s) = \sum_{i=1}^{m} \frac{A_i}{s+a_i}

Laplace Transform Properties
Property	Signal	Laplace Transform	ROC
Linearity	$\,\! ax_1(t) + bx_2(t)$	$\,\! aX_1(s)+bX_2(s)$	At least $R_1 \cap R_2$
Time Shifting	$\,\! x(t-t_0)$	$e^{-st_0}X(s)$	R
Shifting in the s-Domain	$e^{s_0 t}x(t)$	$\,\! X(s-s_0)$	Shifted version of R (i.e., s is in the ROC if $$ s - s_0 $$ is in R)
Time scaling	$\,\! x(at)$	$\frac{1}{\|a\|}X\Bigg( \frac{s}{a} \Bigg)$	Scaled ROC (i.e., s is in the ROC if s/a is in R)
Conjugation	$\,\! x^{*}(t)$	$\,\! X^{}(s^{})$	R
Convolution	$\,\! x_1(t)*x_2(t)$	$\,\! X_1(s)X_2(s)$	At least $R_1 \cap R_2$
Differentiation in the Time Domain	$\frac{d}{dt}x(t)$	$\,\! sX(s)$	At least R
Differentiation in the s-Domain	$\,\! -tx(t)$	$\frac{d}{ds}X(s)$	R
Integration in the Time Domain	$\int_{-\infty}^{t}x(\tau)\,d\tau$	$\frac{1}{s}X(s)$	At least $R \cap \lbrace \mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0 \rbrace$
$\,\!\mbox{Initial- and Final-Value Theorem}$ $\,\!\mbox{If } x(t) = 0 \mbox{ for } t < 0 \mbox{ and } x(t)\mbox{ contains}$ $\,\! \mbox{no impulses or higher-order singularities at }t = 0\mbox{, then}$ $x(0^{+}) = \lim_{x\rightarrow \infty} sX(s)$ $\lim_{t\rightarrow \infty} x(t) = \lim_{s\rightarrow 0}sX(s)$

Laplace Transform Pairs
Transform Pair	Signal	Transform	ROC
1	$\,\!\delta(t)$	$$ 1 $$	$All\,\, s$
2	$\,\! u(t)$	$\frac{1}{s}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0$
3	$\,\! -u(-t)$	$\frac{1}{s}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace < 0$
4	$\frac{t^{n-1}}{(n-1)!}u(t)$	$\frac{1}{s^{n}}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0$
5	$-\frac{t^{n-1}}{(n-1)!}u(-t)$	$\frac{1}{s^{n}}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace < 0$
6	$\,\!e^{-\alpha t}u(t)$	$\frac{1}{s+\alpha}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace > -\alpha$
7	$\,\! -e^{-\alpha t}u(-t)$	$\frac{1}{s+\alpha}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace < -\alpha$
8	$\frac{t^{n-1}}{(n-1)!}e^{-\alpha t}u(t)$	$\frac{1}{(s+\alpha )^{n}}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace > -\alpha$
9	$-\frac{t^{n-1}}{(n-1)!}e^{-\alpha t}u(-t)$	$\frac{1}{(s+\alpha )^{n}}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace < -\alpha$
10	$\,\!\delta (t - T)$	$\,\! e^{-sT}$	$All\,\, s$
11	$\,\![cos( \omega_0 t)]u(t)$	$\frac{s}{s^2+\omega_0^{2}}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0$
12	$\,\![sin( \omega_0 t)]u(t)$	$\frac{\omega_0}{s^2+\omega_0^{2}}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0$
13	$\,\![e^{-\alpha t}cos( \omega_0 t)]u(t)$	$\frac{s+\alpha}{(s+\alpha)^{2}+\omega_0^{2}}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace > -\alpha$
14	$\,\![e^{-\alpha t}sin( \omega_0 t)]u(t)$	$\frac{\omega_0}{(s+\alpha)^{2}+\omega_0^{2}}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace > -\alpha$
15	$u_n(t) = \frac{d^{n}\delta (t)}{dt^{n}}$	$\,\!s^{n}$	$All\,\, s$
16	$u_{-n}(t) = \underbrace{u(t) \dots u(t)}_{n\,\,times}$	$\frac{1}{s^{n}}$	$\mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0$

Recommended Exercises: 9.2, 9.3, 9.4, 9.6, 9.8, 9.9, 9.21, 9.22

Chapter 10

1. The z-Transform The z-Transform is the more general case of the discrete-time Fourier transform. For the DT Fourier transform $z = e^{j\omega }$ with $\omega$ real $\Rightarrow |z| = 1$ . When z is not restricted to 1, it has the form $re^{j\omega}$ . This can be developed into the more general case of transform called z-Transform. The development of the z-Transform is outlined in Chapter 10.1 of the Oppenheim and Wilsky text.

X(z) = \sum_{n = -\infty}^{+\infty}x[n]z^{-n}

2. Region of Convergence for the z-Transform

3. The Inverse z-Transform

The derivation of the inverse z-Transform equation is outlined in chapter 10.3 of the text (pg 757-8).

x[n] = \frac{1}{2\pi j} \oint X(z)z^{n-1}\,dz

This is a closed loop integral around a CCW rotation centered at the origin with radius r. r can be any value for which X(z) converges.

z-Transform Properties
Property	Signal	z-Transform	ROC
Linearity	$\,\! ax_1[n] + bx_2[n]$	$\,\! aX_1(z)+bX_2(z)$	At least $R_1 \cap R_2$
Time Shifting	$\,\! x[n-n_0]$	$z^{-n_0}X(z)$	R, except for the possible addition or deletion of the origin
Scaling in the z-Domain	$e^{j\omega_0 n}x[n]$	$X(e^{-j\omega_0}z)$	R
	$$ z_0^nx[n] $$	$X\Bigg( \frac{z}{z_0} \Bigg)$	$$ z_0 R $$
	$\,\! a^nx[n]$	$\,\! X(a^{-1}z)$	a\|R= the set of point {\|a\|z} for z in R
Time Reversal	$\,\! x[-n]$	$\,\! X(z^{-1})$	Inverted R (i.e., R^-1= the set of point z^-1, where z is in R)
Time Expansion	$x_{(k)}[n] = \begin{cases} x[r], & \mbox{if }n=rk \mbox{ for }r\in \mathbb{Z}\\ 0, &\mbox{if }n\neq rk \mbox{ for } r\in \mathbb{Z}\end{cases}$	$\,\! X(z^k)$	$R^{1/k}$ (i.e., the set of points $z^{1/k}$ , where z is in R)
Conjugation	$\,\! x^{*}[n]$	$\,\! X^{}(z^{})$	R
Convolution	$\,\! x_1[n] * x_2[n]$	$\,\! X_1(z)X_2(z)$	At least the intersection of R_1 and R_2
First Difference	$\,\! x[n] - x[n-1]$	$\,\! (1-z^{-1})X(z)$	At least the intersection of R and $$ \|z\| > 0 $$
Accumulation	$\sum_{k = -\infty}^{n}x[k]$	$\frac{1}{1-z^{-1}}X(z)$	At least the intersection of R and $$ \|z\| > 1 $$
Differentiation in the z-Domain	$\,\! nx[n]$	$-z\frac{dX(z)}{dz}$	R
Initial-Value Theorem If x[n] = 0 for n < 0, then $x[0] = \lim_{z\rightarrow \infty} X(z)$

z-Transform Pairs

z-Transform Pairs
#	Signal	Transform	ROC
1	$\,\!\delta[n]$	$\,\! 1$	All $\,\! z$
2	$\,\!u[n]$	$\,\!\frac{1}{1-z^{-1}}$	$\,\! \|z\| > 1$
3	$\,\!-u[-n-1]$	$\,\!\frac{1}{1-z^{-1}}$	$\,\! \|z\| < 1$
4	$\,\!\delta [n-m]$	$\,\! z^{-m}$	All $\,\!z$ except 0 (if $\,\! m > 0$ ) or $\,\!\infty\mbox{(if } m < 0 \mbox{)}$
5	$\,\!\alpha^{n}u[n]$	$\,\! \frac{1}{1-\alpha z^{-1}}$	$\,\! \|z\| > \|\alpha\|$
6	$\,\! -\alpha^{n}u[-n-1]$	$\,\!\frac{1}{1-\alpha z^{-1}}$	$\,\! \|z\| < \|\alpha\|$
7	$\,\! n\alpha^{n}u[n]$	$\,\! \frac{\alpha z^{-1}}{(1-\alpha z^{-1})^{2}}$	$\,\! \|z\| > \|\alpha\|$
8	$\,\! -n\alpha^{n}u[-n-1]$	$\,\! \frac{\alpha z^{-1}}{(1-\alpha z^{-1})^{2}}$	$\,\! \|z\| < \|\alpha\|$
9	$\,\! [cos(\omega_0 n)]u[n]$	$\,\! \frac{1-[cos(\omega_0)]z^{-1}}{1-[2cos(\omega_0)]z^{-1}+z^{-2}}$	$\,\! \|z\| > 1$
10	$\,\! [sin(\omega_0 n)]u[n]$	$\,\! \frac{1-[cos(\omega_0)]z^{-1}}{1-[2cos(\omega_0)]z^{-1}+z^{-2}}$	$\,\! \|z\| > 1$
11	$\,\! [r^{n}cos(\omega_0 n)]u[n]$	$\,\! \frac{1-[rcos(\omega_0)]z^{-1}}{1-[2rcos(\omega_0)]z^{-1}+r^{2}z^{-2}}$	$\,\! \|z\| > r$
12	$\,\! [r^{n}sin(\omega_0 n)]u[n]$	$\,\! \frac{1-[rcos(\omega_0)]z^{-1}}{1-[2rcos(\omega_0)]z^{-1}+r^{2}z^{-2}}$	$\,\! \|z\| > r$

Recommended Exercises: 10.1, 10.2, 10.3, 10.4, 10.6, 10.7, 10.8, 10.9, 10.10, 10.11, 10.13, 10.15, 10.21, 10.22, 10.23, 10.24, 10.25, 10.26, 10.27, 10.30, 10.31, 10.32, 10.33, 10.43, 10.44.

Note: If a problem states that you should use “long division”, feel free to use the geometric series formula instead.

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