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27
Courses/EE1M2_Calculus_and_Linear_Algebra.typ
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@ -196,7 +196,7 @@ are referred to as *grid curves*.
$ $
] ]
= Lecture 2 - Surface Integrals == Lecture 2 - Surface Integrals
*Orientable surfaces* are surfaces which have a "top" and a "bottom". *Orientable surfaces* are surfaces which have a "top" and a "bottom".
@ -209,28 +209,3 @@ are referred to as *grid curves*.
A surface integral over a *closed surface* $cal(S)$ is denoted as $integral.surf_cal(S) f d cal(S)$. A surface integral over a *closed surface* $cal(S)$ is denoted as $integral.surf_cal(S) f d cal(S)$.
] ]
= 3 - Curl and Divergence
= Lecture 4 - Stokes' theorem
#definition[
Revisiting green's theorem using curl:
$
attach(limits(integral.cont), b: delta D) arrow(F) dot d arrow(r) = attach(limits(integral.double), b: D) (nabla times arrow(F)) dot hat(k) d A
$
]
#theorem[
Let $cal(S)$ be a piecewise-smooth, oriented surface in $RR^3$ bounded by a
simple, closed, piecewise-smooth, positively oriented boundary curve $cal(C)$. if
$arrow(F)$ is a vector field whose components have continuous partial derivatives on a
open region in $RR^3$ that contains $cal(S)$, then
$
attach(limits(integral.cont), b: cal(C)) arrow(F) dot d arrow(r) = attach(limits(integral.double), b: cal(S)) (nabla times arrow(F)) dot hat(n) d S
$
]

106
Courses/EE1P1_Electricity_and_Magnetism.typ
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@ -62,109 +62,3 @@
=== Spherical coordinates === Spherical coordinates
+ EE1M1: $r, theta, phi$, EE1P1: $rho, phi, theta$ + EE1M1: $r, theta, phi$, EE1P1: $rho, phi, theta$
= Lecture 2
= Seminar 1 - Coulomb force
== Coulomb force
Force between two charges:
$
&arrow(F)_(S O)(arrow(r)_(S O), Q_S, Q_O) = k (Q_S Q_O) / r_(S O)^2 hat(r)_(S O) \
&arrow(r)_(S O) = arrow(r)_O - arrow(r)_S \
&r_(S O) = |arrow(r)_(S O)| = |arrow(r)_O - arrow(r)_S| \
&hat(r)_(S O) = arrow(r)_(S O) / r_(S O) \
&k = 10^(-7) c_o^2 approx 8.99 dot 10^9 "N sym.dot m"
$
= Lecture 3
*polarisation* is when an electrical field is applied to a dielectric, which
causes the dipoles within the dielectric to orient to the same point, along
the external electric field.
= Lecture 4 - Gauss's law
== Gauss's law in integral form
Based on the idea of *flux*:
- *flux* is the idea of how much of "something" crosses a given surface
- The *electrostatic field* does not cross anything (it was originally believed to)
The electric flux can be calculated using the surface integral of the electric field.
#definition[
electric flux:
$
Phi = integral_cal(S) arrow(E) dot d arrow(A) = integral_S E cos(theta) d A
$
for a uniform field and a flat surface:
$
Phi = integral_cal(S) arrow(E) dot d arrow(A) = E A cos(theta)
$
]
#definition[
Gauss's law:
$
integral.cont_cal(C) arrow(E) dot d arrow(A) = q_("enclosed") / epsilon_0
$
Gauss's law in a medium:
$
integral.cont_cal(C) dot d arrow(A) = q_("enclosed") / (epsilon_0 epsilon_r)
$
]
assuming that homogenisation is applicable. \
Gauss's law is useful for calculating the electric field in symmetric configurations:
- spherical symmetry
- cylindrical symmetry
- symmetry with respect to a plane
$
arrow(E)(arrow(r)) = arrow(E)(r, theta, phi) \
arrow(E)(arrow(r)) = E_r hat(r) \
$
*Two distinct cases*:
+ Observation point is inside the surface.
with a sphere, with a uniform charge density the total charge is:
$
Q_("tot") = rho (4/3) pi R^3
$
+ Observation point is outside the surface.
=== Examples
==== Calculating the electric field in a sphere with a single charge
==== uniformly charged rod
== Conditions at the surface of perfect conductors
== Local form of Gauss's law
- The function describing $arrow(E)$ is continuously differentiable.
- the medium can be taken as "continuous" (homogenisation).
#definition[
Gauss's law in local-form:
$
arrow(nabla) dot arrow(E)(arrow(r)) = rho(arrow(r)) / epsilon_0
$
]

274
Courses/EE1P1_GHA.typ
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@ -1,274 +0,0 @@
#import "../template/lib.typ": *
#import "@preview/cetz:0.4.2"
#import "@preview/cetz-plot:0.1.3"
#import "@preview/plotsy-3d:0.2.1": *
#set page(paper: "a4")
#show: notes.with(
title: "EE1P1",
subtitle: "Graded Homework Assignment 1",
author: "Folkert Kevelam"
)
== Exercise 1
=== Part a
The value of the electric field $arrow(E)_cal(p)$ at point $cal(P)(cal(l), cal(l), 0)$
can be calculated by the vector sum of the individual electric fields of $Q_1$, $Q_2$, and
$Q_3$.
The individual vector fields are as follows:
$
arrow(E)_(1, cal(P)) &= k q / r_1^2 hat(r_1) = k q / (sqrt(cal(l)^2 + cal(l)^2 + 0))^2 1 / (sqrt(cal(l)^2 + cal(l)^2 + 0)) vec(cal(l), cal(l), 0) \
&= k q / (cal(l)^2) vec(1/(2sqrt(2)), 1/(2sqrt(2)), 0) space [N/C] \
arrow(E)_(2, cal(P)) &= k (2q) / r_2^2 hat(r_2) = 2 k q / (sqrt(0 + cal(l)^2 + 0))^2 1 / sqrt(0 + cal(l)^2 + 0) vec(0, cal(l), 0) \
&= k q / cal(l)^2 vec(0,2,0) [N/C]\
arrow(E)_(2, cal(P)) &= k (-3q) / r_3^2 hat(r_3) = -3 k q / sqrt(cal(l)^2 + 0 + 0)^2 1 / sqrt(cal(l)^2 + 0 + 0) vec(cal(l), 0, 0) \
&= k q / cal(l)^2 vec(-3,0,0) [N/C]
$
The total electric field at point $cal(P)$ is:
$
arrow(E)_cal(P) &= arrow(E)_(1, cal(P)) + arrow(E)_(2, cal(P)) + arrow(E)_(3, cal(P)) = k q / cal(l)^2 dot (vec(1/(2sqrt(2)), 1/(2sqrt(2)), 0) + vec(0,2,0) + vec(-3,0,0)) \
&= k q / cal(l)^2 vec(1/(2sqrt(2)) - 3, 1/(2sqrt(2)) + 2, 0) [N/C]
$
#cetz.canvas({
import cetz.draw: *
import cetz-plot: *
plot.plot(
size: (10,10),
axis-style: "school-book",
x-tick-step: none,
y-tick-step: none,
plot-style: (stroke: 2pt),
{
plot.add(((0,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
plot.add(((1,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
plot.add(((0,1),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: blue))
plot.add(((1,1),), mark: "o", mark-size: 0.25, mark-style: (stroke: none, fill: black))
plot.add-anchor("pt", (1,1))
let vec(data) = {
plot.add(data, mark: "<>", mark-size: 2)
}
let calc_e(charge, distance, v) = {
let x = (charge / distance) * v.at(0)
let y = (charge / distance) * v.at(1)
(x,y)
}
let charges = (
1, 2, -3
)
let distances = (
2, 1, 1
)
let dir_vectors = (
(1/calc.sqrt(2), 1/calc.sqrt(2)),
(0, 1),
(1, 0)
)
let summed_vector = (0,0)
let n = 0
let scale = 0.1
while n < charges.len() {
let field = calc_e(charges.at(n), distances.at(n), dir_vectors.at(n))
let temp_x = field.at(0) + summed_vector.at(0)
let temp_y = field.at(1) + summed_vector.at(1)
summed_vector.at(0) = temp_x
summed_vector.at(1) = temp_y
n += 1
}
vec(((1,1),(1 + scale * summed_vector.at(0), 1 + scale * summed_vector.at(1))))
}
)
})
=== Part b
In order to nullify the x-component of the electric field with the charge
$Q_a$, the value of the charge $Q_a$ should be such that:
$
arrow(E)_(a, cal(P)) hat(x) = - arrow(E)_cal(P) hat(x) \
k Q_a / r_a^2 hat(r)_a hat(x) = - k q / cal(l)^2 (1/(2sqrt(2)) - 3)
$
For the fourth charge, the vector $r_a$ pointing from charge $Q_a$ to point
$cal(P)$ is equal to:
$
r_a = vec(3/2 cal(l), 3/2 cal(l), 0) \
|r_a| = sqrt((3/2 cal(l))^2 + (3/2 cal(l))^2 + 0) = 3/sqrt(2) cal(l) \
hat(r)_a = vec(1/sqrt(2), 1/sqrt(2))
$
Charge $Q_a$ has the same direction vector as charge $Q_1$, which makes sense
since $Q_1$ lies on the line between $Q_a$ and $cal(P)$. We can plug the vector
parameters into our equality:
$
k Q_a / (3/sqrt(2) cal(l))^2 vec(1/sqrt(2), 1/sqrt(2), 0) hat(x) &= - k q / cal(l)^2 (1/(2sqrt(2)) - 3) \
Q_a 2 / 9 1/sqrt(2) &= -q (1/(2sqrt(2)) - 3) \
Q_a &= -q(9/4 - (27 sqrt(2))/2 ) \
Q_a &= ((27 sqrt(2)) / 2 - 9/4) q [C]
$
#cetz.canvas({
import cetz.draw: *
import cetz-plot: *
plot.plot(
size: (10,10),
axis-style: "school-book",
x-tick-step: none,
y-tick-step: none,
plot-style: (stroke: 2pt),
{
plot.add(((0,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
plot.add(((1,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
plot.add(((0,1),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: blue))
plot.add(((1,1),), mark: "o", mark-size: 0.25, mark-style: (stroke: none, fill: black))
plot.add-anchor("pt", (1,1))
let vec(data) = {
plot.add(data, mark: "<>", mark-size: 2)
}
let calc_e(charge, distance, v) = {
let x = (charge / distance) * v.at(0)
let y = (charge / distance) * v.at(1)
(x,y)
}
let charges = (
1, 2, -3, (1/4) * (54 * calc.sqrt(2) - 9)
)
let distances = (
2, 1, 1, 9/2
)
let dir_vectors = (
(1/calc.sqrt(2), 1/calc.sqrt(2)),
(0, 1),
(1, 0),
(1/calc.sqrt(2), 1/calc.sqrt(2))
)
let summed_vector = (0,0)
let n = 0
let scale = 0.1
while n < charges.len() {
let field = calc_e(charges.at(n), distances.at(n), dir_vectors.at(n))
let temp_x = field.at(0) + summed_vector.at(0)
let temp_y = field.at(1) + summed_vector.at(1)
summed_vector.at(0) = temp_x
summed_vector.at(1) = temp_y
n += 1
}
vec(((1,1),(1 + scale * summed_vector.at(0), 1 + scale * summed_vector.at(1))))
}
)
})
== Exercise 2
=== Part a
Since the cross-sections of the charge distributions are negligible, we can
assume that the charges can be calculated using a line integral:
Finding charge 1:
$
Q_1 &= integral_cal(l) lambda_1 (arrow(r)) d arrow(r) \
Q_1 &= integral_a^(a+L) lambda_1 (r(t)) |r^(')(t)| d t
$
with $r(t) = t dot hat(x) $, $a <= t <= a + L$, $|r^(')(t)| = 1$
$
Q_1 &= integral_(a)^(a+L) lambda_0 (t-a) / L d t \
Q_1 &= lambda_0 / L integral_(a)^(a + L) t - a d t \
Q_1 &= lambda_0 / L [1/2 t^2 - a t]^(a + L)_a \
Q_1 &= lambda_0 / L ((1/2 (a + L)^2 - a^2 - a L) - (1/2 a^2 - a^2)) \
Q_1 &= lambda_0 / L ((1/2 a ^2 + a L + 1/2 L^2 - a^2 - a L - 1/2 a^2 + a^2)) \
Q_1 &= lambda_0 / L (1/2 L^2) = 1/2 lambda_0 L space [C]
$
Finding charge 2:
$
Q_2 &= integral_cal(l) lambda_2 (arrow(r)) d arrow(r) \
Q_2 &= integral_a^(a+L) lambda_2 (r(t)) |r^(')(t)| d t
$
with $r(t) = t dot hat(y)$, $a <= t <= a + L$, $|r^(')(t)| = 1$
$
Q_2 &= integral_(a)^(a+L) Q_0 / t d t \
Q_2 &= Q_0 integral_(a)^(a+L) 1 / t d t \
Q_2 &= Q_0 [ln(t)]^(a+L)_a = Q_0 ln((a + L)/a) \
Q_2 &= Q_0 ln(1 + L/a) space [C]
$
=== Part b
=== Part c
== Exercise 3
#let sigma_0 = 1
#let p_0 = 1
#let a = 0.5
#let xfunc(u,v) = u*calc.cos(v)
#let yfunc(u,v) = u*calc.sin(v)
#let zfunc(u,v) = sigma_0*((u - p_0)/(2*a))*calc.sin(v)
#let scale-factor = 0.25
#let (xscale, yscale, zscale) = (1, 1, 1)
#let scale-dim = (xscale*scale-factor, yscale*scale-factor, zscale*scale-factor)
#plot-3d-parametric-surface(
xfunc,
yfunc,
zfunc,
xaxis: (0,2),
yaxis: (0,2),
zaxis: (0,1),
subdivisions: 10,
scale-dim: scale-dim,
udomain: (p_0, p_0 + 2 * a),
vdomain: (0, calc.pi / 2),
axis-step: (1,1,1),
dot-thickness: 0.05em,
front-axis-thickness: 0.1em,
front-axis-dot-scale: (0.04, 0.04),
rear-axis-dot-scale: (0.08, 0.08),
axis-label-size: 1.5em,
xyz-colors: (red, green, blue),
rotation-matrix: ((-2, 2, 4), (0, -1, 0))
)

607
Courses/EE2T1_Telecommunication_and_Sensing.typ
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@ -114,610 +114,3 @@
A signal waveform $w(t)$ cannot both be an *energy waveform* and a *power waveform*. A signal waveform $w(t)$ cannot both be an *energy waveform* and a *power waveform*.
Practical waveforms are always energy waveforms. Practical waveforms are always energy waveforms.
= Lecture 2
== Distortion-free transmission
Requirements for *distortion-free transmission*:
+ $y(t) = A dot x(t - T_d) space |A| > 0, T_d >=0$
+ $Y(f) = A dot X(f)e^(-2 pi j f T_d)$
So
$H(f) = (Y(f))/(X(f)) = A e^(-2 pi j f T_d) arrow phi(f) = -2 pi f T_d$
Over the frequency band which contains the signal
- the response should be flat: $|H(f)| = |A| > 0$
- the phase decreses linear with frequency => constant delay
for all frequency components:
$T_d = -1/(2 pi f) "ang"{H(f)} = - 1/(2 pi f) phi(f)$
=== Time-variant systems: mobile multipath channel
$h(t) = A_0 delta(t- T_0) - A_1 delta(t-T_1)$\
$H(f) = A_0 e^(-2 pi j f T_0) - A_1 e^(-2 pi j f T_1)$
When moving, $T_0$ and $T_1$ change and so does $H(f)$.
== Bandwidths definitions
- *absolute bandwidth* - the bandiwdth between the frequencies with a gain
of $-infinity$.
- *-3dB bandiwdth* - frequencies between -3 dB Gain
- *Equivalent noise bandwidth* -
$
B_eq = integral_0^infinity (|H(f)|^2)/(|H(0)|^2) d f
$
- *Null- or null-to-null bandwidth* - the frequency bandwidth between
the first $-infinity$ gain points.
- *Bounded spectrum bandwidth* -
$
(P(f))/(P(0)) = 10^(-x/10)
$
- *Power bandwidth* -
$
integral_0^(B_(99 %)) P(f) d f = 0.99 integral_0^(infinity) P(f)d f
$
== Band limited signals and noise
+ band-limited signals allow for multiplexing in the frequency domain
+ band-limited signals can be completely represented by a set of discrete-time
sample values
A waveform $w(t)$ is said to be absolutely band-limited, if:
$W(f) = cal(F){w(t)} = 0 space "for" space |f| >= B_0$ \
and absolutely time-limited, if:
$w(t) = 0 space "for" space |t| > T_0$ \
*absolutely time-limited signals cannot be also absolutely band-limited
and vice versa*.
Uncertainty relation for the *time-bandwidth product*:
$
B_0 dot T_0 >= 1/2 space "or" \
B_0 = alpha / T_0 space "with" alpha >= 1/2
$
=== Sampling theorem
Every physical signal $w(t)$ can be expressed as:
$
w(t) = sum_(n=-infinity)^infinity alpha_n phi_n(t)
$
with sample function:
$
phi_n(t) = (sin(pi f_s (t- n T_s)))/(pi f_s (t - n T_s)) = (sin(pi f_s ((t-n)/f_s)))/(pi f_s ((t-n)/f_s)) = sinc(f_s (t-n T_s))
$
and coefficients:
$
&alpha_n = f_s integral_(-infinity)^infinity w(t)phi_n(t) d t space "and" \
&f_s integral_(-infinity)^infinity phi_m(t) phi_n(t) d t = cases(
1 space "for" m=n,
0 space "for" m eq.not n)
$
If the signal $w(t)$ is band-limited in $B$ [Hz] and the sample frequency
$
f_s >= 2 B
$
then the set ${a_k}$ is a complete representation of the signal $w(t)$,
with
$
a_k = w(k / f_s) = w(k T_s)
$
The lowest possible sample frequency: $f_s = 2B$ is called the *Nyquist frequency*
The minimum number of samples required to represent a time-continuous signal $w(t)$
with a bandwidth of $B$ [Hz] over a period $T_0$ is equal to:
$
N = 2B dot t_0
$
where $N$ is the number of dimensions needed to describe the waveform $w(t)$
with a bandwidth $B$ over the period $T_0$.
In practice we need more samples due to the unvertainty relation.
=== Ideal sampling
In ideal sampling we use the $delta$-function
$
w_s(t) = w(t) sum_(n=-infinity)^infinity 1/T_s e^(2 pi j n f_s t) \
W_s(f) = f_s sum_(n=-infinity)^infinity W(f- n f_s)
$
= Lecture 3 - Received signal power in a wireless link
== Travelling EM wave - parameters
*Wavelength* : $lambda = c / f$ with $c approx 3 dot 10^8$ m/s
*Radiation intensity* : $U = P_t / (4 pi R^2)$
with $P_t$ is the total transmit power and $R$ is the distance to the observer.
=== Antenna gain definition
$
G(theta, phi) = 94 pi^(U(theta, phi))) / P_in
$
+ Radiated power intensity at a distance $R$ for an isotropic radiator
$
U = P_t / (4 pi R^2)
$
+ Radiated power intensity at a distance $R$ for a radiator with gain $G$:
$
U = P_t / (4 pi R^2) G_t
$
+ Effective Isotropic Radiated Power:
$
"EIRP" = P_t dot G_t
$
=== Antenna's effective area
*effective area* $A_e$:
$
A_e = P_L / U_(i n)
$
$P_L$ - Power delivered to the load (W) \
$U_(i n)$ - Power intensity of the incident wave (W/m2)
=== Received power
$
P_r = P_t / (4 pi R^2) G_t dot A_e
$
== Reciprocity
*effective area* is related to the antenna *gain*:
$
A_e = G (lambda^2)/(4 pi) arrow G = 4 pi A_e / (lambda^2)
$
Where\
$G$ - antenna gain
$lambda$ - wavelength
$A_e = eta A$
- Isotropic - Effective Area: $lambda^2 / (4 pi)$
- Infinitesimal dipole or loop - Effective Area: $(1.5 lambda^2) / (4 pi)$
- Half-wave dipole - Effective Area: $(1.64 lambda^2)/(4 pi)$
- Horn - Effective Area: $(eta = 0.81) arrow 0.81 A$
- Parabola - Effective Area: $(eta = 0.56) arrow 0.56 A$
== Wireless link budget
$
P_r / P_t = (lambda/(4 pi R))^2 G_t dot G_r
$
Loss free space-
$
L_(F S) = ((4 pi R)/lambda)^2
$
== Transmission line propagation losses
Loss transmission line:
$
L_(T L) = (P(z=0))/(P(z=l)) = e^(2 gamma l)
$
== Radar equation
- An omnidirectional transmitted power $P_t$, induces a power intensity at range
$R$ of:
$
U = P_t / (4 pi R^2)
$
- The radar transmit antenna has a gain $G_t$, therefore the power intensity is:
$
U = (P_t G_t) / (4 pi R^2)
$
- At range $R$ a target with the radar cross section (RCS) $sigma$ reflects a
small portion of the power backward to the radar. Re-radiated power:
$
P = (P_t G_t sigma) / (4 pi R^2)
$
- Received power is
$
P_r = (P_t G_t G_r lambda^2 sigma)/((4 pi)^3 R^4)
$
= Lecture 4
== Thermal Noise
$
V_V(f) approx sqrt(4 R k_b T) \
I_V(f) approx sqrt((4 k_b T)/R) \
V_(r m s) = sqrt(4 R k_b T B_n)
$
The delivered power to a matched ($R_L = R$) load noise power will be based on
half of the resistor noise voltage:
$
V_(L r m s) = sqrt(k_B T B_n R)
$
The *available* load noise power
$
P_a = k_b T B_n
$
The *noise power spectral density* equals
$
P_a(f) = (V_L^2 (f))/R = k_B T
$
== Equivalent noise temperature
$
T_n = P_a / (k_B B_n)
$
== Noise Characterization of a linear device
$
P_(a o) = G_a P_(n_(i n)) + P_(e x) \
P_(a o) = G_a (P_(n_(i n)) + P_e) \
P_e = P_(e x) / G_a
$
== Noise Figure
$
F = ("output noise power actual device")/("output power ideal component")
$
at $T_0 = 290 k$
$
F = (k_b (T_0 + T_e)G_a B_n)/(k_B T_0 G_a B_n) = (T_0 + T_e) / T_0 = 1 + T_e / T_0 >= 1 \
F_(d B) = 10 dot log_(10)(1 + T_e / T_0) >= 0 d B
$
== Noise Figure of a transmission line
$
L_(T L ) = (P(z=0))/(P(z=l)) = e^(2 gamma l) \
L_(T L, d B) = 20 gamma l log_10(e) = alpha l
$
$
T_e = (L-1)T_0
$
== Cascaded devices
$
T_e = T_(e 1) + (T_(e 2))/G_(a 1) + (T_(e 3))/(G_(a 1) G_(a 2)) + (T_(e 4))/(G_(a 1) G_(a 2) G_(a 3)) \
F = F_1 + (F_2 - 1)/G_(a 1) + (F_3 - 1)/(G_(a 1)G_(a 2))
$
= Lecture 5
== Environmental noise received by antenna
Total Noise at the receive antenna due to environmental noise:
$
T_(A E) &= (integral_(Omega=0)^(Omega=4 pi) T_b(Omega) G(Omega d Omega))/(integral_(Omega=0)^(Omega=4 pi)G(Omega) d Omega) \
&= (integral_(phi=0)^(phi=2 pi) integral_(theta=0)^(theta=pi) T_b (theta, phi)G(theta,phi)sin(theta)d theta d phi)/(integral_(phi=0)^(phi=2 pi) integral_(theta=0)^(theta=pi) G(theta, phi) sin(theta) d theta d phi)
$
Where:
- $G$ - antenna gain
- $T_b$ - brightness temperature of different environmental sources
== Total Antenna noise
Total *antenna noise temperature* at the terminal is
$
T_a = e_A T_(A E) + (1-e_A) T_p
$
where:
- $T_(A E)$ is the total captured brightness temperatures of different sources
- $T_p$ is the physical temperature (290K)
- $e_A$ is the thermal efficiency of the antenna
The total *Noise figure* of the antenna is:
$
F_a = 1 + T_a / T_0 \
F_(a, d B) = 10 log_10 (1 + T_a / T_o)
$
== SNR in cable and in free-space
SNR in cable:
$
"SNR"_("cable") = P_("out")/ N_("out") = P_("in") / (k_b B_n (L_(T L) - 1)T_0)
$
SNR in free space:
$
"SNR"_("FS") = P_("FS","out") / N_("FS","out") = (P_("in") lambda G_(T X) G_(R X)) / (k_B B_n (4 pi d)^2 T_a)
$
== Link Budget
$
(S/N)_("Det") = (S/N)_("RX") &= (P_(E I R P) dot G_(F S) dot G_(A R))/(k_B dot T_(s y s) dot B_n) = (P_(E I R P) dot G_(F S) dot G_(A R))/(k_B dot (F-1)T_0 dot B_n) \
&= (P_(T X) dot G_(A T) dot G_(A R)) / (L_(F S) dot k_B (T_(A R) + T_e) B_n)
$
== Communication range equation
$
R_("max") = ((P_t G_t G_r lambda^2)/((4 pi)^2 P_(r "min")))^(1/2)
$
where
- $P_t$ is the transmit power
- $G_t$ is the transmit antenna gain
- $G_r$ is the receive antenna gain
- $lambda$ is the wavelength
$
"SNR" = S_0 / N = P_r / (F-1) k_B T_0 B_n \
P_(r "min") = (F-1) k_b T_0 B_n "SNR"_("min")
$
The maximum operation range can be determined by combining the two:
$
R_("max") = ((P_t G_t G_r lambda^2)/((4 pi)^2 "SNR"_("min")(F-1)k_B T_0 B_n))^(1/2)
$
== Radar range equation
$
R_("max") = ((P_t G_t G_r lambda^2 sigma)/((4 pi)^3 P_(r "min")))^(1/4) \
R_("max") = ((P_t G_t G_r lambda^2 sigma)/((4 pi)^3 "SNR"_("min")(F-1)k_B T_0 B_n))^(1/4)
$
= Pulse amplitude and pulse code modulation
== Sampling theorem
every physical signal can be expressed as:
$
w(t) = sum_(n=-infinity)^infinity a_n phi_n (t)
$
with sample function
$
phi_n(t) = sinc(f_s(t - n T_s))
$
and coefficients
$
a_n = f_s integral_(-infinity)^infinity w(t) phi_n (t) d t \
f_s integral_(-infinity)^infinity phi_m (t) phi_n (t) d t = cases(1 space "for" m=n, 0 space "for" m eq.not n)
$
=== Ideal sampling
$
w_s(t) = w(t) sum_(k=-infinity)^infinity delta (t - k T_s) \
= sum_(k=-infinity)^infinity w(k T_s)delta (t - k T_s) = w(t) sum_(n=-infinity)^infinity f_s e^( 2 pi j n f_s t)
$
with $T_s = 1/f_s, f_s >= 2B$
$
W_s(f) = cal(F){w_s(t)} = f_s sum_(n=-infinity)^infinity W(f - n f_s)
$
=== Natural sampling
$f_(s, "min") = 2 B space "and" space f_s >= 2B$
$
w_s (t) = w(t) dot s(t) \
s(t) = sum_(k=-infinity)^infinity Pi((t-k T_s)/tau) = sum_(n=-infinity)^infinity c_n e^(j 2 pi n f_s t)\
W_s(f) = sum_(n=-infinity)^infinity c_n W(f- n f_s) \
c_n = d (sin(n pi d))/(n pi d) = f_s tau sinc(n f_s tau) \
d = tau / T_s = f_s tau \
S(f) = cal(F){s(t)} = sum_(n=-infinity)^infinity c_n delta(f - n f_s)
$
=== Signal recovery
+ lowpass filter with $B < f_("cut-off") < f_s - B$
+ down-converting the spectrum $W(f-n f_s)$ to baseband ($f=0$) by multiplying
with $cos(2 pi n f_s t)$ using a mixer, followed by an LPF.
$
w_s(t) = w(t) dot sum_(n=-infinity)^infinity c_n e^(j n omega_s t) = w(t) dot sum_(n=-infinity)^infinity c_n (cos(n omega_s t) + j sin(n omega_s t)) \
= w(t)(c_0 + 2 sum_(n=1)^infinity c_n cos(n omega_s t))
$
Multiplying by $cos(k omega_s t)$
$
w_s(t)cos(k omega_s t) = w(t)(c_0 cos(k omega_s t) + 2 sum_(n=1)^infinity c_n cos(n omega_s t)cos(k omega_s t)) \
= w(t)(c_0 cos(k omega_s t) + sum_(n=1)^infinity c_n (cos((n-k)omega_s t) + cos((n+k)omega_s t))) \
= c_k w(t) space "with other components at higher frequencies removed due to LPF"
$
=== Instantaneous sampling (flat-top PAM)
*sample and hold circuit*
$
w_s(t) = sum_(k=-infinity)^infinity w(k T_s) h(t - k T_s) \
h(t) = Pi (t/tau) = cases(1"," space "for" |t| < tau/2, 0"," space "for" |t| > tau/2) \
tau <= T_s \
W_s(f) = H(f) dot f_s sum_(n=-infinity)^infinity W(f - n f_s)
$
+ for *natural sampling*, the individual spectral components *do have a flat
frequency response*. The coefficients are only a function on $n$, $f_s$ and $tau$.
+ *flat-top PAM*, the individual spectral components undergo frequency depndend filtering.
Complicated signal recovery. The filtering results in linear distortion.
The ideal equalizer is $H^(-1)(f)$.
+ The bandwidth required for PAM signal transmission is much larger than needed for
the baseband signal with bandwidth $B$ because of the narrow pulses for $tau/T_s << 1$.
Thus, a *larger receiver bandwidth* is needed, which will pass more noise.
== Time division multiplexing
PAM pulses can be multiplexed in time. but the receiver has to select the correct
pulses, meaning accurate synchronization is required.
== Pulse Code modulation
PCM consists of three basic operations:
+ Signal samling -> discrete-time analog pulses
+ Quantization of the amplitude -> discrete-tie and discrete-amplitude pulses
+ Coding -> digital words are assigned to the discrete-time discrete-amplitude levels.
=== Quantization errors
during quatnization the following error is introduced:
$|epsilon| <= delta / 2$
where $delta = V_(p p) / M$ is the step size or the distance between the
successive quantization levels.
Quantization errors result in quantization noise.
The reconstructed value $Q(x_k)$ for sample $x_k$ is given by:
with polar signaling $a_(k j) in {-1, +1}$
$
Q(x_k) = V sum_(j=1)^n a_(k j) (1/2)^j = delta/2 sum_(j=1)^n a_(k j) 2^(n-j) \
$
=== Noise in a PCM system
In a PCM communication system, the reconstructed signal $y+k = x_k + n_k$ suffers
from three souces of noise:
+ Quantization noise: $e_q = Q(x_k) - x_k$
+ Bit error noise: $e_b = y_k - Q(x_k)$
Reconstruction errors due to *detection errors*
+ Overload noise: The input signal of the ADC is outside the conversion range.
==== Quantization noise
Quantization noise is uniformly distributed between $(-delta/2, delta/2)$.
The PDF $f_(e_q) (e_q)$ is an uniform with height $1/delta$.
The quantization noise power:
$
bar(e^2_q) = integral_(-infinity)^infinity e_q^2 f_(e_q) (e_q) d e_q = \
integral_(-delta/2)^(delta/2) e_q^2 1/delta d e_q = delta^2 / 12 = V^2 / (3M^2)
$
with $delta = (2 dot V)/ 2^n = (2 dot V) / M$.
Signal-to-Noise ratio at maximum signal level *due to quantization errors*.
$
(S/N)_("max") = (P_("signal-max"))/(P_("noise")) = V^2 / bar(e^2_q) = V^2 / (V^2 / (3 M^2)) = 3 M^2
$
with $M$ being the number of quantization levels $M = 2^n$.
Signal-to-Noise ratio for uniformly distributed amplitudes *due to quantization errors*
$
(S/N)_("uniform") = (P_("uniform"))/(P_("noise")) = (V^2/3)/(bar(e_q^2)) = (V^2 / 3)/(V^2 / (3 M^2)) = M^2
$
==== Bit error noise
The probability of a *single error* in a PCM-word with bit error prob of $P_e$
and $n$ the length of the word.
$
P_(e w) = vec(n, 1) P_e (1 - P_e)^(n-1) = n P_e (1- P_e)^(n-1) approx n dot P_e
$
An error in the *jth* bit results in an *error voltage* of:
$
e_j = 2^(n-j) dot delta = 2^(-(j-1)) V = 2 V/ (2^j)
$
The value of $bar(e^2_j)$ averaged over all $n$ bit positions gives the noise power
*given* an error at a random bit location.
$
bar(e_j^2) = 1/n sum_(j=1)^n (delta dot n^(n-j))^2 = 1/n sum_(k=0)^(n-1) (delta dot 2^k)^2 = delta^2 / n sum_(k=0)^(n-1) 4^k \
= delta^2 / n (4^n - 1)/(4-1) = 4/3 V^2 / n (M^2 - 1) / M^2
$
The average *bit error noise power* for bit error probability $P_e$ follows as:
$
bar(e_b^2) = bar([y_k - Q(x_k)]^2) = P_(e w) bar(e_j^2) = n P_e bar(e_j^2) \
= 4/3 V^2 P_e (M^2 - 1) / M^2
$
==== Total noise power and SNR
Combining the results for the quantization noise and bit error noise, we get
$
"SNR"_("pk out") = (3M^2) / (1+4 P_e (M^2 - 1))
$
if only quantized noise is considered:
$
"SNR"_("pk out") = 3 M^2 = 10 dot log_10(3 M^2) space [d B]
$
$
(S/N)_("uniform") = M^2 / (1 + 4(M^2 - 1)P_e) = M^2
$
The maximum possible SNR due to bit error noise is:
$
"SNR" approx 3/4 1/P_e
$