folkert/Notes
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity
b8bdc10fc1
Notes/Courses/EE2T1_Telecommunication_and_Sensing.typ
Folkert Kevelam b8bdc10fc1 Update lecture notes
2026-02-17 12:10:55 +01:00

724 lines
18 KiB
Plaintext
Raw Blame History

#import "../template/lib.typ": *
#set page(paper: "a4")
#show: notes.with(
title: [EE2T1],
subtitle: [Telecommunication and Sensing],
author: "Folkert Kevelam"
)
= Lecture 1 - Introduction
== Information
#definition[
Information content:
Information is related to probability: a less probable message contains
more information
$
I_j = log_2(1/P_j) = - log_2(P_j) space "[bit]"
$
Information is additive:
$
I_(i j) &= log_2(1/(P_i P_J)) = -log_2(P_i)-log_2(P_j) \
&= I_i + I_j
$
iff the messages are independent.
]
#definition[
Source entropy, the average amount of information per message generated
by a source:
$
H = sum_(j=1)^M P_j I_j = sum_(j=1)^M P_j log_2(1/P_j) space "[bit/symbol]"
$
In a binary system, the maximum source entropy will be when $P_1 = P_0 = 0.5$.
The speed of a source:
$
R = H/T space "[bit/s]"
$
]
#theorem[
Shannon-Hartley theorem:
$
C = B dot log_2( 1 + S/N) space "[bit/s]"
$
- $C = "capacity [bit/s]"$
- $B = "bandwidth [Hz]"$
- $S/N = "ratio of signal power to the noise power"$
]
== Principles of range measurement
#definition[
The transmitter "fires" a signal and the receiver measures the time delay
$tau$ between the moments of transmission and reception of the echo.
$
2R = c dot tau arrow R = (c dot tau) / 2
$
with $c$ being the speed of light.
]
#definition[
The ability of a radar to resolve two targets with a range difference
$delta R$ is called *range resolution*.
$
delta R = (c dot tau_p) / 2 approx c/(2 B) space "[m]"
$
]
== Modulation
#definition[
Modulation: *manipulation of a signal waveform* to carry information, in
order to transmit the signal at a specified frequency in the spectrum.
$
s(t) = R(t)cos(2 pi f_c t + phi(t))
$
with
$
R(t) = L{m(t)} space "linear modulation" \
phi(t) = L{m(t)} space "angle modulation"
$
]
== Practical signal waveforms
+ DC-value, mean value: $ w_(D C) = <w(t)> = lim_(T arrow infinity) 1/T integral_(-T/2)^(T/2) w(t) d t $
+ Instantaneous power: $ p(t) = v(t) dot i(t) $
+ Average power: $ P=<p(t)> = <v(t) dot i(t)> $
+ RMS-value: root-mean-square: $ w_(r m s) = sqrt(<w^2(t)>) $
For a resistive load: $ P = v_(r m s) i_(r m s) = (<v^2(t)>)/R = <i^2(t)>R $
+ Normalized power = power delivered to a $1 omega$ load.
$ P = <w^2(t)> = lim_(T arrow infinity) 1/T integral_(-T/2)^(T/2) w^2(t) d t space "[W] = [J/s]" $
$w(t)$ is a *power waveform* iff $0 < P < infinity$.
+ Normalized energy = energy dissipated in a $1 omega$ load.
$ E = lim_(T arrow infinity) integral_(-T/2)^(T/2) w^2(t) d t space "[J]" $
$w(t)$ is an *energy waveform* iff $0 < E < infinity$.
A signal waveform $w(t)$ cannot both be an *energy waveform* and a *power waveform*.
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
$
Reference in New Issue View Git Blame Copy Permalink