diff --git a/Courses/EE1M2_Calculus_and_Linear_Algebra.typ b/Courses/EE1M2_Calculus_and_Linear_Algebra.typ
index fe2f062..3a4de9d 100644
--- a/Courses/EE1M2_Calculus_and_Linear_Algebra.typ
+++ b/Courses/EE1M2_Calculus_and_Linear_Algebra.typ
@@ -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".
 
@@ -209,3 +209,28 @@ are referred to as *grid curves*.
 
   A surface integral over a *closed surface* $cal(S)$ is denoted as $integral.surf_cal(S) f d cal(S)$.
 ]
+
+=  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
+  $
+]
+
+
diff --git a/Courses/EE1P1_GHA.typ b/Courses/EE1P1_GHA.typ
new file mode 100644
index 0000000..dc0a103
--- /dev/null
+++ b/Courses/EE1P1_GHA.typ
@@ -0,0 +1,274 @@
+#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))
+)
diff --git a/Courses/EE2T1_Telecommunication_and_Sensing.typ b/Courses/EE2T1_Telecommunication_and_Sensing.typ
index f201ba8..4c56d2b 100644
--- a/Courses/EE2T1_Telecommunication_and_Sensing.typ
+++ b/Courses/EE2T1_Telecommunication_and_Sensing.typ
@@ -114,3 +114,610 @@
 
 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
+$