- Mobile signals propagate through EM waves.
- The transmission may vary from simple LOS to the one that has lots of obstruction.
- These obstructions can cause reflecting, defraction or scattering of the desired signal.
- Hence the radio channels are extremely random and do not offer early analysis.
- Hence we model them.
- These models predict the power received at the MS or BS under given conditions.

Rare case

$P_r(d) = \frac{P_t \ G_t \ Gr \ \lambda^2}{(4 \pi^2) \ d^2 \ L}$

Gain of any antenna

$G = \frac{4 \pi A^2}{\lambda}$

$\lambda = \frac{c}{f} = \frac{c}{w/ 2 \pi} = \frac{2 \pi c}{w}$

To find path loss in dB

$P_r (dB) = 10 \ log \ (\frac{Pt}{Pr})$

$= \ 10 \ log \ (\frac{P_t (4 \pi)^2 \ d^2 \ L}{P_t \ G_t \ G_r \ \lambda^2})$

$= \ 10 \ log \ (\frac{(4 \pi)^2 \ d^2 L \ w^2}{G_t \ G_r \ 4 \pi \lambda^2})$

$= \ 10 \ log [ \frac{(4 \pi)^2 \ d^2 L}{G_t \ G_r \lambda^2}]$

$= \ -10 \ log (\frac{G_t G_r \lambda^2}{(4 \pi)^2 \ d^2 L^2})$

Assuming lossless system with L = 1 and having unity gain

$P_L (dB) \ = \ -10 \ log (\frac{\lambda^2}{(4 \pi)^2 \ d^2})$

The model is applicable only if the mobile lies in the far field region of the ant i.e. outside the ? distance.

$d \gt d_f$

$d_f : \frac{2 \ D^2}{ \lambda}$

Usually large scale propagation models use a close in reference distance $d_o$ and the received power $P_o$ at this point. The power at any other point is given by

$P_r(d) = \ Po \ (\frac{d_o}{d})^2$

Expressing the above equation

$Pr \ (d) \ dB_m = \ 10 \ log (\frac{P_o}{0.001}) + 20 \ log \ (\frac{d_o}{d})$

1) Find the far field distance for an antenna with a max dimension of 2.5 m and operating at a frequency of 900 MHz.

$\rightarrow$ $d_f = \frac{2D^2}{ \lambda}$

$d_f = 37.5 \ m$

$dBw + 30 = dBm$

2) If a tx produces 65W of power express the transmitted power in units of $dB_m$ and $dB_w$. If 50W is applied to a unity gain antenna with a 900 MHz carrier frequency find the received power in $dB_m$ at a free space distance of 100 m from the antenna. What is $P_r$ at 10 km for the same case. Assume unity gain for the receiving antenna also.

$\rightarrow$ $P_t = 65 w$

$P_t \ (dBw) = \ 10 \ log \ 65 = \ 18.13 \ dBw$

$P_t \ (dBm) = \ 10 \ log \ (65 \times 10^3) = 48.13 dB_m$

$P_r \ (d) \ = \frac{P_t \ G_t \ G_r \ \lambda^2}{(4 \pi)^2 \ d^2 \ L}$

$= \ \frac{50 \times 1 \times c^2}{(900 M)^2 \times (4 \pi)^2 \times 100^2 \times 1}$

$P_r (d) = 3.51 mw$

$P_r (d) = -24.56 dB_m$

$P_r(d) = P_o (\frac{d_o}{d})^2$

$P_r(101 \ cm) = 3.5 \times 10^{-6} \ (\frac{100}{10 \times 10^3})^2$

=$ 0.351 nw$

The mobile signal when propagating will undergo one of the following

**1) Reflection**

It occurs when the EM wave strikes an object that has large dimensions as compared to the wave length of the propagating wave. Example: walls of building, clear ground.

**2) Diffraction**

It occurs by obstructions with sharp irregularities. Secondary waves resulting from the obstructing surfaces are present all over the space, behind the obstacle also. Example: Edges of walls, pyramids, etc.

**3) Scattering**

It occurs when the dimension of the obstruction are smaller than the wave length of the signal.example: Lamp posts, small ?, street lights, etc.

**2 - Ray ground reflection model**

**Part 1:**

To derive an expression $E_{TOT}$ we proceed with following rules.

$E_g = \Gamma \ E_i$

$E_T = (1 + \Gamma) E_i$

And $Q_i = Q_o$

For simplicity we assume the following that $E_T = 0$ and the E component is perpendicularly polarized.

$E (d, t) = \frac{E_o \ d_o }{d} \ cos [ W_c ( t - \frac{d}{c})]$

$E_{Los} = \frac{E_o \ d_o }{d^{'}} \ cos [ W_c ( t - \frac{d^{'}}{c})]$

$E_g = \Gamma \ E_i$

Assuming reflection in near receiver,

Hence,

$E_g = \Gamma \ \frac{E_o \ d_o}{d^{”}} \ cos \ [ W_c (t - \frac{d^{”}}{c})]$

$R_{TOT} = \ E_{cos} + E_g$

$= \frac{E_o \ d_o}{d^{'}} \ cos \ [ \ W_c (t - \frac{ d^{"}}{c})]$

$= \frac{E_o \ d^o}{d^{”}} \ cos \ [ W_c ( t - \frac{ d^{”}}{c})]$ ------ [1]

As $\Gamma$ is perpendicularly polarized $\Gamma = -1$

**Part 2:**

To find out the path difference and phase difference, time delay between the two components.

Taylor series:

$(1 + r)^n = 1 + n r + \frac{n(n-1) r^2}{2!} p - - - -$

Path difference.

$\triangle = d” – d’$

$= \sqrt{ d^2 + (h_t + h_r)^2} - \sqrt{ d^2 + (h_t – h_r)^2}$

$= [d^2 + (h_t + h_r)^2]^{1/2} – [d^2 + (h_t – h_r)^2]^{1/2}$

$= d[ 1 + \frac{(h_t + h_r)^2}{d^2}]^{1/2} – d [ 1 + \frac{(h_t – h_r)^2}{d^2}]^{1/2}$

Using Taylor series expansion.

$\triangle = d [1 + \frac{(h_t + h_r)^2}{2d^2}] – d [1 + \frac{(h_t – h_r)^2}{2d^2}]$

$= d [ \frac{(h_t + h_r)^2}{2d^2} - \frac{ (h_t + h_r)^2}{2d^2}]$

$= \frac{1}{2d} [ 4h_t \ h_r]$

$\triangle = \frac{2(h_t \ h_r)}{d}$

The path difference $\theta_n$ between the 2E field component.

$\theta_{\triangle} = \frac{2 \pi \triangle}{ \lambda} = \frac{2 \pi \triangle}{ c/f} = \frac{ 2 \pi f \triangle}{c} = \frac{w_c \triangle}{c}$

$\theta_{\triangle} = \frac{w_c}{c} \times \frac{2 h_t \ h_r}{d}$

The time delay between arrival of waves

$\zeta_d = \frac{ \triangle}{c} = \frac{\theta_{\triangle}}{w_c}$

**Part 3:**

To find the expression E_{to + at} a particular time internet.

$t = \frac{d^{”}}{c}$

Substituting in [1]

$E_{TOT} = \frac{E_o d_o}{d^{’}} \ cos \ [W_c(\frac{d^{”} – d’}{c})] - \frac{E_o \ d_o}{d^{”}} \ cos \ 0^°$

$= \ \frac{E_o \ d_o}{d_o^{’}} \ cos \ \theta_{\triangle} = \frac{E_o \ d_o}{d^{”}} $

$= \ \frac{E_o \ d_o}{d} (cos \ \theta_{\triangle} – 1)$

$ [ \because \ \frac{E_o \ d_o}{d^{”}} \approx \ \frac{E_o \ d^o}{d^{'}} \approx \ \frac{E_o \ d_o}{d}] $

**Part 4:**

To find expression of $E_{Tot}$ using vector algebra.

$E_{Los} = A \ cos \theta_{ \triangle } \ \ \hat x + A \ sin \theta_{ \triangle} \ \hat y$

$E_g = - A \hat x$

$E_{TOT} = E_{Los} + E_g$

$= A \ cos \theta_{\triangle} \ \hat x^ + A \ sin \ \theta_{\triangle} \ \hat y - A \ \hat x$

$E_{TOT} = A \ \hat x (cos \ \theta_{\triangle} – 1) + A sin \ \theta_{\triangle} \ \hat y$

$| E_{TOT} | = \sqrt{A^2 (cos \theta_{\triangle} – 1)^2 + A^2 sin^2 \theta_{\triangle}} $

$= A \sqrt{cos^2 \theta_{\triangle} – 2 cos \theta_{\triangle} + 1 + sin^2 \theta_{\triangle}} $

= $ A \sqrt{ 2 – 2 \ cos \theta_{\triangle}}$

= $A \sqrt{2(1 – cos \theta_{\triangle})}$

= $A \sqrt{2.2 \ sin^2 \ \theta_{\triangle}/2}$

= $2A sin \ \frac{\theta_{\triangle}}{2}$

= $2 \frac{E_o \ d_o}{d} \ sin \frac{\theta_{\triangle} }{2^{’}}$

Whenever d is very large $\theta_{\triangle}$ is very small hence,

$sin \ \frac{\theta_{\triangle} }{2} \approx \ \frac{\theta_{\triangle} }{2}$

Hence

$|E_{TOT}| = \frac{2 \ E_o d_o}{d} . \frac{\theta_{\triangle}}{2}$

= $ \frac{2 E_o \ d_o}{d} \ \frac{2 \pi \triangle}{2 \lambda}$

= $ \frac{2 E_o d_o}{d} . \frac{2 \pi \ h_t \ h_r}{ \lambda d}$

= $\frac{4 \pi \ E_o \ d_o \ h_t \ h_r}{ \lambda} \times \frac{1}{d^2}$

$|E_{TOT}| \approx \ \frac{k}{d^2} \ v/m$

From this expression we get,

$\lambda = \frac{4 \pi \ E_o \ d_o \ h_t \ h_r}{ |E_{tot}|} \times \frac{1}{d^2}$

$\because$ There is a line of sight component substituting the expression of $\lambda$ into friss free space equation.

$P_r (d) = \frac{P_t \ G_t \ G_r \ \lambda^2}{(4 \pi)^2 \ d^2 \ (L)} $

= $\frac{P_t \ G_t \ G_r}{(4 \pi)^2 \ d^2} \frac{(4 \pi)^2 E_o^2 d_o^2 (h_t \ h_r)^2}{ |E_{tot}|^2}$ $\times \frac{1}{d^h}$

= $\frac{P_t \ G_t \ G_r}{d^2} \frac{E_o^2 \ d_o^2 (h_t \ h_r)^2}{ |E_{tot}|^2} \times \frac{1}{d^4}$

$|E_{tot}|^2 = \frac{E_o^2 \ d_o^2}{d^2}$

$P_r(d) = P_t \frac{G_t \ G_r (h_t \ h_r)^2}{d^4}$

The Perpendicular Polarization was assumed because Brewster angle exists parallel polarized waves.

** Definition:** The angle of incidence at which
$\Gamma = 0$ is called as Brewster, this exists only for parallel polarized waves.

Numerical:

1] Given that $P_t$ = 50w, $G_r$ = 2, $h_t$ = 50 m, $h_r$ = 1.5m, d = 10km.

Find the power received at a distance of 10 km to 2 ray ground reflection mode 1.

$P_r(d) = \frac{P_t \ G_t \ G_r (h_t \ h_r)^2}{d^4}$

= 56.25 PW.

$P_r(d)$ = -102.49 dBw

= - 72.49 dBw

2] A mobile is located 5 km away from the base station and uses a vertical $\lambda/4$ monopole antenna with a gain of 2.55 dB, to receive cellular signal. The E field at 1 km away from the transmitter is measured to be $10^{-3}$ v/m, the carrier frequency is 900 MHz.

A] Find the length and the effective aperture of the receiving antenna.

B] Find the received power at the mobile using 2 ray ground reflection model assuming height of ? is 50 m and receiving ant. is 1.5m above the ground.

$\rightarrow$ a) d = 5 km

$l = \frac{\lambda}{4}$

$G = \frac{4 \pi A_e}{ \lambda^2}$

G = 1.79

L = 8.33 cm

$1.79 = \frac{4 \ \pi A_e}{ \lambda^2}$

$A_e = 0.01 59 m^2$

???????

= $5.7 \times 10^{-5}$

The phenomena of diffraction is totally dependent on Huygens’s principles signals normally propagate through n number of Fresnel zones which are concentric circles around the LOS path.

- Any diffract or in the path will $\uparrow$ the number of Fresnel zones.
- Signals passing through even number of Fresnel zones are constructive in nature.
- Signals passing through odd number of zones are destructive in nature.
- The path difference $\triangle$ of any Fresnel zones can be calculated as

$$\triangle = \frac{x \lambda}{2}$$

- Radius of the nth Fresnel zone can be calculated as follows i.e.

$$r_n = \sqrt{ \frac{n \ \lambda d_1 d_2}{d_1 + d_2}}$$

**Knife Edge Diffraction Model**

The model aims to find the path difference, phase difference angle $\alpha$.The equivalent can be drawn as

$\triangle = (p_1 + p_2) – (d_1 +d_2)$

$p_1 = \sqrt{d_1^2 + h^2}$ $\quad$ $p_2 = \sqrt{d_2^2 + h^2}$

$p_1 = (d_1^2 + h^2)^{1/2}$

$P_1 = d_1 [ 1 + \frac{h^2}{d_1^2}]^{1/2}$

= $ d_1 + \frac{h^2}{2d_1}$

Similarly,

$p_2 = d_2 + \frac{h^2}{2d^2}$

$\triangle = (p_1 + p_2) – (d_1 + d_2)$

= $d_1 + \frac{h^2}{2d_1} + d_2 + \frac{h^2}{2d_2} – d_1 – d_2$

= $\frac{h^2}{2d_1} + \frac{h^2}{2d_2}$

$\triangle = \frac{h^2}{2} ( \frac{1}{d_1} + \frac{1}{d_2})$

$\theta_{ \triangle} = \frac{2 \pi \triangle }{\lambda}$

$\theta_{ \triangle} = \frac{ \Gamma h^2}{ \lambda} ( \frac{d_1 + d_2}{ d_1 d_2}) $

Diffraction angle $\alpha$ depends on $\beta$ and $\gamma$

By exterior angle property,

$\alpha = \beta + \gamma$

$\beta = tan^{-1} (\frac{h}{d_1}) \quad \gamma = \ tan^{-1} (\frac{h}{d_2})$

$\alpha = \ tan^{-1} ( \frac{h}{d_1}) + \ tan^{-1} \ (\frac{h}{d_2})$

$\beta$ and $\gamma$ are very small.

$\alpha = \frac{h}{d_1} + \frac{h}{d_2} = h \ ( \frac{d_1 + d_2}{ d_1 d_2})$

The parameter V i.e. Fresnel Kirchhoff diffraction parameter is a constant for a particular frequency and a given setup and this helps us to predict the losses using this model.

$v = h \sqrt{ 2 \frac{(d_1 + d_2)}{ \lambda \ d_1 d_2}} = \alpha = \sqrt{ \frac{2 d_1 d_2}{ \lambda (d_1 + d_2)}}$

We have a set of empirical equations which depends on parameters.

$L(d B) = 0 \quad \quad \quad V \leq -1$

$L(dB) = 20 \ log \ (0.5 – 0.62 v) \quad \quad -1 \ \angle \ V \ \leq \ \zeta $

$L (dB) = 20 \ log \ (0.5 e^{-0.95} V)$

$L (dB) = 20 \ log \ (0.4 - \sqrt{ 0.1184 – (0.38 – 0.1)^2?} \quad \quad 1 \leq V \leq 2.4$

$L(dB) = 20 \ log \ (\frac{0.225}{V}) \quad \quad V \gt 2.4$

Compute the diffraction loss for the 3 cases when: case a) h = 25 m, case b) h = 0, case © = -25 m Assume $\lambda = 1/3 \ m, \ d_1 = 1 \ km, \ d_2 = 1 \ km$ for each of these cases identify the Fresnel zone within which the tip of the obstruction lies If Frequency correlation function is taken to be greater than 0.9, then

$B_c = \frac{1}{50 \sigma \zeta }$

If frequency correlation function is from 0.5 to 0.9, then

$B_c = \frac{1}{5 \sigma \zeta}$

Doppler shift.

- It is the apparent change in received carrier frequency whenever the t x or r x? are moving.
- Whenever the ? are moving towards each other there is a apparent $\uparrow$ in the frequency expressed as $f_R = f_c + f_d$
- Whenever the ?, are moving away from each other there is apparent $ \downarrow$ in the frequency expressed as: $F_p = f_c – f_d$

*Exp. For Doppler shift*

Assuming S is the source and the rx moves from point A to B in the time interval $\triangle t$. Let d be the distance between A and B, starting with the exp. of the path difference. $\triangle t$ we proceed as follows:

$\triangle l = d cos \theta$

$\theta \lambda = \frac{2 \pi }{ \lambda} \triangle l $

$\theta \triangle = \frac{2 \pi \ d \ cos \ \theta}{ \lambda}$

$\theta \triangle = \frac{ 2 \pi \ V \triangle t \ cos \theta}{ \lambda} $

$\frac{ \theta \triangle }{ \triangle t} = \frac{ 2 \pi }{ \lambda } V \ cos \ \theta$

$w_d = \frac{v}{ \lambda} 2 \ \pi \ cos \ \theta$

$f_d = \frac{v}{ \lambda} \ cos \theta$

**Doppler spread ($B_D$)**

It is the spectral broadening of the received signal due to channel movement or the mobile station movement.

**Coherence time**

The time duration over within the channel response is non-changing.

** Definition:** It is the time duration over which 2 received signals have a strong potential for amplitude correlation.

$T_c \ \approx \ \frac{1}{B_p}$

$T_c = \sqrt{ \frac{9}{16 \ \pi f^2 \ d}} = \frac{0.423}{f_d}$

Calculate the mean excess delay $\zeta_r$ rms delay spread $\sigma_{\zeta}$, for the profile given in figure. Estimate $50 \%$ of coherence between of channel will this channel be suitable for AMPS, GSM service without the use of the equalizer.

$\bar{\zeta} = \frac{0.01 (0) + 0.1 (1) + (0.2) + 1 (5)}{0.01 + 0.1 \times 2 + 1}$

$\bar{\zeta} = \frac{5.3}{1.21} = 4.38 \mu s$

$ \bar{\zeta^2} = \frac{0.01(0)^2 + 0.1 (1)^2 + 0.1 (2)^2 + 1(5)^5}{0.01 + 0.1 \times 2 +1 }$

$= \frac{25.5}{1.21} = 21.07 \mu s^2$

$\sigma_{\zeta} = \sqrt{ \bar{\zeta^2} – ( \bar{\zeta)^2}}$

$\sigma_{\zeta} = 1.373 \mu s$

Since $50 \%$ correlation is accepted

$B_c = \frac{1}{ 5 \ \sigma_{\zeta}}$

= 145. 66 KHz.

For an AMPS signal $B_s$ = 30 KHz here $B_s \lt \lt B_c$ hence it will be a faded signal. Hence use of equalizer is not necessary.

For GSM signal $B_c$ = 200 KHz. Here $B_s \gt \gt B_c$ hence it will be a frequency selective faded signal. Hence equalizer is required.

Consider a ? which radiates a sinusoidal ratio frequency of 1850 MHz. for a vehicle moving 60 miles/hr compute the received carrier frequency if the mobile is moving.

A] Directly towards the t ?

B] Directly away from the t?

C] In a direction which is ? to the direction of arrival of the tx ? signal.

$\rightarrow \ f_c = 185 \ MHz$

$\lambda = 0.162 \ m$

$V = \frac{60 \ miles \ \times \ 1609.344}{3600}$

= 26.82 m/s

a] $f_d = \frac{V}{ \lambda} \ cos \ \theta \quad \quad cos \ \theta = 0$

= 165.55 Hz.

Received carrier frequency = $f_c + f_d = 1850.000165 \ MHz$

b] $\theta$ = 180

$F_d$ = - 26.82 m/s

Received carrier frequency = fc – fd

= 1849.999834 MHz

c] $\theta = 90°$

$f_d$ = 0

$f_r$ = f 1850 MHz

**Models based on probability**

- Most of the received mobile signals follow a Rayleigh distribution.
- The probability of receiving a power level r is given as

$ P(r) = \begin{cases} \frac{r}{\sigma^2} exp (\frac{-r^2}{2 \sigma^2}), & 0 \leq r \leq \infty \\ 0, &r \lt 0 \end{cases}$

Whenever there is no COS it will be Rayleigh distribution.

CDF

$P(R) = P_r (r \leq R) = 1 – exp \ (\frac{-R^2}{2 \sigma^2})$

Mean value

$r_{mean} = E [r] = \int_0^{\infty} \ r P(R) \ dr = \sigma \sqrt{ \pi/2}$

=$ 1.2533 \ \sigma$

Variance

$E [r^2] = E [x^2] + E [y^2] = 2 \sigma^2$

$\sigma \ \sigma^2 E [r^2] – ( E [r] )^2 = 2 \sigma^2 . \sigma (\pi / 2) $

= $0.4292 \sigma^2$

Rician channel when there is LOS

A = maximum amplitude signal ( LOS component)

$I_o$ = Bessel function of $0^{th}$ order.

r = instantaneous power received

$ P(r) = \begin{cases} \frac{r}{\sigma^2} exp (\frac{-r^2+A^2}{2 \sigma^2}) I_o (\frac{A_r}{\sigma^2}), & A \geq 0 \\ 0, &r \lt 0 \end{cases}$

**Risk factor undefinable**

$K (dB) = \ 10 \ log \ [ A^2 \ (2 \sigma^2) ] \ dB$

Definition: The ratio of the LOS component A to the variance of the variable is called as the risk factor. If there is a strong LOS component the variance is less and K increases.