• 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.