A link is described by a 2-ray ground reflection model, where r >> ht, hr and Pt = 50 W, fc = 1800 MHz, Gt = 20 dBi and Gt = 3 dBi.<\/p>\n\n\n\n
a. If the received LOS signal power was measured to be Pr = 1\u03bcW at a range of r = 1 km, determine the path loss and the corresponding path-loss exponent.<\/p>\n\n\n\n
b. Consider the 2-ray ground reflection model with the following system values: ht = 40 m, hr = 3 m and Gr = 3 dBi. Calculate the received power level at r = 2 km.<\/p>\n\n\n\n
c. Consider the 2-ray ground reflection model of (b). Suppose the receiver antenna height hr is adjustable (but all other system values are fixed). Findthe value of hr which maximizes the received power.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve this, we use principles from wireless communication theory, specifically the 2-ray ground reflection model<\/strong>. Here’s the detailed solution:<\/p>\n\n\n\n Given:<\/strong><\/p>\n\n\n\n [ Substitute:<\/p>\n\n\n\n [ In free-space and many environments:<\/p>\n\n\n\n [ At r = 1 km (base), we define this as reference. To get n<\/strong>, we need another measurement at a different r. But since this is from a 2-ray model<\/strong>, at large r:<\/p>\n\n\n\n [ \u2714\ufe0f Path loss = 100 dB<\/strong> [ Substitute:<\/p>\n\n\n\n [ From 2-ray model:<\/p>\n\n\n\n [ But 2-ray model has destructive interference at some ranges<\/strong>, especially when:<\/p>\n\n\n\n [ At large distances, interference diminishes and power becomes proportional to ( h_r^2 ). Hence, maximum Pr occurs when hr is as large as possible<\/strong>.<\/p>\n\n\n\n However, exact maximum<\/strong> occurs when:<\/p>\n\n\n\n [ A link is described by a 2-ray ground reflection model, where r >> ht, hr and Pt = 50 W, fc = 1800 MHz, Gt = 20 dBi and Gt = 3 dBi. a. If the received LOS signal power was measured to be Pr = 1\u03bcW at a range of r = 1 km, […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-210288","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/210288","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=210288"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/210288\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=210288"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=210288"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=210288"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\na. Path Loss and Path-Loss Exponent<\/strong><\/h3>\n\n\n\n
\n
i. Path Loss (PL) in dB<\/strong><\/h4>\n\n\n\n
PL = 10 \\log_{10} \\left( \\frac{P_t \\cdot G_t \\cdot G_r}{P_r} \\right)
]<\/p>\n\n\n\n
PL = 10 \\log_{10} \\left( \\frac{50 \\cdot 100 \\cdot 2}{1 \\times 10^{-6}} \\right) = 10 \\log_{10} (10^4 \\cdot 10^6) = 10 \\log_{10}(10^{10}) = 100\\ \\text{dB}
]<\/p>\n\n\n\nii. Path-Loss Exponent (n)<\/strong><\/h4>\n\n\n\n
P_r \\propto \\frac{1}{r^n} \\Rightarrow 10 \\log_{10} P_r = -10n \\log_{10} r + C
]<\/p>\n\n\n\n
P_r = \\frac{P_t G_t G_r h_t^2 h_r^2}{r^4}
\\Rightarrow n = 4
]<\/p>\n\n\n\n
\u2714\ufe0f Path-loss exponent = 4<\/strong><\/p>\n\n\n\n
\n\n\n\nb. Received Power at r = 2 km using 2-Ray Model<\/strong><\/h3>\n\n\n\n
P_r = \\frac{P_t G_t G_r h_t^2 h_r^2}{r^4}
]<\/p>\n\n\n\n\n
P_r = \\frac{50 \\cdot 100 \\cdot 2 \\cdot (40)^2 \\cdot (3)^2}{(2000)^4}
= \\frac{50 \\cdot 100 \\cdot 2 \\cdot 1600 \\cdot 9}{(16 \\times 10^{12})}
= \\frac{144 \\times 10^6}{1.6 \\times 10^{13}} = 9 \\times 10^{-6} = \\boxed{9\\ \\text{\u03bcW}}
]<\/p>\n\n\n\n
\n\n\n\nc. Maximize Received Power by Optimizing hr<\/strong><\/h3>\n\n\n\n
P_r \\propto h_t^2 h_r^2 \\Rightarrow \\text{Maximize } h_r
]<\/p>\n\n\n\n
r_c = \\frac{4 h_t h_r}{\\lambda}
]<\/p>\n\n\n\n
h_r = \\frac{r \\lambda}{4 h_t}
= \\frac{2000 \\cdot 0.1667}{4 \\cdot 40}
= \\frac{333.4}{160} \u2248 \\boxed{2.08\\ \\text{m}}
]<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/h3>\n\n\n\n
\n
\n","protected":false},"excerpt":{"rendered":"