# عندي أسئلة هامه بأحدى المواد اللي أدرسها

#### Mr-ALDE7WI

##### عضو جديد
بسم الله الرحمن الرحيم

أرجووو منكم يا إخواني المهندسين الكرام الماده هذي مسببه لي عقده .. :55::82::82:

وهذ1 نصه الواجب ... مع الشرح Please أبي أفهم وش يصير :80::80:

[FONT=&quot]1.
[/FONT][FONT=&quot] Find thermal noise power density at T=290 0[/FONT]

[FONT=&quot]

[/FONT]

[FONT=&quot]2.
[/FONT][FONT=&quot] Given a receiver with noise tempertaure of 21 0C and a 10 MHz bandwidth. Find thermal noise at receiver’s output. )[/FONT]

[FONT=&quot]3. [/FONT][FONT=&quot]
[/FONT]
[FONT=&quot][/FONT]
[FONT=&quot])If a signal with a power of 5 W is inserted onto e transmission line and the measured power some distance away is 50 mW. Find the loss of the signal power. )[/FONT]

[FONT=&quot]4. [/FONT][FONT=&quot]
[/FONT]

[FONT=&quot]Consider a series of transmission elements in which the input is at a power level of 4 mW, the first element is a transmission line with 12 dB loss, the second element is an amplifier with 35 dB gain, and the third element is a transmission line with a 10 dB loss. [/FONT]
[FONT=&quot]Find the output power Pout. [/FONT]

:56::56: شكراً مقدماً :56::56:

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#### Mr-ALDE7WI

##### عضو جديد
هذي المعطيات ...

[FONT=&quot]Solution:[/FONT]
[FONT=&quot]1. [/FONT][FONT=&quot]C = 2B = 6200 bit/s[/FONT][FONT=&quot][/FONT]
[FONT=&quot]If we use signal with 8 voltage levels, what will be the capacity C ?[/FONT][FONT=&quot][/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]C = 2B log2M = 18.600 bit/s[/FONT][FONT=&quot][/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]2. [/FONT][FONT=&quot]Amount of thermal noise in a bandwidth of 1 Hz in any device or conductor is:[/FONT][FONT=&quot][/FONT]
[FONT=&quot]N0 = KT (W/Hz)[/FONT]
[FONT=&quot]N0 - noise power density in watts per 1 Hz bandwidth [/FONT]
[FONT=&quot]k - Boltzmann‘s constant - 1.38 x 10-23 J/K[/FONT]
[FONT=&quot]T – temperature, in kelvins (absolute temperature)[/FONT]
[FONT=&quot]At T = 17 0 C, T in kelvin is 273 +17 =300 0 K[/FONT]
[FONT=&quot]N0= 1.38 x 10-23 [/FONT][FONT=&quot]x[/FONT][FONT=&quot] 300 = 4.14 x 10-21[/FONT]
[FONT=&quot]10 [/FONT][FONT=&quot]og10N = -204 dBW/Hz[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]3.Thermal noise in a bandwidth of B can be expressed as: N = kTB [/FONT][FONT=&quot][/FONT]
[FONT=&quot] or in decibel-watts: Ndb = 10logk + 10logT + 10logB[/FONT][FONT=&quot][/FONT]
[FONT=&quot] = -228,6 dBW + 10logT + 10logB[/FONT][FONT=&quot][/FONT]
[FONT=&quot]k - Boltzmann‘s constant 1.38 x 10-23 J/K[/FONT]
[FONT=&quot]T – temperature, in kelvins (absolute temperature)[/FONT]
[FONT=&quot]Sor given T= 294 0 K and B= 10 MHz thermal noise at receiver‘s output will be:[/FONT]
[FONT=&quot]N = KTB = 1.38 x 10-23 [/FONT][FONT=&quot]x [/FONT][FONT=&quot]294 [/FONT][FONT=&quot]x [/FONT][FONT=&quot]10 x 106 = 4057 x 10-17[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]10log10 N = -133,9 dBW[/FONT]