100 W वाहक शक्ति वाले 100% मॉडुलित AM सिग्नल में, निचले साइडबैंड में शक्ति क्या है?

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RRB JE ECE 22 Apr 2025 Shift 2 CBT 2 Official Paper
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  1. 50 W
  2. 25 W
  3. 15 W
  4. 150 W

Answer (Detailed Solution Below)

Option 2 : 25 W
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अवधारणा:

एक AM प्रणाली के लिए कुल संचरित शक्ति इस प्रकार दी गई है:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pc = वाहक शक्ति

μ = मॉडुलन सूचकांक

उपरोक्त व्यंजक को इस प्रकार विस्तारित किया जा सकता है:

\({P_t} = {P_c} + P_c\frac{{{μ^2}}}{2}\)

कुल शक्ति वाहक शक्ति और साइडबैंड शक्ति का योग है, अर्थात्

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

गणना:

दिया गया: Pc = 100 W और μ = 1

हम लिख सकते हैं:

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

\(= 100 \times \frac{1^2}{2}\)

\(P_s=50~W\)

कुल साइडबैंड शक्ति = 50 W

ऊपरी साइडबैंड में शक्ति + निचले साइडबैंड में शक्ति = 50 W

ऊपरी साइडबैंड में शक्ति = निचले साइडबैंड में शक्ति = 25 W

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