Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 10 (Q.No. 7)
7.
The noise figure of individual stages of a two stage amplifier is 2.03 and 1.54 respectively. If gain of first stage is 62, the overall noise figure is
Discussion:
3 comments Page 1 of 1.
Angelo said:
2 years ago
Solving Noise Factor,
F_1 = 10^(NF_1/10).
= 10^(2.03/10).
= 1.60.
F_2 = 10^(1.54/10).
= 1.43.
Solving Total NF,
F_T = F_1 + (F_2-1)/A_1.
= 1.60 + (1.43-1)/62.
= 1.6069.
NF_T = 10log(F_T).
= 2.056.
F_1 = 10^(NF_1/10).
= 10^(2.03/10).
= 1.60.
F_2 = 10^(1.54/10).
= 1.43.
Solving Total NF,
F_T = F_1 + (F_2-1)/A_1.
= 1.60 + (1.43-1)/62.
= 1.6069.
NF_T = 10log(F_T).
= 2.056.
OujaMori said:
9 years ago
The answer should be A.
Using Friis formula:.
NF = 2.03 + (1.54 - 1)/62 is 2.0387.
Using Friis formula:.
NF = 2.03 + (1.54 - 1)/62 is 2.0387.
Arvin Maclang said:
10 years ago
Noise figure is different from the noise factor. The total noise factor can be calculated using the Friis formula for cascaded amplifiers.
Total noise factor = f1+((f2-1)/A1), where f1 and f2 are the respective noise factors for the individual stages.
From that, the total noise figure can be solved by: Total noise figure = 10 log (Total noise factor).
Total noise factor = f1+((f2-1)/A1), where f1 and f2 are the respective noise factors for the individual stages.
From that, the total noise figure can be solved by: Total noise figure = 10 log (Total noise factor).
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers