## User: Sayali Bagwe

Sayali Bagwe1.9k
Reputation:
1,940
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Cotton Green
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http://www.ques10.com/
Last seen:
3 months, 3 weeks ago
Joined:
2 years, 11 months ago
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#### Posts by Sayali Bagwe

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... B= 2 kHz and (S/N) = 24dB $∴24=10 \log_10⁡ \big(\frac{S}{N}\big) \\ ∴\frac{S}{N}=251$ **Channel capacity:** $C= B \log_2⁡ \big[1+\frac{S}{N}\big] \\ =2 \times 10^3 \log_2⁡[1+251] \\ C = 15.95 \times 10^3 bits/sec$ **B is halved:** $∴ b.w, B_2=1kHz, old b.w., B_1=2 kHz \\ ∴N=N_o B,∴N ... written 3 months ago by Sayali Bagwe1.9k 1 answer 136 views 1 answer ... 1. Calculate max capacity of this channel. 2. Assuming constant transmitting power, calculate max. capacity when channel b.w is 1 halved 2, reduced to quarter of its original value. **Subject**: Computer Engineering **Topic**: Electronic Circuits and Communication Fundamentals **Difficulty**: Me ... written 3 months ago by Sayali Bagwe1.9k 1 answer 148 views 1 answers ... Given: Bit rate (n) = 8000 bps Band rate = 1000 band To find: 1. Data elements carried by each signal element (R) 2. Total signal element (L) R=? Bit rate = numbers of data elements per (R) signal × band rate ∴$R=\frac{\text{bit rate(n)}}{\text{Band rate}} =\frac{8000}{1000}=8 Bit/ ...
written 3 months ago by Sayali Bagwe1.9k
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... **Subject**: Computer Engineering **Topic**: Electronic Circuits and Communication Fundamentals **Difficulty**: Medium / High ...
written 3 months ago by Sayali Bagwe1.9k
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... Given: - B = 3100 Hz $\big(\frac{S}{N}\big)dB=10$ $∴10=10 log_{10}⁡big(\frac{S}{N}\big) \\ ∴\frac{S}{N}=10$ ∴ Maximum bit rate $= R_{\max} = B \log_2⁡ \big[1+\frac{S}{N}\big] \\ = 3100 \log_2⁡(1+10) \\ =\frac{3100 \log_{10}⁡11}{\log_{10}⁡2} \\ = 10,724 bits/sec$ ...
written 3 months ago by Sayali Bagwe1.9k
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... **Subject**: Computer Engineering **Topic**: Electronic Circuits and Communication Fundamentals **Difficulty**: Medium / High ...
written 3 months ago by Sayali Bagwe1.9k
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... Given B = 3100 Hz S/N = 20 dB But 20 dB = 10 log (S/N) ∴ S/N = 100 The maximum bit rate: - $R_{\max⁡ }=B \log_2⁡ \big[1+\frac{S}{N}\big] \\ =3100 log_2⁡[1+100 ] \\ =\frac{(3100 \log_{10}⁡101 )}{log_{10}⁡2 } =20,640 bits/sec$ ...
written 3 months ago by Sayali Bagwe1.9k
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113
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... **Subject**: Computer Engineering **Topic**: Electronic Circuits and Communication Fundamentals **Difficulty**: Medium / High ...
written 3 months ago by Sayali Bagwe1.9k
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... **Channel capacity (C):** The channel capacity is defined as the maximum data rate at which the digital date can be transmitted over the channel reliably. The various concepts related to channel capacity are: **Data Rate:** It is defined as no. of bits transmitted by the transmitted per second ...
written 3 months ago by Sayali Bagwe1.9k
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... **Subject**: Computer Engineering **Topic**: Electronic Circuits and Communication Fundamentals **Difficulty**: Medium / High ...
written 3 months ago by Sayali Bagwe1.9k

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