User: Sayali Bagwe

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Sayali Bagwe1.9k
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Posts by Sayali Bagwe

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Answer: A: A 2kHz channel has signal to noise ratio of 24 dB
... 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:** $∴ b.w, B_2=1kHz, old b.w., B_1=2 kHz \\ ∴N=N_o B,∴N ...
written 3 months ago by Sayali Bagwe1.9k
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A 2kHz channel has signal to noise ratio of 24 dB
... 1. Calculate max capacity of this channel. 2. Assuming constant transmitting power, calculate max. capacity when channel b.w is 1 halved 2, reduced to quarter of its original value. **Subject**: Computer Engineering **Topic**: Electronic Circuits and Communication Fundamentals **Difficulty**: Me ...
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Answer: A: An analog signal has a bit rate of 8000 bps and a band rate of 1000 band. How ma
... Given: Bit rate (n) = 8000 bps Band rate = 1000 band To find: 1. Data elements carried by each signal element (R) 2. Total signal element (L) R=? Bit rate = numbers of data elements per (R) signal × band rate ∴$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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An analog signal has a bit rate of 8000 bps and a band rate of 1000 band. How many data elements are carried by each signal element? How many signal elements do we need?
... **Subject**: Computer Engineering **Topic**: Electronic Circuits and Communication Fundamentals **Difficulty**: Medium / High ...
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Answer: A: Calculate the minimum bit rate for a channel having b.w. 3100 Hz and S/N ratio 1
... 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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Calculate the minimum bit rate for a channel having b.w. 3100 Hz and S/N ratio 10 dB
... **Subject**: Computer Engineering **Topic**: Electronic Circuits and Communication Fundamentals **Difficulty**: Medium / High ...
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Answer: A: Calculate the maximum bit rate for a channel having b.w. 3100 Hz and S/N ratio 2
... 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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Calculate the maximum bit rate for a channel having b.w. 3100 Hz and S/N ratio 20 dB
... **Subject**: Computer Engineering **Topic**: Electronic Circuits and Communication Fundamentals **Difficulty**: Medium / High ...
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Answer: A: Define. i) Channel capacity ii) Data rate iii) Channel bandwidth
... **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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Define. i) Channel capacity ii) Data rate iii) Channel bandwidth
... **Subject**: Computer Engineering **Topic**: Electronic Circuits and Communication Fundamentals **Difficulty**: Medium / High ...

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