Thermal Noise & Data Rate Examples

Thermal Noise

Thermal Noise in watts present in a bw(?) of B Hz:

N=k.T.B

indecibel watts:

\begin{align*}
N&=10log(k)+10log(T)+10log(B)\\
N&=-228.6dBW+10log(T)+10log(B)
\end{align*}

Example: Give a receiver with an effective noise temp of 294K and a 10MHz bandwith what is the thermal noise level at this receiver’s output?

\begin{align*}
N&=-228.6~dBW+10log(294)+10log(10^7)\\
N&=-228.6~dBW+24.7+70\\
N&=-133.9~dBW
\end{align*}

DATA RATE

Bir saniyede ne kadar bit yollanabildiği 3 faktöre bağlı

1. Available Bandwith
2. The level of the signal we use
3. The quality of the channel (the level of noise)

Nyquist Channel Capacity (noiseless channel)

C=2B~log_2(M)

Bit~rate=2*bandwith*log_2(L)\\
R=2B*log_2(L)

Shannon Channel Capacity (noisy channel)

C=B*log_2(1+SNR)

• Shannon upper limiti hakkında fikir verir.
• Nyquist kaç level gerektiği hakkında fikir verir.

Ex1

Consider a noiseless channel with bandwith of 3000Hz transmitting a signal with four levels. What is C?

C=2*3000*log_2(4)=12'000~bps

Ex2

We need to send 265 kbps over a noiseless channel with a bandwith of 20kHz. How many signal levels do we need?

256'000~bps=2*20'000*log_2(M)\\
log_2(M)=6'625\\
M=98.7~levels \rightarrow 128~levels(280kbps)

Ex3

We want to calculate the theoritical highest bit rate of a telephone line. A telephone line is type has a bandwith of 3000Hz. The SNR is usually 3162. What is the channel capacity

\begin{align*}
C&=B*log_2(1+SNR)\\
C&=3000*log_2(3163)\\
C&\cong34'860~bps
\end{align*}

Ex4

The SNR is 36 dB in a channel which has a bandwith of 2 MHz. What is channel capacity?

SNRdb=10*log(SNR)\\
26=10*log(SNR) \cong 3981\\
~~\\
C=B*log_2(1+SNR)\\
C=2*10^6*log_2(3981)\\
C=24Mbps

Ex5

Assume that we have a channel with 1 MHz bandwith. The SNR is 63. What is the bitrate and the signal level that we need?

\begin{align*}
C&=B*log_2(1+SNR)\\
C&=10^6*log_2(64)\\
C&=6~Mbps\\
4~Mbps&=2*1~MHz*log_2(M) \Rightarrow M=4
\end{align*}