Shannons Formula for maximum capacity in bps The maximum capacity of a channel is given by : C = B log2(1 + SNR) this is the , bit rate that could be transmitted with no errors So Capacity can be increased by:

Increasing Bandwidth Increasing SNR (note: capacity linear in SNR(dB))

Remark: for an infinite SNR we can send an infinite number of bits/ second whatever the allocated bandwidth. The formula tells us also that if we reduce the power (SNR) we have to increase the bandwidth for same channel capacity and vice- versa. (Power, Bandwidth trade off). C = B log2(1 + SNR) bits/s SHANNON ) 1( r=2.B symbols/s 2.B.log2(M) bit/s NYQUIST (2)

2.B.log2(M) = B log2(1 + SNR) SNRM += 1 (3) M is the number if levels that can be used. log2(M) = The number of bits/symbol that can be used over this channel = 0,5.log2(1+SNR) Special conditions: If log2(1 + SNR)=2 (1 + SNR)=4 M=2. 1 bit binary transmission only. So we can transmit 2B symbols/s 1 bit/symbol= 2B bit/s (C=2B). If log2(1 + SNR) < 2 (1 + SNR) < 4 M< 2 C = B log2(1 + SNR) < 2 B. Here we can not send data in binary form @ nyquist rate over this channel (the limit is C bits/s). Binary data could be sent but a rate smaller than 2.B symbols/s

Channel capacity= Max number of symbols per second * Max number of bits /symbol Max number of symbols per second : limited by the bandwidth

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FirasSticky NoteC : capacity : bits/sec

B : bandwidth

FirasSticky NoteM : L : number of levels = 2^nn : bits / symbol

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Max number of bits /symbol: Limited by the power. C/B function of Eb/N0.

).0

.1(2log.)1(2log.BN

RbEbBSNRBC +=+=

If we are transmitting @ full channel capacity Rb=C.

).0.1(2log.BNCEbBC += )

.0.1(2log/BNCEbBC += BCN

Eb BC

/12

0

/

=

This the theoretical limit for the relation between the spectral efficiency(C/B b/s/Hz) and the power efficiency (Eb/N0)

Maximum spectral efficiency

C/B is the spectral efficiency (Number of bits per second per HZ) Question : What is the Eb/N0 ratio that requires an infinite bandwidth to send data with 0 error probability ??

2ln.0.

.

2ln

).0.1ln(

.).0.1(2log. BN

CEb

BBNCEb

BBNCEbBC =

+=+=

--> Eb/N0= ln2

If x

Exercise 1: A Telephone circuit has a bandwidth of 3000 Hz and a S/N ratio of 30dB a)What is the channel capacity b) Maximum symbol rate for full use of bandwidth (Nyquist limit) c) Whats the corresponding number of bits / symbol d) What is The effective number of distinguishable levels M Solution:

a) Channel capacity : 3000*log2( 1+ 1000)= 30000 *10= 30 000 b/s max b) Maximum symbol rate= 6 Ksym/s = 2*3000.

c) SNRM += 1 , SNR= 1000 in linear, M=32 5 bits/ symbol. d) Number of levels= 32 levels. That means we can transform our Binary

(PCM signal) into a PAM (5 bits/symbol) and we will need less bandwidth.

Exercise 2: We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need? Solution

We can use the Nyquist formula as shown:

In 20 KHz bandwidth we can send a max symbol rate : 2B symbols/s= 40 Ksymbols/s. Noiseless channel M has no limits.

We need to send 265 Kbp/s )(2log..2 MB >= 265 Kbps 40 .log2(M) > 265 log2(M) >=6,625 M

>=98,7 Levels M #2n

M=128 levels, the maximum allowed bit rate with 128 levels is 280 kbps > 265 Kbps

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Remark: in this case the bit rate is fixed but the channel capacity is infinity. Exercise 3:

1- The SNR is 30 dB in the bandwidth of 32 KHz. What is the channel capacity for this channel.

2- How many bits/symbol are allowed 3- A digital signal has a bit rate of 64 Kbps/s . We transmit this signal with

M levels . (M calculated in part 2). Could we use an 8 KHz channel to transmit this signal

Solution:

1- SNR= 30 1000 . KbpsKSNRBC 320~)1001(2log*32)1(2log. =+=

2- SNRM += 1 = 32 levels 3- For binary: 64 Kbps 64 /5 symbols/s ~ 12,8 Ksymbols/s

Exercise 4:

What is the ratio between the bandwidth of an analog signal and a digital signal. We assume fs= 2fmax and digital signal is sent in a binary format M=2. Solution: BW of analog= fmax

BW of the digital: 2.fmax samples/s * n bits/ sample 2n.fmax bits/s. Binary format : 2n.fmax bits /s 2n. fmax symbols/s 2n.fmax symbols/s requires (Nyquist) a bandwidth of n.fmax. This result is interesting because it tells us that the digital signal in its binary format requires a lot of bandwidth when compared with the original signal.

- Exercise 5: What is the required SNR to send a digital signal (n bits/ sample, n=8) in a bandwidth= to the bandwidth of the original analog signal. Solution:

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BW digital= n.BW analog We need to reduce the bandwidth of the digital signal by a factor of n=8 So we should be able to transform 8 bits into 1 symbol M= 256

SNRM += 1 1+SNR= M2= 2^16 SNR~ 2^16 SNR dB= 48 dB.

Exercise 6: We would like to transmit 100 Mb/s in a bandwidth of 10 MHz.

1- What is the required Eb/N0. 2- We consider N0= -174 dBm/Hz What is the required power for the transmission in dBm Solution 1- C/B=10

BCNEb BC

/12

0

/

= =10

2^10=1024/10=102.4 Eb/N0)dB ~ 20 dB

3- EbdB= -174+20=-154 dBm P= Eb.C PdBm= EbdBm+ C dB=-154+ 10.log(100.106)=-154+80= -74 dBm This is Shannon limit , in reality systems try to reach Shannon limit.

Exercise 7

A modem is operating @ 56 Kbps uses an 8 levels modulation (3 bits per symbol)

The modem has an ADC with a quantization step of q=1mV and a resolution N= 8 bits

1- Calculate quantization noise. 2- Calculate the SNR of the ADC 3- Deduce the power of the signal (S) corresponding to the full scale

operation The channel has a frequency bandwidth of 2400 Hz 4- Calculate the channel SNR for the good operation of the modem. 5- Whats the maximum allowed frequency at the input of the ADC.

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lecture 3 channel capacity

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