Bandwidth and Channel Capacity
 |
| Bandwidth of a channel |
Bandwidth of a medium decides the quality of the signal at the other end. A digital signal (usually aperiodic) requires a bandwidth from 0 to infinity. So, it needs a low-pass channel characteristic as shown in Fig(Low-pass channel characteristic required for the transmission of digital signals) On the other hand, a band-pass channel characteristic is required for the transmission of analog signals, as shown in Fig
 |
| Low-pass channel characteristic required for the transmission of digital signals |
 |
| Band-pass channel characteristic required for the transmission of analog signals |
Nyquist Bit Rate:
The maximum rate at which data can be correctly communicated over a channel in presence of noise and distortion is known as its channel capacity. Consider first a noise-free channel of Bandwidth B. Based on Nyquist formulation it is known that given a bandwidth B of a channel, the maximum data rate that can be carried is 2B. This limitation arises due to the effect of intersymbol interference caused by the frequency components higher than B. If the signal consists of m discrete levels, then Nyquist theorem states:
Maximum data rate C = 2 B log2 m bits/sec,
where
C is known as the channel capacity,
B is the bandwidth of the channel
and m is the number of signal levels used.
Baud Rate: The baud rate or signaling rate is defined as the number of distinct symbols transmitted per second, irrespective of the form of encoding. For baseband digital transmission m = 2. So, the maximum baud rate = 1/Element width (in Seconds) = 2B
Bit Rate: The bit rate or information rate I is the actual equivalent number of bits transmitted per second. I = Baud Rate × Bits per Baud
= Baud Rate × N = Baud Rate × log2m
For binary encoding, the bit rate and the baud rate are the same; i.e., I = Baud Rate.
Example: Let us consider the telephone channel having bandwidth B = 4 kHz. Assuming there is no noise, determine channel capacity for the following encoding levels:
(i) 2, and (ii) 128.
Ans: (i) C = 2B = 2×4000 = 8 Kbits/s
(ii) C = 2×4000×log2128 = 8000×7 = 56 Kbits/s
Effects of Noise
When there is noise present in the medium, the limitations of both bandwidth and noise must be considered. A noise spike may cause a given level to be interpreted as a signal of greater level, if it is in positive phase or a smaller level, if it is negative phase. Noise becomes more problematic as the number of levels increases.
Shannon Capacity formulae:
In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacityC = B log2 (1 + S/N),
where S and N are the signal and noise power, respectively, at the output of the channel. This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel.
Example: Suppose we have a channel of 3000 Hz bandwidth, we need an S/N ratio (i.e. signal to noise ration, SNR) of 30 dB to have an acceptable bit-error rate. Then, the maximum data rate that we can transmit is 30,000 bps. In practice, because of the presence of different types of noises, attenuation and delay distortions, actual (practical) upper limit will be much lower.
Between the Nyquist Bit Rate and the Shannon limit, the result providing the smallest channel capacity is the one that establishes the limit.
Example: A channel has B = 4 KHz. Determine the channel capacity for each of the following signal-to-noise ratios: (a) 20 dB, (b) 30 dB, (c) 40 dB.
Answer:
(a) C= B log2 (1 + S/N) = 4×103×log2 (1+100) = 4×103×3.32×2.004 = 26.6 kbits/s
b) C= B log2 (1 + S/N) = 4×103×log2 (1+1000) = 4×103×3.32×3.0 = 39.8 kbits/s
(c) C= B log2 (1 + S/N) = 4×103×log2 (1+10000) = 4×103×3.32×4.0 = 53.1 kbits/s
Example:
A channel has B = 4 KHz and a signal-to-noise ratio of 30 dB. Determine maximum information rate for 4-level encoding.
Answer:
For B = 4 KHz and 4-level encoding the Nyquist Bit Rate is 16 Kbps. Again for B = 4 KHz and S/N of 30 dB the Shannon capacity is 39.8 Kbps. The smallest of the two values has to be taken as the Information capacity I = 16 Kbps.
Example: A channel has B = 4 kHz and a signal-to-noise ratio of 30 dB. Determine maximum information rate for 128-level encoding.
Answer:
The Nyquist Bit Rate for B = 4 kHz and M = 128 levels is 56 kbits/s. Again the Shannon capacity for B = 4 kHz and S/N of 30 dB is 39.8 Kbps. The smallest of the two values decides the channel capacity C = 39.8 kbps.
Answer:
The Nyquist Bit Rate for B = 4 kHz and M = 128 levels is 56 kbits/s. Again the Shannon capacity for B = 4 kHz and S/N of 30 dB is 39.8 Kbps. The smallest of the two values decides the channel capacity C = 39.8 kbps.
Example: The digital signal is to be designed to permit 160 kbps for a bandwidth of 20 KHz. Determine (a) number of levels and (b) S/N ratio.
(a) Apply Nyquist Bit Rate to determine number of levels.
(a) Apply Nyquist Bit Rate to determine number of levels.
C = 2B log2 (M),
or 160×103 = 2×20×103 log2 (M),
or M = 24, which means 4bits/baud.
(b) Apply Shannon capacity to determine the S/N ratio
C = B log2 (1+S/N),
or 160×103 = 20×103 log2 (1+S/N) ×103 log2 (M) ,
or S/N = 28 - 1,
or S/N = 255,
or S/N = 24.07 dB.
Post a Comment
have you a any doubt then tell and if you want some topic then please tell .