Chapter 4 Digital Transmission. 4-1 DIGITAL-TO-DIGITAL CONVERSION. line coding , block coding , and scrambling . Line coding is always needed; block coding and scrambling may or may not be needed. Figure 4.2 Signal element versus data element.
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line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed.
r = number of data elements / number of signal elements
Baseline: running average of the received signal power
Constant digital signal creates low frequencies
Receiver Setting the clock matching the sender’s
Bipolar schemes: AMI (Alternate Mark Inversion) and pseudoternary
The tendency today is to change an analog signal to digital data.
In this section we describe two techniques,
pulse code modulationanddelta modulation.
According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency contained in the signal.
What can we get from this:
1. we can sample a signal only if the signal is
2. the sampling rate must be at least 2 times the highest frequency, not the bandwidth
SNRdb= 6.02nb + 1.76 dB
nb: bits per sample (related to the number of level L)
What is the SNRdB in the example of Figure 4.26?
We have eight levels and 3 bits per sample, so
SNRdB = 6.02 x 3 + 1.76 = 19.82 dB
Increasing the number of levels increases the SNR.
The minimum bandwidth of the digital signal is nb times greater than the bandwidth of the analog signal.
Bmin= nb x Banalog
We have a low-pass analog signal of 4 kHz. If we send the analog signal, we need a channel with a minimum bandwidth of 4 kHz. If we digitize the signal and send 8 bits per sample, we need a channel with a minimum bandwidth of 8 × 4 kHz = 32 kHz.
Next bit is 1, if amplitude of the analog signal is larger
Next bit is 0, if amplitude of the analog signal is smaller
2. Bit rate is the number of bits per second.
2. Baud rate is the number of signal elements per second. 3. In the analog transmission of digital data, the baud rate is less than or equal to the bit rate.
S = N x 1/r baud r = log2L
B = (1+d) x S = (1+d) x N x 1/r
The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BAM = 2B.
The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal:BPM = 2(1 + β)B.
Multiplexing and Spreading
FDM is an analog multiplexing technique that combines analog signals.
WDM is an analog multiplexing technique to combine optical signals.
Figure 6.17 shows the output for four arbitrary inputs. The link carries 50,000 frames per second. The frame duration is therefore 1/50,000 s or 20 μs. The frame rate is 50,000 frames per second, and each frame carries 8 bits; the bit rate is 50,000 × 8 = 400,000 bits or 400 kbps. The bit duration is 1/400,000 s, or 2.5 μs.
Synchronous TDM is not always efficient
Bss >> B
1 Wrap message in a protective envelope for a more secure transmission.
2 the expanding must be done independently
3 two types: frequency hopping spread spectrum (FHSS) and direct sequence spread spectrum (DSSS)
Direct sequence spread spectrum
Replace each data bit with n bits using a spreading code