UNIT 1 (ANALOG COMMUNICATION)
1. Define noise Noise is an unwanted electrical signal which gets added tom a transmitted signal when it is travelling towards the receiver 2. Define noise figure Noise figure is a figure of merit and used to indicate how much the signal to noise ratio gets degraded as a signal passes through a series of circuits 3. What is meant by analog communication system? The modulation systems or techniques in which one of the characteristics of the carrier is changed in proportion with the instantaneous value of modulating signal is called analog communication system. 4. Define modulationModulation is defined as changing the characteristics of the carrier signal with respect to the instantaneous change in message signal. 5. What are the needs for modulation In order to carry the low frequency message signal to a longer distance, the high frequency carrier signal is combined with it. a) Reduction in antenna height b) Long distance communication c) Ease of radiation d) Multiplexing e) Improve the quality of reception f) Avoid mixing up of other signals 6. What are the advantages of Analog communications Transmitters and Receivers are simple Low bandwidth requirement FDM can be used 7. What are the disadvantages of analog communication? Noise affects the signal quality It is not possible to separate noise and signal Repeaters can’t be used between transmitters and receivers Coding is not possible It is not suitable for the transmission of secret information 8. Define Amplitude modulation. Amplitude Modulation is defined as changing the amplitude of the carrier signal with respect to the instantaneous change in message signal. 9. Define Frequency modulation. Frequency Modulation is defined as changing the frequency of the carrier signal with respect to the instantaneous change in message signal. 10. Define Phase modulation.
Phase Modulation is defined as changing the phase of the carrier signal with respect to the instantaneous change in message signal. 11. Define modulation index It is defined as ratio of amplitude of the message signal to the amplitude of the carrier signal 12. Define percentage modulation m=Em/Ec It s defined as the percentage change in the amplitude of the output wave when the carrier is acted on by a modulating signal. M=(Em/Ec)*100 13. State Carson’s rule. Carson’s rule states that the bandwidth required to transmit an angle modulated wave as twice the sum of the peak frequency deviation and the highest modulating signal frequency. Mathematically carson’s rule is B=2( f +fm) Hz. 14. Define Deviation ratio. Deviation ratio is the worst case modulation index and is equal to the maximum peak frequency deviation divided by the maximum modulating signal frequency. Mathematically, the deviation ratio is DR= f (max)/fm(max) PARTB Explain Amplitude Modulation and demodulation in detail. Time domain An AM signal is made up of a carrier (with constant frequency) in which its amplitude is changed (modulated) with respect to the signal (modulating signal) we wish to transmit (voice, music, data, binary). In the example below the carrier (a high frequency sine wave) is being modulated by a lower frequency sine wave. The modulating signal causes the carriers amplitude to change with time. This resulting shape of the carrier is called the envelope. Note the envelope has the shape of a sine wave. AM signal
Modulating signal (sine wave) and modulated carrier
Derivation A carrier is described by v = Vc Sin ( 2 fπ c t + ) To amplitude modulate the carrier its amplitude is changed in accordance with the level of the audio signal, which is described by v = Vm Sin ( 2 fπm t ) The amplitude of the carrier varies sinusoidally about a mean of Vc. When the carrier is modulated its amplitude is varied with the instantaneous value of the modulating signal. The amplitude of the variation of the carrier amplitude is Vm and the angular frequency of the rate at which the amplitude varies is 2 fπm. The amplitude of the
carrier is then:
Carrier amplitude = Vc + Vm Sin ( 2 fπm t )
and the instantaneous value (value at any instant in time) is
v = {Vc + Vm Sin ( 2 fπ m t )} * Sin ( 2 fπ c t ) Eqn. 1
= Vc Sin ( 2 fπ c t ) + Vm Sin ( 2 fπ m t ) * Sin ( 2 fπc t )
Using Sin A * Sin B = ½ Cos (A B) ½ Cos (A + B) this becomes
v = Vc Sin ( 2 fπ c t ) + ½ Vm Cos ( (2 fπc 2 fπm) t ) ½ Vm Cos ((2 fπ c + 2 fπ m)t) Eqn. 2
This is a signal made up of 3 signal components
carrier at 2 fπ c (rad/s) Frequency is fc = 2 fπc/2 Hz
upper side frequency 2 fπ c + 2 fπm (rad/s) Frequency is (2 fπc + 2 fπ m)/2 = fm + fc Hz
lower side frequency 2 fπ c 2 fπm (rad/s) Frequency is (2 fπc 2 fπm)/2 = fm fc Hz
The bandwidth (the difference between the highest and the lowest frequency) is
BW = (2 fπc + 2 fπ m ) (2 fπ c 2 fπm) = 2 * 2 fπ m Rad/s ( = 2 fπm/ Hz)
The spectrum of these signals is shown. This is described as the signal in the frequency domain, as opposed to the signal in the time domain. In this case the audio signal is made up of a single frequency.
In this example the angular frequencies (expressed in Radians/sec, or kRad/sec, or Mrad/sec) are show. In most cases however the frequency is shown (expressed in Hz, or kHz, or MHz).
Amplitude (V) Angular Frequency c - m c c + m Lower side frequency Carrier Upper side frequency Bandwidth = 2 * m If the audio signal is made up of a range of frequencies from f1 to f2 (as is normally the case) rather than a single frequency the output signal will be a band of frequencies, contained in the upper side band (USB), inverted and the lower side band (LSB), erect. A broadcast AM station in the Medium Wave band is usually allocated a frequency slot 9 kHz wide. This means that the carriers of stations in this band are spaced 9 kHz apart. The maximum amplitude in an AM signal is Vc + Vm .The minimum amplitude is Vc Vm. Frequency Domain view of Double Sideband – Full Carrier Modulation Index (or Modulation Factor or Depth of Modulation) This is defined as m = In AM, this quantity, also called modulation depth, indicates by how much the modulated signal varies around its 'original' level. For AM, it relates to the variations in the carrier amplitude. Spectrum of audio signal Carrier Upper Sideband Erect Lower Sideband Inverted fc f1 f2 fc- f2 fc- f1 fc+ f1 fc+ f2
So if m = 0.5, the carrier amplitude varies by 50% above and below its unmodulated level, and for
m = 1.0 it varies by 100%. Modulation depth greater than 100% is generally to be avoided as it creates distortion.
Using this Eqn. 2 can be rewritten as
v = Vc Sin ( 2 fπc t ) + ½ (Vm Cos ( (2 fπ c 2 fπm) t ) Vm Cos ((2 fπc + 2 fπ m)t) ) * Vc /Vc
v = Vc { Sin ( 2 fπc t ) + ½ m [ Cos ( (2 fπ c 2 fπ m) t ) + Cos ((2 fπ c + 2 fπ m)t) ] } Eqn. 3
The maximum allowed value of m is 1.0. If this is exceed the envelope of the output waveform is distorted. This is known as Overmodulation and should never occur in practice, because the distorted envelope will result in a distorted output sound signal in the radio receiver. The effect of overmodulation can be examined in the laboratory. 50% Modulation -1.5 -1 -0.5 0 0.5 1 1.5 1 9 17 25 33 41 49 57 65 100% modulation -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 1 9 17 25 33 41 49 57 65 150% Modulation -3 -2 -1 0 1 2 3 1 9 1 7 25 33 41 49 57 65 Carrier -1 -0.5 0 0.5 1 1 9 17 25 33 41 49 57 65 Modulating Signal -1 -0.5 0 0.5 1 1 9 17 25 33 41 49 57 65
Variations of modulated signal with percentage modulation are shown below. In each image, the maximum amplitude is higher than in the previous image. Note that the scale changes from one image to the next. Power in an AM waveform Assume that the AM signal is dissipated in a load of R . The total power dissipated will be the sum of the powers in all of the components of the signal. The power in the carrier will be Pc = Watts The power in each of the frequencies is Ps = = = Pc The total power is Pt = Pc + Ps + Ps = Pc + 2 Ps = Pc ( 1 + 2 ) = Pc ( 1 + ) Watts The fraction of the power in the carrier is = The maximum value for m is 1.0. This means that at most only 1/3 of the power in the signal will be contained in the sidebands. All of the audio information is contained in either one of the sidebands, so that, in effect, only one sixth of the power (16.7%) is used to carry information.
Alternative form for modulation index
If an AM signal is being displayed on an oscilloscope it can be difficult to read Vm and Vc. Instead
the form for expressing m can be modified to make it easier to read. Modulation index : m= = = = = It is possible to read the maximum and minimum amplitude of the signal from the oscilloscope display. Peak Instantaneous Power The maximum signal voltage is Vc + Vm = Vc (1 + m) so that the maximum instantaneous output power is (1 + m)2 = Pc (1 + m)2 . If the modulation index is 1.0 the maximum output power will be 4 Pc. The transmitter must be designed to carry this level of output power.
AM Demodulators
Diode Detector This is the most commonly used AM demodulator. It is cheap and reasonably accurate. It is also used as an integral part of many designs of (older) FM detector. The basic circuit is as below It has limitations. If the time constant R1 *C1 in the envelope detector is too long relative to the period of the highest frequency modulating signal it will not be it will not be able to follow the peaks and troughs of the envelope giving rise to diagonal clipping. It is required that R1*C1 < [(1 m2)1/2] / (m2 fπ m) where 2 fπm is the highest frequency component of the modulating signal and m is the modulation index. This is derived below. If R1*C1 is too short than there will excessive RF ripple and the output power will be reduced. Because the diode is a non linear device there will be some distortion in the demodulated signal. In general R1 C1 must be a lot longer than the period of the carrier and a lot shorter than the period of the modulating signal. R1 must be a lot larger than the forward resistance of the diode to maintain detector efficiency. It must also provide matching to the next (audio) stage. The Diode detector output signal consists of three components R1 C1 C2 R2 C3 AM in Envelope Detector DC Block LPF Audio Out1. The wanted demodulated audio signal 2. A DC component proportional to the peak amplitude of the RF signal. This is removed by sending the signal through a capacitor C2 (high pass filter). It is also used to provide an input into Automatic Gain Control. 3. An unwanted ripple at the carrier frequency and its harmonics. This is blocked from later stages by using an RC low pass filter (R2 and C3 in this circuit). A 1 MHz carrier is amplitude modulated by an audio signal which contains all frequencies in the range 300 Hz to 5 kHz. What are the frequency bands which are output? What is the output bandwidth? Draw the spectral diagram of these signals. Answer: The carrier is 1 MHz The Upper Side Band is all frequencies in the range 1,000,300 to 1,005,000 Hz The Lower Side Band is all frequencies in the range 995,000 to 999, 700 Hz The Bandwidth is 1,005,000 995,000 = 10,000 Hz = 10 kHz. A 1.5 MHz carrier is amplitude modulated by three sinusoidal signals of frequency 500 Hz, 800 Hz and 1,400 Hz. What are the frequencies in the AM spectrum? Answer: Convert all the frequencies to kHz. 1.5 MHz is 1500 kHz. 500 Hz is 0.5 kHz. 800 Hz is 0.800 kHz. 1400 Hz is 1.4 kHz. The output frequencies are: 1500 kHz, 1500 0.5 kHz, 1500 0.8 kHz 1500 1.4 kHz or 1500, 1500.5 , 1499.5, 1500.8, 1499.2, 1501.4 , 1498.6 kHz An AM signal is represented by the equation v =
( 15 + 3 Sin( 2 * 5 * 10 3 t) ) * Sin( 2 * 0.5 * 10 6 t) volts
(i) What are the values of the carrier and modulating frequencies? (ii) What are the amplitudes of the carrier and of the upper and lower side frequencies? (iii) What is the modulation index? (iv) What is the bandwidth of this signal? Answer: This looks the same as Eqn. 1 above with: 2 fπ c (= 2 fc) = 2 * 0.5 * 106 2 fπ m (= 2 fm) = 2 * 5 * 103 V c = 15 V Vm = 3 V (i) Therefore the carrier frequency fc is 0.5 * 106 = 0.5 MHz and the modulating frequency fm is 5 * 103 = 5 kHz (iii) The bandwidth BW = 2 fm = 10 kHz (ii) The modulation index m = Vm/Vc = 3/15 = 0.2
From Eqn. 3 the amplitude of each side frequency is m* Vc /2 = 0.2 * 15 /2 = 1.5 V A transmitter puts out a total power of 25 Watts of 30% AM signal. How much power is contained in the carrier and each of the sidebands? Answer: Total power = 25 = Pc ( 1 + ) = Pc ( 1 + ) = Pc * 1.045 Therefore the carrier power is Pc = 25/ 1.045 = 23.92 Watts The total power in the 2 sidebands is 25 23.92 = 1.08 W The power in each sideband is 1.08/2 = .54 W The fraction of the power in the carrier is 23.92/25 = 0.957, or 95.7% With neat block diagram explain Generation and detection of SSB signals DSBSC Filtering Method: Since SSB modulation is the transmission of the upper or lower side bands, SSB modulation can be generated by filtering the undesired side band of a DSBSC signal and retaining the desired one using a bandpass filter with bandwidth equal that of the message signal (not twice its bandwidth) and a center frequency equal to the center frequency of the desired side band (not the carrier). The PROBLEM with this modulation method is that it is suitable only for message signals that have a small guard–band (no signal components) around zero frequency, as it is the case for voice signals. The important components of human voice start from a frequency around 300 Hz. The reason is that ideal filters with sharp edges do not exist and therefore, filters with non–sharp edges must be used. Any non– zero components of the message signals close to zero frequency may be lost or distorted because of the filtering process.
The following signal is a representation of a human voice signal. It is seen that this signal has a guard band that separates the two halves of the signal. Crating a SSB signal from this message signal is possible since a guard band of 2*300 Hz is obtained, which means that the transition of the edge of the BPF can take place in this 600 Hz. Phase–Shifting Method: The basic idea of equations (12) above can be used where the Hilbert transform of the message signal is simply obtained using a a device that shifts the message signal by –/2 as shown in the following block diagram.
In fact, even the phase shifter shown in the block diagram below for a general signal is an unrealizable block. It only can be approximated. Therefore, this modulation technique is only a theoretical modulation method.
The above block diagram can also be described in frequency–domain. The following figure show the different signals in frequency–domain at points (a), (b), (c), and (d) in the block diagram.
Demodulation of SSB signals
For demodulation, the same block diagram of a simple DSBSC demodulator can be used. The sideband at the positive and negative frequencies merge (recombine) at zero frequency when the SSB signal is multiplied by the carrier. (Try the exercise of finding the output of the DSBSC demodulator in time– and frequency–domain when its input is a either an USB or a LSB signal).
If the SSB signal includes a LARGE carrier, it can be demodulated using an envelope detector similar to that used for full AM signals .
Explain the Bandwidth of FM and PM Signals
The bandwidth of the different AM modulation techniques ranges from the bandwidth of the message signal (for SSB) to twice the bandwidth of the message signal (for DSBSC and Full AM). When FM signals were first proposed, it was thought that their bandwidth can be reduced to an arbitrarily small value. Compared to the bandwidth of different AM modulation techniques, this would in theory be a big advantage. It was assumed that a signal with an instantaneous frequency that changes over of range of f Hz would have a bandwidth of f Hz. When experiments were done, it was discovered that this was not the case. It was discovered that the bandwidth of FM signals for a specific message signal was at least equal to the bandwidth of the corresponding AM signal. In fact, FM signals can be classified into two types: Narrowband and Wideband FM signals depending on the bandwidth of each of these signalsNarrowband FM and PM
The general form of an FM signal that results when modulating a signals m(t) is ( ) cos ( ) t FM c f g t A t k m d
. A narrow band FM or PM signal satisfies the condition ( ) 1 f k a t = For FM and ( ) 1 p k m t = For PM, where( ) ( ) t a t m d
, such that a change in the message signal does not results in a lot of change in the instantaneous frequency of the FM signal. Now, we can write the above as
( ) cos ( ) FM c f g t A t k a t . Starting with FM, to evaluate the bandwidth of this signal, we need to expand it using a power series expansion. So, we will define a slightly different signal ( ) ( ) ˆ ( ) j ct k a tf j ct jk a tf FM g t A e A e e . Remember that ( )
ˆ ( ) j ct k a tf cos ( ) sin ( ) FM c f c f g t A e A t k a t jA t k a t , so
ˆ
( ) Re ( ) FM FM g t g t . Now we can expand the term ( ) f jk a t e in ˆFM ( ) g t , which gives 2 2 2 3 3 3 4 4 4 2 2 3 3 4 4 ( ) ( ) ( ) ˆ ( ) 1 ( ) 2! 3! 4! ( ) ( ) ( ) ( ) 2! 3! 4! f f f c f f f c c c c c j t FM f j t j t j t j t j t f j k a t j k a t j k a t g t A e jk a t k a t jk a t k a t A e jk a t e e e e K K Since kf and a(t) are real (a(t) is real because it is the integral of a real function m(t)), and since Re{ej2 fπct} =cos(2 fπct) and Re{ jej2 fπct} = –sin(2 fπct), then
2 2 3 3 4 4
ˆ
( ) Re ( )
( ) ( ) ( )
cos( ) ( )sin( ) cos( ) sin( ) cos( )
2! 3! 4! f f f FM FM c f c c c c g t g t k a t k a t k a t A t k a t t t t t K
The assumption we made for narrowband FM is ( ( ) 1 f k a t = ). This assumption will result in making all the terms with powers of ( ) f k a t greater than 1 to be small compared to the first two terms. So, the following is a reasonable approximation for ( ) FM g t
( )( ) cos( ) ( )sin( ) FM Narrowband c f c g t A t k a t t , when ( ) 1 f k a t = . It must be stressed that the above approximation is only valid for narrowband FM signals that satisfy the condition ( ( ) 1 f k a t = ). The above signal is simply the addition (or actually the subtraction) of a cosine (the carrier) with a DSBSC signal (but using a sine as the carrier). The message signal that modulates the DSBSC signal is not m(t) but its integration a(t). One of the properties of the Fourier transform informs us that the bandwidth of a signal m(t) and its integration a(t) (and its derivative too) are the same (verify this). Therefore, the bandwidth of the narrowband FM signal is ( ) 2 ( ) FM Narrowband DSBSC m t BW BW BW . We will see later that when the condition (kf << 1) is not satisfied, the bandwidth of the FM signal becomes higher that twice the bandwidth of the message signal. Similar relationships hold for PM signals. That is ( )( ) cos( ) ( )sin( ) PM Narrowband c p c g t A t k m t t , when ( ) 1 p k m t = , and ( ) 2 ( ) PM Narrowband DSBSC m t BW BW BW .Construction of Narrowband Frequency and Phase Modulators
The above approximations for narrowband FM and PM can be easily used to construct modulators for both types of signals.Narrowband FM Modulator
UNIT 2 (DIGITAL COMMUNICATION)
1. Define Digital modulationDigital Modulation is defined as changing the amplitude of the carrier signal with respect to the binary information or digital signal. 2. What are the advantages of Digital communications It has a better noise immunity Repeaters can be used between transmitters and receivers It becomes simpler and cheaper as compared to the analog communication 3. What are the disadvantages of Digital communications It requires a larger channel bandwidth Delta modulation needs synchronization incase of synchronous modulation 4. Define bit rate Bit rate is defined as number of bits transmitted during one second between the transmitter and receiver. 5. Define baud rate. Baud rate is defined as the rate of change of signal on transmission medium after encoding and modulation have occurred. 6. Define Amplitude Shift Keying (ASK). Amplitude Shift Keying is defined as changing the amplitude of the carrier signal with respect to the binary information or digital signal. 7. Define Frequency Shift Keying (FSK). Frequency Shift Keying is defined as changing the amplitude of the carrier signal with respect to the binary information or digital signal. 8. Define Phase Shift Keying (PSK). Phase Shift Keying is defined as changing the amplitude of the carrier signal with respect to the binary information or digital signal. 9. Define Minimum Shift Keying (MSK). The minimum frequency space that allows the 2 FSK representing symbols 0s and 1s. Thus CP (Continuous Phase) FSK signal with a deviation ratio if one half is defined as MSK. 10. Define Quadrature Amplitude modulation (QAM). QAM is a form of digital modulation similar to PSK except the digital information is contained in both the amplitude and phase of the transmitted carrier. (or) QAM is defined as changing the amplitude as well as the frequency of the carrier signal with respect to the binary information or digital signal. 11. Define bandwidth efficiency Bandwidth efficiency is the ratio of the transmission bit rate to the minimum bandwidth required for a particular modulation
12. What are the advantages of MSK? MSK baseband waveform are smoother compared with QPSK MSK signals have continuous phase It does not have any amplitude variation 13. What are the advantages of QPSK? Very good noise immunity Effective utilization of available bandwidth Low error probability Very high bit rate data transmission PARTB Explain Amplitude Shift Keying (or) Digital amplitude modulation (or) ON – OFF keying The simplest digital modulation technique is Amplitude Shift Keying (ASK), where a binary information signal directly modulates the amplitude of an analog carrier. ASK is similar to standard amplitude modulation except there are only two output amplitudes possible. ASK is sometimes called digital amplitude modulation (DAM). Mathematically, ASK is v(ask)(t) = [1+vm(t)][A/2 cos (ωct)] (1) v(ask)(t) = ASK wave vm(t) =digital information (modulating) signal (volts) A/2 =unmodulatd carrier amplitude (volts) ωc =analog carrier radian frequency(radian per second, 2fct) In equ.(1),the modulating signal (vm(t)) is a normalized binary waveform, where +1V = logic 1and 1 V =logic 0.therefore , For a logic 1 input, vm(t) = +1 V ,Equ.(1) reduces to
v(ask)(t) = [1+1][A/2 cos (ωct)] = A cos (ωct)
For a logic 0 input , vm(t) = 1 V ,Equ.(1) reduces to
v(ask)(t) = [11][A/2 cos (ωct)]= 0
Thus, the modulated wave v(ask)(t),is either A cos (ωct) or 0.hence, the carrier is either ‘on’or ‘off’,
which why ASK is sometimes referred to as onoff keying(OOK). Figure shows the input and output waveforms from an ASK modulator. From the figure, it can be seen that for every change in the input binary data stream, there is one change in the ASK waveform, and the time of one bit (tb) equals the time of one analog signaling element (ts). The bit time is the reciprocal of the bit rate and the time of one signaling element is the reciprocal of the baud. Therefore, the rate of change of the ASK waveform (baud) is the same as the rate of change of the binary input (bps); thus, the bit rate equals the baud. With ASK, the bit rate is also equal to the minimum Nyquist bandwidth. B = fb / 1 = fb baud = fb/1 = fb
Explain Phase shift keying (or) phase reversal keying (or) biphase modulation Binary symbol (0 or 1) modulates the phase of the carrier. Let the carrier be s(t) = A cos 2f0t A peak value of sinusoidal carrier. In the standard 1 load resistor, the power dissipated will be P = 1 / 2 A2 A = P
When the symbol is changed , the phase of the carrier changed by 180 ( radians)
Symbol 1 s (t) =2P cos 2f0t (1) Symbol 0 s(t) =2P cos (2f0t +) (2) cos (2f0 t + ) = cos (2) s(t) = 2P cos 2f0t PSK signal s (t) = b(t) 2P cos 2f0 t Explain BPSK (or) phase reversal keying (or) biphase modulation Simplest form of PSK BPSK N =1 ,M = 2
O N number of bits encoded M number of discrete signals. Two phases (21 = 2) are possible for the carrier
One phase logic 1 Another phase logic 0 As input changes state (0 to 1 (or) 1 to 0), the phase of the output carrier shift between two angles that are separated by 180◦. It is a form of square wave modulation of continuous wave signal. BPSK Transmitter Balanced modulator acts as phase reversing switch. Depending on the logic condition of digital input, the carrier is transferred to the output either in phase or 180 0 out of phase with the reference carrier oscillator.
Internal structure balanced ring modulator The balanced modulator has two inputs. 1. Carrier that is in phase with reference oscillator 2. Binary digital data For the balanced modulator to operate properly the digital input voltage greater than the peak carrier voltage. Digital input input control the ON/ OFF state of diode D1 to D4. Condition 1 :Binary input ‘1’ Diode D1 and D2 forward bias (ON). Diode D3 and D4 reverse bias (OFF). With the polarities shown, the
carrier voltage is developed across transformer T2 in phase with the carrier voltage across T1. Output signal is inphase with the reference oscillator. Condition 2: Binary input ‘0’ Diode D1 and D2 reverse bias (OFF). Diode D3 and D4 forward bias (ON).
With the polarities
voltage is developed across transformer T2
180
◦ out of phase with the carrier voltage across T1.
Output signal is 180◦out of phase with the reference oscillator. Truth table
A constellation diagram (or) signal state space diagram similar to phasor diagram except that the entire phase is not drawn only the peak position of phasors are shown. Bandwidth consideration BPSK Balanced modulator (product modulator) :ouput signal is the product of two input signal. In BPSK modulator, the carrier input signal is multiplied by binary data. Positive logic 1 ; negative logic 0, input carrier (sin 2 fπ c t) multiplied by either 0 (ve) or 1 (+ve). Multiplied by (ve) sin 2 fπ c t 180◦ ( out of phase) Multiplied by (+ve) sin 2 fπ c t 0◦ (inphase). Baud rate = bps (bit rate )
Widest bandwidth occurs when the binary input data are an alternating input
sequence.
The fundamental frequency (fa) of an alternative sequence = one half of bit
rate (fb / 2).
The output (BPSK modulator)
proportional to BPSK output = [sin
(2fat)] [sin (2fct)] fa maximum fundamental frequency of binary input (Hz) fc reference carrier frequency (Hz) product of two sine functions ½ cos [2(fc fa) t] ½ cos [2(fc + fa) t] Minimum double side nyquist
bandwidth (B) fc + fa or fc fa (fc fa) fc + fa / 2fa fa=fb / 2 fb bit rate B = 2fb /2 Time of one BPSK signaling element (ts) = time of one information bit (tn) . Indicates bit rate = baud rate
BPSK receiver The input signal may be ± sin2 fπ ct. Coherent carrier recovery circuit detects and regenerates the carrier that is both frequency and phase coherent with original transmit signal carrier. The balanced modulator, the output is the product of the two inputs (the BPSK and recovered carrier). The low pass filter (LPF) separates the recovered binary data from the complex demodulated signal. For BPSK signal + sin2 fπ ct (logic 1) , the output of balanced modulator (sin2 fπ ct)(sin2 fπ ct) = sin22 fπ ct
Or
sin22 fπ ct= ½ (1 cos 22 fπ ct)
=½ ½ cos 22 fπ ct½ cos 22 fπ ct filtered by using low pass filter (<22 fπ c).
Output of the balanced modulator
contains a
+ve voltage (+ ½ ) and cosine wave at twice the carrier frequency (2 fπ c).
A positive component (voltage) represent a demodulated logic 1
(sin2 fπ ct)(sin2 fπ ct) = sin22 fπ ct sin22 fπ ct = ½ (1cos 22 fπ ct)
= ½ + ½ cos 22 fπ ct ½ cos 22 fπ ct filtered by using low pass filter (<2 fπ c). The output of balanced modulator contains ve voltage ( ½ ) and cosine wave at twice the carrier frequency (2 fπ c). A negative voltage represents a demodulated logic 0.
Probability of error
The general expression for the bit error probability of an Mphase PSK system Explain 8 Phase Shift Keying with neat diagram N = 3 ; M = 2n = 23 = 8 (possible o/p phases) group of three bits tribits 8 PSK transmitter Incoming serial bit stream enters the bit splitter converted to a parallel. Bit rate (three channel ) = fb = 3 Bit I & C to I channel , Bit Q & C bar to Q channel. The I or Q bit determine o/p polarity Logic 1 = + v Logic 0 = v C & C bar determine Logic 1 = 1.307 v Logic 0 = 0.541 v. Tribit code between any two adjacent phase changes by only one bit gray code (or) maximum distance code. For use to reduce the number of transmission errors.
Bandwidth considerations Bit rate I & Q or C channel = 1/3 of binary i/p data rate (fb /3). Highest fundamental frequency in the I , Q , C channel = 1/6 of bit rate (I /p). There is only one change in phase at the o/p for every data i/p bits. Baud rate =fb/2 . O/p of the balanced modulator
Ө = (Xsin ωat)( sin ωct) ωat= 2Пfb/6 t {modulating signal} ωat=2Пfc{carrier}
and X=±1.307 or ±0.541
Thus,
Ө = (Xsin 2Пfb/6 t)( sin 2Пfc t) = X cos 2П(fc fb/6) t X cos 2П(fc +fb/6) t
2 2
The output frequency spectrum extends from fc +fb/4 to fc fb/4, and the minimum
Power splitter directs the i/p 8PSk signal to the I & Q product detector and carrier recovery circuits. The carrier recovery circuit reproduces the original reference oscillator signal. The incoming 8PSK signal is mixed with the recovered carrier in the I product detector and with quadrature carrier in the Q product detector. The o/p of product detector are 4 level PAM o/p from I channel I & C o/p from Q channel Q & C bar. Parallel to serial converter converts the I/C and Q / C bar pairs to serial I,Q& C data streams. Explain 16 Phase Shift keying in detail n = 4 ; M = 2n = 24 = 16 (output phases). 4 bits quad bits. Minimum bandwidth and baud equal one fourth the bit rate. N , M o/p phases . Angular separation between adjacent o/p phases is 22.5.
16 PSK undergoes 11.25 phase shift during transmission and still retain its integrity. Limitation in the level of encoding possible with PSK, as a point is eventually reached where receivers can’t discern the phase of the received signaling element. Phase impairment destroys its integrity producing errors. For 64 PSK n = 6 ; M = 26 = 64 o/p phases , Angular separation between adjacent phases is only 5.6.
Explain Quaternary Phase Shift Keying (Or) Quadrature PSK (Or) Constant Amplitude Digital Modulation (16 mark/10 mark) QPSK is an M ary encoding scheme where N=2 and M = 4 Four phases (o/p) are possible for a single carrier frequency. Four o/p phases four i/p conditions Product modulator requires more than single i/p bit to determine the o/p condition . Four possible conditions : 00,01,10,11 The binary i/p data are combined into group of two bits called dibits. In the modulator, each dibit code generates one of the four possible o/p phases (+45, +135, 45 , 135).
For each two dibit clocked into the modulator single o/p change occurs. Baud rate = ½ bit i/p rate (two i/p bit produce one o/p phase change). QPSK transmitter Two bits (dibit) are clocked into the bit splitter. Serially inputted simultaneously parallel outputted One bit to I channel ; other bit to Q channel The I bit modulates a carrier that is inphase with the reference oscillator. The Qbit modulates a carrier that is 90 out of phase with the reference oscillator. The operation of QPSK same as BPSK i.e two BPSK modulator combined in parallel. Logic 1 = +ve Logic 0 = ve I + sin2 fπ ct sin2 fπ c t Q cos2 fπ ct
cos2 fπc
t
Linear summer o/p = +sin2 fπ ct +cos2 fπ ct, + sin2 fπ ct cos2 fπ ct, sin2 fπ ct + cos2 fπ ct& sin2 fπ ct cos2 fπ ct. Each of the four possible o/p phasors has exactly the same amplitude. Binary information must be encoded entirely in the phase of the o/p signal. Binary i/p QPSK o/p phase Q I 0 0 135 0 1 45 1 0 + 1 1 + 45 (a)
Explain the bandwidth considerations of QPSK system . [ Nov 2010] (6mark) Bit rate (I or Q) equal to i/p data rate (fb / 2). Highest fundamental frequency present at the data i/p to I or Q = ¼ of data rate . Twice (I & Q) fb/ 4 = fb / 2 (nyquist BW). Bandwidth compression is realized in QPSK (min BW < incoming bit rate). QPSK o/p signal doesnot change phase until two bits have been clocked int bit splitter.o/p rate (baud) = ¼ of i/p bit rate.
O/p of the balanced modulator Output = (sin ωat)( sin ωct) ωat= 2Пfb/4 t {modulating signal} ωat=2Пfc{carrier} Thus, Output = (sin 2Пfb/4 t)( sin 2Пfc t) ½ cos 2П(fc fb/4) t ½ cos 2П(fc +fb/4) t
The output frequency spectrum extends from fc +fb/4 to fc fb/4, and the minimum bandwidth (fN) is (fc +fb/4) –( fc fb/4) = 2 fb/4 = fb/2 Explain QPSK receiver with necessary block diagram • The power splitter directs the i/p QPSK signal to the I & Q product detectors and carrier recovery circuit. • Carrier recovery circuit recover / reproduce the original transmit carrier. • QPSK signal is demodulated in I& Q product detector generate original I & Q data. • Product detectors (o/p) Ł converted parallel I& Q data channels to a single binary o/p data stream. • Four possible o/p phases (+45 , +135 , 45 , 135 ). I product detector
The received QPSK signal (sin2 fπ ct+ cos2 fπ ct) is one of the input to the I product
I = (sin2 fπ ct+ cos2 fπ ct)( sin2 fπ ct)
= (sin2 fπ ct) ( sin2 fπ ct)+ (cos2 fπ ct)( sin2 fπ ct)
= sin22 fπ ct + (cos2 fπ ct)( sin2 fπ ct)
= ½ (1 cos22 fπ ct) + ½ sin (2 fπ c + 2 fπ c)t + ½ sin (2 fπ c 2 fπ c)t
= ½ +½ cos22 fπ ct + ½ sin 22 fπ ct + ½ sin 0
Filtered out equals 0 = ½ v (logic 0) Q product detector Again, the receive QPSK signal is one of the input to the Q product detector .the other input is recovered carrier shifted 90 in phase.
I = (sin2 fπ ct + cos2 fπ ct)( cos2 fπ ct)
= cos22 fπ ct (sin2 fπ ct)( cos2 fπ ct)
= ½ (1 cos22 fπ ct) ½ sin (2 fπ c + 2 fπ c)t ½ sin (2 fπ c 2 fπ c)t
= ½ +½ cos22 fπ ct ½ sin 22 fπ ct ½ sin 0
Filtered out equals 0 = ½ v (logic 1) With block diagram explain Offset QPSK (or) offset keyed QPSK • The bit waveforms on the I& Q channel are offset / shifted inphase from each other by one half of a bit time.
• There is never more than 90 shift in o/p phase. • In QPSK, change in i/p dibit from 00 to 11 or 01 to 10 cause 180 shift. Adv: limited phase shift. Disadvantage: Data rate twice (o/p). Bandwidth & baud rate high twice than QPSK. Frequency Shift Keying (FSK) v Form of constant amplitude angle modulation similar to standard frequency modulation except the modulating signal is a binary signal that varies between two discrete voltage levels. v FSK sometimes called binary FSK. Vfsk(t) = Vccos { 2P (fc + Vm(t)D f) t} Vfsk(t) binary FSK Vc peak analog carrier amplitude (volts) fc analog carrier Centre frequency (hertz) Vm(t) binary i/p D f peak carrier change (hertz) v D f Vm(t) direction determined by the polarity.
v Logic 1 = +1 volt Ł Vm(t) = + 1 v Logic 0 = 1 volt Ł Vm(t) = 1. Logic 1 Logic 0 Vfsk(t) = Vccos { 2P (fc + D f) t} Vfsk(t) = Vccos { 2 (fc f) t} FSK in frequency domain v With binary FSK, the carrier centre frequency (fc) is shifted up and down in the frequency domain by the binary i/p (changes from logic 0 to 1or vice versa). v o/p frequency Logic 1 Ł mark frequency Logic 0 Ł space frequency.
v Mark and space frequencies separated from carrier frequency by the peak frequency deviation (D f). v Frequency deviation D f = |fm fs| / 2 |fm fs| absolute difference between mark and space frequency (hertz). FSK in time domain Space frequency (lower frequency) = fc D f Time › frequency ›
Mark frequency (high frequency) = fc + D f Time fl frequency › Related by equation f = 1/t f frequency t time FSK bit rate, baud rate and bandwidth Bit rate = baud rate tb=ts i/p = o/p v baud rate of BFSK can be determined by substituting N =1 Baud rate =fb / N = fb / 1 write a note on bandwidth consideration in FSK[ April 2014] (6marks) B = |(fs fb)(fm fb)| = |fs fm| + 2fb |fs fm| = 2D f using Carson’s rule (get min bandwidth) B minimum bandwidth. D f frequency deviation (|fs fm|) fb i/p bit rate
v The fastest rates of change (highest fundamental frequency) in NRZ binary signal occur when alternate 1’s and 0’s are occurring (square wave). v It take high and low to produce a cycle, the highest fundamental frequency present in a square wave equals the repetition rate of the square wave. v Fundamental frequency binary signal = half of the bit rate fa =fb / 2 v Modulation index h = D f / fa h FM modulation index (h factor). v Worst case modulation index yields widest BW 1. D f / fm at maximum value 2. fa = fb / 2 h = |fm – f s | / 2 f b / 2 h = |fm – fs| / fb Discuss the principle of operation of FSK transmitter. [April 2014][Nov 2011][Nov 2009] (8marks) Vco voltage controlled oscillator v logic 1 Ł shifts the vco o/p to the mark frequency. v logic 0Ł shifts the vco o/p to the space frequency. v Binary i/p changes back and forth between logic 1 and logic 0
v Vco o/p shift / deviates back and forth between mark and space frequency.
v D f is the difference between carrier rest frequency and either mark or space frequency (or difference between carrier rest frequency and mark and space frequency ).
v D f = Vm(t) kl kl deviation sensitivity
v Peak voltage same for logic 1 and 0, magnitude (D f) same for logic 1 and 0. Discuss the principle of operation of FSK receiver. [April 2013] Nov 2009] (8marks) Non coherent FSK v FSK i/p signal to BPF (Band Pass Filter) through power splitter. v BPF allows only mark or space frequency to envelope detector. v Envelope detector find total power in each pass band. v Comparator responds to largest power. v No frequency involvement. Coherent FSK v FSK is multiplied by a recovered carrier signal that has the exact same frequency & phase as the transmitter reference.
v Two transmitted frequencies (fm and fs) are not generally continuous, it is not
practical to reproduce a local reference that is coherent with both of them.
PLL –Phase Locked Loop
v Input to PLL shifts between fmand fs, the dc error voltage at the o/p of the
phase comparator follows the frequency shifts .
v Two o/p error voltage Ł one represents logic 1 another represent logic 0 v Natural frequency (PLL) = frequency (FSK).
v Two frequency matches Ł dc error voltage ‘0’.
Probability of error
The probability of error for noncoherent FSK The probability of error for coherent FSK
Phase continuous FSK (CP FSk)
v With CPFSK ,fm and fs are separated by an exact multiple of one half of the bit
rate fm and fs= n fb / 2.
v Better bit error performance.
v Disadvantages of CPFSk: requires synchronization circuit, more expensive.
Explain minimum shift keying in detail:
• Minimumshift keying (MSK) is a type of continuousphase frequencyshift keying,With a frequency separation of onehalf the bit rate(0r) With h = (1/2) the frequency deviation is half the bit rate.
• This minimum frequency spacing allows the two FSK signals corresponding to symbols 1 and 0 to be coherently orthogonal,
• So That they do not interfere with each other in the process of detection.
• Cpfsk signal with the deviation ratio as (1/2) is commonly referred to as minimum shift keying. • The MSKϕ signalis expressed as S(t) = S1 =
Where
1(t) =
T
b≤t≤T
b ?2(t) = 0≤t≤Tb S1 Tb≤t≤Tb S2 0≤t≤2Tb SIGNAL SPACE REPRSENTATION OF MSK:Transmitted Phase states message binary symbol 0≤t≤Tb S1 S2 1 0 + 2 f/2π + 0 2 fπ + 2 f/2π 1 2 fπ 2 f/2π + 0 0 2 f/2π +
The MSK can signal may assume one of the four possible states depending on the values of (0)θ and (Tb).θ MSK TRANSMITTER: • Two sine i/ps, one of frequency = and the other of frequency , are applied to a product modulator.
• This produces two phase coherent sine waves are separated at frequencies and . • The top most BPF allows only the signal f1. • The bottommost BPF allows only the signal f2. • Then the resulting filter output s are summed to produce and .
• Then and are multiplied by message signal and , with a bit rate of . This produces MSK waveforms. MSK RECEIVER • The received signal x is correlated with locally generated carrier, and . • Then it is integrated over Tb seconds in inphase channel
2 seconds in quadrature channel .
• Then the correlator
outputs, and
, are compared with a threshold of zero volts,
• The estimates of the phase and are obtained.
• Then this phase estimates is given to the logic circuit and original binary waves are reconstructed. Probability of Error Pe = (1/2) erfc (√Eb/N0) Explain QAM Quadrature Amplitude Modulation (Amplitude and PSK are combined ) 8QAM M= 8 possible Mary encoding. Not a constant amplitude signal. 8QAM transmitter v The only difference between the 8QAM transmitter and 8PSK transmitter, the omission of the inverter between C channel and Q product modulator. v Incoming data are divided into groups of three bit (tribits), the I , Q and C bit stream .
v Each with a bit rate equal to 1/3 of the incoming data rate. v The I and Q bits determine the polarity of the PAM signal at the o/p of 2 to 4 level converters and C channel determines the magnitude . v Magnitude equal, logic conditions different. Bandwidth considerations of 8QAM v In 8QAM, the bit rate in the I and Q channel is one third of the i/p binary rate same as 8PSK. v The highest fundamental modulating frequency and fastest o/p rate of change in the 8QA M are same as the 8PSk. v Minimum bandwidth= fb / 3. 8QAM receiver v The differences are the PAM levels at the o/p of the product detector and binary signals at the o/p of the analog to digital converter. v Modulated o/p Ł 8PSK = 8 QAM. Demodulated o/p Ł 8 PSK „ 8 QAM. v Binary o/p from I channel I & C bits. v Binary o/p from Q channel Q & C bits.
(8/10 mark)
Explain 16 Quadrature Amplitude Modulation n = 4; M = 2n = 24 = 16 possible combinations. QAM transmitter v The input binary data are divided into four channels I, I¢ , Q and Q¢ . v The bit rate in each channel is equal to one fourth of i/p bit rate (fb / 4). v Four bits are serially clocked into bit splitter they are outputted simultaneously and in parallel with I, I¢ , Q and Q¢ channel.
v I and Q bit polarity (logic 1 == +ve , logic 0 = ve)
v I¢ and Q¢ bit magnitude (I¢ (logic 1) = 0.821 v and Q¢ (logic 0) = 0.22 v). v 2 to 4 level converters generates a 4 level PAM levels (amplitude) signals.
v Two polarities and two magnitudes ( ± 0.22 v & ± 0.821 v).
v PAM signal modulates the inphase and quadrature carrier in the product modulator.
v Four o/p possible for each product modulator. I product modulator ±0.821 sin ωct , ± 0.22 sin ωct Q product modulator ±0.821
v The linear summer combines the output from I& Q product modulator. v Produces the 16 o/p conditions necessary for 16 QAM.
Bandwidth considerations of 16QAM
v The i/p data are divided into four channels.
v The bit rate in I, I¢ , Q and Q¢ channel is equal to one fourth of the binary input data rate (fb / 4).
v The bit splitter stretches the I, I¢ , Q and Q¢ bits to four times their input bit length.
v Because the I, I¢ , Q and Q¢ bits are outputted simultaneously and in parallel, the 2 to 4
level converters change in their i/p and o/p’s at a rate equal to one fourth of the i/p data rate.
v The bit timing relationship between the binary i/p data, the I, I¢ , Q and Q¢ channel data and the I PAM signal.
v The highest fundamental frequency in the I, I¢ , Q and Q¢ channel is equal to one eighth of the bit rate of the binary i/p data.
v One cycle in the I, I¢ , Q and Q¢ channel take the same amount of the time as eight input bits. v Also the highest fundamental frequency of either PAM signal is equal to one eighth of binary i/p bit rate. v With 16QAM modulator, there is one change in the o/p signal (phase or amplitude) for every four i/p data bits.
Baud rate = fb / 4 as same as minimum bandwidth. The balanced
modulator are product modulator and their outputs can be represented mathematically as Output = (X sin ωat) (sin ωct) ωat = 2P fb/ 8t ωct = 2fct X = ± 0.22 or ± 0.821 Output = (X sin 2P fb/ 8t) (sin 2P fct)
Output =X / 2 cos 2P (fc –(fb / 8))t X / 2 cos 2P (fc + (fb / 8))t
v The output frequency spectrum extends from fc + (fb / 8) to(fc –(fb / 8) and minimum
Compare various modulation techniques ASK,FSK,PSK,QPSK,QAM
S.NO Parameter ASK FSK PSK QPSK QAM
Variable Amplitude of Frequency of Phase of the Phase of the Amplitude and 1. characteristic carrier is carrier is carrier is varied carrier is phase of the
s varied varied varied carrier is varied
Freq shifted 0 (no phase 00> 135 deg For N inputs±
Logic 1 Carrier ON to fm shift) 01>45 deg
2. Amplitude(Volt
Logic 0 Carrier OFF Freq shifted 180 deg phase 10>+135 deg s)
to fs shift 11>+45 deg Phase Degrees
3. o/pwaveform
4 Bit Rate 1 1 1 2 N
5 Detectionmethod Coherent Non Coherent Coherent Coherent
Coherent 6 Symbolduration (T Tb Tb Tb 2Tb Tb /N b) 7. Baud rate fb fb fb fb/2 fb/N 8 Bandwidth fb 2(Δf + fb) fb fb/2 2fb/N 9 No. of o/p 2 2 2 4 2N 10 Bandwidthefficiency 1 1 1 2 N Relation BW = 2(Δf + 11 between BW = fb BW = 2 fb BW = fb/2 BW = fb/N bandwidth & fb) bit rate
Easy to noise Noise & BW Noise & Error efficiency of
12 Advantages generate & immunity probability transmission
detect
Noise & Low High Generation & Generation & more susceptible
13 Disadvantage bit rate bandwidth detection is detection is to noise
requirement complex complex
14 Applications communication Audio transmitter & communications
UNIT 3 (DATA AND PULSE COMMUNICATION)
1. What is meant by data communication Data communication can be defined as two personal computers connected through a Public Telecommunication Network (PTN) 2. What are the applications of data communication? Used in Automatic Teller Machine (ATM) Internet Airline and Hotel reservation system Mass media NEWS network 3. What are the advantages and disadvantages of Parallel communication Advantages: Parallel transmission is speed Used for short distance communication Disadvantages: Require more lines between source and destination More cost 4. What are the advantages and disadvantages of series communication Advantages: Number of transmission lines is less Used for long distance communication Low cost Disadvantages: Speed is low 5. What is meant by point to point communication? Point to point communication is the link between two stations A and B ie)., information is transferred between a main frame computer and a remote computer terminal 6. What is multipoint communication? A multipoint line configuration is one in which more than two specific devices share a single link 7. Define Morse code. It is used to send messages A key which turned the carrier of a transmitter ON and OFFto produce the dots and dashes These dots and dashes were detected at the receiver and it is converter back into letters and numbers makes the original message 8. What are the different types of error detection techniques? a. Redundancy b. Echoplex c. Exact count encoding d. Parity e. Check sum f. Vertical Redundancy Check g. Horizontal Redundancy Check h. Cyclic Redundancy Check 9. What is meant by Forward Error Correction (FEC)?FEC, a receiver can use an error correcting code, which automatically correct certain errors without any retransmissions In FEC, bits are added to the message before the transmission Purpose of FEC code is to reduce the wasted time of retransmission 10. Define Pulse Amplitude modulation The amplitude of a carrier pulse is altered in accordance to that of amplitude of message signal to make it accommodate the information signal. 11. Define Pulse width modulation In PWM system, the message signals are used to vary the duration of carrier pulse. The message signal may vary either the trailing edge or leading edge or both of the carrier pulses n order to accommodate the intelligence of information system. 12. Define Pulse position modulation The position of a carrier pulse is altered in accordance with information contained in sampled waveform. 13. Define Pulse code modulation Pulse code modulation refers a form of source coding. It is a form of digital modulation techniques in which the code refers a binary word that represent digital data.With PCM, the pulses are of fixed length and fixed amplitude. 14. Define sampling rate The sampling rate fs must be atleast two times the highest frequency component of the original signal to be accurately represented fs>=2fm
PARTB
Explain digital Pulse-code modulation in detail.
A form of digital pulse modulation where a message signal is represented in discrete form in both time and amplitude. This form of signal representation permits the transmission of the message signal as a sequence of coded binary pulses.
- PCM is essentially analog-to-digital conversion of a special type where the information contained in the instantaneous samples of an analog signal is represented by digital words in a serial bit stream. Hence, if we assume that each of the digital words has n binary digits, there are M = 2n unique code words that are possible.
- Two fundamental processes are involved in the generation of a binary PCM wave: sampling and quantization.
Advantages of PCM:
1. Relatively inexpensive digital circuitry may be used extensively in the system.
2. PCM signals derived from all types of analog sources (audio, video, etc.) may be merged with data signals (like from digital computers) and transmitted over a common high-speed digital communication system.
3. In long-distance digital telephone systems requiring repeaters, a clean PCM waveform can be regenerated at the output of each repeater, where the input consists of a noisy PCM waveform. 4. The noise performance of a digital system can be superior to that of an analog system.
Main disadvantage: A much wider bandwidth than that of the corresponding analog signal Quantization
It is not necessary to transmit the exact amplitude of the samples. The receiver (like the human ear) can detect only finite intensity differences. This means that the original continuous signal may be approximated by a signal constructed of discrete amplitudes detected on a minimum error basis from an available set. The existence of a finite number discrete amplitude levels is a basic condition of PCM.
Amplitude quantization - The process of transforming the sample amplitude of a message
signal at a time into discrete amplitude taken from a finite set of possible values.
Decision levels (or decision thresholds) – discrete amplitudes at the quantizer input.
Representation levels or reconstruction levels – discrete amplitudes at the quantizer output. Quantum or step size – spacing between two adjacent representation levels.
- Quantizers can be of a uniform or nonuniform type. In a uniform quantizer, the representation levels are uniformly spaced; otherwise the quantizer is nonuniform.
quantization noise – an error defined as the difference between the input signal and the output
signal.
Elements of a PCM system
Basic operations performed in the transmitter: sampling, quantizing, encoding
LPF prior to sampling is included to prevent aliasing of the message signal; the
quantizing and encoding operations are usually performed by a circuit known as analog-to-digital converter..
Basic operations in the receiver are regeneration of impaired signals, decoding and reconstruction of the train of quantized pulses. Regeneration also occurs at intermediate points along the transmission path as necessary
Transmitter:
Low –pass aliasing filter – used at the front-end of the sampler to exclude frequencies greater than B before sampling.
Sampler – permits the reduction of the continuously varying message signal (of some finite
duration) to a limited number of discrete values per second.
Quantizer – provides a new representation of the signal that is discrete in both time and
amplitude.
In telephonic communication, it is preferable to use a variable separation between separation levels
The use of a nonuniform quantizer is equivalent to passing the baseband signal through a compressor and then applying the compressed signal to a uniform quantizer.
