# qpsk full form | QUADRATURE PHASE SHIFT KEYING | GENERATION OF QPSK detection

By   July 22, 2020

GENERATION OF QPSK detection , qpsk full form | QUADRATURE PHASE SHIFT KEYING ?
As a matter of fact, in communication systems, we have two main resources. These are the transmission power and the channel bandwidth. The channel bandwidth depends upon the bit rate or signaling rate fb. In digital bandpass transmission, we use a carrier for transmission. This carrier is transmitted over a channel. If two or more bits are combined in some symbols, then the signaling rate will be reduced. Thus, the frequency of the carrier needed is also reduced. This reduces the transmission channel bandwidth. Hence, because of grouping of bits in symbols, the transmission channel bandwidth can be reduced. In quadrature phase shift keying (QPSK), two successive bits in the data sequence are grouped together. This reduces the bits rate or signaling rate (i.e., fb) and thus reduces the bandwidth of the channel.
In case of BPSK, we know that when symbol changes the level, the phase of the carrier is changed by 180°. Because, there were only two symbols in BPSK, the phase shift occurs in two levels only. However, in QPSK, two successive bits are combined. Infact, this combination of two bits forms four distinct symbols. When the symbol is changed to next symbol, then the phase of the carrier is changed by 45° (p/4 radians). Table 8.7 shows these symbols and their phase shifts.
Table 8.7. Symbol and corresponding phase shifts in QPSK

 S.No. Inputs successive bits Symbol Phase shift in carrier 1. 1(1 V) 0(-1 V) S1 p/4 2. 0(-1 V) 0(-1 V) S2 3p/4 3. 0(-1 V) 1(1 V) S3 5p/4 4. 1(1 V) 1(1 V) S4 7p/4

Hence as shown in Table 8.7, there are four symbols and the phase is shifted by p/4 for each symbol.
8.13 GENERATION OF QPSK
Figure 8.24 shows the block diagram of QPSK transmitter. Here, the input binary sequence is first converted to a bipolar NRZ type of signal. This signal is denoted by b(t). It represents binary ‘1’ by + 1 V and binary `0′ by – 1 V. This signal has been shown in figure 8.25(a). The demultiplexer divides b(t) into two separate bit streams of the odd numbered and even numbered bits. Here, be(t) represents even numbered sequence and b0(t) represents odd numbered sequence. The symbol duration of both of these odd and even numbered sequences is 2Tb. Hence, each symbol consists of two bits. Figure 8.25(b) and (c) illustrate the waveform of be(t) and b0(t).
diagram
FIGURE 8.24 Generation of QPSK.
It may be observed that the first even bit occurs after the first odd bit. Hence, even numbered bit sequence be(t) starts with the delay of one bit period due to first odd bit. Thus, first symbol of be(t) is delayed by one bit period `Tb‘ with respect to first symbol of b0(t). This delay of Tb, is known as offset. This shows that the change in levels of be(t) and b0(t) cannot occur at the same time due to offset or staggering.

 DO YOU KNOW? In QPSK, each symbol represents two bits and the bit rate is talk the baud rate. This is called a dibit system.

Also, the bit steam be(t) modulates carrier  cos(2pfct) and b0(t) modulates  sin(2pfct). These modulators are the balanced modulators. The two carriers cos(2pfct) and  sin(2pfct) have been shown in figure 8.25(d) and (e). There carriers are also known as quadrature carriers.
The two modulated sigals can be written as,
se(t) = be(t)   sin (2pfct)                                …(8.50)
and                              s0(t) = b0(t)   cos (2pfct)                              …(8.51)
Hence, se(t) and s0(t) are basically BPSK signals. The only difference is that T = 2Tb here; The value of be(t) and b0(t) would be + 1V or – 1V. Figure 8.25 (f) and (g) shows the waveforms of se(t) and s0(t). The adder in figure 8.24 adds these two signals be(t) and b0(t).
DIAGRAM
figure 8.25 qpsk waveforms, (a) Input sequence and its corresponding NRZ waveform, (b) Odd numbered bit sequence and its corresponding waveform (c) Even numbered bit sequence and its NRZ waveform (d) Basis function f1(t) (e) Basis function f2(t) (f) Binary PSK waveform for odd numbered channel (g) Binary PSK waveform for even numbered channel (h) Final QPSK waveform.
The output of the adder is QPSK signal and it is given by,
s(t) = s0(t) + se(f)
or                     s(t) = b0(t)  cos (2pfct) + be(t)   sin (2pfct)       …(8.52)
Figure 8.25(h) shows the QPSK signal represented by equation (8.52). In QPSK signal in figure 8.25(h), if there is any phase change, it occurs at minimum duration of Tb. This is because the two signals se(t) and s0(t) have an offset of `Tb‘. Due to this offset, the phase shift in QPSK signal is  .
equation and diagram
FIGURE 8.26 The Phasor diagram of QPSK signal.
8.13.1. Reception of QPSK (i.e. Detection of QPSK)
Figure 8.27 shows the QPSK receiver. This is synchronous reception. Hence, the coherent carrier is to be recovered from the received signal s(t). The received signal s(t) is first raised to its 4th power, i.e., s4(t). After that, it is allowed to pass through a bandpass filter (BPF) which is centred around 4fc. The output of the bandpass filter is a coherent carrier of frequency 4fc. This is divided by 4 and it provides two coherent quadrature carriers, i.e., cos (2pfct) and sin (2pfct). These coherent carriers are applied to two synchronous demodulators. These synchronous demodulators consist of multiplier and an integrator.
diagram
FIGURE 8.27 Reception of QPSK.
The incoming signal is applied to both the multipliers. Here, the integrator integrates; the product signal over two bit interval (i.e., Ts = 2Tb). At the end of this period. the output of integrator is sampled. The outputs of the two integrators are sampled at the offset of one bit period, Tb. Hence, the output of the multiplexer is the signal b(t). This means that the odd and even sequences are combined by multiplexer.
Now, let us consider the product signal at the output of upper multiplier, i.e.,
s(t) sin (2pfct) = b0(t)  cos (2pfct) sin(2pfct) + be(t)  sin2 (2pfct)   …(8.56)
This signal is integrated by the upper integrator in figure 8.27.
Therefore, we have
equation
Now, since  sin (2x) = sin x . cos x
and sine2 (x) =  [1-cos(2x)]
Therefore, using these two trigonometric identities in equation (8.53), we get
equation
In this equation, the first and third integration terms involve integration of sinusoidal carriers over two bit period. They have full (integral number of) cycles over two bit periods and thus integration will be zero, i.e.,
equation
Hence, the upper integrator responds to even sequence only. Similarly, we can obtain the output of lower integrator as b0(t)
NOTE Even though bit synchronizer has not been shown in figure 8.27, it is assumed to be present with the integrator to locate starting and ending times of integration. The multiplexer is also operated by bit synchronizer. The amplitudes of voltage marked in figure 8.27 are arbitrary. They can change depending upon the gains of the integrator.
8.13.2. Concept of Carrier Synchronization in QPSK
Both the carriers are to be synchronized properly in coherent detection in QPSK. Figure 8.28 shows the PLL system for carrier synchronization in QPSK.
The fourth power of the input signal consists of discrete frequency component at 4fc.
We know that,
cos4(2pfc t) = cos(8pfct + 2p N)
where ‘N‘ is the number of cycles over the bit period. It is always an integer value. When the frequency division by four takes place, the RHS of this equation becomes cos. This indicates that the output has a fixed phase error of  . Differential encoding can be used to nullify the phase error events. The PLL remains locked with the phase of ‘4fc‘ and then output of PLL is divided by 4. This provides a coherent carrier. A 90° phase shift is added to this carrier to produce a quadrature carrier. Quadrative carrier.
diagram
FIGURE 8.28 PLL system used for carrier synchronization in QPSK.
8.13.3. Signal Space Representation in QPSK Signals
Figure 8.29 shows the phasor diagram of QPSK signal. Depending upon the combination of two successive bits, the phase shift occurs in carrier. This means that the QPSK signal can be written as,
s(t) =  cos   m = 0, 1, 2, 3    …(8.55)
Here, above expression takes four values and they represent the phasors of figure 8.29. This equation can be expanded as under:
equation
Let us rearrange the above equation as under:
equation
Again, let                                f1(t) =  cos (2pfc t)                                       …(8.57)
and                                          f2(t) =  sin (2pfc t)                                       …(8.58)
These two signals are known as orthogonal signals and they are used as carriers in modulator.
Let                               b0(t) =  cos                                       …(5.59)
and                              be(t) =  sin                                        …(5.60)
With the use of equations (8.57) to (8.60) we can write equation (8.56) as under:
EQUATION
or                                               EQUATION
TS = symbol duration and Ts = 2Tb,
or                                 Tb =                                                                 …(8.61)
Then the above equation becomes,
s(t) =  b0(t) f1(t) +  (t) f2(t)                         …(8.62)
Since bit energy          Eb = Ps Tb
Therefore,                    s(t) =  b0(t) f1(t) +  (t) f2 (t)          …(8.63)
This equation gives signal space representation of QPSK signal. The two orthogonal signals f1(t) and f2(t) form the two axes of the signal space. Figure 8.29 shows the signal space representation. The possible 4 signal points have been shown by small circles on f1 f2-plane.
From each signal point, we obtain two bits. For example, from point ‘A’, we obtain two bits as (1, 1) and from ‘B’ we obtain bits as (- 1, 1). The distance of any signal point from origin ‘0’, given as,
Distance ‘OB’ =  =  =     [⸫ 2Tb = Ts]  …(8.64)
or                                 ‘OB’ =                  [⸫PsTs = Ts]                    …(8.65)
equation and diagram
FIGURE 8.29 The signal space representation for QPSK signals.
Hence, the length of each signal point from origin is . We know that be(t) and b0(t) represent two successive bits. There is an offset of `Tb‘ between be(t) and b0(t). Therefore, be(t) and b0(t) both cannot change their levels simultaneously. Hence, either be(t) or b0(t) can change at a time.
Let us say that be(t) = b0(t) = 1 is representing signal point ‘A’ in figure 8.29. In the next hit interval, if b0(t) = -1, then signal point will be ‘D‘ . Otherwise, if be(t) changes its level [i.e., be(t) = -1], then next signal point will be ‘B’ . Hence, from signal point ‘A’ , then next signal points will be either ‘D’ or ‘B‘.