# MINIMUM SHIFT KEYING (MSK) | minimum shift keying is similar to in digital communication

By   July 23, 2020

minimum shift keying is similar to in digital communication , what is MINIMUM SHIFT KEYING (MSK) signal , bandwidth , transmitter and receiver block diagram formula.
Distance Between Signal Points
As a matter of fact, the ability to determine a bit without error is measured by the distance between two nearest possible signal points in the signal space. Such points differed in signal bit. For example, signal points ‘A’ and ‘B’ are two nearest points since they differ by a signal bit be(t). As ‘A’ and ‘B’ become closer to each other, the possibility of error increases. Therefore, this distance must be as large as possible. This distance is denoted by `d‘. In figure 8.29, the distance between signal points ‘A’ and ‘B’ can be given by,
d2 =  =                           …(8.66)
or                                 d2 = 2                                               …(8.67)
NOTE If we compare this distance with the distance of BPSK signals, then this shows that the distance for QPSK is the same as that for BPSK. Because, this distance repre-sents noise immunity of the system, it shows that noise immunities of BPSK and QPSK are same.
8.13.4. Spectrum of QPSK Signal
The input sequence b(t) is of bit duration Tb. Also, it is a NRZ bipolar waveform. Recall, the power spectral density of such waveform can be given as,
S(f) =
Also, Vb = , then this equation becomes,
S(f) =                                           …(8.68)
This equation gives power spectral density (psd) of signal b(t). This signal is divided into be(t) and b0(t) each of bit period 2Tb. If we consider that symbols 1 and 0 are equally likely, then we can write power spectral densities (pads) of be(t) and b0(t) as,
EQUATION
and                                            equation
In these two equations, we have just replaced Tb by Ts and Ts is the period of bit in be(t) and b0(t). Because, inphase and quadrature components [be(t) and b0(t)] are statistically independent, the baseband power spectral density of QPSK signal equals the sum of the individual power spectral densities of be(t) and b0(t) i.e.,
SB(f) = Se(f) + S0(f)
or                                             SB(f) = 2PsTs                                …(8.71)
This equation gives baseband power spectral density of QPSK signal. Upon modulation of carrier of frequency fc, the spectral density given by above equation is shifted at ± fc. Thus plots of power spectral density of QPSK will be similar to that BPSK.
8.13.5. Bandwidth of QPSK Signal
We have observed that the bandwidth of BPSK signal is equal of 2fb. Here, Tb =  is the one bit period. In QPSK, the two waveforms be(t) and b0(t) form the baseband signals. One bit period for both of these signals is equal to 2Tb. Therefore, bandwidth of QPSK signal will be
BW = 2 x                           …(8.72)
Hence, the bandwidth of QPSK signal is half of the bandwidth of BPSK signal. Earlier, we have observed that noise immunity of QPSK and BPSK is same. This shows that inspite of the reduction in bandwidth in QPSK, the noise immunity remains same as compared to BPSK. BW of QPSK can also be obtained from of figure 8.30.
diagram
FIGURE 8.30 Plot of power spectral density (psd) of QPSK signal.
BW = Highest frequency – Lowest frequency in main lobe
BW =  –  since carrier frequency fc cancels our
BW =
We know that Ts = 2Tb
or         BW =                                                                          …(8.73)
QPSK has some certain advantages as compared to BPSK and DPSK as under:
(i)         For the same bit error rate, the bandwidth required by QPSK is reduced to half as compared to BPSK.
(ii)        Because of reduced bandwidth, the information transmission rate of QPSK is higher.
8.14     MINIMUM SHIFT KEYING (MSK)
(U.P. Tech., Sem., Examination 2003-2004) (10 marks)
We have discussed QPSK technique in last article. The bandwidth requirement of QPSK is high Filters or other methods can overcome these problems, but they have other side effects. For example, filters alter the amplitude of the waveform.
MSK overcomes these problems. In MSK, the output waveform is continuous in phase hence there are no abrupt changes in amplitude. The sidelobes of MSK are very small hence bandpass filtering is not required to avoid interchannel interference. Figure 8.31 shows the waveform of MSK. The binary bit sequence is shown at the top. Figure 8.31(a) shows the corresponding NRZ
diagram
FIGURE 8.31            (a) Bipolar NRZ waveform representing bit sequence (b) Odd bit sequence waveforms b0(t) (c) Even bit sequence waveform be(t) (d) Waveforms of frequency fb/4 used for smoothing of be(t) and b0(t) (e) Modulating waveform of even sequence (f) Modulating waveform of odd sequence (g) Waveform of frequency fH (h) Waveform of frequency fL (i) MSK waveform.
waveform h(t). From b(t), two waveforms are generated for odd even bits. b0(t) represents odd bits and be(t) represents even bit Figure 8.31(b) and (c) shows the waveform of b0(t) and be(t). As shown in those waveforms b1, b3, b5 etc. are represented by odd waveform i.e., b0(t).
The duration of each bit in b0(t) or be(t) is 2Tb, whereas it is Tb in b(t) i.e.,
Ts = 2Tb                                                           …(8.79)
The waveforms b0(t) and be(t) have an offset of Tb. This offset is essential in MSK. Two waveforms sin 2p(t/Tb) and 2p(t/4Tb) are generated as shown in figure 8.31(d). The waveform of sin 2p (t4Tb) passes through zero at the end of symbol time in bo(t). Hence, one symbol duration of bo(t) consists of complete half cycle of cos 2p(t/4Tb). This means that similarly, one symbol duration of be(t) contains half cycle of sin 2p(t/Tb). Thus there is a phase shift of ‘T’b in sine and consine waveforms. be(f) is multiplied by sin 2p (t/4Tb) and bo(t) is multiplied by cons 2p(t/4Tb). These product waveforms are shown in figure 8.31 (e) and (f). The transmitted MSK signal is represented as under:
s(t)=[be(t) sin(2p/4Tb)] cos(2pfct) + [b0(t) cos(2pt/4Tb)] sin (2pfct) …(8.80)
This means that the product signal be(t) sin (2pt/4Tb) and b0(t) cos (2pt/4Tb) modulate the quadrature carriers of frequency fc. We can write last equation as,
equation
We know that fb =  then the last equation (8.7,1) becomes,
equation
Let                                           CH(t) =
and                                          CL(t) =
and let                         fH = fc +                                                          …(8.76)
and                                          fL = fc                                                         …(8.77)
with these substitutions, equation (8.75) becomes,
s(t) =  CH(t) sin (2pfHt) +  CL(t) sin (2pfLt)         …(8.78)
If b0(t) = be(t) then CL(t) = 0 and CH(t) = ± 1, then last equation becomes,
s(t) =  CH(t) sin (2pfHt)                          …(8.79)
Hence, the transmitted frequency is fH.
If b0(t) = -be(t), then CL(t) = 0 and CL(t) = ± 1. Then equation (8.79) becomes,
s(t) =  CH(t) sin (2pfHt)                          …(8.80)
Hence, the transmitted frequency is fL.
The frequencies fH and fL are chosen such that cos(2pfHt) and sin (2pfLt) are orthogonal over the interval Tb. For orthogonality following relation must be satisfied i.e.,
…(8.81)
This relation will be satisfied if we have integers ‘m’ and ‘n’ such that,
2p(fH – fL) Tb = np                                           …(8.82)
and                                             2p(fH + fL) Tb = mp                             …(8.83)
Let us substitue values of fH and fL from equations (8.74) and (8.75) in above relations. From equation (8.80), we get
equation
or                                                   fbTb. = n
or                                             fb x           Þ   n = 1                               …(8.84)         Similarly from equation (8.83), we get
equation
or                                                   4fbTb. = m
or                                             4fb x      Þ   fc =                       …(8.85)
with n = 1 in equation (8.82), we get
2p (fH – fL) Tb = 1 x p
or                                             (fH – fL) =                                           …(8.86)
Here, n = 1 means the difference between fH and fL, is minimum and at the same time, (MSK) they are orthogonal. Therefore, this technique is called minimum shift keying (MSK). This minimum difference is given by equation (8.86) above. From equation (8.85), we know that fc = . This shows that carrier frequency ‘fc‘ is integer multiple of  . Substituting, this value of fc in equation (8.76), we get
fH = fc +
or                                             fH = (m + 1)                                                    …(8.87)
Similarly, substituting fb = m  is equation (8.77), we get
fL = (m – 1)                                  …(8.88)
Figure 8.31(g) and (h) shows the waveforms of sin(2p fH t) and sin (2ptfLt). For these waveforms m = 5. Using equations (8.87) and (8.88), fH and fL are calculated with m = 5. Figure 8.31(i) shows the final MSK waveform. From equation (8.79), we know that if bo(t) = be(t) then transmitted waveform is of frequency fH. And if b0(t)  = – be(t) then the transmitted waveform is given by equation (8.75), which has frequency of fL. This shows that MSK is basically FSK with reduced bandwidth and continuous phase.
8.14.1. Signal Space Representation of MSK and Distance between the Signal Points (i.e., Geometrical Representation of MSK)
Let us rearrange equation (8.73) as follows
s(t) = CH(t)   sin(2pfHt) + CL(t)   sin(2pfLt)   …(8.90)
Here let ,                                 fH(t) = sin(2pfHt)                                  …(8.91)
fL(t) = sin(2pfLt)                                   …(8.92)
The carriers fH(t) and fL(t) are in quadrature. They are in quadrature because their frequencies are in quadrature. In QPSK the carriers are in quadrature because of phase shift. Depending on the values of CH(t) and CL(t), there will be four signal points in fH fL plane. This has benn illustrated in figure 8.32. The distance of each signal point from the origin is
diagram
FIGURE 8.32 Geometrical (Signal Space) representation of MSK signals.
Distance Between Signal Points
Since the points are symmetric, the distance between any two nearest points is same, i.e.,
d2 =   +
or                                 d =                                         .                          …(8.93)
or                                 d =        (since PsTs = Es)                                     …(8.94)
or                                 d =        (since Es = 2Es) = 2)                      …(8.95)
These relations give distance between signal points in MSK. This distance is same as in QPSK.
8.14.2. Power Spectral Density (psd) and Bandwidth of MSK
Let us consider the baseband signal of equation (8.78). The waveform which modulates sin(2pfct)
p(t) =  [b0(t) cos (2pt/4Tb)]                    …(8.96)
or                                             p(t) =  [b0(t) cos (pfbt/2)]                       …(8.97)
The power spectral density (psd) of above waveform is expressed as,
equation                                      …(8.98)
when this signal modulates the carrier ‘fc‘ then the total power spectral density (psd) of baseband signal is divided by ‘4’ and is placed at ± fc, i.e.,
equation
The above equation gives power spectral density (psd) of MSK signal. Figure 8.33 show the normalized spectral densities of MSK and QPSK. Normalization means maximum amplitudes signals are scaled with respect to ‘1’.
DIAGRAM
FIGURE 8.33 Power spectral densities (psd) of MSK and QPSK.
The above plots show that the main lobe in MSK is wider than QPSK. The sides lobes in MSK are very small compared to QPSK.
Bandwidth Calculation of MSK
From figure 8.33, we observe that the width of main lobe in MSK is ± 0.75 i.e.,
fTb = ± 0.75
or                                                  f  = ± 0.75 fb
Hence, bandwidth will be equal to width of the main lobe i.e.,
BW = 0.75 fb –  (- 0.75 fb) = 1.5 fb                        …(8.100)
Thus, the BW of MSK is higher than that of QPSK.
8.14.3. Generation of MSK
Figure 8.34 shows the block diagram of MSK transmitter. The two sinusoidal signals sin(2πfct) and (2π/4Tb) are mixed (i.e., multiplied). The bandpass filters then pass only sum and
diagram
FIGUPE 8.34 MSK transmitter block diagram.
difference components fc +   and fc . The outputs of bandpass filters (BPFs) are then added and subtracted such that two signals x(t) and y(t) are generated. Signal x(t) is multiplied by  b0(t), and y(t) is multiplied by  be(t), The outputs of the multiplies are then added to give final MSK signal. Thus the block diagram of figure 8.34 is the step to step implementation of equation (8.73).
8.14.4. Reception of MSK (i.e. Detection of MSK)
Figure 8.35 shows the block diagram of MSK receiver. MSK uses synchronous detection. The signals x(t) and y(t) are multiplied with the received MSK signal. Here x(t) and y(t) have same values as shown in transmitter block diagram of figure 8.35. The outputs of the multipliers are be(t) and b0(t). The integrators integrate over the period of 2Tb. For the upper correlator, the sampling switch samples output of integrator at t = (2k + 1)Tb. Then the decision device decides whether b0(t) is + 1 or -1. Similarly, lower correlator output is be(t). The outputs of two decision devices are staggered by Tb. The switch S3 operates at t = kTb and simply multiplexes the two correlator outputs.
diagram
FIGURE 8.35 MSK receiver block diagram.
8.14.5. Advantages and Drawbacks of MSK as Compared to QPSK
From the discussion of MSK, we can now compare the advantages of MSK over QPSK.

1. The MSK baseband waveforms are smoother compared to QPSK.
2. MSK signal have continuous phase in all the cases, whereas QPSK has abrupt phase shift of or π.
3. MSK waveform does not have amplitude variations, whereas QPSK signals have abrupt amplitude variations.
4. The main lobe of MSK is wider than that of QPSK. Main lobe of MSK contains around 99% of signal energy whereas QPSK main lobe contains around 90% signal energy.
5. Side lobes of MSK are smaller compared to that of QPSK. Hence, interchannel interference because of side lobes is significantly large in QPSK.
6. To avoid interchannel interference due to sidelobes, QPSK needs bandpass filtering, wherea it is not required in MSK.
7. Bandpass filtering changes the amplitude waveform of QPSK because of abrupt changes in phase. This problem does not exist in MSK.

The distance between signal points is same in QPSK as will as in MSK. Hence, the probability of error is also same. However, there are some drawback of MSK.
(ii)        Drawbacks

1. The bandwidth requirement of MSK is 1.5 fb, whereas it is fb QPSK. Actually, this cannot be said serious drawback of MSK. Because power to bandwidth ratio of MSK is more In fact, 99% of signal power can be transmitted within the bandwidth of 1.2 fb in MSK. While QPSK needs around 8 fb to transmit the same power.

2.        The generation and detection of MSK is slightly complex. Because of incorrect synchronization, phase jitter can be present in MSK. This degrades the performance of MSK