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 b_{e}(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,

d_{2} = = …(8.66)

or d_{2} = 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 T_{b}. Also, it is a NRZ bipolar waveform. Recall, the power spectral density of such waveform can be given as,

S(f) =

Also, V_{b} = , then this equation becomes,

S(f) = …(8.68)

This equation gives power spectral density (psd) of signal b(t). This signal is divided into b_{e}(t) and b_{0}(t) each of bit period 2T* _{b}*. If we consider that symbols 1 and 0 are equally likely, then we can write power spectral densities (pads) of b

_{e}(t) and b

*(t) as,*

_{0}**EQUATION**

and

**equation**

In these two equations, we have just replaced T

*by T*

_{b}*s*and T

*is the period of bit in b*

_{s}_{e}(t) and b

_{0}(t). Because, inphase and quadrature components [b

_{e}(t) and b

_{0}(t)] are statistically independent, the baseband power spectral density of QPSK signal equals the sum of the individual power spectral densities of b

_{e}(t) and b

_{0}(t) i.e.,

S

_{B}(f) = S

_{e}(f) + S

_{0}(f)

or SB(f) = 2PsTs …(8.71)

This equation gives baseband power spectral density of QPSK signal. Upon modulation of carrier of frequency f

*, the spectral density given by above equation is shifted at ± f*

_{c}*. Thus plots of power spectral density of QPSK will be similar to that BPSK.*

_{c}**8.13.5. Bandwidth of QPSK Signal**

We have observed that the bandwidth of BPSK signal is equal of 2f

*. Here, Tb = is the one bit period. In QPSK, the two waveforms b*

_{b}_{e}(t) and b

_{0}(t) form the baseband signals. One bit period for both of these signals is equal to 2T

*. Therefore, bandwidth of QPSK signal will be*

_{b}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 f

*cancels our*

_{c}BW =

We know that T

*= 2T*

_{s}

_{b}*or BW = …(8.73)*

_{ }**8.13.6. Advantages of QPSK**

**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 b

_{0}(t) (c) Even bit sequence waveform b

_{e}(t) (d) Waveforms of frequency f

*/4 used for smoothing of b*

_{b}*(t) and b*

_{e}_{0}(t) (e) Modulating waveform of even sequence (

*f*) Modulating waveform of odd sequence (g) Waveform of frequency f

*(h) Waveform of frequency f*

_{H}*(i) MSK waveform.*

_{L}waveform h(t). From b(t), two waveforms are generated for odd even bits. b

*(t) represents odd bits and b*

_{0}_{e}(t) represents even bit Figure 8.31(b) and (c) shows the waveform of b

_{0}(t) and b

*(t). As shown in those waveforms b*

_{e}_{1}, b

_{3}, b

_{5}etc. are represented by odd waveform i.e., b

_{0}(t).

The duration of each bit in b

_{0}(t) or b

_{e}(t) is 2T

*, whereas it is T*

_{b}*in b(t) i.e.,*

_{b}T

*= 2T*

_{s}*…(8.79)*

_{b}The waveforms b

_{0}(t) and b

*(t) have an offset of T*

_{e}_{b}. This offset is essential in MSK. Two waveforms sin 2p(t/T

*) and 2p(t/4T*

_{b}*) are generated as shown in figure 8.31(d). The waveform of sin 2p (t4T*

_{b}*) passes through zero at the end of symbol time in b*

_{b}_{o}(t). Hence, one symbol duration of b

_{o}(t) consists of complete half cycle of cos 2p(t/4T

_{b}). This means that similarly, one symbol duration of b

_{e}(t) contains half cycle of sin 2p(t/T

_{b}). Thus there is a phase shift of ‘T’

_{b}in sine and consine waveforms. b

_{e}(f) is multiplied by sin 2p (t/4T

_{b}) and b

_{o}(t) is multiplied by cons 2p(t/4T

_{b}). These product waveforms are shown in figure 8.31 (e) and (f). The transmitted MSK signal is represented as under:

s(t)=[b

_{e}(t) sin(2p/4T

_{b})] cos(2pf

_{c}t) + [b

_{0}(t) cos(2pt/4T

_{b})] sin (2pfct) …(8.80)

This means that the product signal b

_{e}(t) sin (2pt/4T

_{b}) and b0(t) cos (2pt/4T

_{b}) modulate the quadrature carriers of frequency f

*. We can write last equation as,*

_{c}**equation**

We know that f

*= then the last equation (8.7,1) becomes,*

_{b}**equation**

Let C

*(t) =*

_{H}and C

*(t) =*

_{L}and let f

*= f*

_{H}*+ …(8.76)*

_{c}and f

*= f*

_{L}*…(8.77)*

_{c}with these substitutions, equation (8.75) becomes,

s(t) = C

*(t) sin (2pf*

_{H}*t) + C*

_{H}*(t) sin (2pf*

_{L}*t) …(8.78)*

_{L}If b

_{0}(t) = b

_{e}(t) then C

*(t) = 0 and C*

_{L}*(t) = ± 1, then last equation becomes,*

_{H}s(t) = C

*(t) sin (2pf*

_{H}*t) …(8.79)*

_{H}Hence, the transmitted frequency is f

*.*

_{H}If b

_{0}(t) = -b

_{e}(t), then C

*(t) = 0 and C*

_{L}*(t) = ± 1. Then equation (8.79) becomes,*

_{L}s(t) = C

*(t) sin (2pf*

_{H}*t) …(8.80)*

_{H}Hence, the transmitted frequency is f

*.*

_{L}The frequencies fH and fL are chosen such that cos(2pf

*t) and sin (2pf*

_{H}*t) are orthogonal over the interval T*

_{L}*. For orthogonality following relation must be satisfied i.e.,*

_{b}…(8.81)

This relation will be satisfied if we have integers ‘m’ and ‘n’ such that,

2p(f

*– f*

_{H }*) T*

_{L}*= np …(8.82)*

_{b}and 2p(f

*+ f*

_{H }*) T*

_{L}*= mp …(8.83)*

_{b}Let us substitue values of f

*and f*

_{H}*from equations (8.74) and (8.75) in above relations. From equation (8.80), we get*

_{L}**equation**

or f

_{b}T

_{b}. = n

or f

_{b}x Þ n = 1 …(8.84) Similarly from equation (8.83), we get

**equation**

or 4f

_{b}T

_{b}. = m

or 4f

_{b}x Þ f

*= …(8.85)*

_{c}with n = 1 in equation (8.82), we get

2p (f

*– f*

_{H}*) T*

_{L}*= 1 x p*

_{b}or (f

*– f*

_{H}*) = …(8.86)*

_{L}Here, n = 1 means the difference between f

*and f*

_{H}*, is minimum and at the same time, (MSK) they are orthogonal. Therefore, this technique is called*

_{L}**minimum shift keying (MSK)**. This minimum difference is given by equation (8.86) above. From equation (8.85), we know that f

_{c}= . This shows that carrier frequency ‘f

*‘ is integer multiple of . Substituting, this value of f*

_{c}*in equation (8.76), we get*

_{c}fH = fc +

or f

*= (m + 1) …(8.87)*

_{H}Similarly, substituting f

*= m is equation (8.77), we get*

_{b}f

*= (m – 1) …(8.88)*

_{L}Figure 8.31(g) and (h) shows the waveforms of sin(2p f

*t) and sin (2ptf*

_{H}*t). For these waveforms m = 5. Using equations (8.87) and (8.88), f*

_{L}*and f*

_{H}*are calculated with m = 5. Figure 8.31(i) shows the final MSK waveform. From equation (8.79), we know that if b*

_{L}_{o}(t) = b

*(t) then transmitted waveform is of frequency f*

_{e}*. And if b*

_{H}*(t) = – b*

_{0}*(t) then the transmitted waveform is given by equation (8.75), which has frequency of f*

_{e}*. This shows that MSK is basically FSK with reduced bandwidth and continuous phase.*

_{L}**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) = C

*(t) sin(2pf*

_{H}*t) + C*

_{H}*(t) sin(2pf*

_{L}*t) …(8.90)*

_{L}Here let , f

*(t) = sin(2pf*

_{H}*t) …(8.91)*

_{H}f

*(t) = sin(2pf*

_{L}*t) …(8.92)*

_{L}The carriers f

*(t) and f*

_{H}*(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 C*

_{L}*(t) and C*

_{H}*(t), there will be four signal points in f*

_{L}*f*

_{H}*plane. This has benn illustrated in figure 8.32. The distance of each signal point from the origin is*

_{L}**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.,**

d

^{2}= +

or d = . …(8.93)

or d = (since P

_{s}T

_{s}= E

_{s}) …(8.94)

or d = (since E

_{s}= 2E

_{s}) = 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(2pf**

*t)*

_{c}p(t) = [b

_{0}(t) cos (2pt/4T

_{b})] …(8.96)

or p(t) = [b

_{0}(t) cos (pf

_{b}t/2)] …(8.97)

The power spectral density (psd) of above waveform is expressed as,

**equation**…(8.98)

when this signal modulates the carrier ‘f

_{c}‘ then the total power spectral density (psd) of baseband signal is divided by ‘4’ and is placed at ± f

_{c}, 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.,**

fT

*= ± 0.75*

_{b}or f = ± 0.75 f

_{b}Hence, bandwidth will be equal to width of the main lobe i.e.,

BW = 0.75 f

*– (- 0.75 f*

_{b}*) = 1.5 f*

_{b}*…(8.100)*

_{b }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(2πf**

*t) and (2π/4T*

_{c}*) are mixed (i.e., multiplied). The bandpass filters then pass only sum and*

_{b}**diagram**

**FIGUPE 8.34**

*MSK transmitter block diagram.*

difference components f

*+ and f*

_{c}_{c}. 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 b

*(t), and y(t) is multiplied by b*

_{0}*(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).*

_{e}**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 b

_{e}(t) and b

_{0}(t). The integrators integrate over the period of 2T

*. For the upper correlator, the sampling switch samples output of integrator at t = (2k + 1)T*

_{b}*. Then the decision device decides whether b*

_{b}_{0}(t) is + 1 or -1. Similarly, lower correlator output is b

*(t). The outputs of two decision devices are staggered by T*

_{e}*. The switch S*

_{b}_{3}operates at t = kT

*and simply multiplexes the two correlator outputs.*

_{b}**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.**

**Advantages:**

- The MSK baseband waveforms are smoother compared to QPSK.
- MSK signal have continuous phase in all the cases, whereas QPSK has abrupt phase shift of or π.
- MSK waveform does not have amplitude variations, whereas QPSK signals have abrupt amplitude variations.
- 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.
- Side lobes of MSK are smaller compared to that of QPSK. Hence, interchannel interference because of side lobes is significantly large in QPSK.
- To avoid interchannel interference due to sidelobes, QPSK needs bandpass filtering, wherea it is not required in MSK.
- 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**

- The bandwidth requirement of MSK is 1.5 f
_{b}, whereas it is f_{b}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 fin MSK. While QPSK needs around 8 f_{b}to transmit the same power._{b}

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