Unit - 2
Digital Modulation-I
Q1) Determine the baud and minimum bandwidth necessary to pass a 10-kbps binary signal using amplitude shift keying.
A1) For ASK, N = 1, and the baud and minimum bandwidth are determined from
B=fb/N
Where N is the number of bits encoded into each signaling element.
B = 10,000 / 1 = 10,000
Baud = 10, 000 /1 = 10,000
Q2) Determine (a) the peak frequency deviation, (b) minimum bandwidth, and (c) baud for a binary FSK signal with a mark frequency of 49 kHz, a space frequency of 51 kHz, and an input bit rate of 2 kbps.
A2)
a) Frequency deviation is ∆f = |fm – fs | / 2
Where ∆f = frequency deviation (hertz) |fm – fs | = absolute difference between the mark and space frequencies (hertz)
∆f = |149kHz - 51 kHz| / 2 =1 kHz
b. The minimum bandwidth is B = 2(∆f + fb)
Where B= minimum Nyquist bandwidth (hertz)
∆f= frequency deviation |(fm– fs)| (hertz)
Fb = input bit rate (bps)
B = 2(1000 + 2000) = 6 kHz c. For FSK, N = 1, and the baud is determined from Equation 2.11 as baud = 2000 / 1 = 2000
c. For FSK, N = 1, and the baud is determined from Equation 2.11 as baud = 2000 / 1 = 2000
Q3) For a BPSK modulator with a carrier frequency of 70 MHz and an input bit rate of 10 Mbps, determine the maximum and minimum upper and lower side frequencies, draw the output spectrum, determine the minimum Nyquist bandwidth, and calculate the baud.
A3) Substituting into Equation
BPSK output = [sin (2πfa t)] x [sin (2πfc t)]
Fa = fb / 2 = 5 MHz = [sin 2π(5MHz)t)] x [sin 2π(70MHz)t)] = 0.5cos[2π(70MHz – 5MHz)t] – 0.5cos[2π(70MHz + 5MHz)t]
Minimum lower side frequency (LSF): LSF=70MHz - 5MHz = 65MHz
Maximum upper side frequency (USF): USF = 70 MHz + 5 MHz = 75 MHz
Therefore, the output spectrum for the worst-case binary input conditions is as follows: The minimum Nyquist bandwidth (B) is
B = 75 MHz - 65 MHz = 10 MHz
And the baud = fb or 10 megabaud.
Q4) For the QPSK modulator shown in Figure, construct the truth table, phasor diagram, and constellation diagram.
A4) For a binary data input of Q = O and I= 0, the two inputs to the I balanced modulator are -1 and sin ωc t, and the two inputs to the Q balanced modulator are -1 and cos ωc t.
Consequently, the outputs are
I balanced modulator =(-1)(sin ωc t) = -1 sin ωc t
Q balanced modulator =(-1)(cos ωc t) = -1 cos ωc t and
The output of the linear summer is -1 cos ωc t - 1 sin ωc t = 1.414 sin(ωc t - 135°)
For the remaining dibit codes (01, 10, and 11), the procedure is the same.
Fig: QPSK modulator: (a) truth table; (b) phasor diagram; (c) constellation diagram
In Figures b and c, it can be seen that with QPSK each of the four possible output phasors has exactly the same amplitude. Therefore, the binary information must be encoded entirely in the phase of the output signal.
Figure b, it can be seen that the angular separation between any two adjacent phasors in QPSK is 90°. Therefore, a QPSK signal can undergo almost a+45° or -45° shift in phase during transmission and still retain the correct encoded information when demodulated at the receiver. Figure shows the output phase-versus-time relationship for a QPSK modulator.
Fig: Output phase-versus-time relationship for a PSK modulator.
Q5) For a QPSK modulator with an input data rate (fb) equal to 10 Mbps and a carrier frequency 70 MHz, determine the minimum double-sided Nyquist bandwidth (fN) and the baud.
A5) The bit rate in both the I and Q channels is equal to one-half of the transmission bit rate, or
fbQ = fb1 = fb / 2 = 10 Mbps / 2 = 5 Mbps
The highest fundamental frequency presented to either balanced modulator is
Fa= fbQ / 2 = 5 Mbps / 2 = 2.5 MHz
The output wave from each balanced modulator is
(sin 2πfa t)(sin 2πfc t)
0.5 cos 2π(fc – fa)t – 0.5 cos 2π(fc + fa)t
0.5 cos 2π[(70 – 2.5)MHz]t – 0.5 cos 2π[(70 – 2.5)MHz]t
0.5 cos 2π(67.5MHz)t - 0.5 cos 2π(72.5MHz)t
The minimum Nyquist bandwidth is B=(72.5-67.5)MHz = 5MHz
The symbol rate equals the bandwidth: thus, symbol rate = 5 megabaud
Q6) For an 8-PSK system, operating with an information bit rate of 24 kbps, determine (a) baud, (b) minimum bandwidth, and (c) bandwidth efficiency.
A6)
a. Baud is determined by
Baud = 24 kbps / 3 = 8000
b. Bandwidth is determined by
B = 24 kbps / 3 = 8000
c. Bandwidth efficiency is calculated
Bη = 24, 000 / 8000 = 3 bits per second per cycle of bandwidth
Q7) For a QPSK system and the given parameters, determine a. Carrier power in dBm. b. Noise power in dBm. c. Noise power density in dBm. d. Energy per bit in dBJ. e. Carrier-to-noise power ratio in dB. f. EblNo ratio. C = 10-12 W, Fb = 60 kbps, N = 1.2 x 10 -14W, B = 120 kHz
A7) a. The carrier power in dBm is determined by
C = 10 log (10 -12 / 0.001) = - 90 dBm
b. The noise power in dBm is determined by
N = 10 log [(1.2x10-14) / 0.001] = -109.2 dBm
c. The noise power density is determined by
No = -109.2 dBm – 10 log 120 kHz = -160 dBm d.
The energy per bit is determined by Eb = 10 log (10-12 / 60 kbps) = -167.8 dBJ
e. The carrier-to-noise power ratio is determined by
C / N = 10 log (10-12 / 1.2 x 10-14) = 19.2 dB
f. The energy per bit-to-noise density ratio is determined by
Eb / No = 19.2 + 10 log 120 kHz / 60 kbps = 22.2 dB
Q8) Determine the minimum bandwidth required to achieve a P(e) of 10-7 for an 8-PSK system operating at 10 Mbps with a carrier-to noise power ratio of 11.7 dB.
A8) The minimum bandwidth is
B / fb = Eb / No = C / N = 14.7 dB – 11.7 dB = 3 dB
B / fb = antilog 3 = 2
B = 2 x 10 Mbps = 20 MHz
Q9) We need to send data 3 bits at a time at 9 bit rate of 3mbps. The carrier frequency is 10MHz. Calculate the number of levels the baud rate and bandwidth?
A9) n= 3 biits
R= 3Mbps
fc= 10MHz
L=2n = 23=8
The number of levels =8
Baud rate = Bit rate/n= 3Mpbs/3 =1Mbaud
Bandwidth = L*r=8*1Mbaud= 8MHz
Q10) We have an available bandwidth of 100kHz which spans from 200- 300kHz. What should be the carrier frequency and the bit rate if we modulated the data by using FSK?
A10)
Q11) A voice signal is sampled at the rate of 8000 samples/sec and each sample is encoded in to 8 bits using PCM. The binary data is transmitted into free space after modulation. Determine the bandwidth of modulated signal, if the modulation technique is ASK, FSK, PSK for f1 =10MHz, f2= 8MHz?
A11)
Bit rate 
Baud rate=8000 samples/sec
No. Of bits per sample =8
Band width of A
Band width of 
Band width of 
Q12) Find the maximum bit rates of an FSK signal in bps, if the bandwidth of the medium is 12 kHz and the difference between the two carriers is 2 kHz(given that transmission mode is full duplex)
A12) As transmission is full duplex only 6kHz is allotted for each direction.
Bandwidth = 2Bit Rate + fc1-fc0
2Bit Rate= Bandwidth – (fc1-fc0)=6000-2000=4000
Bit rate= 2000bps
Q13) A BPSK modulator has a carrier frequency of 70MHz and an input bit rate of 10Mbps. Calculate the maximum upper side band frequency in MHz?
A13)
is maximum fundamental frequency of binary input 
is input bit rate
is reference carrier frequency
The output of a BPSK modulator with carrier frequency of 70 MHz and bit rate of 10 Mbps is
Therefore, the maximum upper sideband frequency
Q14) In digital communication channel employing FSK, the 0 and 1 bits are represented by sine waves of 10kHz and 25kHz respectively. These waveforms will be orthogonal for a bit interval of?
A14)
For orthogonality of two sine waves in 
duration there should be integral multiple of cycles of both sine wave.
Time period of first sine wave
Time period of the second sine wave
Therefore, 200s is the integral multiple of both 
and 
. Hence the two given waveforms will be orthogonal for a bit interval of 200s.
Q15) Find the minimum bandwidth (in kHz) required for transmitting an ASK signal at 2 kbps in half duplex transmission mode.
A15) In ASK, the baud rate and bit rate are the same. Therefore, baud rate is 2000 bauds. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, minimum bandwidth = 2000 Hz = 2 kHz
Q16) A communication system employs ASK to transmit a 10-kbps binary signal. Find the baud rate required in bauds.
A16) For an ASK system, the minimum bandwidth is the same as the bit rate of the signal. Therefore, minimum bandwidth = 10000 Hz
Q17) For the data given in the above Question, find the minimum bandwidth required in hertz
A17) For an ASK system, the minimum bandwidth is the same as the bit rate of the signal. Therefore, minimum bandwidth = 10000 Hz
Q18) Determine the peak frequency deviation, for a binary FSK signal with a mark frequency of 51 kHz, a space frequency of 49 kHz, and an input bit rate of 2.
A18) The peak frequency deviation is ∆f = (mark frequency-space frequency)/2 =|49kHz - 51 kHz| / 2 =1 kHz kbps.
Q19) Determine minimum bandwidth for a binary FSK signal with a mark frequency of 51 kHz, a space frequency of 49 kHz, and an input bit rate of 2.
A19) Minimum bandwidth = 2( 
+ 2Rb) = 2(1+2) =6K
Q20) Explain the generation and detection of QPSK?
A20) GENERATION AND COHERENT DETECTION OF QPSK SIGNALS
(i) Generation
The QPSK Modulator uses a bit-splitter, two multipliers with local oscillator, a 2-bit serial to parallel converter, and a summer circuit.
Fig: QPSK generation
At the modulator’s input, the message signal’s even bits (i.e., 2nd bit, 4th bit, 6th bit, etc.) and odd bits (i.e., 1st bit, 3rd bit, 5th bit, etc.) are separated by the bits splitter and are multiplied with the same carrier to generate odd BPSK (called as PSKI) and even BPSK (called as PSKQ). The PSKQ signal is anyhow phase shifted by 90° before being modulated.
The QPSK waveform for two-bits input is as follows, which shows the modulated result for different instances of binary inputs.
(ii) Detection
The QPSK Demodulator uses two product demodulator circuits with local oscillator, two band pass filters, two integrator circuits, and a 2-bit parallel to serial converter.
Fig: QPSK detectors
The two product detectors at the input of demodulator simultaneously demodulate the two BPSK signals. The pair of bits are recovered here from the original data. These signals after processing are passed to the parallel to serial converter.
Fig: QPSK output
