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A PAM signal is demodulated using an integrator. Signals having same width but different amplitudes produce different values at the output of an integrator.
Which filter is used to detect the PAM signal?
PAM (Pulse Amplitude Modulation):
A speech signal is sampled a rate of 20% above the Nyquist rate. The signal has a bandwidth of 10 kHz. The sample is quantized into 1024 levels and then transmitted through 8level PAM over an AWGN baseband channel. The bandwidth required for transmission is ______
Concept:
Where,
Rb = nf_{s}
n = no. of bits
M = no. of levels of PAM
Calculation:
fm = 10 kHz
Nyquist sampling rate = 2f_{m} = 20 kHz.
The signal is sampled at 20% above Nyquist rate:
∴ fs = (1.2 × 20) = 24 kHz.
No. of Quantization levels
= 1024
No. of bits required = log_{2} (1024)
= log_{2} 2^{10} = 10 bits.
= 80 kHz.
Which of the following Pulse time Modulation does not exist in practice?
PFM does not exist in practice
Additional Information
PWM
The amplitude and position of pulses are constant in modulated signal (PWM), but the width (or duration) of pulses varies proportionally with the amplitude of an analog useful signal.
The carrier signal is produced by a clock.
PPM
The amplitude and width of pulses are constant in a modulated signal (PPM), but the direction of pulses varies proportionally with the amplitude of analogical useful signal.
PAM
The width and location of pulses in a modulated signal (PAM) are constant, while the amplitude of pulses varies proportionally with the amplitude of an analogical useful signal.
Study the given input signal and match the columns.
PFM does not exist in practice
Additional Information
PWM
The amplitude and position of pulses are constant in modulated signal (PWM), but the width (or duration) of pulses varies proportionally with the amplitude of an analog useful signal.
The carrier signal is produced by a clock.
PPM
The amplitude and width of pulses are constant in a modulated signal (PPM), but the direction of pulses varies proportionally with the amplitude of analogical useful signal.
PAM
The width and location of pulses in a modulated signal (PAM) are constant, while the amplitude of pulses varies proportionally with the amplitude of an analogical useful signal.
Hence from the above option (d) is correct
A PAM source generates four symbols 3 V, 1 V, 1 V and 3 V with probability of p(3) = p(3) = 0.2 and p(1) = p(1) = 0.3 respectively. The variance for the source will be
Concept:
For discrete random variable x:
Mean = E[x] = ∑I xi p (xi)
Mean square value (MSQ) = E[x^{2}] = ∑_{I} x_{i}^{2} p (x_{i})
Variance (σ^{2}) = E[x^{2}] – {E(x)}^{2}
Standard deviation (σ) =
Calculation:
Given data: for the given PAM source
Mean = E[x] = ∑i xi p (xi)
= (3)(0.2) + (1)(0.3) + (1)(0.3) + (3)(0.2)
= 0
MSQ = E[x^{2}] = ∑I x_{i}^{2} p (x_{i})
= (3)^{2}(0.2) + (1)^{2}(0.3) + (1)^{2}(0.3) + (3)^{2}(0.2) V
= 4.2 V
Variance = E[x^{2}] – {E[x]}^{2}
= 4.2 – 0
= 4.2 V
Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is white with power spectral density S_{N}(f) = N_{0}/2 = 10^{20} W/Hz. The lowpass filter is ideal with unity gain and cutoff frequency 1 MHz. Let Y_{t}, represent the random variable y(t_{1})
Yt = Ns if transmitted bit bk = 0
YK = a + N_{k} if transmitted bit bk = 1
where N_{k} represents the noise sample value. The noise sample has a probability density function, PN (n) = 0.5αe (This has mean zero and variance 2/α^{2}), Assume transmitted bits lo be equiprobable and threshold z is set to a/^{2} = 10^{6} V
‘The probability of bit error is
Concept:
Consider the block diagram as shown below to represent the noise which is treated as error:
The received signal is:
r(t) = S_{i}(t) + n(t)
The output of the filter is:
z(t) = ai(t) + n0(t)
ai(t): Signal component
n0(t): Noise component
The probability of error is represented in terms of the Q function:
Pe = Q[x]
When the pdf's are not having an equal width then the probability of error will not be the same for both symbols. In this case, the Bit error rate is calculated.
BER = P(1) × Pe_{1} + P(0) × Pe_{0}
P(1): Probability when 1 is transmitted
P(0): Probability when 0 transmitted
Pe0: Probability of error when 0 transmitted
Pe1: Probability of error when 1 transmitted.
Calculation:
The system is defined as shown:
The given mean is 0 and variance is 2/α^{2}
V[X] = E[X^{2}]  (E[X])^{2}
2/α^{2} = E[X^{2}]
The area under the curve is:
E[X^{2}] = 10^{20} × 2 × 10^{6}
2/α^{2} = 10^{14} × 2
α^{2} = 10^{14}
α = 10^{7}
Given Y_{t}, represent the random variable y(t_{1})
Y_{t} = N_{s} if transmitted bit b_{k} = 0
Y_{K} = a + N_{k} if transmitted bit b_{k} = 1 and threshold z is set to a/2 = 10^{6} V
P(0) = P(1) = 1/2
P_{e} = P(1) × Pe_{1} + P(0) × Pe_{0}
Pe_{1} calculation
We know that when the output signal is less than the threshold it is considered as 0 and there is an error when 1 is transmitted.
Y_{k }< z ⇒ error
a + N_{k} < 10^{6}
N_{k} < 10^{6}
Pe_{0} calculation
We know that when the output signal is more than the threshold it is considered as 1 and there is an error when 0 is transmitted.
Y_{k} > z ⇒ error
N_{s} > 10^{6}
N_{s} > 10^{6}
The probability of error is:
Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is white with power spectral density S_{N}(f) = N_{0}/2 = 10^{20} W/Hz. The lowpass filter is ideal with unity gain and cutoff frequency 1 MHz. Let Y_{t}, represent the random variable y(t_{1})
Yt = Ns if transmitted bit bk = 0
YK = a + N_{k} if transmitted bit bk = 1
where N_{k} represents the noise sample value. The noise sample has a probability density function, PN (n) = 0.5αe (This has mean zero and variance 2/α^{2}), Assume transmitted bits lo be equiprobable and threshold z is set to a/^{2} = 10^{6} V
Q. The value of the parameter α (in V1) is
Concept:
The statistical averages Mean and Variance are called as 'first order moment' and 'secondorder moment' respectively.
Variance represents the total power of the random signal.
Total power can be determined as the area of the curve.
When the input is passed through a system the output is shown by:
S_{N}(f): Input noise PSD
S_{N0}(f): Output noise PSD
H(f): system response
Calculation:
The system is defined as shown:
The given mean is 0 and variance is 2/α^{2}
V[X] = E[X^{2}]  (E[X])^{2}
2/α^{2} = E[X^{2}]
The area under the curve is:
E[X^{2}] = 10^{20} × 2 × 10^{6}
2/α^{2} = 10^{14} × 2
α^{2} = 10^{14 }
α = 10^{7}
Which of the following is NOT an advantage of Pulse Duration Modulation (PDM) recording?
Concept:
Pulse Width Modulation (PWM):
As mentioned PWM is short form of Pulse Width Modulation in which width of pulse is proportional to amplitude of the modulating signal.
Following are the benefits or advantages of PWM:
Following are the disadvantages of PWM:
From the given statements, 'It has a less complex electronic circuitry and therefore the reliability of such system is high' is NOT an advantage of Pulse Duration Modulation (PDM) recording.
21 docs263 tests

21 docs263 tests
