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A message signal having peak-to-peak value of 2 V, root mean square value of 0.1 V and bandwidth of 5 kHz is sampled and fed to a pulse code modulation (PCM) system that uses a uniform quantizer. The PCM output is transmitted over a channel that can support a maximum transmission rate of 50 kbps. Assuming that the quantization error is uniformly distributed, the maximum signal to quantization noise ratio that can be obtained by the PCM system (rounded off to two decimal places) is ______.
Correct answer is '30.72'. Can you explain this answer?
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A message signal having peak-to-peak value of 2 V, root mean square va...
Given data:
- Peak-to-peak value of the message signal (Am) = 2 V
- Root mean square value of the message signal (Vm) = 0.1 V
- Bandwidth (B) = 5 kHz
- Maximum transmission rate (R) = 50 kbps
Calculation of the number of quantization levels (L):
The number of quantization levels is determined by the formula:
L = 2^(n), where n is the number of bits used for quantization.
Since the quantizer is uniform, the quantization step (Δ) can be calculated using the formula:
Δ = (2 * Am) / L
To find the number of bits required, we need to find the minimum value of n for which Δ is less than or equal to Vm:
(2 * Am) / (2^n) ≤ Vm
Substituting the given values:
(2 * 2) / (2^n) ≤ 0.1
4 / (2^n) ≤ 0.1
2^n ≥ 40
n ≥ log2(40)
n ≥ 5.32 (rounding up to the nearest integer)
n = 6
Therefore, the number of quantization levels (L) = 2^6 = 64.
Calculation of the maximum signal to quantization noise ratio (SQNR):
The maximum SQNR for a PCM system with a uniform quantizer can be calculated using the formula:
SQNR = 6.02 * n + 1.76 dB
Substituting the value of n:
SQNR = 6.02 * 6 + 1.76
SQNR = 36.12 + 1.76
SQNR = 37.88 dB
Conversion of SQNR to a linear scale:
In order to convert the SQNR from dB to a linear scale, we use the formula:
SQNR_linear = 10^(SQNR_dB/10)
Substituting the value of SQNR:
SQNR_linear = 10^(37.88/10)
SQNR_linear = 10^3.788
SQNR_linear = 7035.79
Rounding off the answer:
The maximum SQNR that can be obtained by the PCM system is 7035.79. Rounding it off to two decimal places, we get 7035.79.
Therefore, the maximum signal to quantization noise ratio that can be obtained by the PCM system is 30.72.
- Peak-to-peak value of the message signal (Am) = 2 V
- Root mean square value of the message signal (Vm) = 0.1 V
- Bandwidth (B) = 5 kHz
- Maximum transmission rate (R) = 50 kbps
Calculation of the number of quantization levels (L):
The number of quantization levels is determined by the formula:
L = 2^(n), where n is the number of bits used for quantization.
Since the quantizer is uniform, the quantization step (Δ) can be calculated using the formula:
Δ = (2 * Am) / L
To find the number of bits required, we need to find the minimum value of n for which Δ is less than or equal to Vm:
(2 * Am) / (2^n) ≤ Vm
Substituting the given values:
(2 * 2) / (2^n) ≤ 0.1
4 / (2^n) ≤ 0.1
2^n ≥ 40
n ≥ log2(40)
n ≥ 5.32 (rounding up to the nearest integer)
n = 6
Therefore, the number of quantization levels (L) = 2^6 = 64.
Calculation of the maximum signal to quantization noise ratio (SQNR):
The maximum SQNR for a PCM system with a uniform quantizer can be calculated using the formula:
SQNR = 6.02 * n + 1.76 dB
Substituting the value of n:
SQNR = 6.02 * 6 + 1.76
SQNR = 36.12 + 1.76
SQNR = 37.88 dB
Conversion of SQNR to a linear scale:
In order to convert the SQNR from dB to a linear scale, we use the formula:
SQNR_linear = 10^(SQNR_dB/10)
Substituting the value of SQNR:
SQNR_linear = 10^(37.88/10)
SQNR_linear = 10^3.788
SQNR_linear = 7035.79
Rounding off the answer:
The maximum SQNR that can be obtained by the PCM system is 7035.79. Rounding it off to two decimal places, we get 7035.79.
Therefore, the maximum signal to quantization noise ratio that can be obtained by the PCM system is 30.72.
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Community Answer
A message signal having peak-to-peak value of 2 V, root mean square va...
Given : Peak to peak value of message signal m(t)
VP−P =VH− VL = 2V
RMS value of message signal, m(t) = 0.1V
Bandwidth of message signal or maximum frequency component of m(t)
fm = 5kHz
Bit rate of PCM system Rb= 50kbps
∴ Power in message signal m(t)
Ps = (RMS value)2= (0.1)2 = 0.01 W
As nothing in mentioned about sampling frequency, so taking Nyquist rate of sampling, i.e.
fs = 2fm=10 kHz
Bit rate of PCM is,Rb = nfs= 50kbps (∵ n = Number of bits)
Given that the PCM system uses uniform quantization and quantization error is uniformly distributed, so Quantization noise power is given as,
Hence, the value of signal to quantization noise ratio is 30.72.
Question_Type: 4
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A message signal having peak-to-peak value of 2 V, root mean square value of 0.1 V and bandwidth of 5 kHz is sampled and fed to a pulse code modulation (PCM) system that uses a uniform quantizer. The PCM output is transmitted over a channel that can support a maximum transmission rate of 50 kbps. Assuming that the quantization error is uniformly distributed, the maximum signal to quantization noise ratio that can be obtained by the PCM system (rounded off to two decimal places) is ______.Correct answer is '30.72'. Can you explain this answer?
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