# Consider an audio signal with spectral components limited to the frequency band of 300 to 3300 Hz.

By   November 29, 2019

EXAMPLE 4.21. Consider an audio signal with spectral components limited to the frequency band of 300 to 3300 Hz. A PCM signal is generated with a sampling rate of 8000 samples/s. The required output signal-to-quantizing-noise ratio is 30 dB.
(i)         What is the minimum number of uniform quantizing levels needed, and what is the minimum number of bits per sample needed?
(ii)        Calculate the minimum system bandwidth required.
(iii)       Repeat parts (i) and (ii) when a -law compander is used with  = 255.                                                             (Karnataka University-1997)
Solution: (i) Here, we have
= 1.76 + 20 log q  ≥ 30
log q ≥  (30-1.76) = 1.412
or                                             q ≥ 25.82
Thus, the minimum number of uniform quatizing levels required is 26.
v = [log2 q] = [log2 26] = [4.7]  5 bits per sample
The minimum number of bits per sample is 5.
(ii) The minimum required system bandwidth will be
fPCM =
fPCM = 20,000 Hz = 20 kHz
we have                                    = 20 log q – 10.1 ≥ 30
log q ≥  (30 + 10.1) = 20.005
or                                             q ≥ 101.2
Thus, the minimum number of quantizing levels needed is 102.
Also,                                        v = [ log2 q] = [6.67]  7
The minimum number of bits per sample is 7.
The minimum bandwidth required for this case will be
fpcm = fs =  (8000) = 28000 Hz = 28 kHz    Ans.
EXAMPLE 4.22. Consider a sinusoidal signal m(t) = A cos  applied to a delta modulator with step size D. Show that the slope overload distortion will occur if
A >  =
where fs = 1/Ts  is the sampling frequency.
Solution: We have
so that                                     m(t) = A cos
EQUATION
To avoid the slope overload, we require that
EQUATION
or                                                         A
Thus, if A >D/(), slope overload distortion will occur. Hence Proved.
EXAMPLE 4.23. For a sinusoidal modulating signal
m(t) = A cos mt
m = 2fm
Prove that the maximum output signal-to-quantizing-noise ration in a DM system under the assumption of no slope overload is given by
EQUATION
where fs = 1/Ts is the sampling rate and fM is the cutoff frequency of a low-pass filter at the output end of the recover.
Solution: For no-slope-overload, we must have
A
Thus, the maximum permissible value of the output signal power equals
EQUATION
We know that the mean quantizing error, or the quantizing noise power,  = D2/3. Let the bandwidth of a postreconstruction low-pass filter at the output end of the reciever be fM ≥ fm and fM << fs. Then, assuming that the quantizing noise power Pq is uniformly distributed over the frequency band up to fs, the output quantizing noise power within the bandwidth fM is
…(iii)
Combining equation (ii) and (iii), we see that the maximum output signal-to-quantizing-noise ratio is
Equation                                              Hence Proved.
EXAMPLE 4.24. A DM system is designed to operate at 3 times the Nyquist rate for a signal with a 3 kHz bandwidth. The quantizing step size is 250 mV.
(i)         Determine the maximum amplitude of a 1-kHz input siinusoid for which the delta modulator does not show slope overload.
(ii)        Determine the posfiltered output signal-to-quantizing-noise ratio for the signal of part (i)
Solution: We have
m(t) = A cos t = A cos  (103)t
The maximum allowable amplitude of the input sinusoid is
Amax =
fs =  = 71.2 mV                        Ans.
(ii) Assuming that the cutoff frequency of the low-pass filter is fm, we have
EQUATION
EXAMPLE 4.25. The pulse rate in a DM system is 56,000 per sec. The input signal is 5 cos (2r 1000t) + 2 cos (2 2000 t) V, with t in sec. Find the minimum value of step size which will avoid slope overload distortion. What would be the disadvantages of choosing a value of larger than the minimum?
(GATE Examination-1998)
Solution: Input signal,
m(t) = 5 cos (2000 t) + 2 cos (4000 t) = m1 (t) + m2(t)
To avoid slope overloading, we have
EQUATION
where D is step size and fs is sampling rate.
or                                                          EQUITATION
D1max (minimum value of step size) =              EQUITATION
Also,                                        EQUITATION
or                                             EQUITATION
Hence, larger step size out of two will be the required step size. i.e., = 0.56 V.
If a value larger than the minimum will be choosen, then granular noise will occur.
EXAMPLE 4.26. Bandwidth of the input to pulse code modulator is restricted to 4 kHz. The input varies from – 3.8 V to 3.8 V and has the average power of 30 mW, the required signal to quantization noise power ratio is 20 dB. The modulator produces binary output. Assume uniform quantization Calculate the number of bits required per sample.
Solution: Given that
or                                 10log
or
Quantizer step size,
where L = 2n, n is the number of bianry digits.
then, average quantizing power is,
Nq =
then,
or                                             100 =
or                                             L =
or                                             2n = 128
Hence, n = 7 = number of bits required per sample.
EXAMPLE 4.27. A low pass signal of 3 kHz bandwidth and amplitude over – 5 Volts to + 5 Volts range is sampled at Nyquist rate and converted to 8-bit PCM using uniform quantization. The mean squared value of message signal is 2 Volt-squared. Determine the following :
(i)         The normalized power for quantization noise.
(ii)        The bit transmission rate.
(iii)       The signal to quantization noise ratio in dB.
(iv)       Derive the expressions used in (i) and (iii).
Solution : Given that
fm= 3 kHz, v = 8
It is given that uniform quantization is used.
Also,                             = Mean square value of message signal is 2 Volt2
(i)         Normalized power for quantization noise (Nq) is given by
Nq where A = Step size
But                                          D =
Therefore,                                D =  = 39.06 mV
Substituting above value of D in equation (i), we obtain
Nq =   = 127.15 x 10-6 W             Ans.
Now, let us calculate the bit transmission rate (r).
The bit transmission rate or signaling rate is the number of bits transmitted by the PCM system per second.
Therefore,                                            r = vfs
As the signal is sampled at Nyquist rate, fs = 2fm.
r = 8 x 2 fm
we have
r = 16 x 3 kHz = 4 K bits/sec. Ans.
(iii)       The signal to quantization noise ratio in dB may be calculated as under:
The normalized signal power P =
P =                             ….(iii)
Therefore,                                (SNR)q =
and                                          (SNR)q in dB = 10 log10 (15728.64)
or                                             (SNR)q = 41.96 dB.                           Ans.
EXAMPLE 4.28. For a full scale sinusoidal modulating signal with peak value A, show that, output signal to quantization noise ratio in binary PCM system is given by,
log M dB
where
A compact disc recording system samples each of the two stereo signals with a 16 bit A/D converter at 44.1 Kb/sec.
(i)         Determine output S/N ratio for a full scale sinusoid.
(ii)        The bit stream of digitized data is augmented by addition of error correcting bits, clock extraction bits etc. and these additional bits represent 100% overhead. Determine output bit rate of CD system.
(iii)       The CD can record an hour’s worth of music. Determine number of bits recorded on CD.
Solution:         There are two stereo channels.
v = 16, fs = 44.1 Kbits/sec.
(i) Output signal to noise ratio for full scale sinusoid is given by
(ii)        Now, let us evaluate the output bit rate of the CD system.
The bit rate for each of two stereo channels  = vfs
Therefore, the bit rate of two channels = 2 vfs
= 2 x 16 x 44.1 x 103 = 1.4112 Mbits/sec.
Including the additional 100% overhead, the output, the output bit rate will be
2 x 1.4112 x 106 b/s = 2.822 Mbits/sec.                                 Ans.
(iii)       Next, we calculate the number of recorded on CD.
The CD can record an hour’s worth of music.
Therefore, the number of bits recorded on CD = bit rate x Number of second/hour
= 2.822 x 106 x 3600 = 10.16 x 109 bits or 10.16 gigabits      Ans.
EXAMPLE 4.29. Determine the output SNR in a DM system for 1 kHz sinusoid, sampled at 32 kHz without slope overload and followed by a 4 kHz post construction filter. Derive the formula used.
Solution : Given that, fm = 1 kHz, fs = 32 kHz, BW = 4 kHz
It is given that there is no slope overload.
The output signal to noise ratio in a DM system is expressed as
(SNR)0 =
Therefore,                                        (SNR)0 =
or                                                     (SNR)0 = 311.25 or 24.93 dB.                Ans.
EXAMPLE 4.30. In a DM system, the voice signal is sampled at a rate of 64,000 samples/sec. The maximum signal amplitude Amax = 1.
(i)         Determine minimum value of step size to avoid slope overload.
(ii)        Determine quantization noise power if voice signal bandwidth is 3.5 kHz.
(iii)       Assuming voice signal to be a sine wave, determine S0 and the SNR.
Solution : Given that fs = 64,000 samples/sec.
Amax = 1.
(i)         First, we determine the minimum value of step size (Dmin).
The condition to avoid slope overload error is expressed as

or                                                         Amax =
Therefore,                                            Dmin =
Since the input to DM system is a voice signal, therefore, let us assume fm= 3.5 kHz.
Hence,                                                 Dmin =         Ans.
(ii)        Quantization noise power Nq‘ can be calculated as under :
The quantization noise power Nq for DM system is given by,
Nq =
However, as the voice bandwidth is only 3.5 kHz, the normalized quantization noise power at the receiver output is given by,
N’q = Nq x
or                                             N’q = 2.15 mW
(iii)       Signal power S0 and SNR can be calculated as under :
Output signal power =                                                                    Ans.
Output signal to noise ratio  =
or                                             (SNR)0 = 232.3           or  23.66 dB.   Ans.
EXAMPLE 4.31. The bandwidth of TV video plus audio signal is 4.5 MHz. If this signal is converted into PCM bit stream with 1024 quantization levels, determine number of bits/sec of the resulting signal. Assume that the signal is sampled at the rate 20% above Nyquist rate.
Solution : Given that fm = 4.5 MHz,             q = 1024
fm = 20% above the Nyquist rate = 1.2 x 2fm = 1.2 x 2 x 4.5 MHz.
or                     fs = 10.8 MHz.
Let us calculate the number of bits/sec.
r = v fs.
But, we do not know the value of v.
We know that                         q = 2v
2v = 1024
v = 10
Therefore,                    bits/sec = 10 x 10.8 MHz = 108 M bits/sec.               Ans.
EXAMPLE 4.32. If a voice frequency signal is sampled at the rate of 32,000 samples/sec and characterized by peak value of 2 Volts, determine the value of step size to avoid slope overload. What is quantization noise power Nq and corresponding SNR? Assume bandwidth of signal as 4 kHz.
Solution : Given that              fs = 32,000 samples/sec.
Peak value of the signal A = 2 V.
Bandwidth BW = 4 kHz.
(i)         Step size A to avoid slope overload can be calculated as under :
To avoid slope overload the following condition must be satisfied :
A £
Substituting the values, we obtain
2 £
or
or                                                         D ³ 1.57 Volt                          Ans.
(ii) Next, we find the quantization noise power (Nq).
The quantization noise power for a delta modulator is given by
Nq =  W
(iii)       We know that the signal to noise ratio is given by
EQUATION
EXAMPLE 4.33. A compact disc (CD) records audio signals digitally by PCM. Assume audio signal’s bandwidth to be 15 kHz. If signals are sampled at a rate 20% above Nyquist rate for practical reasons and the samples are quantized into 65,536 levels. Determine bits/sec required to encode the signal and minimum bandwidth required to transmit encoded signal.
Solution :                    Given that fm = 15 kHz           fs = 1.2 x 2 fm
= 2.4 x 1.5 kHz = 36 kHz
q = 65,536.
Signaling rate (r) can be calculated as under :
We know that                         q = 2v
v = log2 q
or                                             v =
Now, signaling rate r = vfs = 16 x 36 kHz = 576 Kbits/sec.    Ans.
Hence, the signaling rate r is 576 Kbits/sec.
(ii)        Minimum bandwidth can be calculated as under :
BW =  (signaling rate) =  Kbits/sec.
Therefore, minimum bandwidth, BWmin = 288 kHz.              Ans.
EXAMPLE 4.34. In a single integration DM scheme, the voice signal is sampled at a rate of 64 kHz. The maximum signal amplitude is 1 Volt.
(i)         Determine the minimum value of step size to avoid slope overload.
(ii)        Determine granular noise power N0, if the voice signal bandwidth is 3.5 kHz.
(iii)       Assuming signal to be sinusoidal, calculate signal power S0 and signal to noise ratio (SNR).
(iv)       Assuming that the voice signal amplitude is uniformly distributed in the range (- 1, 1), determine So and SNR.
Solution : Given that              fs = 64 kHz A= 1 Volt
(i)         Minimum step size to avoid slope overload is given by
A £
or                                                         Dmin =
or                                                         Dmin = 0.3436 Volt.                 Ans.
(ii)        Granular noise power is expressed as
Nq =
Solving, we get
Nq = 2.15 x 10-3 W                                          Ans.