Communications Test 3

g(t) = 10 cos (50πt) cos² (150πt)

Using trigonometric identity, the above expression can be written as:

\( = 10\cos \left( {50\pi t} \right)\left[ {\frac{{1 + \cos \left( {2 \times 150\pi t} \right)}}{2}} \right]\)

\( = \frac{{10}}{2}\cos \left( {50\pi t} \right)\left[ {1 + \cos \left( {300\pi t} \right)} \right]\)

= 5 cos (50πt) + 5 cos (50πt) cos (300πt)

\( = 5\cos \left( {50\pi t} \right) + \frac{5}{2}\left[ {\cos \left( {250\pi t} \right) + \cos \left( {350\pi t} \right)} \right]\)

So, the message signal is band-limited to:

\({f_m} = \frac{{350\pi }}{{2\pi }} = 175Hz\)

Therefore, the Nyquist sampling rate for the signal will be:

f_N = 2f_m

= 2 × 175 = 350 Hz

Concept:

The transmission rate of a PCM signal is given by:

R_b = nf_s

n = number of bits to encode a sample

f_s = Sampling frequency.

Analysis:

Given the transmission bandwidth capacity of the binary channel is:

B_ch = 36 k bits / sec

The bandwidth of the message signal is f_m = 3.2 kHz.

Sampling frequency must be at least:

f_s ≥ 2 f_m

f_s ≥ 2 × 3.2 kHz

f_s ≥ 6.4 kHz

Let the quantization level be q, we can write:

q = 2ⁿ

n = number of bits required to encode the sample.

Since the channel must support the data rate of the PCM signal, we can write:

R_b ≤ R_ch

n_fs ≤ 36        ---(1)

From the given options, the sampling frequency will be 7.2 kHz, i.e. we can write:

7.2k × n ≤ 36k

n ≤ 5

2ⁿ ≤ 2⁵

q < 32

∴ The appropriate values of the quantizing level and sampling frequencies are q = 32 and f_s = 7.2 kHz.

Concept:

The number of encoded bits is related to the quantization levels by:

q = 2ⁿ

Nyquist rate is the minimum sampling frequency required for a faithful reconstruction and is given by:

f_s = 2f_m

f_m = maximum frequency component present is the message signal.

Calculation:

Given f_m = 15800 Hz.

Nyquist frequency will be:

f_N = 2 × 2 15800 Hz

= 3.16 × 10⁴ Hz

Since the given signal is sampled at a rate 10% higher than the Nyquist rate, the given sampling frequency will be:

f_s = f_N + 0.1 f_N

f_s = 1.1 f_N

f_s = 1.1 × 3.16 × 10⁴ Hz

f_s = 3.476 × 10⁴ Hz

∴ 3.476 × 10⁴ samples/sec are generated.

To store a 3 minutes (3 × 60 sec) performance, the number of samples will be:

Number of samples = f_s × Time duration

= 3.476 × 10⁴ × 3 × 60 samples

= 6.25 × 10⁶ samples

Now, for 128 quantized levels, the required number of bits is obtained as:

2ⁿ = 128

n = 7 bits

Thus, 7 bits per sample are used to quantify the signal. So, the number of bits required to store a time minutes performance will be:

Number of bits = (Number of samples) × (Number of bits per sample)

= 6.25 × 10⁶ × 7 bits

= 4.375 × 10⁷ bits

Concept:

Quantization noise power of a Quantizer is given by

\(Q = \frac{{{\Delta ^2}}}{{12}}\)              

∆ = step size of the Quantizer

\(\Delta = \frac{{{V_{P - P}}}}{{{2^n}}}\)            

V_p-p is the peak to peak voltage

Calculation:

Given, for Quantizer Q_1­,

n = 5 bit,

V_p-p = 2 V

So, \({\Delta _1} = \frac{2}{{{2^5}}} = \frac{1}{{{2^4}}}\)

\(\therefore {Q_1} = \frac{{\Delta _1^2}}{{12}} = {\left( {\frac{1}{{{2^4}}}} \right)^2} \times \frac{1}{{12}}\)

For Quantizer Q₂,

n = 8 bit,

V_p-p = 1V

So, \({\Delta _2} = \frac{1}{{{2^8}}}\)

\(\therefore {Q_2} = \frac{{\Delta _2^2}}{{12}} = {\left( {\frac{1}{{{2^8}}}} \right)^2} \times \frac{1}{{12}}\)

\(\frac{{{Q_1}}}{{{Q_2}}} = \frac{1}{{{2^8} \times 12}} \times {2^{16}} \times 12 \)

\(\frac{Q_1}{Q_2}= {2^8} = 256\)

2 sin² (100 πt) + 3 cos² (100 πt)

⇒ 2 (sin² (100 πt) + cos² 100 πt) + cos² 100 πt

⇒ 2 + cos² 100 πt

V_max = 2 + 1 = 3

 V_mm = 2 + 0 = 0

\(\begin{array}{l}{\rm{\Delta }} = \frac{{{V_{max}} - {V_{min}}}}{L}\\\Rightarrow 0.5 = \frac{{3 - 2}}{L}\end{array}\)

L = 2 = 2ⁿ

n = 1

Bit rate = f_s × n

= 2 (2 f_m) × 1

= 4f_m

2 + cos²100 πt

\(\Rightarrow 2 + \frac{{(1 + \cos 200\;\pi t)}}{2}\)

\(= \frac{5}{2} + \frac{1}{2}\cos \left( {200\;\pi t} \right)\)

f_max = 100 Hz

Bit rate = 4 × 100