ECET310 W5 Assignments HW 5 1 Instructions

Homework 5_1 DIGITAL TRANSMISSION

1.

Determine the Nyquist sample rate for a maximum analog input frequency of: (a) (b)

4 KHz 10 KHz

Fs ≥ 2fa a) Fs = 2(4kHz) = 8 kHz b) Fs = 2(10kHz) = 20 kHz

2.

Determine the dynamic range for a 10-bit sign-magnitude PCM code.

9 bits + 1 sign bit DR = 2n – 1 = 29 – 1 = 511

3.

Determine the minimum number of bits required in a PCM code for a dynamic range of 80 dB. What is the coding efficiency?

DR(db) = 20 log (DR) – 80 = inv log (80/20) = 10000 n = log(DR + 1) / log(2) = log(10000+1)/log(2) = 13.29 bits ≈ 14 bits Max number of bits required = 14 bits + 1 sign bit = 15 bits Coding eff. = (13.29/15) x 100 = 88.6 % 4.

For a resolution of 0.04 V, determine the voltages for the following linear sevenbit sign magnitude PCM codes.

(a)

1 0 0 1 0 1 1 = 11 x 0.04V = 0.44V

(b)

0 1 0 1 1 0 1 = -45 x 0.04V = -1.8V

1

5.

Determine the resolution and quantization error for an eight-bit linear sign magnitude PCM code for a maximum decoded voltage of 1.27 V.

Resolution = Vmax / (2n – 1) = 1.27V/(27 – 1) = 1.27/127 = 0.01 V/step Qe = resolution/2 = 0.01 / 2 = 0.01/2 = 0.005V

6. For a 12-bit linear PCM code with a resolution of 0.02 V, determine the voltage range that would be converted to the following PCM codes.

(a) (b) (c) (d) (e) (f)

100000000001 000000000000 110000000000 01000 0000000 100100000001 101010101010

Qe = ½ (0.02V) = 0.01V a) b) c) d) e) f)

1 x 0.02V = 0.02V range (0.01V ≈ 0.03V) 0 x 0.02V = 0V range (-0.01V ≈ 0.01V) 1024 x 0.02V = 20.48V range (20.47V ≈ 20.49V) -1024 x 0.02V = -20.48 range (-20.49V ≈ 20.47V) 257 x 0.02V = 5.14 range (5.13V ≈ 5.25V) 682 x 0.02V = 13.64V range (13.63V ≈ 13.65V)

7. A µ-law compression encoder has: Vi(max) =

8 V,

vi(max) =

1V,

µ

=

Determine the output voltage for an input voltage of 2 V. Vo = Vo(max) ln (1 + (µVin)/Vi(max)) / (ln (1 + µ)) = 1 ln (1 + 255(2)/8) / (ln (1 + 255)) = 0.75V 8. Repeat problem 7 for µ = 0. Vo = 1 ln (1 + (0x2)/8) / ln (1) = 1 ln (1) – ln (10 = 1(0) – 0 =0V

2

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