IT 420 Homework 2

IT 420

Spring 2015

Homework 2 Solutions

Note:  Where applicable, units of the solution must be correct for full credit.

1. Consider the following composite signal.

f(t) = 4sin(2π2t) + 3sin(2π3t) + 2sin(2π4t) + sin(2π5t)

1. Graph this in the frequency domain, showing amplitude and frequency

1. What is the bandwidth for this signal?

BW = high frequency – low frequency

       = 5 – 2

       = 3 Hz

2. A medium has a bandwidth of 50 KHz and a lower frequency of 90 KHz. What is the upper frequency? 

BW = high frequency – low frequency

High frequency =  BW + low frequency

                          = 50 KHz + 90 KHz

                          = 140 KHz

3. Answer the following questions concerning digital signals:

• A signal has 50 possible levels.  This signal is

1. analog   digital         c.  neither        d.  cannot be determined

• What is a possible unit for a digital bit time?

1. Hz        bits per second       c.  levels per second    d.  seconds

•  _____________ encoding has a transition at the middle of each bit

a.Manchester          b. Differential Manchester     c.  both of these   d.  neither of these

• Sampling is measuring the amplitude of an analog signal _____________

1. when it changes b.  at random intervals            c.  at regular intervals

• Taking too few samples while sampling may lead to

1. too few bits in the quantization
2. using too much bandwidth for PCM
3. c. incorrectly reproducing the data

• Nyquist’s theorem for channel capacity assumes a perfect channel.

1. True   False

4. A digital signal has a bit rate of 2000 bps. What is the time for one bit?

Bit interval = 1 / bit rate

                   = 1 / 2000 bps

                   =  0.0005 sec = 0.5msec

5. A device is sending out data at the rate of 10 Kbps.

1. How long does it take to send out one character (assume one character is represented by 8 bits)?

Time = 8 bits  *   1 sec                        = 0.0008 sec = 0.8 msec

                            10 x10³ bits

1. How long does it take to send a file of 100,000 characters?

Time = 0.0008 sec * 100,000 = 80 sec

6. A device is sending out data at the baud rate of 56000 levels/sec and each level represents 8 bits. What is the data rate in bits/sec?

Data rate = 56000 levels  *   8 bits

                        sec                 level

                 =  448,000 bps

7. Solve the following logarithms:
8. log₂(32) 5
9. log₂(1) 0
10. log₂(256) 8

8. A digital signal has 128 levels. How many bits can be represented by each level?

bits per level = log₂(number of levels)

                       = log₂(128)

                       =7 bits per level

9. Encode the data stream 001110101 using Manchester and Differential Manchester line coding. State any required assumptions (such as the value of the previous bit, if necessary) in your solution.  Please be neat and show bit times.

Manchester (assume starts high)

           

Differential Manchester (assume starts high)

          

10. Calculate the sampling rate according to the Nyquist Theorem for a signal that contains frequencies from 100 KHz to 400 KHz

            Minimum sampling rate = 2 * highest frequency

                                                   = 2 * 400 KHz

                                                   = 800 K samples/sec

11. An analog data signal contains frequencies from 0 Hz to 5000 Hz. We wish to convert this to a digital signal using Pulse Code Modulation encoding. Assume that there will be 256 possible quantized values.

1. What is the minimum sampling rate according to the Nyquist Theorem?

Minimum sampling rate = 2 * highest frequency

                                        = 2 * 5000 Hz

                                        = 10000 samples/sec

1. How many bits are required to represent each value?

log₂(256) =  8, so 8 bits are required.

1. Assuming the bit coding and sampling rate calculated above, what is the bit rate?

            Bit rate = sampling rate * bits/sample

                         = 10000 samples/sec * 8 bits/sample

                         = 80,000 bps

12. What is the maximum channel capacity (in bps) of a perfect channel which accepts frequencies between 1000 Hz and 5000Hz and transmits a signal with 16 signal levels?

            Capacity = 2 * BW * log ₂ L

                            =2 * (5000 – 1000) * log ₂ (16)

                            =2 * (4000) * log ₂ (16)

                            = 2 * 4000 * 4

                            = 32,000 bps

13. You need to send 280 Kbps over a perfect channel with an analog bandwidth of 20 KHz. How many signal levels do you need?

D = 2B log₂ K

            Log₂ K = D/ 2B

            Log₂ K = 280,000/(2 * 20,000)

            Log₂ K = 7

            K = 2⁷ = 128 levels

14. Assume that a channel accepts frequencies between 0 Hz and 10000 Hz and has a signal to noise ratio measured at 15 (not in dB). What is the maximum channel capacity of the channel?

Capacity = BW * log ₂ (1 + SNR)

               = 10000 * log ₂ (1 + 15)

               = 10000* log ₂ (16)

               = 10000 * 4

               = 40000 bps

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