COSC3213W04 Exercise Set 1 - Solutions

1. For each of the following sources of information, identify whether they are analog or digital and explain why:

a) Human voice b) Light switch c) Computer keyboard d) Steering wheel

Answer: a) The human voice produces a sound without discrete frequencies – it can produce

just about any frequency in a given range and at a continuous range of amplitudes so it must be an analog signal.

b) A light switch typically operates in either an “on” or “off” position which means it has discrete values (there’s no “in-between”). Thus a light switch is digital. However, if we are discussing a light dimmer, that allows a continuous variation of the light’s intensity, then that light dimmer would be analog.

c) Each key on a keyboard is distinct (separate) and has a finite set of values – each key may either “up” or “down”. Thus a computer keyboard is digital.

d) A steering wheel may steer to the left or right but to varying degrees. Since there is a continuous set values between “left”, “right” and “forward” a steering wheel is analog.

2.

Given the frequency-domain graph above, answer the following:

a) What is the frequency spectrum? b) What is the bandwidth? c) Is this an analog or digital signal? Why?

Answer: a) The frequency spectrum is the set of frequencies that comprises the signal. In this

case: [3, 6, 12, 14, 16].

b) The bandwidth of the signal is the range of frequencies, that is, the difference between the highest and lowest frequency. The highest frequency is 16, the lowest is 3, so the answer is: 16 – 3 = 13 Hz.

c) As discussed in class, a digital signal is represented as a square wave which requires an infinite number of frequencies to represent. Since there are a finite number of frequencies in this signal, is must be analog.

3. “CAT-5” twisted-pair cable has a bandwidth of roughly 100MHz. We would like to transmit information at a bit rate of 500Mbps. Is a signal-to-noise ratio of 30dB enough to reliably transmit this much information? Why or why not?

Answer: This question requires a simple application of the Shannon Theorem:

)1(log2 SNRBC += In this case we are given C = 500Mbps, B = 100MHz, and SNR = 30dB. In order to answer the question of whether or not the SNR is sufficient, we need to see if the capacity provided by the bandwidth and SNR is >= 500Mbps. As a first step, we need to convert the SNR from decibels to the “unitless” form:

SNRSNRdB 10log10= so

100010

1010/30

10/

==

= dBSNRSNR

We can now apply the Shannon Theorem:

MbpsMHzMbpsMHz

MHzSNRBC

7.996)/1(*967.9*100

)10001(log100)1(log

2

2

=≈

+=+=

Note the use of “Mbps/MHz” to underscore the fact that we are explicitly converting from “Hz” to “bps”. Thus the Shannon Theorem tells us that the theoretical limit for transferring data over this medium at this SNR is approximately 1000Mbps (or 1Gbps). Thus we can reliably transmit 500Mbps over this connection. 4. What is the maximum capacity of a medium with a bandwidth of 750KHz and a signal-to-noise ratio of 30dB?

Answer: This is a straight-forward application of the Shannon Theorem, except that we must first convert the SNR from decibels to its “unitless” form:

SNRSNRdB 10log10= so

100010

1010/30

10/

==

= dBSNRSNR

So, applying the Shannon Theorem:

MbpsKbps

KbpsKHzKbpsKHz

SNRBC

475.725.7475

967.9*750)/1(*)10001(log750

)1(log

2

2

≈≈=

+=+=

So the maximum capacity of a medium with a bandwidth of 750KHz and a signal-to-noise ratio of 30dB is 7.475Mbps. 5. What is the minimum signal-to-noise ratio, in decibels, that must be maintained in order to transmit a 600Kbps signal over a medium with a bandwidth of 20,000Hz?

Answer: Let’s work around the Shannon Theorem to solve for the SNR:

1212

)1(log

)1(log

/

/

2

2

−=

+=

+=

+=

BC

BC

SNRSNR

SNRBC

SNRBC

We can now solve this for the given capacity and bandwidth. Note that our capacity is in Kbps, but our bandwidth is in Hz. Since 20,000Hz = 20Khz, we’ll do the calculations in Kbps/Khz:

107374182312

1230

)20/600(

=−=

−=

SNRSNRSNR

The question however, asks for the SNR in decibels, so we need to convert:

31.901073741823log10

log10

10

10

≈==

dB

dB

dB

SNRSNR

SNRSNR

So we would need at least 90.31dB as a signal-to-noise ratio in order to transmit the desire capacity.

6. We are given a medium that will reliably transmit frequencies between 0 and 25,000Hz. Is it possible to transmit 200Kbps of information along this line? If so, then describe a method and any conditions that must be satisfied. If not, explain why.

Answer: We could apply the Shannon Theorem here but that would just give us a limit on the capacity that could be transmitted by the medium. It would not describe how to do so. If we apply the Nyquist Theorem for two-level signals:

KbpsCC

BC

50)025000(*2

2

=−=

=

we see that with a two-level digital signal, we can only transmit at most 50Kbps. We can apply the reformulated Nyquist Theorem to examine a possible multi-level signal:

162

4log50/200log

log)25(2200log2

42

2

2

2

==

===

=

M

MM

MKHzKbpsMBC

So we can transmit 200Kbps of information along this medium provided we use use a digital signal with at least 16 levels. However this does place some restrictions – the Nyquist Theorem assumes that there is no noise and that transmission is error-free. 7. A triangle wave has the following shape:

This wave can be generally represented by the Fourier series

)2

)12(2sin()12(

11

02

ππ +++∑

∞

=

tfkkk

where f1 is the base frequency of the wave.

a) What is the bandwidth of this signal?

b) If the signal can be closely approximated with the first 3 terms of the series, what is the effective bandwidth of this signal?

c) Sketch the frequency-domain graph of the approximated signal.

Answer:

a) The bandwidth of the signal is the range of frequencies, that is, the difference between the highest and lowest frequency. The lowest frequency is the first term of the series where k = 0:

)2

)1)0(2(2sin()1)0(2(

112

ππ +++

tf

which reduces to:

)2

2sin( 1ππ +tf

The frequency of this term (based upon our standard form of )2sin( φπ +ftA ) is clearly f1. The highest frequency occurs with the highest possible term which is when : ∞=k

)2

)1)(2(2sin()1)(2(

112

ππ ++∞+∞

tf

Note that although the amplitude of this component is infinitesimally small, it is still a component of the signal. The frequency is infinity (∞ ). The bandwidth is thus the difference between the highest and lowest frequencies:

, which is to say, the bandwidth is infinite. ∞=−∞ 1fb) The first three terms would look like:

)2

)1)2(2(2sin()1)2(2(

1

)2

)1)1(2(2sin()1)1(2(

1

)2

)1)0(2(2sin()1)0(2(

1

12

12

12

ππ

ππ

ππ

+++

++++

++++

tf

tf

tf

which simplifies to:

)2

10sin(251)

26sin(

91)

22sin( 111

ππππππ +++++ tftftf

The highest frequency in this approximated signal is the 3rd term, which has a frequency of 5f1. The lowest frequency is in the 1st term which as a frequency of f1. Thus the bandwidth is 5f1 – f1 = 4f1 Hz.

c) The frequency-domain graph would consist of the following amplitude/frequency combinations: (1,1), (1/9,3), (1/25,5)

1 3 5

1/9

1/25

1

8. The Fourier series for a particular signal is represented by the following summation:

1k

sin(2πk −1

kk=1

∞

∑ f0t)

where f0 is a fixed frequency.

a) What is the bandwidth of this signal? b) Assume that we approximate this signal with the first three terms of this series.

What is the bandwidth of this approximation? c) Sketch the frequency-domain graph of the approximated signal.

Answer:

a) The frequency for the k’th term is 01 f

kk − . Look at the values for this frequency for

the first few terms:

00

00

00

0

43

4144

32

3133

21

2122

01

111

ffk

ffk

ffk

fk

=−

→=

=−

→=

=−

→=

=−

→=

Note that although the frequency does increase as k increases, it does not increase

without bound. In fact: 11lim =−

∞→ kk

k. So the lowest frequency component is the first

term (0) and the highest frequency is the “last” term ( ). So the bandwidth of this signal is just .

0f

0f b) We laid out the first three terms in the answer to part (a). The highest frequency

among the first 3 terms is 032 f and the lowest frequency is 0. So the bandwidth of

the approximation is 032 f .

c) The amplitude/frequency pairs will be: (1,0), (1/2,1/2*f0), (1/3,2/3*f0)

1/2f0 2/3*f0 f0

1/2

1/3

1

0

9. For each of the following sine waves identify the amplitude, frequency and phase-shift:

a) )2

52sin(3 ππ +t

b) )4sin( π+t c) )sin(t

Answer: For all of these answers, we need to consider our standard formula for a sine wave:

)2sin( φπ +ftA a) This is the same format as our standard formula, so A = 3, f = 5 and 2/πφ = . b) There is no value in front of “sin” so the amplitude must be 1. Remember that the

frequency component in front of t must be fπ2 so the frequency is ππ /22/4 = . The phase shift in this case is π .

c) The amplitude is 1. The frequency is π2/1 and the phase shift is 0.