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ECE316F Communication Fall 2007 Systems Problem Set 1 (For week of September 17th) 1. Give the following quantities in polar form. (a) (1 + j ) 3 (b) ( 3+ j 3 )(1 - j ) (c) 2-j (6/ 3) 2+j (6/ 3) (d) j (1 + j )e jπ/6 (e) e jπ/3 -1 1+j 3 2. Use Euler’s relation (e = cos(θ)+ j sin(θ)) to show that: (a) sin(θ) sin(φ)= 1 2 cos(θ - φ) - 1 2 cos(θ + φ) (b) sin(θ + φ) = sin(θ) cos(φ) + cos(θ) sin(φ) 3. Haykin and Moher, Drill Problem 2.1 4. Haykin and Moher, Drill Problem 2.2 5. Haykin and Moher, Drill Problem 2.3 6. Haykin and Moher, Drill Problem 2.9 7. Haykin and Moher, Drill Problem 2.13 8. Haykin and Moher, Problem 2.19 9. Haykin and Moher, Problem 2.22 10. Haykin and Moher, Problem 2.25 1

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Page 1: ECE316F Communication Fall 2007 Systems Problem Set 1exams.skule.ca/exams/ECE316H1_20139_641388081078...ECE316F Communication Fall 2007 Systems Problem Set 7 (For week of November

ECE316F Communication

Fall 2007 Systems

Problem Set 1(For week of September 17th)

1. Give the following quantities in polar form.

(a) (1 + j)3

(b) (√

3 + j3)(1− j)

(c) 2−j(6/√

3)

2+j(6/√

3)

(d) j(1 + j)ejπ/6

(e) ejπ/3−11+j

√3

2. Use Euler’s relation (ejθ = cos(θ) + j sin(θ)) to show that:

(a) sin(θ) sin(φ) = 12cos(θ − φ)− 1

2cos(θ + φ)

(b) sin(θ + φ) = sin(θ) cos(φ) + cos(θ) sin(φ)

3. Haykin and Moher, Drill Problem 2.1

4. Haykin and Moher, Drill Problem 2.2

5. Haykin and Moher, Drill Problem 2.3

6. Haykin and Moher, Drill Problem 2.9

7. Haykin and Moher, Drill Problem 2.13

8. Haykin and Moher, Problem 2.19

9. Haykin and Moher, Problem 2.22

10. Haykin and Moher, Problem 2.25

1

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ECE316F Communication

Fall 2007 Systems

Problem Set 2(For week of September 24th)

1. Suppose that g(t)←→ G(f) is a Fourier Transform pair, and let

h(t) = g(t) + G(t) + g(−t) + G(−t).

Show that F [h(t)] = h(f).

2. Show that F [exp(−πt2)] = exp(−πf 2). [This is a tricky problem, more difficult thantypical exam questions.]

3. Using the result of the previous question, find∫∞−∞ exp(−a(x −m)2)dx, where a > 0

and m are real values.

4. Haykin and Moher, Drill Problem 2.10

5. Haykin and Moher, Drill Problem 2.16

6. Haykin and Moher, Drill Problem 2.17

7. Haykin and Moher, Problem 2.20

8. Haykin and Moher, Problem 2.26

9. Haykin and Moher, Problem 2.27

10. Haykin and Moher, Problem 2.30 [Another tricky problem]

1

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ECE316F Communication

Fall 2007 Systems

Problem Set 3

(For week of October 1st)

1. In Example 2.15 of the textbook we see that if x(t) = g(t) cos(2πfct), Sx(f) =(1/4) [Sg(f − fc) + Sg(f + fc)]. Show that the same is true if x(t) = g(t) sin(2πfct),i.e., again Sx(f) = (1/4) [Sg(f − fc) + Sg(f + fc)]. Did you expect this to be true?

2. Haykin and Moher, Problem 2.33.

3. Haykin and Moher, Problem 2.49 (part (b) - periodic square pulse only). Use theapproach suggested in class/the handout on the PSD of a sinusoid that was posted asopposed to the auto correlation function approach.

4. Haykin and Moher, Problem 8.34 (b)

5. Haykin and Moher, Problem 8.36 (b) (c).

6. Haykin and Moher Problem 8.38.

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ECE316F Communication

Fall 2007 Systems

Problem Set 4(For week of October 8th)

1. Haykin and Moher, Drill Problem 3.1 5. Haykin and Moher, Drill Problem 3.5

2. Haykin and Moher, Drill Problem 3.2 6. Haykin and Moher, Drill Problem 3.6

3. Haykin and Moher, Drill Problem 3.3 7. Haykin and Moher, Problem 3.18

4. Haykin and Moher, Drill Problem 3.4 8. Haykin and Moher, Problem 3.19

9. In this question, a(t), b(t), c(t) and d(t) are lowpass energy signals, bandlimited toB Hz, and each having energy E. Let fc � B be a carrier frequency, and let

x(t) = [a(t) cos(2πBt)+b(t) sin(2πBt)] cos(2πfct)+[c(t) cos(2πBt)+d(t) sin(2πBt)] sin(2πfct).

(a) What is the bandwidth of x(t)?

(b) What is the energy of x(t)? You may assume that a(t) cos(2πBt) is orthogonal tob(t) sin(2πBt) and that c(t) cos(2πBt) is orthogonal to d(t) sin(2πBt). Likewise,you may assume that [a(t) cos(2πBt) + b(t) sin(2πBt)] cos(2πfct) is orthogonal to[c(t) cos(2πBt) + d(t) sin(2πBt)] sin(2πfct).

(c) Sketch the block diagram of a demodulator that is able to recover a(t), b(t), c(t)and d(t) from x(t). Indicate clearly the frequency and phase of all local oscillatorsand the bandwidths of all lowpass filters.

10. Throughout this question, the signal m(t) is lowpass with a maximum frequency ofW = 20 KHz.

(a) It is desired to generate the modulated wave y(t) = Am(t) cos(2πfct), wherefc = 1 MHz, by first modulating with a cosinusoidal carrier of frequency f1 =50 KHz and then modulating that wave with a cosinusoidal carrier of frequency f2.Assuming that we have ideal bandpass filters with any desired centre frequencyand bandwidth available, explain carefully what filtering operations are neededand what the frequency f2 should be. Draw a block diagram of the overall system.

(b) Rather than using the technique of part (a) to generate y(t), we ideally samplem(t) with a sampling interval of Ts = 1/(10W ) = 5 × 10−6 seconds to producez(t) =

∑n m(nTs)δ(t − nTs). Explain carefully how to obtain y(t) from z(t) by

ideal filtering, specifying the centre and bandwidths of all filters used.

(c) To demodulate the signal y(t) = Am(t) cos(2πfct) where fc = 1 MHz, we multiplyit by the periodic signal p(t), which is a ±1 square wave with a period of T0 =5× 10−6 secs and Fourier series

p(t) =∞∑

k=0

pk cos(k2πf0t), pk =

{0 if k = 0, 2, 4, 6, . . .

(−1)(k−1)/2 4kπ

if k = 1, 3, 5, 7, . . .

The signal y(t)p(t) is then lowpass-filtered to retrieve m(t). Explain carefullywhat the output of the lowpass filter (with unity gain) is.

1

Page 5: ECE316F Communication Fall 2007 Systems Problem Set 1exams.skule.ca/exams/ECE316H1_20139_641388081078...ECE316F Communication Fall 2007 Systems Problem Set 7 (For week of November

ECE316F Communication

Fall 2007 Systems

Problem Set 5(For week of October 15th)

1. Haykin and Moher, Drill Problem 3.9

2. Haykin and Moher, Drill Problem 3.10

3. Haykin and Moher, Drill Problem 3.11

4. Haykin and Moher, Problem 3.20

5. Haykin and Moher, Problem 3.22

6. Haykin and Moher, Problem 3.23

7. Haykin and Moher, Problem 3.24

8. Haykin and Moher, Problem 3.31

9. In a remote monitoring application, a measurement apparatus deployed in the fieldrecords four low-pass signals: si(t), i = 1, . . . , 4. Two of the signals, s1(t) and s2(t)have B = 100 Hz, while s3(t) and s4(t) have B = 500 Hz. You must design a radiotransmission system to send these signals back to the home office for processing. Youare allocated spectrum in a band of frequencies near 900 MHz. You decide to multiplexthe four signals into a single composite baseband signal x(t) , and then send x(t) usingSSB transmission at fc = 900 MHz. In all cases, it must be possible to recover si(t),i = 1, . . . , 4, from x(t). For the purposes of this problem, we define the spectral extentof x(t) as the smallest value of W such that the spectrum of x(t) is contained in theband [−W, W ].

(a) In one approach, you form the composite signal as follows. First, you create aQAM signal at carrier frequency f1 > 0, modulated by s1(t) and s2(t). To thissignal you add a QAM signal at carrier frequency f2 > f1, modulated by s3(t) ands4(t). How should f1 and f2 be chosen so that the composite signal has minimumspectral extent? What is the bandwidth of the composite signal?

(b) In another approach, you form the composite signal by summing together fourSSB-modulated signals at carrier frequencies f1 < f2 < f3 < f4 where the signalat frequency fi is modulated with signal si. Assuming lower-SSB modulation, howshould f1, . . . , f4 be chosen so that the composite signal has minimum spectralextent? Repeat, assuming upper-SSB modulation. What is the bandwidth of thecomposite signal?

1

Page 6: ECE316F Communication Fall 2007 Systems Problem Set 1exams.skule.ca/exams/ECE316H1_20139_641388081078...ECE316F Communication Fall 2007 Systems Problem Set 7 (For week of November

ECE316F Communication

Fall 2007 Systems

Problem Set 6

(For week of October 22nd)

1. Haykin and Moher, Drill Problem 9.2, pg. 373.

2. Haykin and Moher, Drill Problem 9.3, pg. 377.

3. Haykin and Moher, Drill Problem 9.4, pg. 377. Also, what is the impact of a phaseerror on the output SNR of a DSB-SC system?

4. Haykin and Moher, Problem 9.9.

5. Haykin and Moher, Problem 9.10.

6. Haykin and Moher, Problem 9.13 (a) (b)

1

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ECE316F Communication

Fall 2007 Systems

Problem Set 7

(For week of November 5th)

1. Haykin and Moher, Haykin and Moher, Problem 4.19.

2. The signal m(t) = 10cos(200πt)+40cos(400πt) is frequency modulated with modulatorconstant kf = 10. Estimate the bandwidth of the FM signal using Carsons rule.

3. Instead of applying a message signal m(t) to the input of an FM modulator, it isdecided to apply m(t) + m(t).

(a) Show that the signal being transmitted is

ϕ(t) = A cos(

2πfct + 2πkfm(t) + 2πkf

t

−∞

m(τ)dτ)

.

(b) Specify the impulse response of a filter that recovers in the absence of noise m(t)at the output of an FM demodulator.

4. (a) Write cos5(θ) as a linear combination of 1, cos(θ), cos(2θ), . . . cos(5θ).

(b) For any integer n ≥ 1, show that cosn(θ) can be written as a linear combinationof 1, cos(θ), cos(2θ), . . . cos(nθ). [Hint : use mathematical induction].

5. In this problem, you will design an FM transmission system for a message signal m(t),where m(t) is assumed to be a low-pass signal of bandwidth 10 kHz satisfying, for allt, |m(t)| ≤ 1. You are allocated spectrum of 200 kHz bandwidth, centered at a carrierfrequency of 100 MHz, i.e., extending from 99.9 MHz to 100.1 MHz.

(a) Find the maximum value of the modulation parameter kf for which the resultingFM signal is, according to Carson’s rule, contained within the given bandwidth.

PowerAmplifier

BandpassFilter

FrequencyMultiplier

×50

(kf )NBFM =? fMIX =? Passband = ?

FrequencyMultiplier

×20

NBFMModulator

(fc)NBFM = 225 KHz 2 cos(2πfMIXt)

∼ ∼

fc = 100 MHzkf = 75000

(b) You decide to implement a system with kf = 75000, using Armstrong’s indirectmethod, according to the block diagram shown above. The carrier frequency ofthe first-stage narrowband FM modulator is 225 KHz. Determine the followingparameters: i) the modulation parameter (kf )NBFM for the first-stage narrowband

1

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FM signal signal; ii) the mixing frequency fMIX; iii) the passband of the bandpassfilter (i.e., specify the smallest possible range of (positive) frequencies passed bythe filter). The filter is to be designed with as narrow a passband as possible.

6. “Short snapper” from Final 2006: The Canadian regulatory body the Canadian Radio-television and Telecommunications Commission (CRTC) assigns a local FM radiostation a total of 100 kHz of bandwidth instead of the usual 200 kHz. The stationdecides to set the maximum value of the message signal to 3.5 V. The message signalis bandlimited to 15 kHz.

(a) In units of kHz/V, what is the maximum frequency sensitivity factor that Carsonwould recommend?

(b) If the carrier frequency is fc = 99.3 MHz, and assuming that Carsons rule issufficiently accurate, what frequency range would the transmitted signal occupy?

2

Page 9: ECE316F Communication Fall 2007 Systems Problem Set 1exams.skule.ca/exams/ECE316H1_20139_641388081078...ECE316F Communication Fall 2007 Systems Problem Set 7 (For week of November

ECE316F Communication

Fall 2007 Systems

Problem Set 8

(For week of November 12th)

1. Haykin and Moher, Drill Problem 9.8, page 390.

2. Haykin and Moher, Problem 9.13(c).

3. Haykin and Moher, Problem 9.17.

Even if you haven’t started on Chap 5 in class, please solve these problems using your

knowledge of Fourier transforms, ECE 216/355:

4. Haykin and Moher, Drill problem 5.1, page 192.

5. Haykin and Moher, Drill problem 5.5, page 196 (rate only. The interval is reciprocalof the rate).

6. Consider a signal x(t) whose Fourier transform is shown in Fig. 1(a), where W = 104Hz.This signal is to be sampled using an impulse train, p(t) =

n=−∞δ(t − nTs), with

sampling period Ts = 10−4s.

1

(a) (b)

p(t)

px (t)Anti

Alias

Filter

x(t)

X(f)

-W Wf

Figure 1: X(f), the Fourier transform of x(t) and impulse train sampling for Q. 6.

(a) Design an anti-aliasing filter given this sampling rate, i.e., modify the samplingscheme to that in Fig. 1(b). You need to provide the frequency response of the filter,indicating all parameters that define the filter.

(b) On using the anti-aliasing filter in Part (a), carefully sketch Xp(f), the Fouriertransform of xp(t), in the range f ∈ (−2.5W, 2.5W ).

1

Page 10: ECE316F Communication Fall 2007 Systems Problem Set 1exams.skule.ca/exams/ECE316H1_20139_641388081078...ECE316F Communication Fall 2007 Systems Problem Set 7 (For week of November

ECE316F Communication

Fall 2007 Systems

Problem Set 9

(For week of November 19th)

1. Haykin and Moher, Problem 5.14.

2. Haykin and Moher, Problem 5.15(a).

3. (Final 2006) In wideband speech processing, a speech signal is assumed to be ban-dlimited to 7 kHz.

(a) Assuming that the signal is sampled at the Nyquist rate and quantized using a 512-level quantizer, what is the bit rate of the corresponding pulse code modulation(PCM) system?

(b) The system is found to achieve a signal-to-quantization noise ratio (SQNR) of60 dB. How many levels would the quantizer need to have in order to achieve anSQNR of 72 dB?

4. (Final 2006)

(a) The message signal, m(t), has bandwidth 4 kHz, and is scaled so that −2 ≤m(t) ≤ 2. In the initial design, m(t) is sampled at the minimum rate required tomeet Nyquist’s sampling criterion. What is this minimum sampling rate?

(b) Next, the samples are quantized. What is the minimum number of bits requiredper sample if we require the quantization error ǫ to satisfy −0.25 ≤ ǫ ≤ 0.25?

(c) Using the sampling rate found in (4a) and 8 bits per sample, what bit rate isrequired to communicate this signal?

1

Page 11: ECE316F Communication Fall 2007 Systems Problem Set 1exams.skule.ca/exams/ECE316H1_20139_641388081078...ECE316F Communication Fall 2007 Systems Problem Set 7 (For week of November

ECE316F Communication

Fall 2007 Systems

Problem Set 10

(For Thurs., Nov. 29th, Mon., Dec. 3rd and Tues. Dec. 4th)

1. Haykin and Moher, Problem 6.3.

2. Haykin and Moher, Problem 6.9 (a) (d)

3. Haykin and Moher, Problem 6.10

4. Haykin and Moher, Problem 6.13 (a) (c)

5. Haykin and Moher, Problem 6.17. Use a data rate of 100kbps. Also, repeat the eyediagram with a 10% jitter in timing.

6. (Final 2006): A 4-level baseband pulse transmission system operates in a bandwidthof 10 kHz using bandlimited pulses with 25% excess bandwidth. What bit rate issupported by this system?

7. (Final 2006): Suppose that g(t) is a pulse that satisfies Nyquist’s criterion for zerointersymbol interference assuming a symbol interval of T seconds and that the centralsample appears at time t = 0. Assuming the symbol interval remains at T seconds,which of the following must also satisfy Nyquist’s criterion? (Justify your answer ineach case.)

(a) g(−t)

(b) g(2t)

(c) g(t/2)

8. (Final 2006)

(a) A twisted-pair wireline can support transmission of a baseband signal. Assumethat the bit rate is set to 64 kbps. To avoid ISI, you decide to use a raised-cosinepulse shape with 100% excess bandwidth (α = 1). What transmission bandwidthis needed, assuming binary pulse amplitude modulation?

(b) After some experimentation, you find that the usable bandwidth on the givenchannel is 48 kHz. To increase the efficiency of your system, you decide to use araised-cosine pulse with 25% excess bandwidth, and to change the quantizationscheme to delta modulation. With this scheme, what is the maximum rate atwhich the message signal may be sampled?

1

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9. (Final 2006) In a particular baseband data transmission system the basic transmittedpulse is rectangular, given by g(t) = rect(t−1/2). The channel is ideal, with frequencyresponse H(f) = 1.

(a) The receiver filter is linear time-invariant, with impulse response given by q(t) =rect(t− 1/2). Given that rect(t) ∗ rect(t) = (1− |t|)rect(t/2), show that when g(t)is transmitted, the output of the receiver filter is the triangular pulse

p(t) = (1 − |t − 1|)rect((t − 1)/2).

(b) To transmit N bits, the pulse train

s(t) =N−1∑

k=0

akg(t − k)

is sent, where ak ∈ {+1,−1}. Sketch the transmitted pulse train s(t) correspond-ing to the eight bit sequence a0 = +1, a1 = +1, a2 = −1, a3 = +1, a4 = −1,a5 = −1, a6 = +1. a7 = −1.

(c) Sketch the corresponding output of the receiver filter when s(t) from (9b) istransmitted.

(d) Does this system satisfy Nyquist’s criterion for zero intersymbol interference?Explain.

(e) Sketch the eye pattern for this system, and identify the times at which the receiverfilter output should be sampled in order to minimize ISI.

(f) Suppose, instead of the binary alphabet {+1,−1}, the system were to use thequaternary alphabet {+3, +1,−1,−3}. Sketch the eye pattern for this system.

2

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ECE316F Communication

Fall 2007 Systems

Problem Set 1 Solutions

1. Probably the easiest way to approach this problem is to place individual factors into polarform, and then combine the factors.

(a) We have 1 + j =√

2ejπ/4, thus (1 + j)3 = 23/2ej3π/4.(b) j3 = −j, so

√3 + j3 = 2ejθ, for θ = tan−1(−1/

√3) = −π/6. Also 1 − j =

√2e−jπ/4.

Thus the quantity of interest is 23/2e−jπ/6−jπ/4 = 23/2e−j5π/12.

(c) The numerator is 2 − j(6/√

3) = 4ejθ for θ = tan−1(3/√

3) = −π/3. The denominatoris the complex conjugate, namely 4e−jθ. Thus the quantity of interest is ej2θ = e−2π/3.

(d) j = ejπ/2, 1 + j =√

2ejπ/4, thus the quantity of interest is√

2ej(π/2+π/4+π/6) =√2ej11π/12.

(e) We have

ejπ/3 − 1 = ejπ/6(ejπ/6 − e−jπ/6)= 2jejπ/6 sin(π/6)= jejπ/6

= ejπ/2+jπ/6

= ej2π/3.

(Alternatively, since ejπ/3 = cos(π/3) + j sin(π/3) = 1/2 + j√

3/2, we have ejπ/3 − 1 =−1/2+j

√3/2 = ej2π/3.) Also, 1+j

√3 = 2ejπ/3. Thus, the quantity of interest is 1

2ejπ/3.

2. (a) We have

sin(θ) sin(φ) =

(ejθ − e−jθ

2j

)(ejφ − e−jφ

2j

)

=−14

(ej(θ+φ) + e−j(θ+φ) − ej(θ−φ) − e−j(θ−φ)

)=

12

(ej(θ−φ) + e−j(θ−φ)

2

)− 1

2

(ej(θ+φ) + e−j(θ+φ)

2

)

=12

cos(θ − φ) +12

cos(θ + φ).

(b) We have

sin(θ) cos(φ) + cos(θ) sin(φ) =14j

((y − y∗)(z + z∗) + (y + y∗)(z − z∗))

for y = ejθ and z = ejφ. Expanding, we get14j

(yz + yz∗ − y∗z − y∗z∗ + yz − yz∗ + y∗z − y∗z∗) =14j

(2yz − 2y∗z∗)

=12j

(ej(θ+φ) − e−j(θ+φ))

= sin(θ + φ).

1

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3. Haykin and Moher, Drill Problem 2.1:

We have g(t) = e−t sin(2πfct)u(t) = e−t(ej2πfct − e−j2πfct)u(t)/2j. Thus

G(f) =∫ ∞−∞

g(t)e−j2πftdt

=12j

∫ ∞0

e−t(ej2πfct − e−j2πfct)e−j2πftdt

=12j·∫ ∞0

e−t(1−j2πfc+j2πf)dt− 12j·∫ ∞0

e−t(1+j2πfc+j2πf)dt

=12j· 11 + j2π(f − fc)

− 12j· 11 + j2π(f + fc)

=12j· 1 + j2π(f + fc)− 1− j2π(f − fc)(1 + j2π(f − fc))(1 + j2π(f + fc))

=2πfc

1− 4π2(f2 − f2c ) + j4πf

4. Haykin and Moher, Drill Problem 2.2:

We have

g(t) =∫ ∞−∞

G(f)ej2πftdf

=∫ 0

−Wejπ/2ej2πftdf +

∫ W

0e−jπ/2ej2πftdf

= jej2πft

j2πt

∣∣∣∣∣0

f=−W

− jej2πft

j2πt

∣∣∣∣∣W

f=0

=1

2πt((1− e−j2πWt)− (ej2πWt − 1))

=1πt· 2− ej2πWt − e−j2πWt

2

=1πt

(1− cos(2πWt))

=2 sin2(πWt)

πt

5. Haykin and Moher, Drill Problem 2.3:

We know that G(0) gives the area under g(t). If g(t) is real, then the area under g(t) is real,and so G(0) must be real, i.e., Im[G(0)] = 0.

6. Haykin and Moher, Drill Problem 2.9:

(a) We have g(t) = cos2(2πfct) = 12(1 + cos(4πfct)).

G(f) = F [12

+12

cos(4πfct)]

=12δ(f) +

14δ(f − 2fc) +

14δ(f + 2fc).

2

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(b) Similarly, g(t) = sin2(2πfct) = 12(1− cos(4πfct)), so

G(f) = F [12− 1

2cos(4πfct)]

=12δ(f)− 1

4δ(f − 2fc)−

14δ(f + 2fc).

7. Haykin and Moher, Drill Problem 2.13:

From (2.40) in the text, we know that

exp(−πt2)←→ exp(−πf2).

From the dilation property (2.20), with dilation parameter a = 1/(τ√

2π), we get

exp(−t2/2τ2)←→ τ√

2π exp(−2π2τ2f2).

8. Haykin and Moher, Problem 2.19:

(a) We have ga(t) = A cos(πt/T )rect(t/T ), which is A times the signal of Example 2.5 inthe text, with fc = 1/(2T ). Thus, from (2.29) we get

Ga(f) =AT

2sinc[T (f − 1/2T )] +

AT

2sinc[T (f + 1/2T )]

=AT

2sinc(fT − 1/2) +

AT

2sinc(fT + 1/2)

=2AT cos(πfT )π(1− 4f2T 2)

.

The last expression follows—after some manipulation—by expanding sinc(fT ± 1/2)as ∓ cos(πfT )/(πfT ± π/2). Note that, because g(t) is real-valued and even, G(f) isreal-valued .

(b) The new signal gb(t) = ga(t− T/2) is a T/2 shifted version of ga(t) from part (a); thus

Gb(f) = e−jπfT Ga(f).

(c) The half-sine pulse with duration aT is given by gb(t/a), and hence has Fourier transform

aGb(af) = e−jπafT Ga(af).

(d) The new signal gc(t) = −ga(t + T/2); thus

Gc(f) = −ejπfT Ga(f).

(e) The new signal gd(t) = gb(t) + gc(t); thus

Gd(f) = Ga(f)e−jπfT −Ga(f)e+jπfT

= −2j sin(πfT )Ga(f)

Note that since Ga(f) is purely real, Gd(f) is purely imaginary, which is to be expectedsince gd(t) is real-valued and odd.

3

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9. Haykin and Moher, Problem 2.22:

(a) We know from (2.23) in the text that if g(t)↔ G(f) then

g∗(−t)↔ G∗(f)

Now if g(t) is real and even, then g∗(−t) = g(t), hence G∗(f) = G(f), which impliesthat G(f) is purely real.If g(t) is real and odd, then g∗(−t) = −g(t), hence G∗(f) = −G(f), which implies thatG(f) is purely imaginary.

(b) From (2.35) in the Text, we know that if g(t)←→ G(f), then

tg(t)←→ j

d

dfG(f).

By induction it follows that

tng(t)←→ jn

(2π)n

dn

dfnG(f).

(c) From the area property and the previous result it follows that∫ ∞−∞

tng(t)dt =jn

(2π)n

dn

dfnG(f)

∣∣∣∣f=0

.

(d) Let R12(τ) =∫∞−∞ g1(t)g∗2(t − τ)dt. From the correlation theorem, equation (2.53) in

the text, we have R12(τ) ←→ G1(f)G∗2(f). From the area property, we get R12(0) =∫∞−∞G1(f)G∗2(f)df , i.e., ∫ ∞

−∞g1(t)g∗2(t)dt =

∫ ∞−∞

G1(f)G∗2(f)df.

10. Haykin and Moher, Problem 2.25:

If y(t) = x2(t) = x(t)x(t), then Y (f) = X(f)∗X(f). If X(f) is bandlimited to [−W,W ] thenthe convolution is zero outside [−2W, 2W ], i.e., y(t) is bandlimited to [−2W, 2W ].

4

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ECE316F Communication

Fall 2007 Systems

Problem Set 2 Solutions

1. We are given that F [g(t)] = G(f). From the duality property, we have F [G(t)] =g(−f). From the reflection property, we have F [g(−t)] = G(−f). Combining theduality and reflection properties, we have F [G(−t)] = g(f).

Thus we get

F [h(t)] = F [g(t) +G(t) + g(−t) +G(−t)]= F [g(t)] + F [G(t)] + F [g(−t)] + F [G(−t)]= G(f) + g(−f) +G(−f) + g(f)

= h(f).

2. This is trickier than it looks. Let’s start with

g(t) = exp(−πt2)←→ G(f).

We must find G(f). From the derivative property of the Fourier transform we have

dg(t)

dt= −2πtg(t)←→ j2πfG(f).

On the other hand, from the dual of derivative property we have

−j2πtg(t)←→ dG(f)

df

or

−2πtg(t)←→ −jdG(f)

df.

Equating the two expressions for the Fourier transform of −2πtg(t), we get

j2πfG(f) = −jdG(f)

df

ordG(f)

df+ 2πfG(f) = 0

which is a first-order ordinary differential equation that G(f) must satisfy. (Actually,we have already seen that g(t) satisfies exactly this differential equation. Could G(f) =g(f)?)

This differential equation can be solved by multiplying by an appropriate integratingfactor M(f), i.e., a function M(f) such that M(f)G′(f) + 2πfM(f)G(f) is equal to

1

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the derivative of M(f)G(f). We find that M(f) = eπf2, so the differential equation is

equivalent tod

df

(eπf2

G(f))

= 0.

This equation is easily solved: we find G(f)eπf2= C for some constant C. Thus

G(f) = Ce−πf2. The remaining task is to find the constant C.

Now, we havee−πt2 ←→ Ce−πf2

.

Applying the time-frequency duality property, we get

Ce−πt2 ←→ e−πf2

or

e−πt2 ←→ 1

Ce−πf2

.

Thus C = 1/C, or C2 = 1. Now, since G(0) = C =∫∞−∞ e

−πt2dt > 0, we find thatC = 1 is the desired solution. Thus, we have shown that

e−πt2 ←→ e−πf2

.

(Whew!)

We could also have tried to proceed as follows. We have

F [exp(−πt2)] =∫ ∞

−∞exp(−πt2) exp(−j2πft)dt

=∫ ∞

−∞exp(−π(t+ jf)2 − πf 2)dt

= e−πf2∫ ∞

−∞exp(−π(t+ jf)2)dt

= e−πf2

limA→∞

∫ A

−Aexp(−π(t+ jf)2)dt.

We could substitute u = t+ jf to get

G(f) = e−πf2

limA→∞

∫ A+jf

−A+jfexp(−πu2)du

which is a path integral in the complex plane. To solve this we could resort to Cauchy’sIntegral Theorem. However, the approach taken above avoids complex integration.

3. We haveexp(−πt2)←→ exp(−πf 2).

From the scaling property of the Fourier transform, we have

exp(−π(t√a/π)2)←→

√π/a exp(−π(f

√π/a)2)

2

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From the time shifting property, it follows that

exp(−π((t−m)√

(a/π))2)←→ exp(−j2πfm)√π/a exp(−π(f

√π/a)2).

Finally, from the area property of the Fourier transform (i.e., evaluating the Fouriertransform at f = 0) we get∫ ∞

−∞exp(−π((t−m)

√(a/π))2)dt =

∫ ∞

−∞exp(−a(t−m)2)dt =

√π/a.

4. Haykin and Moher, Drill Problem 2.10

If g(t) = δ(t+ 1/2)− δ(t− 1/2), then G(f) = exp(jπf)− exp(−jπf), with G(0) = 0.Since ∫ t

−∞g(τ)dτ = rect(t)←→ G(f)

j2πf+G(0)δ(f)

2,

we get

rect(t)←→ exp(jπf)− exp(−jπf)

j2πf=

sin(πf)

πf= sinc(f),

as expected.

5. Haykin and Moher, Drill Problem 2.16

The energy spectral density of a signal g(t) with Fourier transform G(f) is given asψg(f) = |G(f)|2. If g(t) = exp(−at)u(t), then G(f) = (a+ j2πf)−1. It follows that

ψg(f) = |G(f)|2 =1

a2 + 4π2f 2.

6. Haykin and Moher, Drill Problem 2.17

The double exponential pulse may be written as g(t) = exp(−at)u(t) + exp(at)u(−t).Thus

G(f) =1

a+ j2πf+

1

a− j2πf=

2a

a2 + 4π2f 2,

and

ψf (g) = |G(f)|2 =4a2

(a2 + 4π2f 2)2.

7. Haykin and Moher, Problem 2.20

0 T

t

g(t)

A

The signal g(t) is shown above.

3

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(a)

ge(t) =1

2(g(t) + g(−t))

=A

2

(rect(

t

T− 1

2) + rect(

−tT− 1

2))

go(t) =1

2(g(t)− g(−t))

=A

2

(rect(

t

T− 1

2)− rect(

−tT− 1

2))

The even and odd parts are sketched below.

0 T

tA/2

−T

ge(t)

T

tA/2

−T 0

go(t)

(b) We haveGe(f) = F [(A/2)rect(t/2T )] = AT sinc(2fT )

and

Go(f) = e−jπfT AT

2sinc(fT )− ejπfT AT

2sinc(fT )

= −jAT sinc(fT ) sin(πfT )

Alternatively, from

G(f) = e−jπfTAT sinc(fT ) = AT sinc(fT )(cos(πfT )− j sin(πfT )),

we get

Ge(f) = Re[G(f)]

= AT sinc(fT ) cos(πfT )

= AT sin(πfT ) cos(πfT )/πfT

= AT sin(2πfT )/2πfT

= AT sinc(2fT )

and

Go(f) = Im[G(f)]

= −jAT sinc(fT ) sin(πfT ).

8. Haykin and Moher, Problem 2.26

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(a) The rectangular pulse of unit area and width T is given as g(t) = 1Trect(t/T ),

which has Fourier transform sinc(fT ). In the limit as T → 0, sinc(fT ) → 1 forany fixed f .

(b) The unit area sinc pulse with first zero crossing at T is given as 1Tsinc(t/T ). The

Fourier transform of this pulse is rect(fT ). In the limit as T → 0, rect(fT )→ 1for any fixed f .

9. Haykin and Moher, Problem 2.27

We have G(f) = u(f). We know that

u(t)←→ 1

j2πf+

1

2δ(f).

From the duality property of the Fourier transform we get

1

j2πt+

1

2δ(t)←→ u(−f).

From the reflection property and using the property that δ(t) is even, we get

j

2πt+

1

2δ(t)←→ u(f).

Thus g(t) = j2πt

+ 12δ(t).

10. Haykin and Moher, Problem 2.30

(Another tricky problem.)

An LTI system with impulse response h(t) is stable if and only if∫ ∞

−∞|h(t)|dt <∞.

Now if h(t) is absolutely integrable, then h(t) is square-integrable, i.e.,∫∞−∞ |h(t)|2dt <

∞, i.e., the impulse response has finite energy.

If x(t) is the input and y(t) the output of this system, we have ψy(f) = |H(f)|2ψx(f),and so

Ey =∫ ∞

−∞ψy(f)df =

∫ ∞

−∞|H(f)|2ψx(f)df.

From the Cauchy-Schwarz inequality (see Appendix 5), we have∫ ∞

−∞|H(f)|2ψx(f)df ≤

(∫ ∞

−∞|H(f)|2df

∫ ∞

−∞ψx(f)df

)1/2

.

However, since both h(t) and x(f) have finite energy, i.e.,∫∞−∞ |H(f)|2df < ∞ and∫∞

−∞ ψx(f)df <∞, we must have(∫ ∞

−∞|H(f)|2df

∫ ∞

−∞ψx(f)df

)1/2

<∞

so y(t) has finite energy.

5

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ECE316F Communication

Fall 2007 Systems

Problem Set 3

(For week of October 1st)

1. We will take an approach similar to that in the book.

sin(2πfct) =exp[j2πfct] − exp[−j2πfct]

2j

, i.e., after forming gT (t) and xT (t), we have

XT (f) =1

2j[GT (f − fc) − GT (f + fc)]

Again, ignoring the overlap between the GT (f − fc) and GT (f + fc),

|XT (f)|2 =1

4

[

|GT (f − fc)|2 + |GT (f + fc)|

2]

i.e.,

Sx(f) =1

4[Sg(f − fc) + Sg(f + fc)]

One should expect this result because multiplying with either cos(2πfct) and sin(2πfct)cause a frequency shift of G(f) to frequency ±fc; sin(·) is only a phase shift of cos(·)while the power spectral density is concerned with the magnitude (rather magnitudesquared).

2. Haykin and Moher, Problem 2.33: First, the pulse in part (b) is a time shift of thepulse in part (a) by T/2. In the frequency domain, Gb(f) = exp [−j2πfT/2] Ga(f).This phase term has unit magnitude and so does not affect the energy distribution over

frequency, i.e., both signals have the energy spectral density.

Now, focusing on the signal in part (a), it is one segment of a cosine function.

g(t) = Arect(

t

T

)

cos(

πt

T

)

,

⇒ G(f) = ATsin(πfT )

πfT⋆

1

2

[

δ(

f −1

2T

)

+ δ(

f +1

2T

)]

,

where ⋆ represents convolution. Therefore,

G(f) =AT

2

[

sin(π(f − 1/2T )T )

π(f − 1/2T )T+

sin(π(f + 1/2T )T )

π(f + 1/2T )T

]

,

1

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=2AT cos(πfT )

π(1 − 2fT )(1 + 2fT ),

=2AT cos(πfT )

π(1 − 4f 2T 2),

⇒ Ψg(f) = |G(f)|2 =4A2T 2 cos2(πfT )

π2(1 − 4f 2T 2)2,

which is the required answer. We have used sin(π/2 − θ) = cos(θ).

3. Haykin and Moher, Problem 2.49 (part (b) - periodic square pulse only). Use theapproach suggested in class/the handout on the PSD of a sinusoid that was posted asopposed to the auto correlation function approach.

Using the definition of the Fourier series

gp(t) =∞∑

n=−∞

cnej2πnf0t, (1)

⇒ Gp(f) =∞∑

n=−∞

cnδ(f − nf0), (2)

where f0 = 1/T0 and

cn =1

T0

T0/2

−T0/2

gp(t)e−j2πnf0tdt

=

{

A

2n = 0

A sin(nπ/2)

nπn 6= 0

.

Note that since sin(kπ) = 0 for integer k, cn = 0 for even n.

Now, the first step in finding the power spectral density (PSD) is to time limit g(t) toobtain gT (t) and then take its Fourier transform

gT (t) =

{

g(t) |t| ≤ T/2

0 |t| > T/2= rect

(

t

T

)

g(t),

⇒ GT (f) = Gp(f) ⋆ T sinc(fT ).

Using Eqn. (2),

GT (f) =∞∑

n=−∞

cnT sinc((f − nf0)T ),

i.e., each δ-function in Gp(f) is replaced with a sinc function with width 1/T . Now,the PSD of the original signal, gp(t) is

Sg(f) = limT→∞

1

T|GT (f)|2

2

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As T → ∞, each sinc gets narrower and one can ignore the overlap between the sinc

functions and

|GT (f)|2 ≃∞∑

n=−∞

|cn|2T 2sinc2((f − nf0)T ),

⇒ Sg(f) =∞∑

n=−∞

|cn|2

[

limT→∞

1

TT 2sinc2((f − nf0)T )

]

,

=∞∑

n=−∞

|cn|2δ(f − nf0),

where we have derived cn earlier. Though, note that this procedure can be used tofind the PSD of any periodic signal, including, as a special case, that of the sinusoid(as posted to the class website).

4. Haykin and Moher, Problem 8.34 (b)

If x(t) ↔ X(f), we know that dx/dt ↔ j2πfX(f), i.e., the derivative of a signal canbe obtained by using a filter with frequency response H(f) = j2πf . We also know thatif a signal x(t) passes through a filter with frequency response H(f) to create outputy(t),

Sy(f) = |H(f)|2Sx(f),

i.e., the power spectral density of dx/dt is 4π2f 2Sx(f).

5. Haykin and Moher, Problem 8.36 (b) (c).

(b) The power spectral density includes a δ-function at f = 0, i.e., it has some dcpower. We know the average power is

P =∫

−∞

Sx(f)df

Think of the dc power as the power concentrated around f = 0

dc power = Pdc = limδ→0

δ

−δ

Sx(f)df = 1

(c) The ac power is the total power minus the dc power

Pac =∫

−∞

Sx(f)df − Pdc = f0.

(this is the area under the triangle).

6. Haykin and Moher Problem 8.38.

We using Eqn. (8.98) of the textbook (the bandwidth B = 2)

SxI(f) = SxQ

(f) =

{

Sx(f − fc) + Sx(f + fc) |f | < B

0 |f | > B=

2 − 3|f |/2 0 < |f | < 1

1 − |f |/2 1 < |f | < 2

0 |f | > 2

3

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ECE316F Communication

Fall 2007 Systems

Problem Set 4 Solutions

1. Haykin and Moher, Drill Problem 3.1

Yes. The following figure shows the result of modulating a sinusoidal carrier with asinusoid modulating wave at 100% modulation. The envelope becomes zero wheneverthe modulating wave reaches it maximum negative value.

2. Haykin and Moher, Drill Problem 3.2

Here s(t) = Ac[1 +µ cos(2πfmt)] cos(2πfct), as in Example 3.1 in the text. The carrierpower is 1

2A2

c while each sideband has power 18µ2A2

c . With µ = 1/5, we get 1200A2

c powerin each sideband.

3. Haykin and Moher, Drill Problem 3.3

In order to avoid spectral overlap, we require the carrier frequency fc to be at least W ,the bandwidth of the message signal.

4. Haykin and Moher, Drill Problem 3.4

(a) We have

v2(t) = a1v1(t) + a2v21(t)

= a1[Ac cos(2πfct) +m(t)] + a2[Ac cos(2πfct) +m(t)]2

= a1[Ac cos(2πfct) +m(t)] + a2[A2c cos2(2πfct) + 2Acm(t) cos(2πfct) +m2(t)]

= a1[Ac cos(2πfct) +m(t)] + a2[A2c

1

2(1 + cos(2π(2fc)t)) + 2Acm(t) cos(2πfct) +m2(t)]

= a2A2c/2 + a1m(t) + a2m

2(t)︸ ︷︷ ︸spurious baseband components

+ [a1Ac + 2a2Acm(t)] cos(2πfct)︸ ︷︷ ︸desired AM wave at fc

+

1

2a2A

2c cos(2π(2fc)t)︸ ︷︷ ︸

spurious carrier at 2fc

(b) To extract the desired AM wave, we require a bandpass filter of bandwidth 2Wcentered at ±fc, passing frequencies in the range [±fc −W,±fc +W ].

(c) The spurious baseband signal components extend to frequency 2W . Thus werequire that fc is at least 3W , so that the lower sideband doesn’t overlap withthese components. The condition is fc > 3W .

1

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5. Haykin and Moher, Drill Problem 3.5

Each sideband contains 50% of the total power in the DSB-SC signal.

6. Haykin and Moher, Drill Problem 3.6

(a) The modulating signal is m(t) = Am cos(2πfmt), so the modulated signal is s(t) =m(t)Ac cos(2πfct). Thus the output of the product modulator, assuming a unitamplitude local oscillator, is

v(t) = s(t) cos(2πfct)

= m(t)Ac cos(2πfct) cos(2πfct)

=Ac

2m(t)[1 + cos(2π(2fc)t)]

(b) Expressed in terms of complex exponential functions, we have

s(t) =AmAc

4

(ej2π(fm+fc)t + ej2π(−fm+fc)t + ej2π(fm−fc)t + ej2π(−fm−fc)t

).

Thus

v(t) = s(t) cos(2πfct)

=AmAc

8

ej2π(fm+fc)t︸ ︷︷ ︸+

+ ej2π(−fm+fc)t︸ ︷︷ ︸+

+ ej2π(fm−fc)t︸ ︷︷ ︸−

+ ej2π(−fm−fc)t︸ ︷︷ ︸−

ej2πfct︸ ︷︷ ︸+

+ e−j2πfct︸ ︷︷ ︸−

=

AmAc

4

(ej2πfmt + e−j2πfmt

)︸ ︷︷ ︸

desired signal

+terms centered at ±2fc.

Terms labelled ‘+’ are at positive frequency and terms labelled ‘−’ are at negativefrequency. The two terms labelled ‘+’ in the first factor combine with the termlabelled ‘−’ in the second factor to contribute half of the total desired signal,while the two terms labelled ‘−’ in the first factor combine with the term labelled‘+’ in the second factor to contribute the other half.

7. Haykin and Moher, Problem 3.18

(a) With 75% modulation, the modulated wave looks like this:

2

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(b) The carrier power is A2c

2÷ 100 = 2500/200 = 12.5 W. The total sideband power

is A2cµ2

4÷ 100 = 3.52 W. Thus the total power is 16.02 W. Note that this power

depends only on the carrier magnitude and µ — not on the message power. (Theefficiency factor is 3.52/16.02 ≈ 22% = µ2/(2 + µ2).)

8. Haykin and Moher, Problem 3.19

The maximum and minimum values of m(t) occur at t = ±1, taking values ±1/2,respectively. The modulated wave is thus given by

s(t) = Ac(1 + 2µm(t)) cos(2πfct),

where µ is the modulation factor. The modulated wave is sketched (for fc = 10 Hz)below.

(a) 50% modulation (b) 100% modulation (c) 125% modulation

9. (a) Let z1(t) = a(t) cos(2πBt) + b(t) sin(2πBt) and let z2(t) = c(t) cos(2πBt) +d(t) sin(2πBt). Both z1(t) and z2(t) are QAM signals of bandwidth 2B Hz atcarrier frequency B Hz. Similarly x(t) = z1(t) cos(2πfct) + z2(t) sin(2πfct) is aQAM signal of bandwidth 4B Hz at carrier frequency fc. Thus the bandwidth ofx(t) is 4B Hz.

(b) The energy spectral density of a(t) cos(2πBt) is given by 14(ψa(f−B)+ψa(f+B)),

where ψa(f) is the energy spectral density of a. It follows that the energy ofa(t) cos(2πBt) is E/2. Likewise the energy of b(t) sin(2πBt) is E/2. Since thetwo components of z1(t) are orthogonal, their energies add, thus the energy ofz1(t) is E. It follows that z1(t) cos(2πfct) has energy E/2. Likwise, the energyof z2(t) sin(2πfct) is E/2. Since the two components of x(t) are orthogonal, theirenergies add, and hence the energy of x(t) is E.

3

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(c) The block diagram is shown in the figure.The first-stage lowpass filters of bandwidth 2Bshown in the figure may be omitted.

LPF

LPF

LPF

LPF

LPF

LPFbandwidth

2B

bandwidth

2B

bandwidth

bandwidth

bandwidth

bandwidth

sin(2πfct)

sin(2πBt)

sin(2πBt)

B

B

B

B

a(t)

b(t)

c(t)

d(t)

x(t)

cos(2πBt)

cos(2πfct)

cos(2πBt)

10. (a) We choose f2 = 950 KHz or f2 = 1050 KHz, and bandpass filter the resultwith bandpass filter having centre frequency 1 MHz and bandwidth 40 KHz (i.e.,passing frequencies in the range 980 to 1020 KHz). The gain of the filter shouldbe 2A in the passband. (This gain factor can also appear elsewhere, for examplemultiplying one of the mixing frequencies.) The block diagram is shown below.

BPFm(t)

cos(2πf1t) cos(2πf2t)

f1 = 50 KHz f2 = 950 or 1050 KHz

Am(t) cos(2πfct)

fc = 1000 KHz

passband = 980 to 1020 KHzpassband gain = 2A

(b) We have z(t) =∑

nm(nTs)δ(t− nTs), which has Fourier transform

Z(f) =1

Ts

∞∑n=−∞

M(f − nfs),

as shown below.

0 200 400 600 800 1000−200−400−600−800−1000

· · ·· · ·

f (KHz)

The desired signal component appears at n = ±5. Thus we bandpass filter in theregion (980, 1020) KHz, with a passband gain of ATs/2.

(c) Let f0 = 1/T0. The signal p(t) has Fourier transform

P (f) = p0δ(f) +1

2

∞∑k=1

pk[δ(f − kf0) + δ(f + kf0)]

=1

2

∞∑k=1

k odd

pk[δ(f − kf0) + δ(f + kf0)].

4

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Thus z(t) = y(t)p(t) has Fourier transform

Z(f) = Y (f) ∗ P (f) =1

2

∞∑k=1

k odd

pk[Y (f − kf0) + Y (f + kf0)]

Now, since fc = 5f0, we have Y (f) = A2[M(f − 5f0) + M(f + 5f0)]. Thus, in

terms of M(f), z(t) has Fourier transform

Z(f) =A

4

∞∑k=1

k odd

pk [M(f − 5f0 − kf0) +M(f + 5f0 − kf0)+

M(f − 5f0 + kf0) +M(f + 5f0 + kf0)] .

The lowpass filter retains only the components centered at zero frequency, whicharise when k = 5, these terms being

A

4p5 [M(f) +M(f)] .

Thus the filter output will be

A

2p5m(t) =

2A

5πm(t).

5

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ECE316F Communication

Fall 2007 Systems

Problem Set 5 Solutions

1. Haykin and Moher, Drill Problem 3.9

We will need the following trigonometric identities:

cos2(θ) =1

2(1 + cos(2θ)); sin2(θ) =

1

2(1− cos(2θ)); cos(θ) sin(θ) =

1

2sin(2θ).

The multiplexed signal is s(t) = Acm1(t) cos(2πfct) + Acm2(t) sin(2πfct). The inputto the low-pass filter in the upper receiver branch (the I-channel) is

s(t)A′c cos(2πfct) = AcA

′cs(t)[m1(t) cos2(2πfct) + m2(t) cos(2πfct) sin(2πfct)]

=1

2AcA

′c[m1(t) + m1(t) cos(4πfct) + m2(t) sin(4πfct).

The high-frequency terms are rejected by the low-pass filter, so the final output is12AcA

′cm1(t). Similarly, the input to the low-pass filter in the lower receiver branch

(the Q-channel) is

s(t)A′c sin(2πfct) =

1

2AcA

′c[m1(t) sin(4πfct) + m2(t)−m2(t) cos(4πfct)]

The high-frequency terms are rejected by the low-pass filter, so the final output is12AcA

′cm2(t).

2. Haykin and Moher, Drill Problem 3.10

(a) We have

s(t) =1

2Acm(t) cos(2πfct)−

1

2Acm(t) sin(2πfct);

thus

S(f) =1

4Ac[M(f − fc) + M(f + fc)]−

1

4jAc[M(f − fc)− M(f + fc)],

where M(f) = F [m(t)] = −jsgn(f)M(f). Therefore we have

S(f) =1

4Ac[M(f−fc)+M(f+fc)]+

1

4Ac[M(f−fc)sgn(f−fc)−M(f+fc)sgn(f+fc)].

For f > 0, M(f + fc) = 0 (assuming the message has bandwidth smaller thanfc); thus, for f > 0, we have

S(f) =1

4AcM(f − fc) +

1

4AcM(f − fc)sgn(f − fc)

=

{14AcM(f − fc) + 1

4AcM(f − fc) f > fc,

14AcM(f − fc)− 1

4AcM(f − fc) 0 < f ≤ fc;

=

{12AcM(f − fc) f > fc,

0 0 < f ≤ fc.

1

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(b) We have

s(t) =1

2Acm(t) cos(2πfct) +

1

2Acm(t) sin(2πfct);

thus

S(f) =1

4Ac[M(f − fc) + M(f + fc)] +

1

4jAc[M(f − fc)− M(f + fc)]

=1

4Ac[M(f − fc) + M(f + fc)]−

1

4Ac[M(f − fc)sgn(f − fc)−M(f + fc)sgn(f + fc)].

For f > 0, we have

S(f) =1

4AcM(f − fc)−

1

4AcM(f − fc)sgn(f − fc)

=

{14AcM(f − fc)− 1

4AcM(f − fc) f > fc,

14AcM(f − fc) + 1

4AcM(f − fc) 0 < f ≤ fc;

=

{0 f > fc,12AcM(f − fc) 0 < f ≤ fc.

3. Haykin and Moher, Drill Problem 3.11

The Fourier transform of m(t) is simply M(f) = −jsgn(f)M(f). Since |M(f)| =|M(f)|, a signal and its Hilbert transform have the same bandwidth.

4. Haykin and Moher, Problem 3.20

Let vi(t) = x(t) + cos(2πf1t), where x(t) is a signal of bandwidth W . Note thatcos2(θ) = 1

2(1 + cos(2θ)) and that cos3(θ) = 1

4cos(3θ) + 3

4cos(θ). Then

io(t) = a1vi(t) + a3v3i (t)

= a1[x(t) + cos(2πf1t)] + a3[x(t) + cos(2πf1t)]3

= a1x(t) + a1 cos(2πf1t) + a3x3(t) + 3a3x(t) cos2(2πf1t) +

3a3x2(t) cos(2πf1t) + a3 cos3(2πf1t)

= a1x(t) + a1 cos(2πf1t) +

a3x3(t) + (3a3/2)x(t) + (3a3/2)x(t) cos(4πf1t) +

3a3x2(t) cos(2πf1t) + (3a3/4) cos(2πf1t) + (a3/4) cos(6πf1t).

We find modulation products centered at dc, at f1, at 2f1 and at 3f1. Only the termcentered at 2f1, namely

s(t) = (3a3/2)x(t) cos(4πf1t)

is the product of x(t) with a cosine. Setting x(t) = 1 + m(t), f1 = fc/2 and filteringout components not centered at fc, gives amplitude modulation.

Can such a device be used as an envelope detector? Let s(t) = Ac(1+kam(t)) cos(2πfct),and let N(x) = a1x + a3x

3. Then

N(s(t)) = a1Ac(1 + kam(t)) cos(2πfct) + a3(A3c(1 + kam(t))3 cos3(2πfct)).

2

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No terms appear that are centered at dc. Even by considering N(N(s(t)) or N(N(N(s(t))),only odd powers of s(t) appear, giving rise to terms at odd harmonics of the carrierfrequency. No such odd harmonic appears at dc. Thus such a device cannot be usedas the first stage of an envelope detector.

5. Haykin and Moher, Problem 3.22

As described in the the solution to 3.20, the modulation product appear at 2f1 isproportional to the product of m(t) and cos(4πf1t). A bandpass filter centered at 2f1

would pass this component and reject components at other multiples of f1 (assumingf1 is larger than the signal bandwidth), and thus a product modulator is obtained.

6. Haykin and Moher, Problem 3.23

In general, the demodulator first produces a signal proportional to M(f) + 12M(f −

2fc) + 12M(f + 2fc); this signal is then low-pass filtered to bandwidth W .

(a) When fc > W , then no spectral overlap occurs; and the signal recovered has aspectrum proportional to M(f).

(b) When fc < W , spectral overlap occurs. In this case, the spectrum of the detectoroutput is proportional to the signal sketched (in bold) below.

W0

I’m not sure what is being asked in the last part of the question. Each componentof s(t) is uniquely determined by m(t). However, the smallest (positive) carrierfrequency for which no spectral overlap occurs is fc = W .

7. Haykin and Moher, Problem 3.24

Let

s(t) = Ac cos(2πfct + φ) + m(t) cos(2πfct)

= (Ac cos(φ) + m(t)) cos(2πfct)− Ac sin(φ) sin(2πfct).

(a) If φ = 0, we see that s(t) is the ideal AM signal (Ac + m(t)) cos(2πfct). Pro-vided that Ac > |m(t)|, so that no overmodulation distortion occurs, the envelopedetector will produce Ac + m(t) at its output.

(b) We can always write a cos(2πfct)− b sin(2πfct) as√

a2 + b2 cos(2πfct + θ), wherecos(θ) = a/

√a2 + b2 and sin(θ) = b/

√a2 + b2 so that θ = tan−1(b/a). Here we

have a(t) = (Ac cos φ + m(t)) and b = (Ac sin(φ)) so

s(t) =√

a2(t) + b2 cos(2πfct + θ(t)).

3

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The ideal envelope detector would output√a2(t) + b2 =

[A2

c cos2(φ) + 2Ac cos(φ)m(t) + m2(t) + A2c sin2(φ)

]1/2

=[A2

c + 2Ac cos(φ)m(t) + m2(t)]1/2

= Ac

[1 + 2 cos(φ)m(t)/Ac + m2(t)/A2

c

]1/2

Now, from the Taylor series expansion, we know that if x � 1,√

1 + x ≈ 1+x/2.Thus, assuming that |m(t)|/Ac � 1 (so that m2(t)/A2

c can be neglected), we findthat the envelope detector output is approximately

Ac(1 + cos(φ)m(t)/Ac) = Ac + m(t) cos(φ).

8. Haykin and Moher, Problem 3.31

If g(t) = m(t) cos(2πfct), then G(f) = 12(M(f − fc) + M(f + fc)). Thus G(f) =

−jsgn(f)12(M(f − fc) + M(f + fc)). If m(t) is low-pass, with bandwidth smaller than

fc, it follows that G(f) = 12j

(M(f − fc) − M(f + fc)). Thus, under the conditionsstated,

[m(t) cos(2πfct)]∧ = m(t) sin(2πfct)

and similarly[m(t) sin(2πfct)]

∧ = −m(t) cos(2πfct)

(a) The upper-SSB signal is given by

su(t) =Ac

2m(t) cos(2πfct)−

Ac

2m(t) sin(2πfct).

Thus

su(t) =Ac

2m(t) sin(2πfct) +

Ac

2m(t) cos(2πfct).

Collecting these two equations in matrix form, we have[su(t)su(t)

]=

Ac

2

[cos(2πfct) − sin(2πfct)sin(2πfct) cos(2πfct)

] [m(t)m(t)

].

Inverting this system of equations yields[m(t)m(t)

]=

2

Ac

[cos(2πfct) sin(2πfct)− sin(2πfct) cos(2πfct)

] [su(t)su(t)

].

(b) The lower-SSB signal is given by

sl(t) =Ac

2m(t) cos(2πfct) +

Ac

2m(t) sin(2πfct).

4

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Following the same approach as in (a), we get[sl(t)sl(t)

]=

Ac

2

[cos(2πfct) sin(2πfct)sin(2πfct) − cos(2πfct)

] [m(t)m(t)

],

hence [m(t)m(t)

]=

2

Ac

[cos(2πfct) sin(2πfct)sin(2πfct) − cos(2πfct)

] [sl(t)sl(t)

].

(NB: the expression for m(t) given in the text is incorrect.)

(c) If s(t) denotes the SSB signal (either upper-SSB or lower-SSB), then

m(t) =2

Ac

[s(t) cos(2πfct) + s(t) sin(2πfct)].

This yields the block diagram shown below.

m(t)

Hilbert

H

Transform

2Ac

sin(2πfct)

2Ac

cos(2πfct)s(t)

Interestingly (ignoring scale factors), this diagram is the same as that of a lower-SSB modulator, with m(t) and s(t) reversed.

9. The best way to approach this problem is to draw pictures.

Here are example baseband spectra:

500−500

f

S3(f), Sf (4)

100−100

f

S1(f), S2(f)

(a) QAM modulated spectrum:

f1 f2

f

s3, s4

s1, s2

To minimize spectral extent, choose f1 = 100 Hz and f2 = 700 Hz. The totalbandwidth is then 1200 Hz.

5

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(b) Lower SSB spectrum:

f

f3 f4

s4s3

f1 f2

s2s1

Upper SSB spectrum:

f

f2 f4

s4s3

f3

s2s1

To minimize spectral extent in the lower-SSB case, choose f1 = 100 Hz, f2 =200 Hz, f3 = 700 Hz, and f4 = 1200 Hz. In the upper-SSB case, choose f1 = 0,f2 = 100 Hz, f3 = 200 Hz, and f4 = 700 Hz. In both cases, the bandwidth is 1200Hz.

6

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ECE316F Communication

Fall 2007 Systems

Problem Set 6

1. Haykin and Moher, Drill Problem 9.3, pg. 377. (Solution to 9.2 on next page)

1

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2. Haykin and Moher, Drill Problem 9.2, pg. 373.

3. Haykin and Moher, Drill Problem 9.4, pg. 377.

A phase shift does not impact on the envelope detector. However, a phase error of θ reducesthe output SNR by a factor of cos2 θ.

4. Haykin and Moher, Problem 9.9.

The frequency response of this filter is given by H(f) = 1/ (1 + j2πfRC), i.e., the noise PSDis given by:

SN (f) =N0

2|H(f)|2 =

N0

2

1

1 + (2πfRC)2,

The total noise power is therefore

Pn =

−∞

SN (f)df =N0

4RC.

Due to the filter, the sinusoid is attenuated by H(fc). Since the average power of the inputsinusoid is A2

c/2, the average output power of the sinusoid is given by:

(

A2c

2

)

1

1 + (2πfcRC)2,

and the SNR is given by:

SNR =

(

A2c

2

)

1

1 + (2πfcRC)2

(

4RC

N0

)

.

5. Haykin and Moher, Problem 9.10.

At a frequency of 200kHz, the noise PSD is approximately 0.5 × 10−18 = 5 × 10−19W/Hz.Setting this to be N0/2 (assuming this to be constant over the small message bandwidth of4kHz) we have N0 = 10−18W/Hz.

Furthermore, the received signal power is A2cP/2 which is given to be -80dBm = 10−11W

(note that in the dBm scale, the ”reference power” is 1mW).

The post-detection SNR is therefore

SNR =A2

cP

2N0W=

10−11

4 × 103 × 10−18= 2.5 × 103 =≃ 34dB.

2

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6. Haykin and Moher, Problem 9.13 (a) (b)

The power in the message signal is

P =

−∞

SM (f)df =

W

−W

a|f |

Wdf = 2

W

0

af

Wdf = aW.

Therefore

(a)

SNRDSB =A2

cP

2N0W=

A2ca

2N0

(b)

SNRAM =A2

ckaP

2N0W= 0.09

A2ca

2N0

3

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ECE316F Communication

Fall 2007 Systems

Problem Set 7

(For week of November 5th)

1. Haykin and Moher, Haykin and Moher, Problem 4.19.

Answer: The given signal may be written as

s(t) = cos(2πfct)− cos(2πfct)rect(t/T ) + cos(2π(fc + ∆f)t)rect(t/T ).

Since

cos(2πfct)←→1

2δ(f − fc) +

1

2δ(f + fc)

and

rect(t/T ) cos(2πf0t)←→T

2sinc((f − f0)T ) +

T

2sinc((f + f0)T )

we have

S(f) =1

2δ(f − fc) +

1

2δ(f + fc)−

T

2sinc((f − fc)T )−

T

2sinc((f + fc)T )

+T

2sinc((f − fc −∆f)T ) +

T

2sinc((f + fc + ∆f)T ).

This function is sketched below for the parameters fc = 20, ∆f = 3.

2320-20-23

As usual in FM, we find that the spectrum occupies all frequencies, but is concentrated nearthe carrier frequency.

1

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2. The signal m(t) = 10cos(200πt) + 40cos(400πt) is frequency modulated with modulator con-stant kf = 10. Estimate the bandwidth of the FM signal using Carsons rule.

Answer: The peak signal value (achieved, for example, at t = 0) is mp = 50. The maximumfrequency deviation (in Hz) is ∆f = kfmp = 500 Hz. The effective bandwidth (highestfrequency component) of m(t) is B = 200 Hz. Hence, by Carson’s rule, the bandwidth of theFM signal is BFM = 2(B + ∆f) = 1400 Hz.

3. Instead of applying a message signal m(t) to the input of an FM modulator, it is decided toapply m1(t) = m(t) + m′(t).

(a) Show that the signal being transmitted is

ϕ(t) = A cos

(

2πfct + 2πkfm(t) + 2πkf

t

−∞

m(τ)dτ

)

.

Answer: In FM, if the message signal is m(t),

s(t) = A cos

(

2πfct + 2πkf

t

−∞

m(τ)dτ

)

,

i.e., the phase function is related to the integral of the message. Here,∫

t

−∞m1(t)dt =

m(t) +∫

t

−∞m(τ)dτ , i.e., the transmitted signal is as above.

(b) Specify the impulse response of a filter that recovers in the absence of noise m(t) atthe output of an FM demodulator.

Answer: Note that M1(f) = M(f)(1 + j2πf), i.e., to invert this operation at thereceiver, pass the demodulated signal through a filter with frequency response H(f) =1/(1 + j2πf), which corresponds to an impulse response of h(t) = e−tu(t).

4. For any integer n, we have

cos(θ) cos(nθ) =1

2[cos((n− 1)θ) + cos((n + 1)θ)].

(a) Write cos5(θ) as a linear combination of 1, cos(θ), cos(2θ), . . . cos(5θ).

Answer: From above,

cos2(θ) =1

2[1 + cos(2θ)]

cos3(θ) =1

4[cos(−θ) + cos(θ) + cos(θ) + cos(3θ)]

=1

4[3 cos(θ) + cos(3θ)]

cos4(θ) =1

8[3 + 3 cos(2θ) + cos(2θ) + cos(4θ)]

=1

8[3 + 4 cos(2θ) + cos(4θ)]

cos5(θ) =1

16[3 cos(−θ) + 3 cos(θ) + 4 cos(θ) + 4 cos(3θ) + cos(3θ) + cos(5θ)]

=1

16[10 cos(θ) + 5 cos(3θ) + cos(5θ)]

2

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(b) For any integer n ≥ 1, show that cosn(θ) can be written as a linear combination of 1,cos(θ), cos(2θ), . . . cos(nθ). [Hint : use mathematical induction].

Answer: Let P (n) denote the proposition that cosn(θ) can be written as a linear com-bination of 1, cos(θ), . . . , cos(nθ). Since cos1(θ) = cos(θ), P (1) is certainly true.

Suppose P (n) is true for some n ≥ 1. This would mean that

cosn(θ) =n

i=0

ai cos(iθ)

for some coefficients a0, a1, a2, . . . , an. Then

cosn+1(θ) = cos(θ) cosn(θ)

= cos(θ)n

i=0

ai cos(iθ)

=n

i=0

ai cos(θ) cos(iθ)

=1

2

n∑

i=0

(ai cos((i− 1)θ) + ai cos((i + 1)θ))

=1

2

n∑

i=0

ai cos((i− 1)θ) +1

2

n∑

i=0

ai cos((i + 1)θ)

=1

2a0 cos(θ) +

1

2

n−1∑

j=0

aj+1 cos(jθ) +1

2

n+1∑

j=1

aj−1 cos(jθ),

which is a linear combination of 1, cos(θ), . . . , cos((n + 1)θ). Thus P (n + 1) is true.

Since P (1) is true and P (n) → P (n + 1) for all n ≥ 1, it follows from the principle ofmathematical induction that P (n) is true for all n ≥ 1.

5. In this problem, you will design an FM transmission system for a message signal m(t), wherem(t) is assumed to be a low-pass signal of bandwidth 10 kHz satisfying, for all t, |m(t)| ≤ 1.You are allocated spectrum of 200 kHz bandwidth, centered at a carrier frequency of 100MHz, i.e., extending from 99.9 MHz to 100.1 MHz.

(a) Find the maximum value of the modulation parameter kf for which the resulting FMsignal is, according to Carson’s rule, contained within the given bandwidth.

Answer: We have B = 10 KHz and mp = 1. According to Carson’s rule, the bandwidthis given as 2(∆f + B) = 200 KHz, from which we obtain ∆f = 90 KHz. Since ∆f =kfmp = kf , we obtain kf ≤ 90000.

PowerAmplifier

BandpassFilter

FrequencyMultiplier×50

(kf )NBFM =? fMIX =? Passband = ?

FrequencyMultiplier×20

NBFMModulator

(fc)NBFM = 225 KHz 2 cos(2πfMIXt)

∼ ∼

fc = 100 MHzkf = 75000

3

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(b) You decide to implement a system with kf = 75000, using Armstrong’s indirect method,according to the block diagram shown above. The carrier frequency of the first-stagenarrowband FM modulator is 225 KHz. Determine the following parameters: i) themodulation parameter (kf )NBFM for the first-stage narrowband FM signal signal; ii) themixing frequency fMIX; iii) the passband of the bandpass filter (i.e., specify the smallestpossible range of (positive) frequencies passed by the filter). The filter is to be designedwith as narrow a passband as possible.

Answer: (i) We see from the figure that kf = 1000(kf )NBFM; hence (kf )NBFM = 75.

(ii) After frequency-multiplying by a factor of 20, the resulting carrier frequency is225 KHz×20 = 4.5 MHz. On the other hand, prior to frequency-multiplying by a factorof 50, the carrier frequency should be 100 MHz/50 = 2 MHz. We need to down-shift from4.5 MHz to 2 MHz, which we can do by choosing fMIX = 6.5 MHz or fMIX = 2.5 MHz,so that |4.5 MHz− fMIX| = 2 MHz.

(iii) The input to the bandpass filter is an FM signal at carrier frequency 2 MHz,with ∆f = 75 × 20 = 1500 Hz. By Carson’s rule, the bandwidth of this FM signal is2(∆f + B) = 2(1500 + 10000) = 23 KHz. The filter passband should therefore extendfrom 2 MHz− 11.5 KHz to 2 MHz + 11.5 KHz, i.e, from 1.9885 MHz to 2.0115 MHz.

6. “Short snapper” from Final 2006: The Canadian regulatory body the Canadian Radio-television and Telecommunications Commission (CRTC) assigns a local FM radio station atotal of 100 kHz of bandwidth instead of the usual 200 kHz. The station decides to set themaximum value of the message signal to 3.5 V. The message signal is bandlimited to 15 kHz.

(a) In units of kHz/V, what is the maximum frequency sensitivity factor that Carson wouldrecommend?

Answer: According to Carson, bandwidth of the transmitted signal, s(t), is

BT = 2 (∆f + W ) , ∆f = kf |m(t)|max ⇒ 100 = 2(3.5kf + 15)⇒ kf = 10kHz/V

(b) If the carrier frequency is fc = 99.3 MHz, and assuming that Carsons rule is sufficientlyaccurate, what frequency range would the transmitted signal occupy?

Answer: The transmitted signal, s(t) occupies the bandwidth fc−BT /2 to fc + BT /2,i.e., the frequencies occupied is 99.25MHz to 99.35MHz.

4

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ECE316F Communication

Fall 2007 Systems

Problem Set 8

(For week of November 12th)

1. Haykin and Moher, Drill Problem 9.8, page 390.

Answer: The post-detection and pre-detection SNRs are related by

SNRFM

post= SNRFM

pre

3

4

(

BT

W

)3

.

Note 3

4

(

BT

W

)3

= (3/4) × (30 × 106/6 × 106)3 = 93.75 ≡ 19.71dB. The post-detectionSNR is therefore, 15 + 19.71 = 34.71dB.

Using the equation at the bottom of page 389,

I =(W/f3dB)2

3 [(W/f3dB) − tan−1 (W/f3dB)].

Here f3dB = 800kHz = 0.8MHz. Plugging the numbers in, I = 23.2 ≡ 13.6dB.

2. Haykin and Moher, Problem 9.13(c).

Answer: The average power of the message is given by

P =∫

−∞

SM(f)df = 2∫

W

0

af

Wdf = aW.

For frequency modulation and detection with kf = 500Hz/V , the post-detection SNRis

SNRFM =3A2

ck2

fP

2N0W 3=

A2

ca

2N0

× 3 ×

[

kf

W

]2

3. Haykin and Moher, Problem 9.17.

Answer: Carson’s rule tells us

BT = 2 (∆f + fm) = 2 (kfAm + fm) = 2(1000(1) + 200) = 2400Hz.

The pre-detection SNR is

SNRFM

pre= 500 =

A2

c

2N0BT

⇒A2

c

2N0

= 500 × 2400 = 1.2 × 106Hz.

1

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Due to the strange post-detection filter, the output noise power is given by

N0

A2c

× 2 ×∫

−100

−300

f 2df =2N0

3A2c

[

3003 − 1003]

=2N0

3A2c

× 2.6 × 107.

The post-detection SNR is therefore,

SNRFM

post=

3A2

ck2

fP

2N0(2.6 × 107)= 69230.8 = 48.4dB,

since kf = 1000Hz/V and P = A2

m/2 = 0.5W.

4. Haykin and Moher, Drill problem 5.1, page 192.

The goal of this problem is to illustrate the fact that sampling in domain (time/frequency)necessarily implies periodicity in the other (frequency/time). In Section 2.5, the text-book introduces the notion that a periodic signal, gT0

(t), with period T0 can be viewedas a periodic repetition (every T0) of a template or generating function g(t) where

g(t) =

{

gT0(t) −T0

2≤ t < T0

2

0 elsewhere⇒ gT0

(t) =∞∑

m=−∞

g(t − mT0), (1)

i.e., g(t) is one period of gT0centered around zero. This section of the textbook then

goes on to show that, in the frequency domain, the Fourier transform of gT0(t), GT0

(f),is G(f) sampled every f0 = 1/T0 Hz.

Consider what this implies: g(t) [with transform G(f)] is not a periodic function; gT0(t)

is its periodic extension (it is g(t) repeated every T0). Creating this periodic extensionin the time domain implies sampling in the frequency domain:

GT0(f) = f0

∞∑

n=−∞

G(nf0)δ(f − nf0). (2)

Note that since the Fourier transform is (for all intents and purposes) a one-to-onemapping, one can reverse this last statement: sampling in the frequency domain im-

plies a periodic extension in the time domain. Now on to answer Q1 (drill problem 5.1):

(a) From Eqn. (2),

gT0(t) ⇀↽ GT0

(f),

⇀↽ f0

∞∑

n=−∞

G(nf0)δ(f − nf0),

⇒ T0gT0(t) ⇀↽

∞∑

n=−∞

G(nf0)δ(f − nf0), (since T0 = 1/f0),

2

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Now using the expansion of gT0(t) in Eqn. (1), we have the required answer:

⇒ T0

∞∑

m=−∞

g(t − mT0) ⇀↽∞∑

n=−∞

G(nf0)δ(f − nf0). (3)

(b) Duality suggests that in any valid sentence one can switch the words “time” and“frequency” with the sentence still valid (For the most part. Formally, one has to switch

“f” with t and “t” with “−f”). Above the solution to part (a) we said “sampling in thefrequency implies a periodic extension in the time domain”. Using duality, sampling

in the time domain implies a periodic extension in the frequency domain. In Eqn. 3,we have:

T0

∞∑

m=−∞

g(t − mT0) ⇀↽∞∑

n=−∞

G(nf0)δ(f − nf0). (4)

Using the duality property and replacing f0 with Ts and T0 with fs, we get

⇒ fs

∞∑

m=−∞

g(−f − mfs) ⇀↽∞∑

n=−∞

G(nTs)δ(t − nTs). (5)

Now duality also tells us that G(t) ⇀↽ g(−f), i.e., g(−f) is the Fourier transform ofG(t). Letting F represent the Fourier transform, F [G(t)] = g(−f). Switching theletters t and f , F [G(f)] = g(−t) we get

fs

∞∑

m=−∞

G(−f − mfs) ⇀↽∞∑

n=−∞

g(−nTs)δ(t − nTs),

⇒ fs

∞∑

m=−∞

G(f − mfs) ⇀↽∞∑

n=−∞

g(nTs)δ(t − nTs), (6)

which is the statement in the bottom right of Table 5.1.

5. Haykin and Moher, Drill problem 5.5, page 196 (rate only. The interval is reciprocalof the rate).

(a) g(t) has bandwidth of 100Hz, hence, the Nyquist rate is 200Hz.

(b) The signal here is the square of the signal in part (a). Hence, its bandwidth is200Hz and the Nyquist rate is 400Hz.

(c) The first term has bandwidth of 100Hz, the second a bandwidth of 200Hz. Hence,total bandwidth is 200Hz and the Nyquist rate is, again, 400Hz.

3

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1

(a) (b)

p(t)

px (t)Anti

Alias

Filter

x(t)

X(f)

-W Wf

Figure 1: X(f), the Fourier transform of x(t) and impulse train sampling for Q. 6.

6. Consider a signal x(t) whose Fourier transform is shown in Fig. 1(a), where W = 104Hz.This signal is to be sampled using an impulse train, p(t) =

n=−∞δ(t − nTs), with

sampling period Ts = 10−4s.

(a) Design an anti-aliasing filter given this sampling rate, i.e., modify the samplingscheme to that in Fig. 1(b). You need to provide the frequency response of the filter,indicating all parameters that define the filter.

In this problem fs = 1/Ts = 104Hz. The anti-aliasing filter ensures that the input tothe sampler meets Nyquist’s criterion. The filter is, therefore, a low-pass filter withcut-off = fs/2 = 5000Hz. The amplification factor is largely irrelevant and may be,conveniently, set to 1.

(b) On using the anti-aliasing filter in Part (a), carefully sketch Xp(f), the Fouriertransform of xp(t), in the range f ∈ (−2.5W, 2.5W ).

Note that W = fs. Denote the output of the anti-aliasing filter as g(t). Its Fouriertransform, G(f) is given in Fig. 2(a). The sampling process causes a periodic repetitionof G(f) every fs. Therefore, the Fourier transform of xp(t), Xp(f) is as given inFig. 2(b) below.

X (f)

f

- f /2s f /2s

p

f s

5f /2s -5f /2s -3f /2s 3f /2s

G(f)

- f /2s f /2s

f s

(a) (b)

Figure 2: Solution to Q3.

4

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ECE316F Communication

Fall 2007 Systems

Problem Set 9

(For week of November 19th)

1. Haykin and Moher, Problem 5.14.

The goal of this problem is to illustrate multiplexing. In this problem, s1(t) and s2(t)have bandwidth of 80Hz (with a Nyquist frequency of 2 × 80 = 160 Hz). s3(t) ands4(t) are sampled at a rate of fs = 2400 samples/second.

(a) In theory, therefore, the sampling rate could be reduced by a factor of 2400/160 =15. Since the sampling rate must be reduced by a factor of 2R (integer R) only, we canset R = 3 and reduce the sampling rate by a factor of 23 = 8. The signals s1(t) ands2(t) are therefore sampled at a rate of 2400/8 = 300 samples/sec.

(b) This problem will be largely “solved” via figures. There is no one answer and thesolution presented here is the simplest. The first step is to combine the PAM signalscorresponding to s1(t) and s2(t) into a single signal s5(t). Figure 1 illustrates thisprocess. The figure uses arbitrary signals. The width of each pulse is chosen sufficientlysmall. Note that because s5(t) combines two signals, its effective bandwidth doubles(a “sample” every 1/600 seconds).

The next step is to combine this signal s5(t) with the other two, s3(t) and s4(t), asshown below. The signals s3(t) and s4(t) are interleaved such that the there are samplesof each signal every 1/2400 seconds. The samples of s5(t) are interleaved every 1/600seconds. For this interleaver to work, we therefore require that the pulse width (T ) beless than 1/(3fs), i.e., less than 1/7200 seconds.

1

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1 1

2

2

5

(1)5

(2)

0 1/300

0

1/300

5

(1)

5

(2)

2/6001/600

3/600

Figure 1: Combining two signals into a single multiplexed signal.

3 5

(1)

01/2400

5

(2)

43

4 3

4

3

4

1/600

3

4

Figure 2: Combining s3(t), s4(t) and s5(t) into a single multiplexed signal.

2

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2. Haykin and Moher, Problem 5.15(a).

Figure 3: Solution to Problem 5.15(a)

3. (Final 2006) In wideband speech processing, a speech signal is assumed to be ban-dlimited to 7 kHz.

(a) Assuming that the signal is sampled at the Nyquist rate and quantized using a 512-level quantizer, what is the bit rate of the corresponding pulse code modulation(PCM) system?

Answer: BW of the message signal is W = 7kHz. The Nyquist sampling rate istherefore 2W = 14kHz. Furthermore, 512 = 29, i.e., each sample creates 9 bitsmaking the bit rate = 14 × 9 = 126kbps.

(b) The system is found to achieve a signal-to-quantization noise ratio (SQNR) of60 dB. How many levels would the quantizer need to have in order to achieve anSQNR of 72 dB?

Answer: Each bit adds 6dB to the SQNR, i.e., if 9 bits provide 60dB of SQNR,we need 11 bits for 72dB of SQNR. 11 bits can address 2048 levels.

3

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4. (Final 2006)

(a) The message signal, m(t), has bandwidth 4 kHz, and is scaled so that −2 ≤m(t) ≤ 2. In the initial design, m(t) is sampled at the minimum rate required tomeet Nyquist’s sampling criterion. What is this minimum sampling rate?

Answer: Since message bandwidth = W = 4kHz ⇒ Nyquist’s sampling rate= 2W = 8kHz.

(b) Next, the samples are quantized. What is the minimum number of bits requiredper sample if we require the quantization error ǫ to satisfy −0.25 ≤ ǫ ≤ 0.25?

Answer: Let ∆ be the spacing between levels. The maximum error is ∆ = 2,i.e., ∆ = 0.5. The levels are, therefore, -1.75, -1.25, -0.75, -0.25, 0.25, 0.75, 1.25and 1.75, i.e., we have 8 levels, requiring 3 bits.

(c) Using the sampling rate found in (4a) and 8 bits per sample, what bit rate isrequired to communicate this signal?

Answer: Each sample creates 8 bits and there are 8k samples per second, i.e.,bit rate = 64kbps.

4

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ECE316F Communication

Fall 2007 Systems

Problem Set 10: Solutions

(For Thurs., Nov. 29th, Mon., Dec. 3rd and Tues. Dec. 4th)

1. Haykin and Moher, Problem 6.3.

For the ideal sinc pulse shape, its Fourier transform is the rect which is clearly discon-tinuous, i.e., the zero-th order derivative is discontinuous ⇒ k = 0. This pulse falls off,as expected, as 1/t.

For the raised cosine with 0 < α ≤ 1, we have

P ′(f) =dP

df=

0 |f | < f1,

E

4B0

π

2(B0−f1)(−sgn(f)) sin

[

π(|f |−f1)

2(B0−f1)

]

f1 ≤ |f | ≤ 2B0 − f1,

0 |f | > 2B0 − f1.

Note that this function is continuous (the sin term is 0 at the end points, f = f1 andf = 2B0 − f1. One can show that the second derivative is discontinuous (the sin termbecomes a cos). Hence the corresponding pulse drops off as 1/t3. Note

p(t) ∝ sinc(2B0t)

(

cos(2παB0t)

1 − 16α2B20t

2

)

∝sin(2πB0t)

2πB0t

(

cos(2παB0t)

1 − 16α2B20t

2

)

,

which gives the 1/t3 behavior.

2. Haykin and Moher, Problem 6.9 (a) (d)

(a) Since the data rate is 56kbps, Tb = 1/56000s and B0 = 1/(2Tb) = 28kHz. Withα = 0.25, B = B0(1 + α) = 35kHz (d) α = 1.0 ⇒ B = B0(1 + α) = 56kHz.

“Points to ponder”: In both cases we’re transmitting data at 56kbps. What did webuy by using 56kHz in part (d) as opposed to only 35kHz in part (a)?

3. Haykin and Moher, Problem 6.10

We have Tb = 10µs ⇒ B0 = 1/(2Tb) = 100kHz/2 = 50kHz. The system uses abandwidth of 75kHz. The excess bandwidth is therefore 25kHz = αB0 ⇒ α = 0.5.

4. Haykin and Moher, Problem 6.13 (a) (c)

It’s rather unfortunate that the textbook uses B0 to denote the parameter of the pulseshapes. In the chapter we used B0 to be something specific; namely 1/(2Tb).

(a) As derived in class, if Tb is the bit period, the criterion for zero ISI in the frequencydomain is

1

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∞∑

k=−∞

P

(

f −k

Tb

)

= constant.

We showed that this implies P(

1

2Tb+ x

)

+P ∗

(

1

2Tb− x

)

= constant, i.e., odd symmetry

about the f = 1/(2Tb) point.

In this particular example,

P (f) =

{

E

2B0

(

1 − |f |

B0

)

|f | ≤ B0,

0 |f | > B0

For this to be odd symmetric about the point f = 1/(2Tb), we need B0 = 1/Tb.

(c) Let’s start with a hint: think in terms of Problem 6.3 whose solution is given above.

Note that while the pulse spectrum in Fig. 6.12(a) looks a LOT like the raised cosinespectrum with α = 1, the first derivative of this pulse is discontinuous, i.e., the pulsefalls off at 1/t2, not 1/t3. (The triangle is a convolution of two rect functions, i.e., thepulse shape is a sinc2).

5. Haykin and Moher, Problem 6.17. Use a data rate of 100kbps. Also, repeat the eyediagram with a 10% jitter in timing.

The following figure plots p(t), the raised cosine pulse with α = 1. Note Tb = 1/105 =10µs.

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

x 10−5

−0.2

0

0.2

0.4

0.6

0.8

Time

Pul

se s

hape

p(t

)

Problem 6.17, the pulse shape with α = 1

Figure 1: The pulse shape with Tb = 10µs.

Therefore the given 6-bit sequence creates the following received signal:

2

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−2 −1 0 1 2 3 4 5 6 7

x 10−5

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Time

Rec

eive

d si

gnal

y(t

)

Problem 6.17, the received signal for the given 6−bit sequence

Figure 2: The received signal for Problem 6.17.

0 0.5 1 1.5 2

x 10−5

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Time

Rec

eive

d si

gnal

s

Problem 6.17, the eye diagram with ideal sampling

Eye opening

Figure 3: The eye diagram with ideal sampling.

The eye diagram plots the signal corresponding to each received bit (generally a verylong random sequence of 1 and 0) one on top of the other. See Fig. 3.

With a 10% error in the sampling times, we get an eye diagram as in Fig. 4:

Note the significantly smaller eye opening.

6. (Final 2006): A 4-level baseband pulse transmission system operates in a bandwidthof 10 kHz using bandlimited pulses with 25% excess bandwidth. What bit rate issupported by this system?

Answer: If Ts is the symbol period, let B0 = 1/2Ts. The bandwidth used is B0(1+α)where α = 0.25. Hence,

1.25

2Ts

= 10 × 103 ⇒1

Ts

= 20 × 1031.25 = 16 × 103symbols per second.

3

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0 0.5 1 1.5 2

x 10−5

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Time

Rec

eive

d si

gnal

s

Problem 6.17, the eye diagram with 10% error in sampling

Eye opening

Figure 4: The eye diagram with 10% sampling jitter.

Given 4 levels, each symbol comprises 2 bits, i.e, the bit rate is 32kbps.

7. (Final 2006): Suppose that g(t) is a pulse that satisfies Nyquist’s criterion for zerointersymbol interference assuming a symbol interval of T seconds and that the centralsample appears at time t = 0. Assuming the symbol interval remains at T seconds,which of the following must also satisfy Nyquist’s criterion? (Justify your answer ineach case.)

(a) g(−t)

Let g1(t) = g(−t). g1(iT ) = g(−iT ) = 0∀i 6= 0, so g(−t) does satisfy Nyquist’s

zero-ISI criterion.

(b) g(2t)

Let g2(t) = g(2t). g2(iT ) = g(2iT ) = 0∀i 6= 0, so g(2t) does satisfy Nyquist’s

zero-ISI criterion.

(c) g(t/2)

Let g3(t) = g(t/2). g3(iT ) = g((i/2)T ) which may not be zero for odd i. So,g(t/2) need not satisfy Nyquist’s zero-ISI criterion.

8. (Final 2006)

(a) A twisted-pair wireline can support transmission of a baseband signal. Assumethat the bit rate is set to 64 kbps. To avoid ISI, you decide to use a raised-cosinepulse shape with 100% excess bandwidth (α = 1). What transmission bandwidthis needed, assuming binary pulse amplitude modulation?

Answer: With α = 1, the bandwidth used by the pulses = 1/Tb. With a bit rateof 64kbps, this bandwidth = 64kHz.

4

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(b) After some experimentation, you find that the usable bandwidth on the givenchannel is 48 kHz. To increase the efficiency of your system, you decide to use araised-cosine pulse with 25% excess bandwidth, and to change the quantizationscheme to delta modulation. With this scheme, what is the maximum rate atwhich the message signal may be sampled?

Answer: With α = 0.25, bandwidth = 1.25/2Tb = 48kHz. Therefore 1/Tb =48 × 2/1.25 = 76.8kbps. Since, in delta modulation, each sample creates a singlebit, the sampling rate can be as high as 76.8k samples per second.

9. (Final 2006) In a particular baseband data transmission system the basic transmittedpulse is rectangular, given by g(t) = rect(t−1/2). The channel is ideal, with frequencyresponse H(f) = 1.

(a) The receiver filter is linear time-invariant, with impulse response given by q(t) =rect(t− 1/2). Given that rect(t) ∗ rect(t) = (1− |t|)rect(t/2), show that when g(t)is transmitted, the output of the receiver filter is the triangular pulse

p(t) = (1 − |t − 1|)rect((t − 1)/2).

Answer: The overall pulse shape p(t) = g(t) ⋆ h(t) ⋆ q(t) = g(t) ⋆ q(t). Sinceg(t) = q(t) are both versions of rect(t) delayed by 0.5, and we are given rect(t) ∗rect(t) = (1 − |t|)rect(t/2),

g(t) ⋆ q(t) = rect(t − 1/2) ⋆ rect(t − 1/2) = (1 − |t − 1|)rect((t − 1)/2).

(b) To transmit N bits, the pulse train

s(t) =N−1∑

k=0

akg(t − k)

is sent, where ak ∈ {+1,−1}. Sketch the transmitted pulse train s(t) correspond-ing to the eight bit sequence a0 = +1, a1 = +1, a2 = −1, a3 = +1, a4 = −1,a5 = −1, a6 = +1. a7 = −1.

$s(t)$

$t$

1 5

$+1$

$−1$

8

Figure 5: Transmit Signal.

(c) Sketch the corresponding output of the receiver filter when s(t) from (9b) istransmitted.

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$t$

1 5

$+1$

$−1$

8

$y(t)$

2 3 4 76

Figure 6: Received Signal after filtering.

(d) Does this system satisfy Nyquist’s criterion for zero intersymbol interference?Explain.

Answer: Yes. The symbol interval is T = 1, and p(kT ) = p(k) = 0 for everynonzero integer k.

(e) Sketch the eye pattern for this system, and identify the times at which the receiverfilter output should be sampled in order to minimize ISI.

Answer: Note that there need not be ISI, at any t = iTb, the signal can take ononly ±1 at these sample points. Hence the possible transitions between samplesare 1 to 1, 1 to -1, -1 to 1 and -1 to -1:

1

-10 T 2T

b b

Figure 7: Eye diagram for the binary case.

The ideal sampling time is clearly at iTb.

(f) Suppose, instead of the binary alphabet {+1,−1}, the system were to use thequaternary alphabet {+3, +1,−1,−3}. Sketch the eye pattern for this system.

Answer: Note that there need not be ISI, at any t = iTb, the signal can takeon only ±1 or ±3 at these sample points. All possible transitions between thesepoints are possible:

The ideal sampling time is clearly at iTb.

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-1

-30 T 2T

b b

1

3

Figure 8: Eye diagram for the quanternary case.

7