ECE316F
Fall 2007
Communication
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)ejπ/6
(e)
ejπ/3√
−1
1+j 3
2. Use Euler’s relation (ejθ = cos(θ) + j sin(θ)) to show that:
(a) sin(θ) sin(φ) = 12 cos(θ − φ) − 12 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
ECE316F
Fall 2007
Communication
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 than
typical exam questions.]
3. Using the result of the previous question, find
and m are real values.
R∞
−∞
exp(−a(x − m)2 )dx, where a > 0
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
ECE316F
Fall 2007
Communication
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(2πfc t), Sx (f ) =
(1/4) [Sg (f − fc ) + Sg (f + fc )]. Show that the same is true if x(t) = g(t) sin(2πfc t),
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 the
approach suggested in class/the handout on the PSD of a sinusoid that was posted as
opposed 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.
1
ECE316F
Fall 2007
Communication
Systems
Problem Set 4
(For week of October 8th)
1.
2.
3.
4.
Haykin
Haykin
Haykin
Haykin
and
and
and
and
Moher,
Moher,
Moher,
Moher,
Drill
Drill
Drill
Drill
Problem
Problem
Problem
Problem
3.1
3.2
3.3
3.4
5.
6.
7.
8.
Haykin
Haykin
Haykin
Haykin
and
and
and
and
Moher,
Moher,
Moher,
Moher,
Drill Problem 3.5
Drill Problem 3.6
Problem 3.18
Problem 3.19
9. In this question, a(t), b(t), c(t) and d(t) are lowpass energy signals, bandlimited to
B 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(2πfc t)+[c(t) cos(2πBt)+d(t) sin(2πBt)] sin(2πfc t).
(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 to
b(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(2πfc t) is orthogonal to
[c(t) cos(2πBt) + d(t) sin(2πBt)] sin(2πfc t).
(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 oscillators
and the bandwidths of all lowpass filters.
10. Throughout this question, the signal m(t) is lowpass with a maximum frequency of
W = 20 KHz.
(a) It is desired to generate the modulated wave y(t) = Am(t) cos(2πfc t), where
fc = 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 frequency
and bandwidth available, explain carefully what filtering operations are needed
and 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 sample
m(t) with a sampling interval of Ts = 1/(10W ) = 5 × 10−6 seconds to produce
P
z(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(2πfc t) where fc = 1 MHz, we multiply
it 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) =
∞
X
(
pk cos(k2πf0 t), pk =
k=0
0
4
(−1)(k−1)/2 kπ
if k = 0, 2, 4, 6, . . .
if k = 1, 3, 5, 7, . . .
The signal y(t)p(t) is then lowpass-filtered to retrieve m(t). Explain carefully
what the output of the lowpass filter (with unity gain) is.
1
ECE316F
Fall 2007
Communication
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 field
records 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 radio
transmission system to send these signals back to the home office for processing. You
are allocated spectrum in a band of frequencies near 900 MHz. You decide to multiplex
the four signals into a single composite baseband signal x(t) , and then send x(t) using
SSB 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 extent
of x(t) as the smallest value of W such that the spectrum of x(t) is contained in the
band [−W, W ].
(a) In one approach, you form the composite signal as follows. First, you create a
QAM signal at carrier frequency f1 > 0, modulated by s1 (t) and s2 (t). To this
signal you add a QAM signal at carrier frequency f2 > f1 , modulated by s3 (t) and
s4 (t). How should f1 and f2 be chosen so that the composite signal has minimum
spectral extent? What is the bandwidth of the composite signal?
(b) In another approach, you form the composite signal by summing together four
SSB-modulated signals at carrier frequencies f1 < f2 < f3 < f4 where the signal
at frequency fi is modulated with signal si . Assuming lower-SSB modulation, how
should f1 , . . . , f4 be chosen so that the composite signal has minimum spectral
extent? Repeat, assuming upper-SSB modulation. What is the bandwidth of the
composite signal?
1
ECE316F
Fall 2007
Communication
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 phase
error 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
ECE316F
Fall 2007
Communication
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 modulator
constant 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 is
decided to apply m(t) + m(t).
(a) Show that the signal being transmitted is
ϕ(t) = A cos 2πfc t + 2πkf m(t) + 2πkf
Z
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 combination
of 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 all
t, |m(t)| ≤ 1. You are allocated spectrum of 200 kHz bandwidth, centered at a carrier
frequency 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 resulting
FM signal is, according to Carson’s rule, contained within the given bandwidth.
(kf )NBFM =?
NBFM
Modulator
fMIX =?
Frequency
Multiplier
×20
Passband = ?
Bandpass
Filter
∼
∼
(fc )NBFM = 225 KHz
2 cos(2πfMIX t)
Frequency
Multiplier
×50
Power
Amplifier
fc = 100 MHz
kf = 75000
(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-stage narrowband FM modulator is 225 KHz. Determine the following
parameters: i) the modulation parameter (kf )NBFM for the first-stage narrowband
1
FM signal signal; ii) the mixing frequency fMIX ; iii) the passband of the bandpass
filter (i.e., specify the smallest possible range of (positive) frequencies passed by
the 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 Radiotelevision and Telecommunications Commission (CRTC) assigns a local FM radio
station a total of 100 kHz of bandwidth instead of the usual 200 kHz. The station
decides to set the maximum 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
would recommend?
(b) If the carrier frequency is fc = 99.3 MHz, and assuming that Carsons rule is
sufficiently accurate, what frequency range would the transmitted signal occupy?
2
ECE316F
Fall 2007
Communication
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 reciprocal
of the rate).
6. Consider a signal x(t) whose Fourier transform is shown in Fig. 1(a), where W = 104 Hz.
P
This signal is to be sampled using an impulse train, p(t) = ∞
n=−∞ δ(t − nTs ), with
−4
sampling period Ts = 10 s.
p(t)
X(f)
1
x(t)
-W
W
f
(a)
Anti
Alias
Filter
x p (t)
(b)
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 sampling
scheme 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 Fourier
transform of xp (t), in the range f ∈ (−2.5W, 2.5W ).
1
ECE316F
Fall 2007
Communication
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 bandlimited to 7 kHz.
(a) Assuming that the signal is sampled at the Nyquist rate and quantized using a 512level 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) of
60 dB. How many levels would the quantizer need to have in order to achieve an
SQNR 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 to
meet Nyquist’s sampling criterion. What is this minimum sampling rate?
(b) Next, the samples are quantized. What is the minimum number of bits required
per 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 is
required to communicate this signal?
1
ECE316F
Fall 2007
Communication
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 eye
diagram with a 10% jitter in timing.
6. (Final 2006): A 4-level baseband pulse transmission system operates in a bandwidth
of 10 kHz using bandlimited pulses with 25% excess bandwidth. What bit rate is
supported by this system?
7. (Final 2006): Suppose that g(t) is a pulse that satisfies Nyquist’s criterion for zero
intersymbol interference assuming a symbol interval of T seconds and that the central
sample 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 in
each 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. Assume
that the bit rate is set to 64 kbps. To avoid ISI, you decide to use a raised-cosine
pulse shape with 100% excess bandwidth (α = 1). What transmission bandwidth
is needed, assuming binary pulse amplitude modulation?
(b) After some experimentation, you find that the usable bandwidth on the given
channel is 48 kHz. To increase the efficiency of your system, you decide to use a
raised-cosine pulse with 25% excess bandwidth, and to change the quantization
scheme to delta modulation. With this scheme, what is the maximum rate at
which the message signal may be sampled?
1
9. (Final 2006) In a particular baseband data transmission system the basic transmitted
pulse is rectangular, given by g(t) = rect(t − 1/2). The channel is ideal, with frequency
response 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
X
ak g(t − k)
k=0
is sent, where ak ∈ {+1, −1}. Sketch the transmitted pulse train s(t) corresponding 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) is
transmitted.
(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 receiver
filter output should be sampled in order to minimize ISI.
(f) Suppose, instead of the binary alphabet {+1, −1}, the system were to use the
quaternary alphabet {+3, +1, −1, −3}. Sketch the eye pattern for this system.
2
ECE316F
Fall 2007
Communication
Systems
Problem Set 1 Solutions
1. Probably the easiest way to approach this problem is to place individual factors into polar
form, and then combine the factors.
√
(a) We have 1 + j = 2ejπ/4 , thus (1 + j)3 = 23/2 ej3π/4 .
√
√
√
(b) j 3 = −j, so 3 + j 3 = 2ejθ , for θ = tan−1 (−1/ 3) = −π/6. Also 1 − j = 2e−jπ/4 .
Thus the quantity of interest is 23/2 e−jπ/6−jπ/4 = 23/2 e−j5π/12 .
√
√
(c) The numerator is 2 − j(6/ 3) = 4ejθ for θ = tan−1 (3/ 3) = −π/3. The denominator
is 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 12 ejπ/3 .
2. (a) We have
!
sin(θ) sin(φ) =
=
=
=
!
ejθ − e−jθ
ejφ − e−jφ
2j
2j
−1 j(θ+φ)
e
+ e−j(θ+φ) − ej(θ−φ) − e−j(θ−φ)
4
!
!
1 ej(θ−φ) + e−j(θ−φ)
1 ej(θ+φ) + e−j(θ+φ)
−
2
2
2
2
1
1
cos(θ − φ) + cos(θ + φ).
2
2
(b) We have
sin(θ) cos(φ) + cos(θ) sin(φ) =
1
((y − y ∗ )(z + z ∗ ) + (y + y ∗ )(z − z ∗ ))
4j
for y = ejθ and z = ejφ . Expanding, we get
1
(yz + yz ∗ − y ∗ z − y ∗ z ∗ + yz − yz ∗ + y ∗ z − y ∗ z ∗ ) =
4j
1
(2yz − 2y ∗ z ∗ )
4j
1 j(θ+φ)
=
(e
− e−j(θ+φ) )
2j
= sin(θ + φ).
1
3. Haykin and Moher, Drill Problem 2.1:
We have g(t) = e−t sin(2πfc t)u(t) = e−t (ej2πfc t − e−j2πfc t )u(t)/2j. Thus
Z
∞
g(t)e−j2πf t dt
G(f ) =
−∞
=
=
=
=
=
1
2j
1
2j
1
2j
1
2j
∞
Z
e−t (ej2πfc t − e−j2πfc t )e−j2πf t dt
0
·
Z
∞
1
·
2j
e−t(1−j2πfc +j2πf ) dt −
0
Z
∞
e−t(1+j2πfc +j2πf ) dt
0
1
1
1
·
−
·
1 + j2π(f − fc ) 2j 1 + j2π(f + fc )
1 + j2π(f + fc ) − 1 − j2π(f − fc )
·
(1 + j2π(f − fc ))(1 + j2π(f + fc ))
2πfc
2
2
1 − 4π (f − fc2 ) + j4πf
4. Haykin and Moher, Drill Problem 2.2:
We have
Z
∞
g(t) =
−∞
0
Z
G(f )ej2πf t df
jπ/2 j2πf t
e
=
e
−W
Z
df +
W
e−jπ/2 ej2πf t df
0
0
W
ej2πf t ej2πf t = j
−j
j2πt f =−W
j2πt f =0
=
=
=
=
1
((1 − e−j2πW t ) − (ej2πW t − 1))
2πt
1 2 − ej2πW t − e−j2πW t
·
πt
2
1
(1 − cos(2πW t))
πt
2 sin2 (πW t)
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 (2πfc t) = 21 (1 + cos(4πfc t)).
1 1
G(f ) = F[ + cos(4πfc t)]
2 2
1
1
1
=
δ(f ) + δ(f − 2fc ) + δ(f + 2fc ).
2
4
4
2
(b) Similarly, g(t) = sin2 (2πfc t) = 12 (1 − cos(4πfc t)), so
1 1
G(f ) = F[ − cos(4πfc t)]
2 2
1
1
1
δ(f ) − δ(f − 2fc ) − δ(f + 2fc ).
=
2
4
4
7. Haykin and Moher, Drill Problem 2.13:
From (2.40) in the text, we know that
exp(−πt2 ) ←→ exp(−πf 2 ).
√
From the dilation property (2.20), with dilation parameter a = 1/(τ 2π), we get
√
exp(−t2 /2τ 2 ) ←→ τ 2π exp(−2π 2 τ 2 f 2 ).
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 in
the text, with fc = 1/(2T ). Thus, from (2.29) we get
Ga (f ) =
=
=
AT
AT
sinc[T (f − 1/2T )] +
sinc[T (f + 1/2T )]
2
2
AT
AT
sinc(f T − 1/2) +
sinc(f T + 1/2)
2
2
2AT cos(πf T )
.
π(1 − 4f 2 T 2 )
The last expression follows—after some manipulation—by expanding sinc(f T ± 1/2)
as ∓ cos(πf T )/(πf T ± π/2). Note that, because g(t) is real-valued and even, G(f ) is
real-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−jπf T Ga (f ).
(c) The half-sine pulse with duration aT is given by gb (t/a), and hence has Fourier transform
aGb (af ) = e−jπaf T Ga (af ).
(d) The new signal gc (t) = −ga (t + T /2); thus
Gc (f ) = −ejπf T Ga (f ).
(e) The new signal gd (t) = gb (t) + gc (t); thus
Gd (f ) = Ga (f )e−jπf T − Ga (f )e+jπf T
= −2j sin(πf T )Ga (f )
Note that since Ga (f ) is purely real, Gd (f ) is purely imaginary, which is to be expected
since gd (t) is real-valued and odd.
3
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 implies
that G(f ) is purely real.
If g(t) is real and odd, then g ∗ (−t) = −g(t), hence G∗ (f ) = −G(f ), which implies that
G(f ) is purely imaginary.
(b) From (2.35) in the Text, we know that if g(t) ←→ G(f ), then
tg(t) ←→
j d
G(f ).
2π df
By induction it follows that
tn g(t) ←→
j n dn
G(f ).
(2π)n df n
(c) From the area property and the previous result it follows that
j n dn
t g(t)dt =
G(f )
.
n
n
(2π) df
−∞
f =0
Z
∞
n
∞
(d) Let R12 (τ ) = −∞
g1 (t)g2∗ (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) =
R∞
∗
−∞ G1 (f )G2 (f )df , i.e.,
R
Z
∞
−∞
g1 (t)g2∗ (t)dt
Z
∞
=
−∞
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 ] then
the convolution is zero outside [−2W, 2W ], i.e., y(t) is bandlimited to [−2W, 2W ].
4
ECE316F
Fall 2007
Communication
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 the
duality 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)
= −2πtg(t) ←→ j2πf G(f ).
dt
On the other hand, from the dual of derivative property we have
−j2πtg(t) ←→
dG(f )
df
or
−2πtg(t) ←→ −j
dG(f )
.
df
Equating the two expressions for the Fourier transform of −2πtg(t), we get
j2πf G(f ) = −j
or
dG(f )
df
dG(f )
+ 2πf G(f ) = 0
df
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 integrating
factor M (f ), i.e., a function M (f ) such that M (f )G0 (f ) + 2πf M (f )G(f ) is equal to
1
2
the derivative of M (f )G(f ). We find that M (f ) = eπf , so the differential equation is
equivalent to
d πf 2
e G(f ) = 0.
df
2
This equation is easily solved: we find G(f )eπf = C for some constant C. Thus
2
G(f ) = Ce−πf . The remaining task is to find the constant C.
Now, we have
2
2
e−πt ←→ Ce−πf .
Applying the time-frequency duality property, we get
2
Ce−πt ←→ e−πf
2
or
1 −πf 2
e
.
C
R ∞ −πt2
Thus C = 1/C, or C 2 = 1. Now, since G(0) = C = −∞
e
dt > 0, we find that
C = 1 is the desired solution. Thus, we have shown that
2
e−πt ←→
2
2
e−πt ←→ e−πf .
(Whew!)
We could also have tried to proceed as follows. We have
2
F[exp(−πt )] =
Z
∞
−∞
∞
=
Z
exp(−πt2 ) exp(−j2πf t)dt
exp(−π(t + jf )2 − πf 2 )dt
−∞
−πf 2
= e
Z
∞
exp(−π(t + jf )2 )dt
−∞
−πf 2
= e
lim
Z
A
A→∞ −A
exp(−π(t + jf )2 )dt.
We could substitute u = t + jf to get
−πf 2
G(f ) = e
lim
Z
A+jf
A→∞ −A+jf
exp(−πu2 )du
which is a path integral in the complex plane. To solve this we could resort to Cauchy’s
Integral Theorem. However, the approach taken above avoids complex integration.
3. We have
exp(−πt2 ) ←→ exp(−πf 2 ).
From the scaling property of the Fourier transform, we have
q
exp(−π(t a/π)2 ) ←→
2
q
q
π/a exp(−π(f π/a)2 )
From the time shifting property, it follows that
q
q
q
exp(−π((t − m) (a/π))2 ) ←→ exp(−j2πf m) π/a exp(−π(f π/a)2 ).
Finally, from the area property of the Fourier transform (i.e., evaluating the Fourier
transform at f = 0) we get
Z
∞
q
exp(−π((t − m) (a/π))2 )dt =
Z
−∞
∞
exp(−a(t − m)2 )dt =
q
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
Z t
G(f ) G(0)δ(f )
g(τ )dτ = rect(t) ←→
+
,
j2πf
2
−∞
we get
sin(πf )
exp(jπf ) − exp(−jπf )
=
= sinc(f ),
rect(t) ←→
j2πf
πf
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 as
ψg (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 + 4π 2 f 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
1
1
2a
G(f ) =
+
= 2
,
a + j2πf
a − j2πf
a + 4π 2 f 2
and
4a2
ψf (g) = |G(f )|2 = 2
.
(a + 4π 2 f 2 )2
7. Haykin and Moher, Problem 2.20
g(t)
A
t
0
The signal g(t) is shown above.
3
T
(a)
1
(g(t) + g(−t))
2 t
1
−t 1
A
rect( − ) + rect(
− )
=
2
T
2
T
2
1
go (t) =
(g(t) − g(−t))
2 A
1
−t 1
t
=
− )
rect( − ) − rect(
2
T
2
T
2
ge (t) =
The even and odd parts are sketched below.
ge (t)
A/2
A/2
−T
t
−T
0
go (t)
t
0
T
T
(b) We have
Ge (f ) = F[(A/2)rect(t/2T )] = AT sinc(2f T )
and
AT
AT
sinc(f T ) − ejπf T
sinc(f T )
2
2
= −jAT sinc(f T ) sin(πf T )
Go (f ) = e−jπf T
Alternatively, from
G(f ) = e−jπf T AT sinc(f T ) = AT sinc(f T )(cos(πf T ) − j sin(πf T )),
we get
Ge (f ) =
=
=
=
=
Re[G(f )]
AT sinc(f T ) cos(πf T )
AT sin(πf T ) cos(πf T )/πf T
AT sin(2πf T )/2πf T
AT sinc(2f T )
and
Go (f ) = Im[G(f )]
= −jAT sinc(f T ) sin(πf T ).
8. Haykin and Moher, Problem 2.26
4
(a) The rectangular pulse of unit area and width T is given as g(t) = T1 rect(t/T ),
which has Fourier transform sinc(f T ). In the limit as T → 0, sinc(f T ) → 1 for
any fixed f .
(b) The unit area sinc pulse with first zero crossing at T is given as T1 sinc(t/T ). The
Fourier transform of this pulse is rect(f T ). In the limit as T → 0, rect(f T ) → 1
for any fixed f .
9. Haykin and Moher, Problem 2.27
We have G(f ) = u(f ). We know that
1
1
+ δ(f ).
j2πf
2
u(t) ←→
From the duality property of the Fourier transform we get
1
1
+ δ(t) ←→ u(−f ).
j2πt 2
From the reflection property and using the property that δ(t) is even, we get
j
1
+ δ(t) ←→ u(f ).
2πt 2
Thus g(t) =
j
2πt
+ 12 δ(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
Z
∞
|h(t)|dt < ∞.
−∞
Now if h(t) is absolutely integrable, then h(t) is square-integrable, i.e.,
∞, i.e., the impulse response has finite energy.
R∞
−∞
|h(t)|2 dt <
If x(t) is the input and y(t) the output of this system, we have ψy (f ) = |H(f )|2 ψx (f ),
and so
Z ∞
Z ∞
Ey =
ψy (f )df =
|H(f )|2 ψx (f )df.
−∞
−∞
From the Cauchy-Schwarz inequality (see Appendix 5), we have
Z
∞
−∞
2
|H(f )| ψx (f )df ≤
Z
∞
2
|H(f )| df
−∞
Z
1/2
∞
−∞
ψx (f )df
However,
since both h(t) and x(f ) have finite energy, i.e.,
R∞
−∞ ψx (f )df < ∞, we must have
Z
∞
|H(f )|2 df
Z
−∞
−∞
so y(t) has finite energy.
5
1/2
∞
ψx (f )df
R∞
<∞
−∞
.
|H(f )|2 df < ∞ and
ECE316F
Fall 2007
Communication
Systems
Problem Set 3
(For week of October 1st)
1. We will take an approach similar to that in the book.
sin(2πfc t) =
exp[j2πfc t] − exp[−j2πfc t]
2j
, i.e., after forming gT (t) and xT (t), we have
XT (f ) =
1
[GT (f − fc ) − GT (f + fc )]
2j
Again, ignoring the overlap between the GT (f − fc ) and GT (f + fc ),
|XT (f )|2 =
i
1h
|GT (f − fc )|2 + |GT (f + fc )|2
4
i.e.,
1
[Sg (f − fc ) + Sg (f + fc )]
4
One should expect this result because multiplying with either cos(2πfc t) and sin(2πfc t)
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 magnitude
squared).
Sx (f ) =
2. Haykin and Moher, Problem 2.33: First, the pulse in part (b) is a time shift of the
pulse in part (a) by T /2. In the frequency domain, Gb (f ) = exp [−j2πf T /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.
t
g(t) = Arect
cos
T
sin(πf T )
⇒ G(f ) = AT
⋆
πf T
πt
,
T
1
1
δ f−
2
2T
+δ f +
1
2T
,
where ⋆ represents convolution. Therefore,
"
#
AT sin(π(f − 1/2T )T ) sin(π(f + 1/2T )T )
,
+
G(f ) =
2
π(f − 1/2T )T
π(f + 1/2T )T
1
2AT cos(πf T )
,
π(1 − 2f T )(1 + 2f T )
2AT cos(πf T )
=
,
π(1 − 4f 2 T 2 )
4A2 T 2 cos2 (πf T )
=
,
π 2 (1 − 4f 2 T 2 )2
=
⇒ Ψg (f ) = |G(f )|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 the
approach suggested in class/the handout on the PSD of a sinusoid that was posted as
opposed to the auto correlation function approach.
Using the definition of the Fourier series
∞
X
gp (t) =
n=−∞
∞
X
⇒ Gp (f ) =
cn ej2πnf0 t ,
(1)
cn δ(f − nf0 ),
(2)
n=−∞
where f0 = 1/T0 and
cn
1 Z T0 /2
gp (t)e−j2πnf0 t dt
=
T0 −T0 /2
( A
n=0
2
=
.
sin(nπ/2)
n 6= 0
A nπ
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) to
obtain gT (t) and then take its Fourier transform
(
t
g(t) |t| ≤ T /2
gT (t) =
= rect
T
0
|t| > T /2
⇒ GT (f ) = Gp (f ) ⋆ T sinc(f T ).
g(t),
Using Eqn. (2),
GT (f ) =
∞
X
cn T sinc((f − nf0 )T ),
n=−∞
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
1
|GT (f )|2
T →∞ T
Sg (f ) = lim
2
As T → ∞, each sinc gets narrower and one can ignore the overlap between the sinc
functions and
|GT (f )|2 ≃
⇒ Sg (f ) =
=
∞
X
|cn |2 T 2 sinc2 ((f − nf0 )T ),
n=−∞
∞
X
2
|cn |
n=−∞
∞
X
1
lim T 2 sinc2 ((f − nf0 )T ) ,
T →∞ T
|cn |2 δ(f − nf0 ),
n=−∞
where we have derived cn earlier. Though, note that this procedure can be used to
find 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 ↔ j2πf X(f ), i.e., the derivative of a signal can
be obtained by using a filter with frequency response H(f ) = j2πf . We also know that
if a signal x(t) passes through a filter with frequency response H(f ) to create output
y(t),
Sy (f ) = |H(f )|2 Sx (f ),
i.e., the power spectral density of dx/dt is 4π 2 f 2 Sx (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 dc
power. We know the average power is
P =
Z
∞
Sx (f )df
−∞
Think of the dc power as the power concentrated around f = 0
dc power = Pdc = lim
Z
δ
Sx (f )df = 1
δ→0 −δ
(c) The ac power is the total power minus the dc power
Pac =
Z
∞
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 ) =
(



 2 − 3|f |/2
Sx (f − fc ) + Sx (f + fc ) |f | < B
=  1 − |f |/2
0
|f | > B

 0
3
0 < |f | < 1
1 < |f | < 2
|f | > 2
ECE316F
Fall 2007
Communication
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 a
sinusoid modulating wave at 100% modulation. The envelope becomes zero whenever
the modulating wave reaches it maximum negative value.
2. Haykin and Moher, Drill Problem 3.2
Here s(t) = Ac [1 + µ cos(2πfm t)] cos(2πfc t), as in Example 3.1 in the text. The carrier
1
power is 12 A2c while each sideband has power 18 µ2 A2c . With µ = 1/5, we get 200
A2c power
in 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) = a1 v1 (t) + a2 v12 (t)
= a1 [Ac cos(2πfc t) + m(t)] + a2 [Ac cos(2πfc t) + m(t)]2
= a1 [Ac cos(2πfc t) + m(t)] + a2 [A2c cos2 (2πfc t) + 2Ac m(t) cos(2πfc t) + m2 (t)]
1
= a1 [Ac cos(2πfc t) + m(t)] + a2 [A2c (1 + cos(2π(2fc )t)) + 2Ac m(t) cos(2πfc t) + m2 (t)]
2
=
a2 A2c /2 + a1 m(t) + a2 m2 (t) + [a1 Ac + 2a2 Ac m(t)] cos(2πfc t) +
|
{z
}
spurious baseband components
1
a2 A2c cos(2π(2fc )t)
2
|
{z
}
spurious carrier at 2fc
|
{z
desired AM wave at fc
}
(b) To extract the desired AM wave, we require a bandpass filter of bandwidth 2W
centered at ±fc , passing frequencies in the range [±fc − W, ±fc + W ].
(c) The spurious baseband signal components extend to frequency 2W . Thus we
require that fc is at least 3W , so that the lower sideband doesn’t overlap with
these components. The condition is fc > 3W .
1
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(2πfm t), so the modulated signal is s(t) =
m(t)Ac cos(2πfc t). Thus the output of the product modulator, assuming a unit
amplitude local oscillator, is
v(t) = s(t) cos(2πfc t)
= m(t)Ac cos(2πfc t) cos(2πfc t)
Ac
m(t)[1 + cos(2π(2fc )t)]
=
2
(b) Expressed in terms of complex exponential functions, we have
s(t) =
Am Ac j2π(fm +fc )t
e
+ ej2π(−fm +fc )t + ej2π(fm −fc )t + ej2π(−fm −fc )t .
4
Thus
v(t) = s(t) cos(2πfc t)



Am Ac  j2π(fm +fc )t
j2π(−fm +fc )t
j2π(fm −fc )t
j2π(−fm −fc )t   j2πfc t
ct
e| {z } + e| −j2πf
=
e|
|
{z
} + |e
{z
}+e
{z
} + |e
{z
}
{z }
8
+
+
−
−
+
−
=
Am Ac j2πfm t
e
+ e−j2πfm t +terms centered at ±2fc .
{z
}
| 4
desired signal
Terms labelled ‘+’ are at positive frequency and terms labelled ‘−’ are at negative
frequency. The two terms labelled ‘+’ in the first factor combine with the term
labelled ‘−’ 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
2
(b) The carrier power is A2c ÷ 100 = 2500/200 = 12.5 W. The total sideband power
2 2
is Ac4µ ÷ 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. (The
efficiency 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(2πfc t),
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 at
carrier frequency B Hz. Similarly x(t) = z1 (t) cos(2πfc t) + z2 (t) sin(2πfc t) is a
QAM signal of bandwidth 4B Hz at carrier frequency fc . Thus the bandwidth of
x(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 of
a(t) cos(2πBt) is E/2. Likewise the energy of b(t) sin(2πBt) is E/2. Since the
two components of z1 (t) are orthogonal, their energies add, thus the energy of
z1 (t) is E. It follows that z1 (t) cos(2πfc t) has energy E/2. Likwise, the energy
of z2 (t) sin(2πfc t) is E/2. Since the two components of x(t) are orthogonal, their
energies add, and hence the energy of x(t) is E.
3
bandwidth
LPF
bandwidth
LPF
(c) The block diagram is shown in the figure.
The first-stage lowpass filters of bandwidth 2B
shown in the figure may be omitted.
cos(2πBt)
B
bandwidth
2B
cos(2πfc t)
LPF
b(t)
B
sin(2πBt)
x(t)
a(t)
bandwidth
LPF
bandwidth
LPF
sin(2πfc t)
cos(2πBt)
c(t)
B
bandwidth
2B
LPF
sin(2πBt)
d(t)
B
10. (a) We choose f2 = 950 KHz or f2 = 1050 KHz, and bandpass filter the result
with 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 should
be 2A in the passband. (This gain factor can also appear elsewhere, for example
multiplying one of the mixing frequencies.) The block diagram is shown below.
passband = 980 to 1020 KHz
passband gain = 2A
m(t)
BPF
Am(t) cos(2πfc t)
fc = 1000 KHz
cos(2πf1 t)
cos(2πf2 t)
f1 = 50 KHz
(b) We have z(t) =
P
n
f2 = 950 or 1050 KHz
m(nTs )δ(t − nTs ), which has Fourier transform
Z(f ) =
∞
1 X
M (f − nfs ),
Ts n=−∞
as shown below.
···
−1000
···
−800
−600
−400
−200
0
200
400
600
800
1000 f (KHz)
The desired signal component appears at n = ±5. Thus we bandpass filter in the
region (980, 1020) KHz, with a passband gain of ATs /2.
(c) Let f0 = 1/T0 . The signal p(t) has Fourier transform
P (f ) = p0 δ(f ) +
=
∞
1X
pk [δ(f − kf0 ) + δ(f + kf0 )]
2 k=1
∞
1 X
pk [δ(f − kf0 ) + δ(f + kf0 )].
2 k=1
k odd
4
Thus z(t) = y(t)p(t) has Fourier transform
Z(f ) = Y (f ) ∗ P (f ) =
∞
1 X
pk [Y (f − kf0 ) + Y (f + kf0 )]
2 k=1
k odd
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
∞
A X
Z(f ) =
pk [M (f − 5f0 − kf0 ) + M (f + 5f0 − kf0 )+
4 k=1
k odd
M (f − 5f0 + kf0 ) + M (f + 5f0 + kf0 )] .
The lowpass filter retains only the components centered at zero frequency, which
arise when k = 5, these terms being
A
p5 [M (f ) + M (f )] .
4
Thus the filter output will be
2A
A
p5 m(t) =
m(t).
2
5π
5
ECE316F
Fall 2007
Communication
Systems
Problem Set 5 Solutions
1. Haykin and Moher, Drill Problem 3.9
We will need the following trigonometric identities:
1
1
1
cos2 (θ) = (1 + cos(2θ)); sin2 (θ) = (1 − cos(2θ)); cos(θ) sin(θ) = sin(2θ).
2
2
2
The multiplexed signal is s(t) = Ac m1 (t) cos(2πfc t) + Ac m2 (t) sin(2πfc t). The input
to the low-pass filter in the upper receiver branch (the I-channel) is
s(t)A0c cos(2πfc t) = Ac A0c s(t)[m1 (t) cos2 (2πfc t) + m2 (t) cos(2πfc t) sin(2πfc t)]
1
=
Ac A0c [m1 (t) + m1 (t) cos(4πfc t) + m2 (t) sin(4πfc t).
2
The high-frequency terms are rejected by the low-pass filter, so the final output is
1
A A0 m (t). Similarly, the input to the low-pass filter in the lower receiver branch
2 c c 1
(the Q-channel) is
1
s(t)A0c sin(2πfc t) = Ac A0c [m1 (t) sin(4πfc t) + m2 (t) − m2 (t) cos(4πfc t)]
2
The high-frequency terms are rejected by the low-pass filter, so the final output is
1
A A0 m (t).
2 c c 2
2. Haykin and Moher, Drill Problem 3.10
(a) We have
1
1
s(t) = Ac m(t) cos(2πfc t) − Ac m̂(t) sin(2πfc t);
2
2
thus
1
1
S(f ) = Ac [M (f − fc ) + M (f + fc )] − Ac [M̂ (f − fc ) − M̂ (f + fc )],
4
4j
where M̂ (f ) = F[m̂(t)] = −jsgn(f )M (f ). Therefore we have
1
1
S(f ) = Ac [M (f −fc )+M (f +fc )]+ Ac [M (f −fc )sgn(f −fc )−M (f +fc )sgn(f +fc )].
4
4
For f > 0, M (f + fc ) = 0 (assuming the message has bandwidth smaller than
fc ); thus, for f > 0, we have
1
1
S(f ) =
Ac M (f − fc ) + Ac M (f − fc )sgn(f − fc )
4
4
(
1
A M (f − fc ) + 41 Ac M (f − fc ) f > fc ,
4 c
=
1
A M (f − fc ) − 14 Ac M (f − fc ) 0 < f ≤ fc ;
4 c
(
=
1
A M (f
2 c
0
− fc ) f > fc ,
0 < f ≤ fc .
1
(b) We have
1
1
s(t) = Ac m(t) cos(2πfc t) + Ac m̂(t) sin(2πfc t);
2
2
thus
1
Ac [M (f − fc ) + M (f + fc )] +
4
1
=
Ac [M (f − fc ) + M (f + fc )] −
4
For f > 0, we have
S(f ) =
1
Ac [M̂ (f − fc ) − M̂ (f + fc )]
4j
1
Ac [M (f − fc )sgn(f − fc ) − M (f + fc )sgn(f + fc )].
4
1
1
Ac M (f − fc ) − Ac M (f − fc )sgn(f − fc )
4
4
(
1
A M (f − fc ) − 14 Ac M (f − fc ) f > fc ,
4 c
=
1
A M (f − fc ) + 41 Ac M (f − fc ) 0 < f ≤ fc ;
4 c
S(f ) =
(
=
0
1
A M (f
2 c
f > fc ,
− fc ) 0 < f ≤ fc .
3. Haykin and Moher, Drill Problem 3.11
The Fourier transform of m̂(t) is simply M̂ (f ) = −jsgn(f )M (f ). Since |M̂ (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(2πf1 t), where x(t) is a signal of bandwidth W . Note that
cos2 (θ) = 12 (1 + cos(2θ)) and that cos3 (θ) = 14 cos(3θ) + 34 cos(θ). Then
io (t) = a1 vi (t) + a3 vi3 (t)
= a1 [x(t) + cos(2πf1 t)] + a3 [x(t) + cos(2πf1 t)]3
= a1 x(t) + a1 cos(2πf1 t) + a3 x3 (t) + 3a3 x(t) cos2 (2πf1 t) +
3a3 x2 (t) cos(2πf1 t) + a3 cos3 (2πf1 t)
= a1 x(t) + a1 cos(2πf1 t) +
a3 x3 (t) + (3a3 /2)x(t) + (3a3 /2)x(t) cos(4πf1 t) +
3a3 x2 (t) cos(2πf1 t) + (3a3 /4) cos(2πf1 t) + (a3 /4) cos(6πf1 t).
We find modulation products centered at dc, at f1 , at 2f1 and at 3f1 . Only the term
centered at 2f1 , namely
s(t) = (3a3 /2)x(t) cos(4πf1 t)
is the product of x(t) with a cosine. Setting x(t) = 1 + m(t), f1 = fc /2 and filtering
out components not centered at fc , gives amplitude modulation.
Can such a device be used as an envelope detector? Let s(t) = Ac (1+ka m(t)) cos(2πfc t),
and let N (x) = a1 x + a3 x3 . Then
N (s(t)) = a1 Ac (1 + ka m(t)) cos(2πfc t) + a3 (A3c (1 + ka m(t))3 cos3 (2πfc t)).
2
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 carrier
frequency. No such odd harmonic appears at dc. Thus such a device cannot be used
as 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 is
proportional to the product of m(t) and cos(4πf1 t). A bandpass filter centered at 2f1
would pass this component and reject components at other multiples of f1 (assuming
f1 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 ) + 12 M (f −
2fc ) + 21 M (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 a
spectrum proportional to M (f ).
(b) When fc < W , spectral overlap occurs. In this case, the spectrum of the detector
output is proportional to the signal sketched (in bold) below.
0
W
I’m not sure what is being asked in the last part of the question. Each component
of s(t) is uniquely determined by m(t). However, the smallest (positive) carrier
frequency for which no spectral overlap occurs is fc = W .
7. Haykin and Moher, Problem 3.24
Let
s(t) = Ac cos(2πfc t + φ) + m(t) cos(2πfc t)
= (Ac cos(φ) + m(t)) cos(2πfc t) − Ac sin(φ) sin(2πfc t).
(a) If φ = 0, we see that s(t) is the ideal AM signal (Ac + m(t)) cos(2πfc t). Provided that Ac > |m(t)|, so that no overmodulation distortion occurs, the envelope
detector will produce Ac + m(t) at its output.
√
(b) We can always
a2 + b2 cos(2πfc t + θ), where
c t) as
√ write a cos(2πfc t) − b sin(2πf
√
cos(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) =
q
a2 (t) + b2 cos(2πfc t + θ(t)).
3
The ideal envelope detector would output
q
i1/2
a2 (t) + b2 =
h
A2c cos2 (φ) + 2Ac cos(φ)m(t) + m2 (t) + A2c sin2 (φ)
=
h
A2c + 2Ac cos(φ)m(t) + m2 (t)
i1/2
h
= Ac 1 + 2 cos(φ)m(t)/Ac + m2 (t)/A2c
i1/2
√
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)/A2c can be neglected), we find
that 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(2πfc t), then G(f ) = 21 (M (f − fc ) + M (f + fc )). Thus Ĝ(f ) =
−jsgn(f ) 12 (M (f − fc ) + M (f + fc )). If m(t) is low-pass, with bandwidth smaller than
fc , it follows that Ĝ(f ) = 2j1 (M (f − fc ) − M (f + fc )). Thus, under the conditions
stated,
[m(t) cos(2πfc t)]∧ = m(t) sin(2πfc t)
and similarly
[m(t) sin(2πfc t)]∧ = −m(t) cos(2πfc t)
(a) The upper-SSB signal is given by
su (t) =
Ac
Ac
m(t) cos(2πfc t) −
m̂(t) sin(2πfc t).
2
2
Thus
Ac
Ac
m(t) sin(2πfc t) +
m̂(t) cos(2πfc t).
2
2
Collecting these two equations in matrix form, we have
ŝu (t) =
"
su (t)
ŝu (t)
#
Ac
=
2
"
cos(2πfc t) − sin(2πfc t)
sin(2πfc t) cos(2πfc t)
#"
m(t)
m̂(t)
#
su (t)
ŝu (t)
#
.
Inverting this system of equations yields
"
m(t)
m̂(t)
#
2
=
Ac
"
cos(2πfc t) sin(2πfc t)
− sin(2πfc t) cos(2πfc t)
#"
(b) The lower-SSB signal is given by
sl (t) =
Ac
Ac
m(t) cos(2πfc t) +
m̂(t) sin(2πfc t).
2
2
4
.
Following the same approach as in (a), we get
"
hence
"
sl (t)
ŝl (t)
#
m(t)
m̂(t)
#
Ac
=
2
"
2
=
Ac
"
cos(2πfc t) sin(2πfc t)
sin(2πfc t) − cos(2πfc t)
#"
cos(2πfc t) sin(2πfc t)
sin(2πfc t) − cos(2πfc t)
#"
m(t)
m̂(t)
#
sl (t)
ŝl (t)
#
,
.
(NB: the expression for m̂(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
[s(t) cos(2πfc t) + ŝ(t) sin(2πfc t)].
Ac
This yields the block diagram shown below.
2
Ac
s(t)
cos(2πfc t)
m(t)
2
Ac
sin(2πfc t)
Hilbert
Transform
H
Interestingly (ignoring scale factors), this diagram is the same as that of a lowerSSB 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:
S1 (f ), S2 (f )
S3 (f ), Sf (4)
f
−100
100
f
−500
500
(a) QAM modulated spectrum:
s1 , s2
s3 , s4
f
f1
f2
To minimize spectral extent, choose f1 = 100 Hz and f2 = 700 Hz. The total
bandwidth is then 1200 Hz.
5
(b) Lower SSB spectrum:
s1 s2
s3
s4
f
f1 f2
f3
f4
Upper SSB spectrum:
s1s2 s3
s4
f
f2 f3
f4
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 1200
Hz.
6
ECE316F
Fall 2007
Communication
Systems
Problem Set 6
1. Haykin and Moher, Drill Problem 9.3, pg. 377. (Solution to 9.2 on next page)
1
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 reduces
the 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 + j2πf RC), i.e., the noise PSD
is given by:
N0
1
N0
SN (f ) =
,
|H(f )|2 =
2
2 1 + (2πf RC)2
The total noise power is therefore
Pn =
Z
∞
SN (f )df =
−∞
N0
.
4RC
Due to the filter, the sinusoid is attenuated by H(fc ). Since the average power of the input
sinusoid is A2c /2, the average output power of the sinusoid is given by:
A2c
2
!
1
,
1 + (2πfc RC)2
A2c
2
!
4RC
1
.
2
N0
1 + (2πfc RC)
and the SNR is given by:
SNR =
5. Haykin and Moher, Problem 9.10.
At a frequency of 200kHz, the noise PSD is approximately 0.5 × 10−18 = 5 × 10−19 W/Hz.
Setting this to be N0 /2 (assuming this to be constant over the small message bandwidth of
4kHz) we have N0 = 10−18 W/Hz.
Furthermore, the received signal power is A2c P/2 which is given to be -80dBm = 10−11 W
(note that in the dBm scale, the ”reference power” is 1mW).
The post-detection SNR is therefore
SNR =
A2c P
10−11
=
= 2.5 × 103 =≃ 34dB.
2N0 W
4 × 103 × 10−18
2
6. Haykin and Moher, Problem 9.13 (a) (b)
The power in the message signal is
P =
Z
∞
−∞
SM (f )df =
Z
W
a
−W
|f |
df = 2
W
Z
W
a
0
Therefore
(a)
SNRDSB =
A2c P
A2 a
= c
2N0 W
2N0
(b)
SNRAM =
A2 a
A2c ka P
= 0.09 c
2N0 W
2N0
3
f
df = aW.
W
ECE316F
Fall 2007
Communication
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(2πfc t) − cos(2πfc t)rect(t/T ) + cos(2π(fc + ∆f )t)rect(t/T ).
Since
1
1
cos(2πfc t) ←→ δ(f − fc ) + δ(f + fc )
2
2
and
rect(t/T ) cos(2πf0 t) ←→
T
T
sinc((f − f0 )T ) + sinc((f + f0 )T )
2
2
we have
S(f ) =
1
1
T
T
δ(f − fc ) + δ(f + fc ) − sinc((f − fc )T ) − sinc((f + fc )T )
2
2
2
2
T
T
+ sinc((f − fc − ∆f )T ) + sinc((f + fc + ∆f )T ).
2
2
This function is sketched below for the parameters fc = 20, ∆f = 3.
-23 -20
20 23
As usual in FM, we find that the spectrum occupies all frequencies, but is concentrated near
the carrier frequency.
1
2. The signal m(t) = 10cos(200πt) + 40cos(400πt) is frequency modulated with modulator constant 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 maximum
frequency deviation (in Hz) is ∆f = kf mp = 500 Hz. The effective bandwidth (highest
frequency component) of m(t) is B = 200 Hz. Hence, by Carson’s rule, the bandwidth of the
FM 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 to
apply m1 (t) = m(t) + m′ (t).
(a) Show that the signal being transmitted is
ϕ(t) = A cos 2πfc t + 2πkf m(t) + 2πkf
Z
t
m(τ )dτ .
−∞
Answer: In FM, if the message signal is m(t),
s(t) = A cos 2πfc t + 2πkf
Z
t
−∞
m(τ )dτ ,
i.e., the phase function is related to the integral of the message. Here,
Rt
m(τ )dτ , i.e., the transmitted signal is as above.
m(t) + −∞
Rt
−∞ m1 (t)dt
=
(b) Specify the impulse response of a filter that recovers in the absence of noise m(t) at
the output of an FM demodulator.
Answer: Note that M1 (f ) = M (f )(1 + j2πf ), i.e., to invert this operation at the
receiver, 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−t u(t).
4. For any integer n, we have
1
cos(θ) cos(nθ) = [cos((n − 1)θ) + cos((n + 1)θ)].
2
(a) Write cos5 (θ) as a linear combination of 1, cos(θ), cos(2θ), . . . cos(5θ).
Answer: From above,
cos2 (θ) =
cos3 (θ) =
=
cos4 (θ) =
=
cos5 (θ) =
=
1
[1 + cos(2θ)]
2
1
[cos(−θ) + cos(θ) + cos(θ) + cos(3θ)]
4
1
[3 cos(θ) + cos(3θ)]
4
1
[3 + 3 cos(2θ) + cos(2θ) + cos(4θ)]
8
1
[3 + 4 cos(2θ) + cos(4θ)]
8
1
[3 cos(−θ) + 3 cos(θ) + 4 cos(θ) + 4 cos(3θ) + cos(3θ) + cos(5θ)]
16
1
[10 cos(θ) + 5 cos(3θ) + cos(5θ)]
16
2
(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 combination 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
X
ai cos(iθ)
i=0
for some coefficients a0 , a1 , a2 , . . . , an . Then
cosn+1 (θ) = cos(θ) cosn (θ)
= cos(θ)
n
X
ai cos(iθ)
i=0
=
n
X
ai cos(θ) cos(iθ)
i=0
=
n
1X
(ai cos((i − 1)θ) + ai cos((i + 1)θ))
2 i=0
=
n
n
1X
1X
ai cos((i − 1)θ) +
ai cos((i + 1)θ)
2 i=0
2 i=0
=
X
X
1 n+1
1 n−1
1
aj+1 cos(jθ) +
aj−1 cos(jθ),
a0 cos(θ) +
2
2 j=0
2 j=1
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 of
mathematical 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), where
m(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 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 resulting FM
signal 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 bandwidth
is given as 2(∆f + B) = 200 KHz, from which we obtain ∆f = 90 KHz. Since ∆f =
kf mp = kf , we obtain kf ≤ 90000.
(kf )NBFM =?
NBFM
Modulator
fMIX =?
Frequency
Multiplier
×20
Passband = ?
Bandpass
Filter
∼
∼
(fc )NBFM = 225 KHz
2 cos(2πfMIX t)
3
Frequency
Multiplier
×50
Power
Amplifier
fc = 100 MHz
kf = 75000
(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-stage
narrowband FM modulator is 225 KHz. Determine the following parameters: i) the
modulation parameter (kf )NBFM for the first-stage narrowband FM signal signal; ii) the
mixing frequency fMIX ; iii) the passband of the bandpass filter (i.e., specify the smallest
possible range of (positive) frequencies passed by the filter). The filter is to be designed
with 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 is
225 KHz × 20 = 4.5 MHz. On the other hand, prior to frequency-multiplying by a factor
of 50, the carrier frequency should be 100 MHz/50 = 2 MHz. We need to down-shift from
4.5 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 is
2(∆f + B) = 2(1500 + 10000) = 23 KHz. The filter passband should therefore extend
from 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 Radiotelevision and Telecommunications Commission (CRTC) assigns a local FM radio station a
total of 100 kHz of bandwidth instead of the usual 200 kHz. The station decides to set the
maximum 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 would
recommend?
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 sufficiently
accurate, 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
ECE316F
Fall 2007
Communication
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
=
3
SNRFM
pre
BT
4 W
3
.
3
Note 34 BWT = (3/4) × (30 × 106 /6 × 106 )3 = 93.75 ≡ 19.71dB. The post-detection
SNR is therefore, 15 + 19.71 = 34.71dB.
Using the equation at the bottom of page 389,
(W/f3dB )2
.
3 [(W/f3dB ) − tan−1 (W/f3dB )]
I=
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 =
Z
∞
SM (f )df = 2
−∞
Z
W
a
0
f
df = aW.
W
For frequency modulation and detection with kf = 500Hz/V , the post-detection SNR
is
"
#2
3A2c kf2 P
A2c a
kf
SNRFM =
=
×3×
2N0 W 3
2N0
W
3. Haykin and Moher, Problem 9.17.
Answer: Carson’s rule tells us
BT = 2 (∆f + fm ) = 2 (kf Am + fm ) = 2(1000(1) + 200) = 2400Hz.
The pre-detection SNR is
SNRFM
pre = 500 =
A2
A2c
⇒ c = 500 × 2400 = 1.2 × 106 Hz.
2N0 BT
2N0
1
Due to the strange post-detection filter, the output noise power is given by
Z −100
i
N0
2N0 h 3
2N0
2
3
×
2
×
× 2.6 × 107 .
f
df
=
300
−
100
=
A2c
3A2c
3A2c
−300
The post-detection SNR is therefore,
SNRFM
post
3A2c kf2 P
=
= 69230.8 = 48.4dB,
2N0 (2.6 × 107 )
since kf = 1000Hz/V and P = A2m /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 textbook introduces the notion that a periodic signal, gT0 (t), with period T0 can be viewed
as a periodic repetition (every T0 ) of a template or generating function g(t) where
g(t) =
(
gT0 (t) − T20 ≤ t <
0
elsewhere
T0
2
⇒ gT0 (t) =
∞
X
g(t − mT0 ),
(1)
m=−∞
i.e., g(t) is one period of gT0 centered 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 extension
in the time domain implies sampling in the frequency domain:
GT0 (f ) = f0
∞
X
G(nf0 )δ(f − nf0 ).
(2)
n=−∞
Note that since the Fourier transform is (for all intents and purposes) a one-to-one
mapping, one can reverse this last statement: sampling in the frequency domain implies 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
⇒ T0 gT0 (t) ⇀
↽
∞
X
G(nf0 )δ(f − nf0 ),
n=−∞
∞
X
G(nf0 )δ(f − nf0 ),
n=−∞
2
(since T0 = 1/f0 ),
Now using the expansion of gT0 (t) in Eqn. (1), we have the required answer:
∞
X
⇒ T0
∞
X
g(t − mT0 ) ⇀
↽
m=−∞
G(nf0 )δ(f − nf0 ).
(3)
n=−∞
(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 the
frequency 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:
∞
X
T0
g(t − mT0 ) ⇀
↽
m=−∞
∞
X
G(nf0 )δ(f − nf0 ).
(4)
n=−∞
Using the duality property and replacing f0 with Ts and T0 with fs , we get
⇒ fs
∞
X
g(−f − mfs ) ⇀
↽
m=−∞
∞
X
G(nTs )δ(t − nTs ).
(5)
n=−∞
Now duality also tells us that G(t) ⇀
↽ g(−f ), i.e., g(−f ) is the Fourier transform of
G(t). Letting F represent the Fourier transform, F [G(t)] = g(−f ). Switching the
letters t and f , F [G(f )] = g(−t) we get
fs
∞
X
G(−f − mfs ) ⇀
↽
m=−∞
⇒ fs
∞
X
∞
X
g(−nTs )δ(t − nTs ),
n=−∞
∞
X
G(f − mfs ) ⇀
↽
m=−∞
g(nTs )δ(t − nTs ),
(6)
n=−∞
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 reciprocal
of 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 is
200Hz 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
p(t)
X(f)
1
x(t)
-W
W
f
Anti
Alias
Filter
x p (t)
(a)
(b)
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 = 104 Hz.
P
This signal is to be sampled using an impulse train, p(t) = ∞
n=−∞ δ(t − nTs ), with
sampling period Ts = 10−4 s.
(a) Design an anti-aliasing filter given this sampling rate, i.e., modify the sampling
scheme 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 = 104 Hz. The anti-aliasing filter ensures that the input to
the sampler meets Nyquist’s criterion. The filter is, therefore, a low-pass filter with
cut-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 Fourier
transform 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 Fourier
transform, G(f ) is given in Fig. 2(a). The sampling process causes a periodic repetition
of G(f ) every fs . Therefore, the Fourier transform of xp (t), Xp (f ) is as given in
Fig. 2(b) below.
G(f)
Xp(f)
fs
fs
- f s /2
- f s /2
f
fs /2
-5f s /2
-3f s /2
fs /2
(b)
(a)
Figure 2: Solution to Q3.
4
3fs /2
5fs /2
ECE316F
Fall 2007
Communication
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) and
s4 (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 can
set R = 3 and reduce the sampling rate by a factor of 23 = 8. The signals s1 (t) and
s2 (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 the
solution presented here is the simplest. The first step is to combine the PAM signals
corresponding to s1 (t) and s2 (t) into a single signal s5 (t). Figure 1 illustrates this
process. The figure uses arbitrary signals. The width of each pulse is chosen sufficiently
small. 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), as
shown below. The signals s3 (t) and s4 (t) are interleaved such that the there are samples
of each signal every 1/2400 seconds. The samples of s5 (t) are interleaved every 1/600
seconds. For this interleaver to work, we therefore require that the pulse width (T ) be
less than 1/(3fs ), i.e., less than 1/7200 seconds.
1
1
1
1/300
0
2
1/300
0
2
5
(1)
5
(1)
5
(2)
3/600
1/600
2/600
5
(2)
Figure 1: Combining two signals into a single multiplexed signal.
3
4
0
3
4
5
(1)
4
4
3
3
3
1/2400
1/600
4
5
(2)
Figure 2: Combining s3 (t), s4 (t) and s5 (t) into a single multiplexed signal.
2
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 bandlimited to 7 kHz.
(a) Assuming that the signal is sampled at the Nyquist rate and quantized using a 512level 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 is
therefore 2W = 14kHz. Furthermore, 512 = 29 , i.e., each sample creates 9 bits
making the bit rate = 14 × 9 = 126kbps.
(b) The system is found to achieve a signal-to-quantization noise ratio (SQNR) of
60 dB. How many levels would the quantizer need to have in order to achieve an
SQNR 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
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 to
meet 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 required
per 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, 1.25
and 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 is
required to communicate this signal?
Answer: Each sample creates 8 bits and there are 8k samples per second, i.e.,
bit rate = 64kbps.
4
ECE316F
Fall 2007
Communication
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 discontinuous, 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 ) =

0






 √
dP
=
df






|f | < f1 ,
h
|−f1 )
E
π
(−sgn(f )) sin π(|f
4B0 2(B0 −f1 )
2(B0 −f1 )
i
0
f1 ≤ |f | ≤ 2B0 − f1 ,
|f | > 2B0 − f1 .
Note that this function is continuous (the sin term is 0 at the end points, f = f1 and
f = 2B0 − f1 . One can show that the second derivative is discontinuous (the sin term
becomes a cos). Hence the corresponding pulse drops off as 1/t3 . Note
cos(2παB0 t)
p(t) ∝ sinc(2B0 t)
1 − 16α2 B02 t2
!
sin(2πB0 t)
∝
2πB0 t
!
cos(2παB0 t)
,
1 − 16α2 B02 t2
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. With
α = 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 we
buy 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 a
bandwidth 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 pulse
shapes. 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 frequency
domain is
1
k
P f−
Tb
k=−∞
∞
X
We showed that this implies P
about the f = 1/(2Tb ) point.
1
2Tb
!
+ x +P ∗
= constant.
1
2Tb
− x = constant, i.e., odd symmetry
In this particular example,
√
P (f ) =
(
E
2B0
0
1 − B|f0|
|f | ≤ B0 ,
|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 cosine
spectrum with α = 1, the first derivative of this pulse is discontinuous, i.e., the pulse
falls off at 1/t2 , not 1/t3 . (The triangle is a convolution of two rect functions, i.e., the
pulse shape is a sinc2 ).
5. Haykin and Moher, Problem 6.17. Use a data rate of 100kbps. Also, repeat the eye
diagram with a 10% jitter in timing.
The following figure plots p(t), the raised cosine pulse with α = 1. Note Tb = 1/105 =
10µs.
Problem 6.17, the pulse shape with α = 1
0.8
Pulse shape p(t)
0.6
0.4
0.2
0
−0.2
−2
−1.5
−1
−0.5
0
Time
0.5
1
1.5
2
−5
x 10
Figure 1: The pulse shape with Tb = 10µs.
Therefore the given 6-bit sequence creates the following received signal:
2
Problem 6.17, the received signal for the given 6−bit sequence
1
0.8
Received signal y(t)
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−2
−1
0
1
2
3
4
5
Time
6
7
−5
x 10
Figure 2: The received signal for Problem 6.17.
Problem 6.17, the eye diagram with ideal sampling
1
0.8
Received signals
0.6
0.4
0.2
Eye opening
0
−0.2
−0.4
−0.6
−0.8
−1
0
0.5
1
Time
1.5
2
−5
x 10
Figure 3: The eye diagram with ideal sampling.
The eye diagram plots the signal corresponding to each received bit (generally a very
long 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 bandwidth
of 10 kHz using bandlimited pulses with 25% excess bandwidth. What bit rate is
supported 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
1
= 10 × 103 ⇒
= 20 × 103 1.25 = 16 × 103 symbols per second.
2Ts
Ts
3
Problem 6.17, the eye diagram with 10% error in sampling
1
0.8
Received signals
0.6
0.4
0.2
Eye opening
0
−0.2
−0.4
−0.6
−0.8
−1
0
0.5
1
Time
1.5
2
−5
x 10
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 zero
intersymbol interference assuming a symbol interval of T seconds and that the central
sample 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 in
each 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. Assume
that the bit rate is set to 64 kbps. To avoid ISI, you decide to use a raised-cosine
pulse shape with 100% excess bandwidth (α = 1). What transmission bandwidth
is needed, assuming binary pulse amplitude modulation?
Answer: With α = 1, the bandwidth used by the pulses = 1/Tb . With a bit rate
of 64kbps, this bandwidth = 64kHz.
4
(b) After some experimentation, you find that the usable bandwidth on the given
channel is 48 kHz. To increase the efficiency of your system, you decide to use a
raised-cosine pulse with 25% excess bandwidth, and to change the quantization
scheme to delta modulation. With this scheme, what is the maximum rate at
which 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 single
bit, 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 transmitted
pulse is rectangular, given by g(t) = rect(t − 1/2). The channel is ideal, with frequency
response 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). Since
g(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
X
ak g(t − k)
k=0
is sent, where ak ∈ {+1, −1}. Sketch the transmitted pulse train s(t) corresponding to the eight bit sequence a0 = +1, a1 = +1, a2 = −1, a3 = +1, a4 = −1,
a5 = −1, a6 = +1. a7 = −1.
$s(t)$
$+1$
$t$
1
5
8
$−1$
Figure 5: Transmit Signal.
(c) Sketch the corresponding output of the receiver filter when s(t) from (9b) is
transmitted.
5
$y(t)$
$+1$
$t$
1
2
3
4
5
6
7
8
$−1$
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 every
nonzero integer k.
(e) Sketch the eye pattern for this system, and identify the times at which the receiver
filter 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 on
only ±1 at these sample points. Hence the possible transitions between samples
are 1 to 1, 1 to -1, -1 to 1 and -1 to -1:
1
0
Tb
2Tb
-1
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 the
quaternary 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 take
on only ±1 or ±3 at these sample points. All possible transitions between these
points are possible:
The ideal sampling time is clearly at iTb .
6
3
1
-1
0
Tb
2Tb
-3
Figure 8: Eye diagram for the quanternary case.
7