Solution of ECE 342 Test 3 S12
1.
A random power signal has a mean of three and a standard deviation of five. Find its numerical total
average signal power.
Signal Power = ____________
P = 32 + 5 2 = 34
2.
⎛ τ − 1⎞
⎛ τ + 1⎞
+ Λ⎜
A random energy signal has an autocorrelation R x (τ ) = Λ ⎜
. Find its numerical
⎝ 4 ⎟⎠
⎝ 4 ⎟⎠
⎧1 − x , x < 1
signal energy. Λ ( x ) = ⎨
, otherwise
⎩0
Signal Energy = ____________
⎛ 0 − 1⎞
⎛ 0 + 1⎞
P = Rx (0) = Λ ⎜
+ Λ⎜
= 3 / 4 + 3 / 4 = 1.5
⎟
⎝ 4 ⎠
⎝ 4 ⎟⎠
3.
A random signal with power spectral density G x ( f ) = 10Λ ( f / 50 ) is the input signal to a filter with
⎡ ⎛ f − 20 ⎞
⎛ f + 20 ⎞ ⎤
frequency response H ( f ) = 5 ⎢Π ⎜
⎟⎠ + Π ⎜⎝
⎟ . Find the numerical signal power of the output
⎝
10
10 ⎠ ⎥⎦
⎣
⎪⎧1 , x < 1 / 2
signal from the filter. Π ( x ) = ⎨
⎪⎩0 , x > 1 / 2
Output Signal Power = ____________
∞
Pout =
∫ G ( f ) H( f )
x
−∞
2
25
df = 2 × 25 × 10 ∫ Λ ( f / 50 ) df = 2 × 25 × 10 ×
15
0.7 + 0.5
× 10 = 3000
2
4.
An AM signal with carrier amplitude Ac = 10 is modulated by
x ( t ) = 2 cos (1000π t ) + 4 sin ( 4000π t )
with a modulation index µ = 1 . The modulated carrier is transmitted through a channel with a loss of 120
dB. It is received by an AM receiver using synchronous detection and DC blocking with noise
temperature T = 4000K . The predection bandpass filter in the AM receiver is practically ideal and just
wide enough to pass the AM signal. The last lowpass filter (after the synchronous detection) is just wide
enough to pass the demodulated signal.
(a)
Find the numerical signal power of the transmitted signal ST .
ST = ____________
{
}
ST = 10 ⎡⎣1 + 2 cos (1000π t ) + 4 sin ( 4000π t ) ⎤⎦ cos (ω ct )
2
ST = 100 (1 + 2 + 8 ) × 1 / 2 = 50 × 11 = 550
(b)
Find the numerical signal power of the received signal SR .
SR = ____________
SR = ST / 1012 = 5.5 × 10 −10
(c)
Find the numerical destination signal power SD .
SD = ____________
This answer depends on the assumptions about the synchronous detector. Do we multiply by
cos (ω ct ) and take the low-frequency part, yielding a factor of 1/2 in amplitude and 1/4 in power as
in my slides? Or do we multiply by 2 cos (ω ct ) , yielding a factor of 1 in amplitude and in power
as in the book? The signal-to noise ratio is the same either way and I accepted either answer if
the signal power and noise power were computed consistently with either assumption.
SD =
(d)
1 Ac2
1 10 2
A2
10 2
Sx =
10 = 2.5 × 10 −10 OR SD = c Sx = 12 10 = 10 −9
12
4 L
4 10
L
10
Find the numerical destination noise power in the receiver's bandwidth N D .
N D = ____________
σ n2 = N 0 BT = kT × 4000 = 1.38 × 10 −23 × 4000 × 4000 = 2.208 × 10 −16
N D = σ n2 / 4 = 2.208 × 10 −16 / 4 = 5.52 × 10 −17 OR N D = σ n2 = 2.208 × 10 −16
(e)
Find the numerical signal-to-noise ratio at the destination ( S / N )D in dB.
( S / N )D
= ____________
( S / N )D = 2.5 × 10 −10 / 5.52 × 10 −17 = 4.529 × 10 6
or 66.56 dB
OR
( S / N )D = 10 −9 / 2.208 × 10 −16 = 4.529 × 10 6
(f)
or 66.56 dB
In dB, how much better could the signal-to-noise ratio be made by using a bandpass filter after the
detection process instead of a lowpass filter?
Improvement in dB = ____________
We could use a bandpass filter between 500 Hz and 2 kHz instead of a lowpass filter between 0
and 2 kHz. That would reduce the noise by a factor of 0.75, realizing an improvement in signal-tonoise ratio of about 1.249 dB.
5.
A signal with signal power Pin = 100 mW is the input signal to a transmission cable with loss of
α = 1.5 dB/km . The cable is 100 km in length. Repeaters are used to avoid the signal's getting lost in
noise. The signal power during transmission cannot be allowed to fall below 1 mW at any point on the
cable and each repeater has a power gain of 20 dB.
(a)
What is the minimum number of repeaters necessary?
Minimum number of repeaters is ____________
The minimum number of repeaters occurs when the signal is allowed to fall to 1 mW at the input
of all the repeaters. The cable length at which the signal power falls from 100 mW to 1mW is
13.3333 km. 100 km/13.3333km is 7.5. So we need a minimum of 7 repeaters.
(b)
Using the minimum number of repeaters what is the output signal power?
Output signal power is ____________
The seventh repeater is at the 93.3333 km point. Its input power is 1 mW and its output power is
100 mW. The distance to the cable end is 6.667 km. The signal loses 10 dB in that distance. So
the output signal power at the end of the cable is 10 mW.
6.
What is the advantage of using an envelope detector compared with using a synchronous detector?
It allows for a very simple and economical demodulation of the signal compared to synchronous
techniques.
7.
What is the purpose of guard bands in frequency-division multiplexing?
Guard bands are used to minimize crosstalk between channels when multiple signals are frequency
multiplexed onto to one signal. Crosstalk occurs because real filters do not have vertical sides and their
stopbands do not have infinite rejection.
8.
Given that the full range of AM carrier frequencies permitted by law is from 530 kHz to 1600 kHz in 10
kHz steps, and that the intermediate frequency used is 455 kHz, identify all the legal AM carrier
frequencies for which an image frequency exists in the AM range. You may do this by simply listing all
these frequencies or by specifying with <, >, ≤ and/or ≥ the ranges of frequencies.
The answer to this question depends on whether one assumes high-side injection or low-side injection of
the local oscillator.
With high-side injection, all carrier frequencies for which the frequency minus 910 kHz are in the AM
range are image frequencies. Those are 1440, 1450, 1460, 1470, 1480, 1490, 1500, 1510, 1520, 1530,
1540, 1550, 1560, 1570, 1580, 1590, 1600 or 1440 ≤ fc ≤ 1600 .
With low-side injection, all carrier frequencies for which the frequency plus 910 kHz are in the AM
range are image frequencies. Those are 530, 540, 550, 560, 570, 580, 590, 600, 610, 620, 630, 640, 650,
660, 670, 680, 690 or 530 ≤ fc ≤ 690 .
9.
If a phase locked loop VCO free runs at a frequency f0 and the incoming frequency to be locked to is at
fc and if f0 = fc and if the loop is locked, describe the exact phase relationship between the VCO output
signal and the incoming signal. (What is the phase difference and which signal is leading or lagging?)
The phase difference is exactly 90° with the VCO output leading the incoming signal.
If the incoming signal increases in size by a factor of two, what happens to this phase relationship?
It is momentarily perturbed but then returns back exactly to the same phase relationship.
10.
What is the effect on a demodulated USSB signal of a local oscillator at a frequency that is too high?
(Be specific.)
The lower edge of the USSB signal, instead of being shifted to zero, will be shifted to a frequency below
zero. So all the frequencies in the demodulated signal will be too low by the error in the local oscillator
frequency and the two replicas of the USSB signal will overlap causing distortion.
Solution of ECE 342 Test 3 S12
1.
A random power signal has a mean of two and a standard deviation of eight. Find its numerical total
average signal power.
Signal Power = ____________
P = 2 2 + 8 2 = 68
2.
⎛ τ − 1⎞
⎛ τ + 1⎞
+ Λ⎜
A random energy signal has an autocorrelation R x (τ ) = Λ ⎜
. Find its numerical
⎝ 6 ⎟⎠
⎝ 6 ⎟⎠
⎧1 − x , x < 1
signal energy. Λ ( x ) = ⎨
, otherwise
⎩0
Signal Energy = ____________
⎛ 0 − 1⎞
⎛ 0 + 1⎞
P = Rx (0) = Λ ⎜
+ Λ⎜
= 5 / 6 + 5 / 6 = 5 / 3 or 1.6667
⎝ 6 ⎟⎠
⎝ 6 ⎟⎠
3.
A random signal with power spectral density G x ( f ) = 4Λ ( f / 50 ) is the input signal to a filter with
⎡ ⎛ f − 25 ⎞
⎛ f + 25 ⎞ ⎤
frequency response H ( f ) = 5 ⎢Π ⎜
⎟⎠ + Π ⎜⎝
⎟ . Find the numerical signal power of the output
⎝
10
10 ⎠ ⎥⎦
⎣
⎪⎧1 , x < 1 / 2
signal from the filter. Π ( x ) = ⎨
⎪⎩0 , x > 1 / 2
Output Signal Power = ____________
∞
Pout =
∫ G ( f ) H( f )
x
−∞
2
30
df = 2 × 25 × 4 ∫ Λ ( f / 50 ) df = 2 × 25 × 4 ×
20
0.6 + 0.4
× 10 = 1000
2
4.
An AM signal with carrier amplitude Ac = 8 is modulated by
x ( t ) = 2 cos ( 2000π t ) + 4 sin ( 4000π t )
with a modulation index µ = 1 . The modulated carrier is transmitted through a channel with a loss of 120
dB. It is received by an AM receiver using synchronous detection and DC blocking with noise
temperature T = 6000K . The predection bandpass filter in the AM receiver is practically ideal and just
wide enough to pass the AM signal. The last lowpass filter (after the synchronous detection) is just wide
enough to pass the demodulated signal.
(a)
Find the numerical signal power of the transmitted signal ST .
ST = ____________
{
}
ST = 8 ⎡⎣1 + 2 cos ( 2000π t ) + 4 sin ( 4000π t ) ⎤⎦ cos (ω ct )
2
ST = 64 (1 + 2 + 8 ) × 1 / 2 = 32 × 11 = 352
(b)
Find the numerical signal power of the received signal SR .
SR = ____________
SR = ST / 1012 = 3.52 × 10 −10
(c)
Find the numerical destination signal power SD .
SD = ____________
This answer depends on the assumptions about the synchronous detector. Do we multiply by
cos (ω ct ) and take the low-frequency part, yielding a factor of 1/2 in amplitude and 1/4 in power as
in my slides? Or do we multiply by 2 cos (ω ct ) , yielding a factor of 1 in amplitude and in power
as in the book? The signal-to noise ratio is the same either way and I accepted either answer if
the signal power and noise power were computed consistently with either assumption.
SD =
(d)
1 Ac2
1 82
Ac2
82
−10
Sx =
10
=
1.6
×
10
OR
S
=
S
=
10 = 6.4 × 10 −10
D
x
4 L
4 1012
L
1012
Find the numerical destination noise power in the receiver's bandwidth N D .
N D = ____________
σ n2 = N 0 BT = kT × 4000 = 1.38 × 10 −23 × 6000 × 4000 = 3.312 × 10 −16
N D = σ n2 / 4 = 3.312 × 10 −16 / 4 = 8.28 × 10 −17 OR N D = σ n2 = 3.312 × 10 −16
(e)
Find the numerical signal-to-noise ratio at the destination ( S / N )D in dB.
( S / N )D
= ____________
( S / N )D = 1.6 × 10 −10 / 8.28 × 10 −17 = 1.9324 × 10 6
or 62.861 dB
OR
( S / N )D = 6.4 × 10 −10 / 3.312 × 10 −16 = 1.9324 × 10 6
or 62.861 dB
(f)
In dB, how much better could the signal-to-noise ratio be made by using a bandpass filter after the
detection process instead of a lowpass filter?
Improvement in dB = ____________
We could use a bandpass filter between 1 kHz and 2 kHz instead of a lowpass filter between 0
and 2 kHz. That would reduce the noise by a factor of 0.5, realizing an improvement in signal-tonoise ratio of about 3 dB.
5.
A signal with signal power Pin = 200 mW is the input signal to a transmission cable with loss of
α = 1.5 dB/km . The cable is 100 km in length. Repeaters are used to avoid the signal's getting lost in
noise. The signal power during transmission cannot be allowed to fall below 1 mW at any point on the
cable and each repeater has a power gain of 23 dB.
(a)
What is the minimum number of repeaters necessary?
Minimum number of repeaters is ____________
The minimum number of repeaters occurs when the signal is allowed to fall to 1 mW at the input
of all the repeaters. The cable length at which the signal power falls from 200 mW to 1mW is
15.3333 km. 100 km/15.3333km is 6.5217. So we need a minimum of 6 repeaters.
(b)
Using the minimum number of repeaters what is the output signal power?
Output signal power is ____________
The sixth repeater is at the 92 km point. Its input power is 1 mW and its output power is
200 mW. The distance to the cable end is 8 km. The signal loses 12 dB in that distance. So
the output signal power at the end of the cable is 12.6191 mW.
6.
What is the effect on a demodulated USSB signal of a local oscillator at a frequency that is too high?
(Be specific.)
The lower edge of the USSB signal, instead of being shifted to zero, will be shifted to a frequency below
zero. So all the frequencies in the demodulated signal will be too low by the error in the local oscillator
frequency and the two replicas of the USSB signal will overlap causing distortion.
7.
What is the advantage of using an envelope detector compared with using a synchronous detector?
It allows for a very simple and economical demodulation of the signal compared to synchronous
techniques.
8.
What is the purpose of guard bands in frequency-division multiplexing?
Guard bands are used to minimize crosstalk between channels when multiple signals are frequency
multiplexed onto to one signal. Crosstalk occurs because real filters do not have vertical sides and their
stopbands do not have infinite rejection.
9.
Given that the full range of AM carrier frequencies permitted by law is from 530 kHz to 1600 kHz in 10
kHz steps, and that the intermediate frequency used is 455 kHz, identify all the legal AM carrier
frequencies for which an image frequency exists in the AM range. You may do this by simply listing all
these frequencies or by specifying with <, >, ≤ and/or ≥ the ranges of frequencies.
The answer to this question depends on whether one assumes high-side injection or low-side injection of
the local oscillator.
With high-side injection, all carrier frequencies for which the frequency minus 910 kHz are in the AM
range are image frequencies. Those are 1440, 1450, 1460, 1470, 1480, 1490, 1500, 1510, 1520, 1530,
1540, 1550, 1560, 1570, 1580, 1590, 1600 or 1440 ≤ fc ≤ 1600 .
With low-side injection, all carrier frequencies for which the frequency plus 910 kHz are in the AM
range are image frequencies. Those are 530, 540, 550, 560, 570, 580, 590, 600, 610, 620, 630, 640, 650,
660, 670, 680, 690 or 530 ≤ fc ≤ 690 .
10.
If a phase locked loop VCO free runs at a frequency f0 and the incoming frequency to be locked to is at
fc and if f0 = fc and if the loop is locked, describe the exact phase relationship between the VCO output
signal and the incoming signal. (What is the phase difference and which signal is leading or lagging?)
The phase difference is exactly 90° with the VCO output leading the incoming signal.
If the incoming signal increases in size by a factor of two, what happens to this phase relationship?
It is momentarily perturbed but then returns back exactly to the same phase relationship.
Solution of ECE 342 Test 3 S12
1.
A random power signal has a mean of four and a standard deviation of twelve. Find its numerical total
average signal power.
Signal Power = ____________
P = 4 2 + 12 2 = 160
2.
⎛ τ − 1⎞
⎛ τ + 1⎞
+ Λ⎜
A random energy signal has an autocorrelation R x (τ ) = Λ ⎜
. Find its numerical
⎝ 8 ⎟⎠
⎝ 8 ⎟⎠
⎧1 − x , x < 1
signal energy. Λ ( x ) = ⎨
, otherwise
⎩0
Signal Energy = ____________
⎛ 0 − 1⎞
⎛ 0 + 1⎞
P = Rx (0) = Λ ⎜
+ Λ⎜
= 7 / 8 + 7 / 8 = 7 / 4 or 1.75
⎝ 8 ⎟⎠
⎝ 8 ⎟⎠
3.
A random signal with power spectral density G x ( f ) = 6Λ ( f / 50 ) is the input signal to a filter with
⎡ ⎛ f − 20 ⎞
⎛ f + 20 ⎞ ⎤
frequency response H ( f ) = 5 ⎢Π ⎜
⎟⎠ + Π ⎜⎝
⎟ . Find the numerical signal power of the output
⎝
20
20 ⎠ ⎥⎦
⎣
⎪⎧1 , x < 1 / 2
signal from the filter. Π ( x ) = ⎨
⎪⎩0 , x > 1 / 2
Output Signal Power = ____________
∞
Pout =
∫ G ( f ) H( f )
x
−∞
2
30
df = 2 × 25 × 6 ∫ Λ ( f / 50 ) df = 2 × 25 × 6 ×
10
0.4 + 0.8
× 20 = 3600
2
4.
An AM signal with carrier amplitude Ac = 12 is modulated by
x ( t ) = 2 cos ( 3000π t ) + 4 sin ( 4000π t )
with a modulation index µ = 1 . The modulated carrier is transmitted through a channel with a loss of 120
dB. It is received by an AM receiver using synchronous detection and DC blocking with noise
temperature T = 5000K . The predection bandpass filter in the AM receiver is practically ideal and just
wide enough to pass the AM signal. The last lowpass filter (after the synchronous detection) is just wide
enough to pass the demodulated signal.
(a)
Find the numerical signal power of the transmitted signal ST .
ST = ____________
{
}
ST = 12 ⎡⎣1 + 2 cos ( 3000π t ) + 4 sin ( 4000π t ) ⎤⎦ cos (ω ct )
2
ST = 144 (1 + 2 + 8 ) × 1 / 2 = 72 × 11 = 792
(b)
Find the numerical signal power of the received signal SR .
SR = ____________
SR = ST / 1012 = 7.92 × 10 −10
(c)
Find the numerical destination signal power SD .
SD = ____________
This answer depends on the assumptions about the synchronous detector. Do we multiply by
cos (ω ct ) and take the low-frequency part, yielding a factor of 1/2 in amplitude and 1/4 in power as
in my slides? Or do we multiply by 2 cos (ω ct ) , yielding a factor of 1 in amplitude and in power
as in the book? The signal-to noise ratio is the same either way and I accepted either answer if
the signal power and noise power were computed consistently with either assumption.
1 Ac2
1 12 2
Ac2
12 2
−10
Sx =
10
=
3.6
×
10
OR
S
=
S
=
10 = 1.44 × 10 −9
D
x
4 L
4 1012
L
1012
SD =
(d)
Find the numerical destination noise power in the receiver's bandwidth N D .
N D = ____________
σ n2 = N 0 BT = kT × 4000 = 1.38 × 10 −23 × 5000 × 4000 = 2.76 × 10 −16
N D = σ n2 / 4 = 2.76 × 10 −16 / 4 = 6.9 × 10 −17 OR N D = σ n2 = 2.76 × 10 −16
(e)
Find the numerical signal-to-noise ratio at the destination ( S / N )D in dB.
( S / N )D
= ____________
( S / N )D = 3.6 × 10 −10 / 6.9 × 10 −17 = 5.22 × 10 6
or 67.175 dB
OR
( S / N )D = 1.44 × 10 −9 / 2.76 × 10 −16 = 5.22 × 10 6
or 67.175 dB
(f)
In dB, how much better could the signal-to-noise ratio be made by using a bandpass filter after the
detection process instead of a lowpass filter?
Improvement in dB = ____________
We could use a bandpass filter between 1.5 Hz and 2 kHz instead of a lowpass filter between 0
and 2 kHz. That would reduce the noise by a factor of 0.25, realizing an improvement in signal-tonoise ratio of about 6 dB.
5.
A signal with signal power Pin = 400 mW is the input signal to a transmission cable with loss of
α = 1.5 dB/km . The cable is 100 km in length. Repeaters are used to avoid the signal's getting lost in
noise. The signal power during transmission cannot be allowed to fall below 1 mW at any point on the
cable and each repeater has a power gain of 26 dB.
(a)
What is the minimum number of repeaters necessary?
Minimum number of repeaters is ____________
The minimum number of repeaters occurs when the signal is allowed to fall to 1 mW at the input
of all the repeaters. The cable length at which the signal power falls from 400 mW to 1mW is
17.3333 km. 100 km/17.3333km is 5.7692. So we need a minimum of 5 repeaters.
(b)
Using the minimum number of repeaters what is the output signal power?
Output signal power is ____________
The fifth repeater is at the 86.6667 km point. Its input power is 1 mW and its output power is 400
mW. The distance to the cable end is 13.3333 km. The signal loses 20 dB in that distance. So the
output signal power at the end of the cable is 4 mW.
6.
What is the purpose of guard bands in frequency-division multiplexing?
Guard bands are used to minimize crosstalk between channels when multiple signals are frequency
multiplexed onto to one signal. Crosstalk occurs because real filters do not have vertical sides and their
stopbands do not have infinite rejection.
7.
Given that the full range of AM carrier frequencies permitted by law is from 530 kHz to 1600 kHz in 10
kHz steps, and that the intermediate frequency used is 455 kHz, identify all the legal AM carrier
frequencies for which an image frequency exists in the AM range. You may do this by simply listing all
these frequencies or by specifying with <, >, ≤ and/or ≥ the ranges of frequencies.
The answer to this question depends on whether one assumes high-side injection or low-side injection of
the local oscillator.
With high-side injection, all carrier frequencies for which the frequency minus 910 kHz are in the AM
range are image frequencies. Those are 1440, 1450, 1460, 1470, 1480, 1490, 1500, 1510, 1520, 1530,
1540, 1550, 1560, 1570, 1580, 1590, 1600 or 1440 ≤ fc ≤ 1600 .
With low-side injection, all carrier frequencies for which the frequency plus 910 kHz are in the AM
range are image frequencies. Those are 530, 540, 550, 560, 570, 580, 590, 600, 610, 620, 630, 640, 650,
660, 670, 680, 690 or 530 ≤ fc ≤ 690 .
8.
If a phase locked loop VCO free runs at a frequency f0 and the incoming frequency to be locked to is at
fc and if f0 = fc and if the loop is locked, describe the exact phase relationship between the VCO output
signal and the incoming signal. (What is the phase difference and which signal is leading or lagging?)
The phase difference is exactly 90° with the VCO output leading the incoming signal.
If the incoming signal increases in size by a factor of two, what happens to this phase relationship?
It is momentarily perturbed but then returns back exactly to the same phase relationship.
9.
What is the effect on a demodulated USSB signal of a local oscillator at a frequency that is too high?
(Be specific.)
The lower edge of the USSB signal, instead of being shifted to zero, will be shifted to a frequency below
zero. So all the frequencies in the demodulated signal will be too low by the error in the local oscillator
frequency and the two replicas of the USSB signal will overlap causing distortion.
10.
What is the advantage of using an envelope detector compared with using a synchronous detector?
It allows for a very simple and economical demodulation of the signal compared to synchronous
techniques.