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my digital communications cheat sheet
Nasser M. Abbasi
January 5, 2019 Compiled on May 23, 2020 at 3:35am
Contents
1 What is the relation between bandpass, baseband ,complex envelop and pre
envelop? 2
2 Some useful Fourier Transforms 3
3 Random process definitions 3
4 How to determine Hilbert transform of a signal? 3
5 How to find Power Spectrum (PSD) of a random signal 𝑥 (𝑡) 3
6 What is the relation between variance and power for a random signal 𝑥 (𝑡)? 4
7 How to find the SNR for sampling quantization? 4
8 How to determine coding of a number from quantization? 5
8.1 sign magnitude.................................... 5
8.2 ones complement................................... 6
8.3 offset binary...................................... 6
8.4 2’s complement.................................... 7
9 How to derive the Phase and Frequency modulation signals? 7
10 How to obtain the phase deviation and the frequency deviation for angle
modulated signal? 8
11 How to quickly determine SNR𝑖 from 𝑆𝑁𝑅𝑐? 8
12 How to determine figure of merit for DSB-SC using coherent detector? 9
13 How to determine figure of merit for AM transmission using coherent de-
tector? 12
14 How to determine figure of merit for AM using envelope detector? 13
15 How to determine figure of merit for SSB using coherent detector? 15
16 How to determine figure of merit for VSB using coherent detector? 17
1 What is the relation between bandpass, baseband
,complex envelop and pre envelop?
BandPass signal
BasePass signal
Shifts frequency To band pass
s t sI t cos ct sQ t sin ct
In phase component (^) quadrature component
sI t s t cos ct ŝ t sin ct sQ t ŝ t cos ct s t sin ct
Hilbert transform of x(t)
s t a t cos ct t
Envelope of s(t) a t sI^2 t sQ^2 t t ^ ^ tan^1
sQ t sI t
Phase of s(t)
x t ^1 j sgn f X f
Complex envelope
-fc fc
X f X
f
X f
Pre-envelope
Bandpass signal^ Complex envelope
f
X f 2 X f U f x t 2 x t t 2 ^
j 2 t x t x t jx t
X ^ f X f fc
x t x t e jct
x t x t jx t e jct x t ejct^ x t ^ jx t Re x t ejct^ x t
x t x t ^1 t
x t xI t jxQ t
2 ways to find Hilbert transform :
Figure 1: bandpass, baseband ,complex envelop
1. Find Fourier Transform 𝑋 �𝑓� of 𝑥 (𝑡)
2. Find the �𝑋 �𝑓��
2 = 𝑋 �𝑓� 𝑋∗^ �𝑓�
6 What is the relation between variance and power for a
random signal 𝑥 (𝑡)?
Variance is the sum of the total average normalized power and the DC power.
𝜎^2 𝑥 =
total Power ���������𝐸 �𝑥 (^2) (𝑡)� +
DC power ���������𝐸 [𝑥 (𝑡)] 2
For the a signal whose mean is zero,
𝜎^2 𝑥 =
total Power ���������𝐸 �𝑥 (^2) (𝑡)�
How to find average, power, PEP, effective value (or the RMS) of a periodic function?
Let 𝑥 (𝑡) be a periodic function, of period 𝑇, then
average of 𝑥 (𝑡) = ⟨𝑥 (𝑡)⟩ =
𝑇 0
The average power is
𝑝𝑎𝑣 = �𝑥^2 (𝑡)� =
𝑇 0
|𝑥 (𝑡)|^2 𝑑𝑡
Effective value, or the RMS value is
𝑥𝑟𝑚𝑠 (𝑡) = ��𝑥^2 (𝑡)� = √𝑝𝑎𝑣 =
𝑇 0
𝑥^2 (𝑡) 𝑑𝑡
For example, for 𝑥 (𝑡) = 𝐴 cos (𝑥) , ⟨𝑥 (𝑡)⟩ = 0, 𝑃𝑎𝑣 = 𝐴
2 2 , 𝑥𝑟𝑚𝑠^ (𝑡) = 0.707𝐴
To find PEP (which is the peak envelope power), find the complex envelope 𝑥 (𝑡)̃ , then find
the average power of it. i.e.
𝑥^2 max (𝑡)
7 How to find the SNR for sampling quantization?
Suppose we have a message 𝑚 (𝑡) that is sampled. Assume we have 𝑛 bits to use for encoding
the sample levels. Hence there are 2 𝑛^ levels of quantizations. We want to find the ration of
the signal to the noise power. Noise here is generated due to quantization (i.e. due to the
rounding off values of 𝑚 (𝑡) during sampling).
This is the algorithm:
Input: 𝑛, the number of bits for encoding, 𝑚𝑝 absolute maximum value of the message
𝑚 (𝑡), the pdf 𝑓𝑋 (𝑡) of the message 𝑚 (𝑡) is 𝑚 (𝑡) is random message or 𝑚 (𝑡) function if it is
deterministic (such as cos (𝑡))
1. Find the quantization step size 𝑆 =
2𝑚𝑝 22
2. Find 𝑃𝑎𝑣 of the error is 121 𝑆^2 where 𝑆 is the step size found in (1), hence 𝑃𝑎𝑣 = 121 𝑆^2 =
1 12 �^
2𝑚𝑝 22 �
2
𝑚^2 𝑝 3×22𝑛
3. If 𝑚 (𝑡) is deterministic find 𝑝𝑎𝑣 = �𝑚^2 (𝑡)� = 𝑇^1 ∫
𝑇 0 |𝑚 (𝑡)|
4. If 𝑚 (𝑡) is random, find 𝑝𝑎𝑣 = 𝐸 (𝑚 (𝑡)) = ∫𝑚^2 (𝑡) 𝑓𝑋 (𝑡) 𝑑𝑡, this is called the second moment
of the pdf
3×22𝑛
Hence find 𝑆𝑁𝑅 for noise quantisation comes down to finding the power in the message
Examples: For sinusoidal message 𝑚 (𝑡), 𝑆𝑁𝑅𝑑𝑏 = 6𝑛 + 1.761. For random 𝑚 (𝑡) with PDF
which is uniform distributed 𝑆𝑁𝑅𝑑𝑏 = 6𝑛, for random 𝑚 (𝑡)^ which is AWGN. Do this later
8 How to determine coding of a number from
quantization?
Given an analog value say 𝑥 and given a maximum absolute possible value to be 𝑚𝑝, and
given the number of bits available for coding to be 𝑁, the following are the algorithm to
generate the quantized version of 𝑥, called 𝑥̂
8.1 sign magnitude
Input: 𝑥, 𝑚𝑝, 𝑁
output: 𝑥̂
Let Δ =
𝑚𝑝
2 𝑁−1^ called the step size
Let 𝑞 = 𝑟𝑜𝑢𝑛𝑑 � 𝑎𝑏𝑠(𝑥)Δ � which is the quantization level
If 𝑞 ≥ 2𝑁−1^ − 1 then 𝑞 = 2𝑁−1^ end if
8.4 2’s complement
Input: 𝑥, 𝑚𝑝, 𝑁
output: 𝑥̂
Let Δ =
𝑚𝑝
2 𝑁−1^ called the step size
Let 𝑞 = 𝑟𝑜𝑢𝑛𝑑 � 𝑎𝑏𝑠(𝑥)Δ � which is the quantization level
If 𝑥 ≥ − Δ 2 then
if 𝑞 ≥ 2𝑁−1^ − 1 then
𝑞 = 2𝑁−1^ − 1
end if
else
if 𝑞 ≥ 2𝑁−1^ − 1 then
end if
𝑐𝑜𝑑𝑒 = 2𝑁^ − 𝑞
end if
return 𝑐𝑜𝑑𝑒 in base 2
9 How to derive the Phase and Frequency modulation
signals?
For any bandpass signal, we can write it as
𝑥 (𝑡) = Re �̃𝑥 (𝑡) 𝑒𝑗𝜔𝑐𝑡�
Where 𝑥 (𝑡)̃ is the complex envelope of 𝑥 (𝑡). For PM and FM, the baseband modulated signal,
𝑥 (𝑡)̃ has the form 𝐴𝑐𝑒𝑗𝜃(𝑡)^ Hence the above becomes
𝑥 (𝑡) = Re �𝐴𝑐𝑒𝑗𝜃(𝑡)𝑒𝑗𝜔𝑐𝑡�
= 𝐴𝑐 (cos 𝜔𝑐𝑡 cos 𝜃 (𝑡) − sin 𝜔𝑐𝑡 sin 𝜃 (𝑡))
But cos (𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵, hence the above becomes
𝑥 (𝑡) = cos (𝜔𝑐𝑡 + 𝜃 (𝑡)) (1)
The above is the general form for PM and FM. Now, for PM, 𝜃 (𝑡) = 𝑘𝑝𝑚 (𝑡) and for FM,
𝑡
0 𝑚 (𝑡^1 ) 𝑑𝑡^1. Hence, substituting in (1) we obtain
𝑥𝐹𝑀 (𝑡) = cos �𝜔𝑐𝑡 + 𝑘𝑓 �
𝑡 0
and
𝑥𝑃𝑀 (𝑡) = cos �𝜔𝑐𝑡 + 𝑘𝑝𝑚 (𝑡)�
10 How to obtain the phase deviation and the frequency
deviation for angle modulated signal?
From the general form for angle modulated signal (see above note)
𝑥 (𝑡) = cos (𝜔𝑐𝑡 + 𝜃 (𝑡))
The phase deviation is 𝜃 (𝑡). And the maximum phase deviation is simply the maximum of
Now, to find the frequency deviation, we need a little bit more work. Start with
Where 𝑓𝑖 is the instantaneous frequency in Hz. But
Δ𝑓 ����������� 1 2𝜋
11 How to quickly determine SNR𝑖 from 𝑆𝑁𝑅𝑐?
First find 𝑆𝑁𝑅𝑐, for to find 𝑆𝑁𝑅𝑖 use the following
𝑇
, where 𝐵𝑇 is the transmission bandwidth, and 𝐵 is the baseband bandwidth.
For 𝐴𝑀, 𝐵𝑇 = 2𝐵. For 𝐷𝑆𝐵 − 𝑆𝐶, 𝐵𝑇 = 2𝐵. For 𝐷𝑆𝐵 − 𝑆𝑆, 𝐵𝑇 = 𝐵.
X
Ac^ ^ cos ct
LPF
Band Pass Filter (center at fc )
Coherent detector for DSB-SC
(SNR)i (SNR)o
+
w t
(SNR)C
(^1 )
3 4 5
s 1 t Acm t cos ct s 2 t Acm t cos ct w t
AWGN
s 3 t Acm t cos ct
narrow band noise n t Acm t cos ct nI t cos ct nQ t sin ct
in phase Acm t nI t cos ct
quadrature nQ t sin ct
s 4 t s 3 t Ac^ ^ cos ct
Acm t nI t cos ct nQ t sin ct Ac^ cos ct
Ac^ Acm t nI t cos^2 ct Ac^ nQ t sin ct cos ct
12 12 cos 2 ct Ac^ Acm t nI t Ac^ nQ t sin 0 sin 2 ct
Ac^ Acm t nI t
2 ^
Ac
2 Acm t ^ ^ nI t ^ cos 2 ct^ ^ Ac
nQ t sin 2 ct
s 5 t
Ac^ ^ Acm t nI t
SNRi
Ac^2 m^2 t ^ cos^2 ct
2 BN 0
Ac^2
2 Pm
2 BN 0
SNRc Ac
(^2) m (^2) t cos (^2) ct BN 0
Ac^2 2 Pm BN 0
Figure of Merit for DSB-SC signal By Nasser Abbasi11/30/ Study_notes_2.vsd
SNRo
Ac^ ^24 Ac 2 m (^2) t E nI 2 t ^2
Ac^ ^24 Ac 2 Pm (^14) E nI (^2) t
Ac^ ^24 Ac 2 Pm (^14 2) BN 0 ^ Ac^ ^2 Ac^2 Pm 2 BN 0
SNRo SNRi
Ac^ ^2 Ac^2 Pm 2 BN 0 Ac^2 22 BN^ Pm 0
2 Ac
2 SNR SNRoc
Ac^ ^2 Ac^2 Pm 2 BN 0 Ac^2 2 Pm BN 0
Ac
2B 2B
Figure 3: figure of merit for DSB-SC
Question: Verify the above.
SNRo
Ac^ ^ Ack a 2 m t
2
E Ac
2 nI t
2
Ac^ ^2 Ac^2 k a^2 4 Pm Ac^ ^2 4 E nI
(^2) t
Ac^ ^2 Ac^2 k a^2 4 Pm Ac^ ^2 4 2 BN^0
Ac^2 ka^2 Pm 2 BN 0
SNRo SNRc
Ac^2 k a^2 Pm 2 BN 0 Ac^2 2 1 k a (^2) Pm BN 0
ka^2 Pm 1 ka^2 Pm
Nasser M. Abbasi11/30/ AM_coherent_2.vsdx
Figure 5: figure of merit for AM coherent (2)
14 How to determine figure of merit for AM using
envelope detector?
Band Pass Filter (center at fc )
Envelope detector for AM
(SNR)i (^) (SNR) o
+
w t (SNR)C
2 3 4 5
AWGN
(^1) Envelope detector
DC blocker
2B 2B
Figure 6: figure of merit for AM using envelope detector
𝑠 1 (𝑡) = 𝐴𝑐 (1 + 𝑘𝑎𝑚 (𝑡)) cos 𝜔𝑐𝑡
𝑠 2 (𝑡) = 𝐴𝑐 (1 + 𝑘𝑎𝑚 (𝑡)) cos 𝜔𝑐𝑡 + 𝑤 (𝑡)
And
�(𝐴𝑐 (1 + 𝑘𝑎𝑚 (𝑡))^ cos 𝜔𝑐𝑡)^2 �
𝐴^2 𝑐 2 �(1 + 𝑘𝑎𝑚 (𝑡))
𝐴^2 𝑐 2 �1 + 𝑘
Now assuming ⟨𝑚 (𝑡)⟩ = 0, the above simplifies to
𝐴^2 𝑐 2 �1 + 𝑘
Hence
𝐴^2 𝑐 2 �1 + 𝑘
𝐴^2 𝑐 2 �1 + 𝑘
Now find 𝑠 3 (𝑡)
𝑠 3 (𝑡) = 𝐴𝑐 (1 + 𝑘𝑎𝑚 (𝑡)) cos 𝜔𝑐𝑡 +
narrow band noise 𝑛 (𝑡)^ �
= 𝐴𝑐 (1 + 𝑘𝑎𝑚 (𝑡)) cos 𝜔𝑐𝑡 + �𝑛𝐼 (𝑡) cos 𝜔𝑐𝑡 − 𝑛𝑄 (𝑡) sin 𝜔𝑐𝑡�
in phase
���������������������������������[𝐴𝑐 (1 + 𝑘𝑎𝑚 (𝑡)) + 𝑛𝐼 (𝑡)] cos 𝜔𝑐𝑡 −
quadrature
𝑛^ �𝑄 (𝑡) sin 𝜔𝑐𝑡
Now, to find 𝑠 4 (𝑡), which is the envelope of 𝑠 3 (𝑡).
𝑠 4 (𝑡) = envelope (𝑠 3 (𝑡))
= �(𝑠 3 )^2 𝐼 + (𝑠 3 )^2 𝑄
= �(𝐴𝑐 (1 + 𝑘𝑎𝑚 (𝑡)) + 𝑛𝐼 (𝑡))^2 + 𝑛^2 𝑄 (𝑡)
Now, assuming 𝐴𝑐 ≫ |𝑛𝐼 (𝑡)| and 𝐴𝑐 ≫ �𝑛𝑄 (𝑡)�, then the above simplifies to
now apply the DC blocker, we obtain
�(𝑘 [𝑚 (𝑡)^ cos^ 𝜔𝑐𝑡 −̂ 𝑚 (𝑡)^ sin^ 𝜔𝑐𝑡])^2 �
𝑘^2 �𝑚^2 (𝑡) cos^2 𝜔𝑐𝑡� + 𝑘^2 �̂𝑚^2 (𝑡) sin^2 𝜔𝑐𝑡� − 2𝑘^2 ⟨𝑚 (𝑡)̂ 𝑚 (𝑡) cos (𝜔𝑐𝑡) sin (𝜔𝑐𝑡)⟩
𝑘^2 �𝑚^2 (𝑡)� �cos^2 𝜔𝑐𝑡� + 𝑘^2 �𝑚^2 (𝑡)� �sin^2 𝜔𝑐𝑡� − 2𝑘^2 ⟨𝑚 (𝑡)̂ 𝑚 (𝑡) cos (𝜔𝑐𝑡) sin (𝜔𝑐𝑡)⟩
Assume ⟨𝑚 (𝑡)⟩ = 0, we obtain
𝑘^2 𝑃 2 𝑚 + 𝑘^2 𝑃 2 𝑚
𝑘^2 𝑃𝑚
𝑠 3 (𝑡) = 𝑘 [𝑚 (𝑡) cos 𝜔𝑐𝑡 −̂ 𝑚 (𝑡) sin 𝜔𝑐𝑡] + 𝑛 (𝑡)
Hence
𝑘^2 𝑃𝑚
𝑘^2 𝑃𝑚
𝑠 4 (𝑡) = [𝑘 [𝑚 (𝑡) cos 𝜔𝑐𝑡 −̂ 𝑚 (𝑡) sin 𝜔𝑐𝑡] + 𝑛 (𝑡)] 𝐴′ 𝑐 cos 𝜔𝑐𝑡
= 𝐴′ 𝑐𝑘𝑚 (𝑡)^ cos^2 𝜔𝑐𝑡 − 𝐴′ 𝑐𝑘̂𝑚 (𝑡)^ sin 𝜔𝑐𝑡 cos 𝜔𝑐𝑡 + 𝐴′ 𝑐 �𝑛𝐼 (𝑡)^ cos 𝜔𝑐𝑡 − 𝑛𝑄 (𝑡)^ sin 𝜔𝑐𝑡�^ cos 𝜔𝑐𝑡
cos 2𝜔𝑐𝑡� − 𝐴′ 𝑐𝑘̂𝑚 (𝑡)
(sin (0) + sin (2𝜔𝑐𝑡))
+ 𝐴′ 𝑐 �𝑛𝐼 (𝑡) cos^2 𝜔𝑐𝑡 − 𝑛𝑄 (𝑡) sin 𝜔𝑐𝑡 cos 𝜔𝑐𝑡�
′ 𝑐𝑘𝑚 (𝑡) +
′
𝑐𝑘𝑚 (𝑡)^ cos^ 2𝜔𝑐𝑡 − 𝐴
′ 𝑐𝑘̂𝑚 (𝑡)
2 sin^
cos 2𝜔𝑐𝑡� − 𝑛𝑄 (𝑡)
(sin (0) + sin (2𝜔𝑐𝑡))�
𝐴′ 𝑐𝑘𝑚 (𝑡) cos 2𝜔𝑐𝑡 −
𝐴′ 𝑐𝑘̂𝑚 (𝑡) sin (2𝜔𝑐𝑡)
𝑛𝐼 (𝑡) cos 2𝜔𝑐𝑡 −
𝑛𝑄 (𝑡) sin 2𝜔𝑐𝑡
After low pass filter, we obtain
′ 𝑐𝑘𝑚 (𝑡) +
2 𝑛𝐼^
Hence,
��^
1 2 𝐴
′ 𝑐𝑘𝑚 (𝑡)�
2 �
𝐸 �� 𝐴
′𝑐 2 𝑛𝐼^ (𝑡)�
2 �
1 4 �𝐴
′ 𝑐�
2 𝑘^2 𝑃𝑚 1 4 �𝐴
′ 𝑐�
2 𝑁 0 𝐵
=
𝑘^2 𝑃𝑚
Hence
𝑘^2 𝑃𝑚 𝑁 0 𝐵 𝑘^2 𝑃𝑚 𝐵𝑁 0 = 1
Hence
𝑘^2 𝑃𝑚 𝑁 0 𝐵 𝑘^2 𝑃𝑚 𝐵𝑁 0 = 1
16 How to determine figure of merit for VSB using
coherent detector?
�𝑚 (𝑡) cos 𝜔𝑐𝑡 ∓ 𝑚𝑄 (𝑡) sin 𝜔𝑐𝑡�
𝑚𝑄 (𝑡) is the output of VSB filter when input is 𝑚 (𝑡)
