DIGITAL COMMUNICATION
Block Diagram
 Functional model of pass-band data transmission system.
2
Digital Modulation Technique :
-ASK : Amplitude shift keying .
-FSK : Frequency shift keying .
-PSK : Phase shift keying .
-QAM: Quadrature amplitude modulation.
(Special case of Hybrid Modulation Techniques)
3
What do we want to study?
 We are going to study and compare different
modulation techniques in terms of
 Power Spectrum.
 Bandwidth efficiency:
Ro

B
Bits/s/Hz
 Probability of errors.
4
Complex Representation of bandpass Signals:
 Let 𝑔(𝑡) be any signal.
 Define 𝑔+ 𝑡 = 𝑔 𝑡 + 𝑗 ∙ 𝑔(𝑡) which called the
analytic signal (pre-envelope) of 𝑔(𝑡).
 𝑔(𝑡) is the Helbert transform of 𝑔(𝑡).
𝑔 𝑡 =
1
𝜋𝑡
⊗ 𝑔(𝑡),
 ℱ 𝑔(𝑡) = ℱ
𝐺 𝑓 =ℱ
𝑗
1 ℱ
𝜋𝑡
1
𝜋𝑡
1
𝜋𝑡
⊗convolution.
∙ ℱ 𝑔(𝑡)
∙ 𝐺(𝑓)
𝑠𝑔𝑛 𝑓 
1 ℱ
𝜋𝑡
− 𝑗 ∙ 𝑠𝑔𝑛 𝑓
5
Cont.
ℱ
1
𝜋𝑡
−𝑗, 𝑓 > 0
= 0, 𝑓 = 0
𝑗, 𝑓 < 0
sgn(f)
+1
f
-1
 𝐺 𝑓 = −𝑗 ∙ 𝑠𝑔𝑛 𝑓 ∙ 𝐺 𝑓

−𝑗𝐺 𝑓 = 𝐺 𝑓 𝑒 −𝑗𝜋 2 , 𝑓 > 0
0, 𝑓 = 0
=
𝑗𝐺 𝑓 = 𝐺 𝑓 𝑒 +𝑗𝜋 2 , 𝑓 < 0
6
Cont.
 ℱ 𝑔+ (𝑡) = 𝐺 𝑓 + 𝑠𝑔𝑛 𝑓 ∙ 𝐺(𝑓)
2𝐺 𝑓 , 𝑓 > 0
 𝐺+ 𝑓 = 𝐺 0 , 𝑓 = 0
0, 𝑓 < 0
 Also define 𝑔− 𝑡 = 𝑔 𝑡 − 𝑗 ∙ 𝑔(𝑡) called the –ve pre-envelope.
 𝐺− 𝑓 = 𝐺 𝑓 − 𝑠𝑔𝑛 𝑓 ∙ 𝐺(𝑓)
0, 𝑓 > 0
 𝐺− 𝑓 = 𝐺 0 , 𝑓 = 0
2𝐺 𝑓 , 𝑓 < 0
 𝑔− 𝑡 = 𝑔+ ∗ (𝑡)
7
Cont.
 Let 𝑔(𝑡) be a bandpass signal.
𝐺(𝑓)
𝐺(𝑓𝑐 )
−𝑓𝑐 − 𝑓𝑜
−𝑓𝑐
𝑓
−𝑓𝑐 + 𝑓𝑜
𝑓𝑐 − 𝑓𝑜
2 𝐺(𝑓𝑐 )
𝑓𝑐
𝑓𝑐 + 𝑓𝑜
𝐺+ (𝑓)
𝑓
𝑓𝑐 − 𝑓𝑜
2 𝐺(𝑓𝑐 )
−𝑓𝑜
𝑓𝑐
𝑓𝑐 + 𝑓𝑜
𝐺 (𝑓)
𝑓𝑜
𝑓
𝑓𝑐 − 𝑓𝑜
𝑓𝑐
𝑓𝑐 + 𝑓𝑜
8
Cont.
 Let 𝐺+ 𝑓 + 𝑓𝑐 = 𝐺 𝑓 = ℱ 𝑔(𝑡)
 𝑔(𝑡) is the equivalent low-pass signal of 𝑔(𝑡) called the




complex envelope of 𝑔(𝑡).
𝑔 𝑡 = ℱ −1 𝐺+ (𝑓 + 𝑓𝑐 )
= 𝑔+ (𝑡)𝑒 −𝑗2𝜋𝑓𝑐𝑡
 𝑔 𝑡 𝑒 𝑗2𝜋𝑓𝑐 𝑡 = 𝑔+ 𝑡
= 𝑔 𝑡 + 𝑗 𝑔(𝑡)
 𝑔 𝑡 = 𝑅𝑒 𝑔+ 𝑡
= 𝑅𝑒 𝑔 𝑡 𝑒 𝑗2𝜋𝑓𝑐 𝑡
9
Cont.
 𝑔 𝑡 in general is complex-valued function.
 ⇒ 𝑔 𝑡 = 𝑔𝐼 𝑡 + 𝑗 𝑔𝑄 (𝑡)
 𝑔𝐼 𝑡 : is real lowpass signal called the In-phase component of
𝑔(𝑡).
 𝑔𝑄 𝑡 : is real lowpass signal called the Quadrature component of
𝑔(𝑡).
 ⇒ 𝑔 𝑡 = 𝑅𝑒 𝑔𝐼 𝑡 + 𝑗 𝑔𝑄 (𝑡) 𝑒 𝑗2𝜋𝑓𝑐 𝑡

= 𝑅𝑒 𝑔𝐼 𝑡 + 𝑗 𝑔𝑄 (𝑡) cos 2𝜋𝑓𝑐 𝑡 + 𝑗 sin(2𝜋𝑓𝑐 𝑡)
 ⇒ 𝑔 𝑡 = 𝑔𝐼 (𝑡) cos 2𝜋𝑓𝑐 𝑡 − 𝑔𝑄 (𝑡) sin(2𝜋𝑓𝑐 𝑡)
 ⇒ 𝑔 𝑡 = 𝐼𝑚 𝑔+ 𝑡
= 𝐼𝑚 𝑔 𝑡 𝑒 𝑗2𝜋𝑓𝑐 𝑡
 ⇒ 𝑔 𝑡 = 𝑔𝐼 (𝑡) sin 2𝜋𝑓𝑐 𝑡 + 𝑔𝑄 (𝑡) cos(2𝜋𝑓𝑐 𝑡)
10
Cont.
 To get 𝑔𝐼 𝑡 and 𝑔𝑄 (𝑡) from 𝑔(𝑡) use the following
system.
LPF
𝑔𝐼 (𝑡)
LPF
𝑔𝑄 (𝑡)
2 cos 𝜔𝑐 𝑡
𝑔(𝑡)
~
π/2
−2 sin 𝜔𝑐 𝑡
11
Cont.
 To get back 𝑔(𝑡) from 𝑔𝐼 𝑡 and 𝑔𝑄 (𝑡) use the following
system.
𝑔𝐼 (𝑡)
cos 𝜔𝑐 𝑡
+
~
∑
-π/2
𝑔𝑄 (𝑡)
𝑔(𝑡)
-
sin 𝜔𝑐 𝑡
12
Cont.
 Since 𝑔 𝑡 = 𝑔𝐼 𝑡 + 𝑗 𝑔𝑄 (𝑡) is complex it can be
written in the polar form as:
 𝑔 𝑡 = a(t)𝑒 𝑗𝜙(𝑡)
 Where 𝑎 𝑡 =
𝑔2 𝐼 𝑡 + 𝑔2 𝑄 𝑡 Called Natural
Envelope of 𝑔(𝑡).
𝜙 𝑡 =
𝑔𝑄 (𝑡)
−1
tan
𝑔𝐼 𝑡
Called the phase of 𝑔(𝑡)
 ⇒ 𝑔 𝑡 = 𝑎 𝑡 cos 2𝜋𝑓𝑐 𝑡 + 𝜙(𝑡)
 𝑔𝐼 𝑡 = 𝑎 𝑡 cos 𝜙 𝑡 and 𝑔𝑄 𝑡 = 𝑎 𝑡 sin 𝜙 𝑡
13
Cont.





∞
𝐺 𝑓 = −∞ 𝑔(𝑡)𝑒 −𝑗2𝜋𝑓𝑡 𝑑𝑡
∞
= −∞ 𝑅𝑒 𝑔(𝑡)𝑒 𝑗2𝜋𝑓𝑐 𝑡 𝑒 −𝑗2𝜋𝑓𝑡 𝑑𝑡
1
But 𝑅𝑒 𝑥 = 𝑥 + 𝑥 ∗
2
1 ∞
⇒ 𝐺 𝑓 = −∞ 𝑔 𝑡 𝑒 𝑗2𝜋𝑓𝑐𝑡 + 𝑔∗ (𝑡)𝑒 −𝑗2𝜋𝑓𝑐 𝑡
2
1
⇒ 𝐺 𝑓 = 𝐺 𝑓 − 𝑓𝑐 + 𝐺 ∗ (−𝑓 − 𝑓𝑐 )
2
𝑒 −𝑗2𝜋𝑓𝑡 𝑑𝑡
14
Power Spectral Density
 Define PSD:
𝑆𝑔 𝑓 ≜ 𝐺 𝑓
 𝑆𝑔 𝑓 =
1
2
2
= 𝐺(𝑓)𝐺 ∗ (𝑓)
𝐺 𝑓 − 𝑓𝑐 + 𝐺 ∗ (−𝑓 − 𝑓𝑐 )
1
2
𝐺 ∗ 𝑓 − 𝑓𝑐 + 𝐺(−𝑓 − 𝑓𝑐 )
 Since 𝐺 𝑓 − 𝑓𝑐 and 𝐺 ∗ (−𝑓 − 𝑓𝑐 ) do not overlap
 ⇒ 𝑆𝑔 𝑓 =
 ⇒ 𝑆𝑔 𝑓 =
1
1 ∗
∗
𝐺 𝑓 − 𝑓𝑐 𝐺 (𝑓 − 𝑓𝑐 ) + 𝐺 (−𝑓
4
4
2
2
1
1
𝐺 𝑓 − 𝑓𝑐 + 𝐺 −𝑓 − 𝑓𝑐
4
4
1
4
− 𝑓𝑐 )𝐺 −𝑓 − 𝑓𝑐
1
4
 ⇒ 𝑆𝑔 𝑓 = 𝑆𝑔 𝑓 − 𝑓𝑐 + 𝑆𝑔 −𝑓 − 𝑓𝑐
15
Representation of linear bandpass system
 Let ℎ(𝑡) be the impulse response of a linear system (filter)
⇒ ℱ ℎ(𝑡) = 𝐻 𝑓 its transfer function.
 Since ℎ(𝑡) is real ⇒ 𝐻 ∗ −𝑓 = 𝐻 𝑓 .
𝐻 𝑓 ,𝑓 > 0
 Define 𝐻 𝑓 − 𝑓𝑐 =
0, 𝑓 < 0
⇒
𝐻∗
0,
𝑓>0
−𝑓 − 𝑓𝑐 =
𝐻 ∗ −𝑓 , 𝑓 < 0
0,
𝑓>0
=
𝐻 𝑓 ,𝑓 < 0
 So 𝐻 𝑓 = 𝐻 𝑓 − 𝑓𝑐 + 𝐻 ∗ −𝑓 − 𝑓𝑐
 ⇒ ℎ 𝑡 = ℎ 𝑡 𝑒 𝑗2𝜋𝑓𝑐𝑡 + ℎ∗ 𝑡 𝑒 −𝑗2𝜋𝑓𝑐𝑡 = 2𝑅𝑒 ℎ(𝑡)𝑒 𝑗2𝜋𝑓𝑐𝑡
16
Response of a bandpass system to a bandpass signal
 𝑔(𝑡) is a bandpass signal, 𝑔(𝑡) is its equivalent lowpass
signal.
 ℎ 𝑡 is an impulse response of bandpass system, ℎ(𝑡)
is its equivalent lowpass impulse response.
 𝑟(𝑡) the output of a bandpass system is also bandpass.
 ⇒ 𝑟 𝑡 = 𝑅𝑒 𝑟(𝑡)𝑒 𝑗2𝜋𝑓𝑐 𝑡
 𝑟 𝑡 =𝑔 𝑡 ⊗ℎ 𝑡 =
∞
𝑔
−∞
𝜏 ℎ 𝑡 − 𝜏 𝑑𝜏
 𝑅 𝑓 = ℱ 𝑟(𝑡) = 𝐺 𝑓 𝐻 𝑓
 ⇒𝑅 𝑓 =
1
2
𝐺 𝑓 − 𝑓𝑐 + 𝐺 ∗ (−𝑓 − 𝑓𝑐 ) 𝐻 𝑓 − 𝑓𝑐 + 𝐻∗ −𝑓 − 𝑓𝑐
𝑔(𝑡)
𝑟(𝑡)
ℎ(𝑡)
17
Cont.
 But for narrowband signal and system:
𝐺 𝑓 − 𝑓𝑐 𝐻 ∗ −𝑓 − 𝑓𝑐 = 0
And 𝐺 ∗ −𝑓 − 𝑓𝑐 𝐻 𝑓 − 𝑓𝑐 = 0
 ⇒𝑅 𝑓 =
1
2
𝐺 𝑓 − 𝑓𝑐 𝐻 𝑓 − 𝑓𝑐 + 𝐺 ∗ (−𝑓 − 𝑓𝑐 )𝐻 ∗ (−𝑓 −
𝑔(𝑡)
𝑟(𝑡)
ℎ(𝑡)
18
Representation of Bandpass Stationary
Random Process
 Let 𝑛(𝑡) be a sample function of a wide-sense stationary
random process with zero mean and PSD of 𝜓𝑛 𝑓 . The PSD
is concentrated about ±𝑓𝑐 and zero outside a certain band.
 ⇒ 𝑛 𝑡 = 𝑎 𝑡 cos 2𝜋𝑓𝑐 𝑡 + 𝜃(𝑡)
= 𝑥 𝑡 cos 2𝜋𝑓𝑐 𝑡 − 𝑦 𝑡 sin 2𝜋𝑓𝑐 𝑡
= 𝑅𝑒 𝑛(𝑡)𝑒 𝑗2𝜋𝑓𝑐𝑡
 𝑛 𝑡 = 𝑥 𝑡 + 𝑗𝑦(𝑡)
 𝑛 𝑡 is the complex envelope (lowpass)
 𝑥 𝑡 is the In phase component (lowpass)
 𝑦(𝑡) is the Quadrature component (lowpass)
19
Cont.

𝑅𝑛𝑛 𝜏 = 𝐸 𝑛 𝑡 𝑛(𝑡 − 𝜏)
= 𝐸 𝑥 𝑡 cos 2𝜋𝑓𝑐 𝑡 − 𝑦 𝑡 sin 2𝜋𝑓𝑐 𝑡
𝑥 𝑡 − 𝜏 cos 2𝜋𝑓𝑐 (𝑡 − 𝜏) − 𝑦 𝑡 − 𝜏 sin(2𝜋𝑓𝑐 (𝑡 −
20
Cont.
𝑅𝑛𝑛 𝜏
⇒ 𝑅𝑛𝑛
𝑅𝑥𝑥 𝜏
𝑅𝑦𝑦 𝜏
= 𝐸 𝑛 𝑡 𝑛(𝑡 − 𝜏)
−𝜏 = 𝐸 𝑛 𝑡 𝑛(𝑡 + 𝜏) = 𝑅𝑛𝑛 𝜏 even function
= 𝐸 𝑥 𝑡 𝑥(𝑡 − 𝜏) even function
= 𝐸 𝑦 𝑡 𝑦(𝑡 − 𝜏) even function
But:
 𝑅𝑦𝑥 −𝜏 = 𝐸 𝑦 𝑡 𝑥(𝑡 + 𝜏)
= 𝐸 𝑥 𝛾 𝑦(𝛾 − 𝜏) = 𝑅𝑥𝑦 𝜏 = −𝑅𝑦𝑥 𝜏
⇒odd function
And 𝑅𝑥𝑦 (𝜏) is also odd.
21
Cont.
 𝑛 𝑡 = 𝑥 𝑡 + 𝑗𝑦(𝑡)
 𝑅𝑛𝑛 𝜏 =
=
=
1
𝐸 𝑛(𝑡)𝑛∗ (𝑡 − 𝜏)
2
1
𝐸 𝑥 𝑡 + 𝑗𝑦(𝑡)
2
1
𝑅𝑥𝑥 𝜏 + 𝑅𝑦𝑦 𝜏
2
𝑥 𝑡 − 𝜏 − 𝑗𝑦(𝑡 − 𝜏)
− 𝑗𝑅𝑥𝑦 𝜏 + 𝑗𝑅𝑦𝑥 (𝜏)
But 𝑅𝑥𝑥 𝜏 = 𝑅𝑦𝑦 𝜏 , 𝑅𝑦𝑥 𝜏 = −𝑅𝑥𝑦 𝜏
⇒ 𝑅𝑛𝑛 𝜏 = 𝑅𝑥𝑥 𝜏 + 𝑗𝑅𝑦𝑥 (𝜏)
𝑅𝑛𝑛 𝜏 = 𝑅𝑥𝑥 𝜏 cos(2𝜋𝑓𝑐 𝜏) − 𝑅𝑦𝑥 𝜏 sin 2𝜋𝑓𝑐 𝜏
⇒ 𝑅𝑛𝑛 𝜏 = 𝑅𝑒 𝑅𝑛𝑛 𝜏 𝑒 𝑗2𝜋𝑓𝑐𝜏
22
Cont.
 𝑅𝑛𝑛 𝜏 = 𝑅𝑛∗ 𝑛 (−𝜏) since 𝑅𝑦𝑥 (𝜏) is odd function.
 PSD of 𝑛(𝑡): 𝑆𝑛 𝑓 = ℱ 𝑅𝑛𝑛 𝜏
=
∞
𝑅 𝜏
−∞ 𝑛𝑛
−𝑗2𝜋𝑓𝜏
∞
∗
𝑅
(−𝜏)𝑒
−∞ 𝑛𝑛
∞
= −∞ 𝑅𝑛𝑛 −𝜏 𝑒 𝑗2𝜋𝑓𝜏
= 𝑆𝑛∗ 𝑓
=
𝑒 −𝑗2𝜋𝑓𝜏 𝑑𝜏
𝑑𝜏
𝑑𝜏
∗
⇒ 𝑆𝑛 𝑓 is real-valued function.
 PSD of 𝑛 𝑡 :
𝑆𝑛 𝑓 = ℱ 𝑅𝑛𝑛 𝜏
=ℱ
1
=
2
1
=
2
1
2
= ℱ 𝑅𝑒 𝑅𝑛𝑛 𝜏 𝑒 𝑗2𝜋𝑓𝑐𝜏
𝑅𝑛𝑛 𝜏 𝑒 𝑗2𝜋𝑓𝑐𝜏 + 𝑅𝑛∗ 𝑛 (𝜏)𝑒 −𝑗2𝜋𝑓𝑐𝜏
𝑆𝑛 𝑓 − 𝑓𝑐 + 𝑆𝑛∗ −𝑓 − 𝑓𝑐
𝑆𝑛 𝑓 − 𝑓𝑐 + 𝑆𝑛 −𝑓 − 𝑓𝑐
23
Power Spectral Density of Digital Carrier Systems:
Φ(𝑡) = 𝑚(𝑡) cos(𝜔𝑐 𝑡)
Where:
Φ(𝑡): Modulated signal.
𝑚(𝑡): baseband message signal (Modulating signal).
cos(𝜔𝑐 𝑡): un modulated carrier.
𝜔𝑐 : carrier frequency.
The PSD is:
SΦ(ω)
Ψ𝑇(𝜔) 2
= lim
𝑇
𝑇→∞
24
PSD of Digital Carrier Systems (Cont.)
Where: Ψ𝑇 ω = ℱ ΦT(t)
𝑡
𝑇
ΦT(t) = Φ(t) ∙ 𝑟𝑒𝑐𝑡( )
𝑡
𝑇
ΦT(t) = m(t) ∙ cos(𝜔𝑐 t) rect( )
ΦT(t)
𝑡
𝑟𝑒𝑐𝑡(𝑇)
Φ(t)
t
-T/2
T/2
t
-T/2
T/2
25
𝑡
ΦT(t) = m(t) rect( ) cos(𝜔𝑐 t)
𝑇
𝑚 𝑇 (𝑡)
ℱ 𝑚 𝑇 (𝑡) = 𝑀𝑇 (𝜔)
1
Ψ𝑇(ω) = [𝑀𝑇 (ω − 𝜔𝑐 ) + 𝑀𝑇 (ω + 𝜔𝑐 )]
2
1 𝑀𝑇 (ω − 𝜔𝑐 ) + 𝑀𝑇 (ω + 𝜔𝑐 )
SΦ(ω) = lim
𝑇→∞ 4
T
2
𝑀 𝜔 : Band limited
−𝜔𝐵
𝜔𝐵
𝜔
26
1 MT(ω−𝜔𝑐 )
SΦ(ω) = lim
𝑇→∞ 4
T
If 𝜔𝑐 ≥𝜔𝐵
2
MT(ω+𝜔𝑐 )
+
T
2
1
SΦ(ω) = [SM(ω−𝜔𝑐 ) + SM(ω+𝜔𝑐 )]
4
𝑤𝑜
2𝜋
1
Bandwidth =
Hz
Bandwidth = fo Hz
−𝜔𝑜
PSD of m(t)
𝜔𝑜
𝜔
PSD of Φ(t)
PΦ =
1
2
1/4
𝑃𝑚
−𝜔𝑐 − 𝜔𝑜 −𝜔𝑐
𝑤
Bandwidth = 2 (2𝜋𝑜) Hz
Bandwidth = 2 fo Hz
−𝜔𝑐 + 𝜔𝑜
𝜔𝑐 − 𝜔𝑜
𝜔𝑐
𝜔
𝜔𝑐 + 𝜔𝑜
27
Binary Digital Modulation
 Binary ASK:
Φ𝐴𝑆𝐾 𝑡 = 𝑦 𝑡 + 𝐷𝐶 cos 𝜔𝑐 𝑡
𝑚(𝑡)
= 𝑦 𝑡 cos 𝜔𝑐 𝑡 +(DC) cos 𝜔𝑐 𝑡
Where 𝑦(𝑡) is polar NRZ.
28
Φ𝐴𝑆𝐾 (𝑡)
A
-A
A+DC
Φ𝐴𝑆𝐾 (𝑡)
DC
-DC
AM signal with m<1
2A
Φ𝑂𝑂𝐾 (𝑡)
-2A
AM signal with m=1
On-Off keying
m: modulation index
29
𝑃(𝜔)
𝑆𝑌 𝜔 =
𝑇
2
𝑃 𝜔
=
𝑇0
2
∞
ℛ𝑛 𝑒 −𝑗𝑛𝜔𝑇
𝑛=−∞
For polar line code
 𝑃 𝜔 = ℱ 𝑝 𝑡 ] could be any shape
 Let 𝑝 𝑡 = 𝐴 ∙
𝑡−𝑇0 2
𝑟𝑒𝑐𝑡(
)
𝑇0
(NRZ rect)
𝑝(𝑡)
𝑃 𝜔
𝜔𝑇0 −𝑗𝜔𝑇0
2
= 𝐴 ∙ 𝑇0 𝑠𝑖𝑛𝑐(
)𝑒
2
𝜔𝑇0
2
2
2
2
𝑃 𝜔 = 𝐴 ∙ 𝑇0 𝑠𝑖𝑛𝑐 (
)
2
𝑆𝑌 𝜔 =
𝐴2
∙ 𝑇0
𝑠𝑖𝑛𝑐 2 (
𝑇0
2
t
𝑇0
𝜔𝑇0
)
2
30
𝑆Φ (𝜔)
−𝜔𝑐 − 2𝜋𝑅𝑜 −𝜔𝑐
−𝜔𝑐 + 2𝜋𝑅𝑜
(𝐷𝐶)2 𝜋
2
𝜔𝑐 − 2𝜋𝑅𝑜 𝜔𝑐
𝐴2 𝑇𝑜
4
𝜔𝑐 + 2𝜋𝑅𝑜
f
𝑓𝑐 − 𝑅𝑜 𝑓𝑐 𝑓𝑐 + 𝑅𝑜
EB = 2Ro
31
OOK:
ΦOOK(t) = y(t) cos(𝜔𝑐 t)
Here, y(t) is Unipolar NRZ.
𝑃(𝜔)
SY(ω) =
4𝑇𝑜
∞
2
2𝜋
1+
𝛿(𝜔 − 2𝜋𝑛𝑅𝑜 )]
𝑇𝑜
𝑛=−∞
𝑡 − 𝑇0 2
𝑝 𝑡 = 2𝐴 ∙ 𝑟𝑒𝑐𝑡(
)
𝑇0
∞
𝜔𝑇𝑜
2𝜋
2𝜋𝑛
2
2
SY(ω) =𝐴 𝑇𝑜 𝑠𝑖𝑛𝑐 (
) 1+
𝛿(𝜔 −
)]
2
𝑇𝑜
𝑇𝑜
𝑛=−∞
𝐴2 𝜋
2
𝑆Φ (𝜔)
𝐴2 𝑇𝑜
4
−𝜔𝑐 − 2𝜋𝑅𝑜 −𝜔𝑐 −𝜔𝑐 + 2𝜋𝑅𝑜
𝜔𝑇
Since sinc( 2 𝑜) is
zero at (2𝜋nR𝑜)
then, one discrete
component at 𝜔𝑐
𝜔𝑐 − 2𝜋𝑅𝑜 𝜔𝑐 𝜔𝑐 + 2𝜋𝑅𝑜
32
Binary ASK:Demodulation
1.
2.
Coherent (Synchronous).
Non-Coherent (Asynchronous) Using Envelope Detector (ED).
 Coherent:
r(t)
ΦASK(t)
LPF
𝟐𝒄𝒐𝒔(𝝎𝒄 𝒕)
ΦASK(t) = m(t) cos(𝜔𝑐 t)
r(t) = m(t) cos(𝜔𝑐 t) ((2)cos(𝜔𝑐 t))
1
r(t) = 2m(t)[ (cos(2𝜔𝑐 ) + cos(0))] = m(t) cos(2𝜔𝑐 t) + m(t)
2
And after the LPF:
l(t) = m(t) = y(t) + DC
l(t)
y(t) + DC
With noise it will replaced
by a Match Filter
(will explained later)
33
 Non-coherant :
 ASK
ENVELOPE
DETECTOR
yt   DC
34
Constellation of Binary ASK
𝑠2
𝑠1
𝜓1 (𝑡)
35
Binary FSK
 𝜑𝐹𝑆𝐾 = 𝐴 cos 𝜔𝑐 𝑡 +
𝑡
𝑘𝑓 −∞ 𝑦
𝛼 𝑑𝛼
⇒ 𝐹𝑀
 Where y(t) is polar NRZ.
36
1
0
0
1
1
0
1
0
0
1
1
0
polar A
NRZ
-A
A
𝜑𝐵𝐹𝑆𝐾
-A
A
cos(𝜔1 𝑡)
𝜑𝑂𝑂𝐾1 𝑡
-A
A
0
1
1
0
0
𝜑𝑂𝑂𝐾2 𝑡
-A
cos(𝜔2 𝑡)
1
 𝜑𝐵𝐹𝑆𝐾 𝑡 = 𝜑𝑂𝑂𝐾1 𝑡 + 𝜑𝑂𝑂𝐾2 (𝑡)
Binary FSK(Cont.)
 "1": 𝐴 𝑝 𝑡 cos 𝜔1 𝑡
, 𝜆𝑇𝑜 < 𝑡 < (𝜆 + 1)𝑇0
 "0": 𝐴 𝑝 𝑡 cos 𝜔2 𝑡
, 𝜆𝑇𝑜 < 𝑡 < 𝜆 + 1 𝑇𝑜
 𝜔1 = 𝜔𝑐 + ∆𝜔
→ mark frequincy.
 𝜔2 = 𝜔𝑐 − ∆𝜔
→ space frequincy.

∆𝜔
2𝜋
= ∆𝑓
 Using Carson's Rule
 𝐵𝐹𝑀 = 2∆𝑓 + 2𝐵 , B : Bandwidth of the base band signal
Binary FSK(Cont.)
 Since FSK is an FM modulation with the base band signal
y(t) which is polar NRZ
 → Bandwidth of y t is 𝐵 = 𝑅𝑜
  𝐵𝑊 𝑜𝑓 𝐹𝑆𝐾 = 2∆𝑓 + 2𝑅𝑜
 PSD of FSK is :
PSD of 𝜑𝑂𝑂𝐾1 𝑡 + PSD of 𝜑𝑂𝑂𝐾2 𝑡 +𝑆12 𝜔 + 𝑆21 (𝜔)
Cross PSDs
𝑆𝐵𝐹𝑆𝐾 𝜔
2𝑅0
2𝑅0
𝜔𝑐 − ∆𝜔 𝜔2 + 2π𝑅𝑜 𝜔𝑐
𝜔𝑐 + ∆𝜔 𝜔1 + 2π𝑅𝑜
𝜔
𝑓
𝑓2
𝑓2 + 𝑅𝑜
𝑓𝑐
𝑓1 − 𝑅𝑜
2∆𝑓 + 2𝑅𝑜
𝑓1
𝑓1 + 𝑅𝑜
Binary FSK(Cont.)
 With choosing Δω , ωc properly we can get rid of the
discrete components at (ωc+ Δω) & (ωc-Δω)
 → (𝜔𝑐 +∆𝜔) − (𝜔𝑐 − ∆𝜔)
 → 2∆𝜔 = 𝑛(2𝜋𝑅𝑜 )
 or 2∆𝑓 = 𝑛𝑅𝑜 → (Multiple of 𝑅𝑜 )
 → ∆𝑓 =
𝑛𝑅𝑜
2
 (No discrete component at (ωc+ Δω) & (ωc-Δω))
 Δf increase → BW increase
 Δf decrease → BW decrease
NonCoherent BFSK
 𝑠1 𝑡 = 𝐴 cos 𝜔1 𝑡 + 𝜃1 , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜 , for “1”
 𝑠2 𝑡 = 𝐴 cos 𝜔2 𝑡 + 𝜃2 , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜 , for “0”
 𝜃1 , 𝜃2 are initial phases at 𝑡 = 0.
 These two signals are not coherent since 𝜃1 and 𝜃2 are not the same in
general. The waveform is not continuous at bit transitions. This form of
FSK is therefore called noncoherent or discontinuous FSK.
 It can be noncoherently demodulated.
cos 𝜔1 𝑡 + 𝜃1
~
1
𝜑𝐵𝐹𝑆𝐾 (𝑡)
2
𝑐𝑜𝑠 𝜔2 𝑡 + 𝜃2
~
𝐼𝑘
43
Coherent BFSK
 The two signals have the same initial phase 𝜃 at t = 0.
 𝑠1 𝑡 = 𝐴 cos 𝜔1 𝑡 + 𝜃 , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜 , for “1”
 𝑠2 𝑡 = 𝐴 cos 𝜔2 𝑡 + 𝜃 , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜 , for “0”
 For coherent demodulation of the coherent FSK signal, the two frequencies
are chosen so that the two signals are orthogonal:
(𝜆+1)𝑇𝑜
𝑠1 (𝑡)𝑠2 (𝑡) 𝑑𝑡 = 0
𝜆𝑇𝑜
𝐴 cos 𝜔1 𝑡 + 𝜃
1
Frequency
Synthesizer
𝜑𝐵𝐹𝑆𝐾 (𝑡)
2
𝐴 cos 𝜔2 𝑡 + 𝜃
44
Cont.
⇒
⇒
(𝜆+1)𝑇𝑜
𝐴𝑐𝑜𝑠
𝜆𝑇𝑜
𝐴2 (𝜆+1)𝑇𝑜
2 𝜆𝑇𝑜
2𝜋𝑓1 𝑡 + 𝜃 𝐴𝑐𝑜𝑠 2𝜋𝑓2 𝑡 + 𝜃 𝑑𝑡 = 0
𝑐𝑜𝑠 2𝜋 𝑓1 + 𝑓2 𝑡 + 2𝜃 + 𝑐𝑜𝑠 2𝜋 𝑓1 − 𝑓2 𝑡 𝑑𝑡 = 0
𝐴2
⇒ 2𝜋 𝑓 +𝑓 cos 2𝜃 𝑠𝑖𝑛 2𝜋 𝑓1 + 𝑓2
1 2
𝐴2
(1+𝜆)𝑇𝑜
𝑠𝑖𝑛
2𝜋
𝑓
−
𝑓
𝑡
|
1
2
𝜆𝑇𝑜
2𝜋 𝑓1 −𝑓2
(1+𝜆)𝑇𝑜
𝑡 + sin 2𝜃 𝑐𝑜𝑠 2𝜋 𝑓1 + 𝑓2 𝑡 |𝜆𝑇𝑜
+
=0
 This requires 2𝜋 𝑓1 + 𝑓2 𝑇𝑜 = 𝑛𝜋 and 2𝜋 𝑓1 − 𝑓2 𝑇𝑜 = 𝑚𝜋, 𝑛&𝑚 are integers.
𝑛+𝑚
4𝑇𝑜
𝑛−𝑚
𝑓2 = 4𝑇
𝑜
 ⇒ 𝑓1 =
 ⇒
 2Δ𝑓 = 𝑓1 − 𝑓2 =
𝑚
2𝑇𝑜
45
Cont.
 Thus we conclude that for orthogonality 𝑓1 and 𝑓2 must be integer multiple
of
1
4𝑇𝑜
and their difference must be integer multiples of
1
.
2𝑇𝑜
 𝑓1 = 𝑓𝑐 + Δ𝑓
 𝑓2 = 𝑓𝑐 − Δ𝑓
 ⇒ 𝑓𝑐 =
𝑓1 +𝑓2
2
=
𝑛
2𝑇𝑜
Where 𝑓𝑐 is the nominal (or apparent) carrier frequency which must be an
integer multiple of 1 2𝑇𝑜 for orthogonality.
 When the separation between 𝑓1 and 𝑓2 is chosen as 1 𝑇𝑜 , then the
phase continuity will be maintained at bit transitions and the FSK is
called Sunde ’s FSK.
46
Cont.
 As a matter of fact, if the separation is 𝛾 𝑇𝑜 , where 𝛾 is an integer, the
phase of the coherent FSK signal is always continuous.
Proof:
at 𝑡 = 𝜆𝑇𝑜 , the phase of 𝑠1 (𝑡) is
2𝜋𝑓1 𝜆𝑇𝑜 + 𝜃 = 2𝜋(𝑓2 + 𝛾/𝑇𝑜 )𝜆𝑇𝑜 + 𝜃
= 2𝜋𝑓2 𝜆𝑇𝑜 + 2𝜋𝜆𝛾 + 𝜃
= 2𝜋𝑓2 𝜆𝑇𝑜 + 𝜃 (𝑀𝑜𝑑𝑢𝑙𝑜 2𝜋)
which is exactly the phase of 𝑠2 (𝑡). Thus at 𝑡 = 𝜆𝑇𝑜 , if the input bit switches
from 1 to 0, the new signal 𝑠2 (𝑡) will start at exactly the same amplitude
where 𝑠1 (𝑡) has ended.
 The minimum separation for orthogonality between 𝑓1 and 𝑓2 is 1 2𝑇𝑜 . As
we have just seen above, this separation cannot guarantee continuous phase.
A particular form of FSK called minimum shift keying (MSK) not only has
the minimum separation but also has continuous phase.
47
Cont.
 Figure (a) is an example of Sunde’s FSK waveform where bit 1 corresponds
to a higher frequency 𝑓1 and bit 0 a lower 𝑓2 . Since 𝑓1 and 𝑓2 are multiples
of 1 𝑇𝑜 , the ending phase of the carrier is the same as the starting phase,
therefore the waveform has continuous phase at the bit boundaries.
 A coherent FSK waveform might have discontinuous phase at bit
boundaries. Figure (b) is an example of such a waveform, where
𝑓1 = 9/4𝑇𝑜 , 𝑓2 = 6/4𝑇𝑜 , and the separation is 2Δ𝑓 = 3/4𝑇𝑜 .
48
PSD of Sunde’s FSK signal
 We expand the Sunde’s FSK signal as:
𝑔 𝑡 = 𝐴 cos 2𝜋 𝑓𝑐 + 𝑎𝜆
1
2𝑇𝑜
𝑡, 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜
Where 𝑎𝜆 = ±1
⇒ 𝑔 𝑡 = 𝐴 cos 𝑎𝜆
⇒ 𝑔 𝑡 = 𝐴 cos
⇒ 𝑔𝐼 𝑡 = 𝐴 cos
𝜋𝑡
𝑇𝑜
𝜋𝑡
𝑇𝑜
⇒ 𝑔𝑄 𝑡 = 𝐴𝑎𝜆 sin
𝜋𝑡
𝑇𝑜
cos 2𝜋𝑓𝑐 𝑡 − 𝐴 sin 𝑎𝜆
cos 2𝜋𝑓𝑐 𝑡 − 𝐴𝑎𝜆 sin
𝜋𝑡
𝑇𝑜
𝜋𝑡
𝑇𝑜
sin 2𝜋𝑓𝑐 𝑡
sin 2𝜋𝑓𝑐 𝑡
is independent of the data
𝜋𝑡
𝑇𝑜
is directly related to the data
So the inphase and quadrature components are independent.
49
Cont.
 ⇒ 𝑔 𝑡 = 𝑔𝐼 (𝑡) cos 2𝜋𝑓𝑐 𝑡 − 𝑔𝑄 (𝑡) sin(2𝜋𝑓𝑐 𝑡)
 𝑔 𝑡 = 𝑅𝑒 𝑔 𝑡 𝑒 𝑗2𝜋𝑓𝑐𝑡
 𝑔 𝑡 = 𝑔𝐼 𝑡 + 𝑗 𝑔𝑄 (𝑡)
1
1
 ⇒ 𝑆𝑔 𝑓 = 𝑆𝑔 𝑓 − 𝑓𝑐 + 𝑆𝑔 −𝑓 − 𝑓𝑐
4
4
 Since the inphase component and the quadrature component of the FSK signal
are independent of each other, the PSD for the complex envelope is the sum of
the PSDs of these two components.
 ⇒ 𝑆𝑔 𝑓 = 𝑆𝐼 𝑓 + 𝑆𝑄 (𝑓)
 𝑆𝐼 𝑓 = ℱ 𝐴 cos
𝜋𝑡
𝑇𝑜
2
=
𝐴2
4
𝛿 𝑓−
1
2𝑇𝑜
+𝛿 𝑓+
1
2𝑇𝑜
50
Cont.
 𝑔𝑄 𝑡 = 𝐴𝑎𝜆 sin
𝜋𝑡
𝑇𝑜
, 𝑎𝜆 = −1, +1 which are equally likely.
 We can look at it as a polar signal with pulse shape 𝑝 𝑡 = 𝐴 sin
 𝑆𝑄 𝑓 =
𝑃(𝜔) 2
𝑇𝑜
 ⇒ 𝑆𝑔 𝑓 =
𝐴2
4
=
1
𝑇𝑜
ℱ 𝐴 sin
𝛿 𝑓−
1
2𝑇𝑜
𝜋𝑡
𝑇𝑜
2
+𝛿 𝑓+
=
1
2𝑇𝑜
1 2𝐴𝑇𝑜 cos 𝜋𝑇𝑜 𝑓
𝑇𝑜 𝜋 1− 2𝑇𝑜 𝑓 2
+
𝜋𝑡
𝑇𝑜
2
2𝐴 cos 𝜋𝑇𝑜 𝑓
𝑇𝑜
𝜋 1− 2𝑇𝑜 𝑓 2
2
51
Orthogonality of Noncoherent BFSK





𝑠1 𝑡 = 𝐴 cos 𝜔1 𝑡 + 𝜃1 , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜 , for “1”
𝑠2 𝑡 = 𝐴 cos 𝜔2 𝑡 + 𝜃2 , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜 , for “0”
𝜃1 , 𝜃2 are initial phases at 𝑡 = 0. To simplify assume 𝜃1 = 0 and 𝜃2 = 𝜃
⇒ 𝑠1 𝑡 = 𝐴 cos 𝜔1 𝑡 , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜 , for “1”
⇒ 𝑠2 𝑡 = 𝐴 cos 𝜔2 𝑡 + 𝜃 , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜 , for “0”

(𝜆+1)𝑇𝑜
𝐴𝑐𝑜𝑠 2𝜋𝑓1 𝑡 𝐴𝑐𝑜𝑠 2𝜋𝑓2 𝑡 + 𝜃 𝑑𝑡 =
𝜆𝑇𝑜
(𝜆+1)𝑇𝑜
𝐴2 cos 𝜃 𝜆𝑇
𝑐𝑜𝑠 2𝜋𝑓1 𝑡 𝑐𝑜𝑠 2𝜋𝑓2 𝑡 𝑑𝑡 −
𝑜
(𝜆+1)𝑇𝑜
𝐴2 sin 𝜃 𝜆𝑇
𝑐𝑜𝑠 2𝜋𝑓1 𝑡 𝑠𝑖𝑛 2𝜋𝑓2 𝑡 𝑑𝑡 = 0
𝑜

𝑠𝑖𝑛 2𝜋 𝑓1 +𝑓2 𝑡
𝐴2 cos 𝜃
2𝜋 𝑓1 +𝑓2 𝑡
+
𝐴2 sin 𝜃
=0
 ⇒
0
𝑠𝑖𝑛 2𝜋 𝑓1 −𝑓2 𝑡 (𝜆+1)𝑇𝑜
+
2𝜋 𝑓1 −𝑓2 𝑡
𝜆𝑇𝑜
𝑐𝑜𝑠 2𝜋 𝑓1 +𝑓2 𝑡
𝑐𝑜𝑠 2𝜋 𝑓1 −𝑓2 𝑡 (𝜆+1)𝑇𝑜
+
2𝜋 𝑓1 +𝑓2 𝑡
2𝜋 𝑓1 −𝑓2 𝑡
𝜆𝑇𝑜
52
Cont.
 For arbitrary 𝜃, this requires that the sums inside the brackets be zero. This,
in turn, requires that 2𝜋(𝑓1 + 𝑓2 )𝑇𝑜 = 𝑘𝜋 for the first term and
2𝜋(𝑓1 − 𝑓2 )𝑇𝑜 = 𝑙𝜋 for the second term in the first bracket, that
2𝜋(𝑓1 + 𝑓2 )𝑇𝑜 = 2𝑚𝜋 for the first term and 2𝜋(𝑓1 − 𝑓2 )𝑇𝑜 = 2𝑛𝜋 for the
second term in the second bracket, where 𝑘, 𝑙, 𝑚, 𝑎𝑛𝑑 𝑛 are integers and 𝑘 >
𝑙, 𝑚 > 𝑛. The 𝑘𝜋 and 𝑙𝜋 cases are included in the 2𝑚𝜋 and 2𝑛𝜋 case,
respectively.
⇒ 2𝜋(𝑓1 − 𝑓2 )𝑇𝑜 = 2𝑛𝜋
⇒ 2𝜋(𝑓1 + 𝑓2 )𝑇𝑜 = 2𝑚𝜋
𝑛+𝑚
2𝑇𝑜
𝑚−𝑛
𝑓2 =
2𝑇𝑜
 ⇒ 𝑓1 =
 ⇒
 2Δ𝑓 = 𝑓1 − 𝑓2 =
𝑛
𝑇𝑜
53
Cont.
 This is to say that for two noncoherent FSK signals to be orthogonal, the
two frequencies must be integer multiples of 1/2𝑇𝑜 and their separation
must be a multiple of 1/𝑇𝑜 . When 𝑛 = 1, the separation is the 1/𝑇𝑜 , which
is the minimum. Comparing with a coherent FSK case, the separation of
noncoherent FSK is double that of FSK. Thus more system bandwidth is
required for noncoherent FSK for the same symbol rate.
54
Continuous phase FSK(CPFSK):
 𝜑 𝑡 = 𝐴 𝑝 𝑡 cos 𝜔𝑐 𝑡 + 𝜃(𝑡) , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜
 Where θ(t) is a continuous function of time
 𝜃 𝑡 = 𝜃 0 + ∆𝜔𝑡
, "1"
, "0"
 𝜃 𝑡 = 𝜃 0 − ∆𝜔𝑡
 Where θ(0) = phase at (t = 0)
 ⇒ 𝜃 𝑡 = 𝜃 0 ± 2𝜋∆𝑓𝑡
 𝜃 𝑡 = 𝜃 0 ± 2𝜋(
𝑛𝑅𝑜
2
)𝑡
 𝜃 𝑡 = 𝜃 0 ± 𝜋𝑛𝑅𝑜 𝑡
𝜃 𝑡 = 𝜃 0 ±
𝜋𝑛
𝑡
𝑇𝑜
, 0 ≤ 𝑡 ≤ 𝑇𝑜
CPFSK (Cont.)
 → 𝜑 𝑡 = 𝐴 𝑝 𝑡 cos 𝜔1 𝑡 + 𝜃(0) → "1“
 & 𝜑 𝑡 = 𝐴 𝑝 𝑡 cos 𝜔2 𝑡 + 𝜃(0) → "0"


𝑛
𝑓1 = 𝑓𝑐 +
&
2𝑇𝑜
1
→ 𝑓𝑐 = (𝑓1 + 𝑓2 )
2
𝑓2 = 𝑓𝑐 −
𝑛
2𝑇𝑜
→ 𝑛 = 𝑇0 (𝑓1 − 𝑓2 )
 Where n is called the deviation ratio
 𝜃 𝑡 −𝜃 0 =±
𝜋
𝑛𝑡
𝑇𝑜
 At 𝑇𝑜 : 𝜃(𝑇𝑜 ) − 𝜃 0 = ±𝜋𝑛
 → sending "1" increase the phase by πn radian
 → sending "0" decrease the phase by πn radian
Example:
Data : 110100
𝜃 𝑡 − 𝜃(0)
2𝜋𝑛
𝜋𝑛
𝑡
2𝑇𝑜
−2𝑇𝑜
4𝑇𝑜
6𝑇𝑜
−𝜋𝑛
−2𝜋𝑛
Phase
Tree
CPFSK (Cont.)
 Let 𝑛 =
1
2
→ 𝜃(𝑡) − 𝜃 0 =
𝜋
±
𝑡
2𝑇𝑜
𝜋
=±
2
 𝑎𝑡 𝑡 = 𝑇𝑜
→
𝜃 𝑡 −𝜃 0
 𝑎𝑡 𝑡 = 2𝑇𝑜
→
𝜃 𝑡 − 𝜃 0 = ±𝜋
 𝑎𝑡 𝑡 = 3𝑇𝑜
→
𝜃 𝑡 −𝜃 0 =
3𝜋
±
2
=
𝜋
∓
2
𝜃 𝑡 − 𝜃(0)
Example:
Data : 1101100
𝜋
𝜋/2
2𝑇𝑜
−2𝑇𝑜
4𝑇𝑜
𝑡
6𝑇𝑜
−𝜋/2
−𝜋
Phase
Trellis
Minimum Shift Keying (MSK)
 what is the min Δf?
 → minimum shift keying (MSK);
 it is a continuous phase FSK → CPFSK
𝑛=
1
2
→ ∆𝑓 =
𝑅𝑜
4
→ ∆𝑓 =
1
4𝑇𝑜
 How?
 Polar min BW is
𝑅𝑜
2
→ (zero ISI or Duobinary pulse)
Duobinary
−
𝑅𝑜
2
𝑅𝑜
2
f
𝐵𝑊 = 1.5𝑅𝑜
𝑅𝑜
2
𝑅𝑜
𝑓2 −
2
𝑓2 = 𝑓𝑐 − ∆𝑓
𝑓𝑐
𝑓1 = 𝑓𝑐 + ∆𝑓
𝑅𝑜
𝑓1 +
2
f
BFSK Demodulation:
1. Coherent:
62
2. Non-coherent:
𝜔1
𝜔2
𝜔
𝜔
63
Constellation of BFSK
𝜓2 (𝑡)
𝑠2
𝑠1
𝜓1 (𝑡)
64
Binary PSK (BPSK):
 𝜑𝐵𝑃𝑆𝐾 𝑡 = 𝑦 𝑡 . cos(𝜔𝑐 𝑡) → DSB-SC
 y(t) : polar NRZ
65
Polar NRZ
1
0
0
1
1
0
A
-A
180°
180°
180°
66
𝑃(𝜔)
𝑆𝑌 𝜔 =
𝑇
2
𝑃 𝜔
=
𝑇0
2
∞
ℛ𝑛 𝑒 −𝑗𝑛𝜔𝑇
𝑛=−∞
For polar line code
𝑃 𝜔 =ℱ 𝑝 𝑡 ]
 𝑝 𝑡 =𝐴∙
𝑡−𝑇0 2
𝑟𝑒𝑐𝑡(
)
𝑇0
(NRZ rect)
𝑝(𝑡)
𝑃 𝜔
𝜔𝑇0 −𝑗𝜔𝑇0
2
= 𝐴 ∙ 𝑇0 𝑠𝑖𝑛𝑐(
)𝑒
2
𝜔𝑇0
2
2
2
2
𝑃 𝜔 = 𝐴 ∙ 𝑇0 𝑠𝑖𝑛𝑐 (
)
2
𝑆𝑌 𝜔 =
𝐴2
∙ 𝑇0
𝑠𝑖𝑛𝑐 2 (
𝑇0
2
t
𝑇0
𝜔𝑇0
)
2
67
𝑆𝐵𝑃𝑆𝐾 𝜔
𝐴2 𝑇𝑜 4
𝜔𝑐 − 2𝜋
𝑓𝑐 − 2𝑅0
𝑓𝑐 − 𝑅0
𝑇0
𝜔𝑐
𝜔
𝜔𝑐 + 2𝜋 𝑇
0
𝑓
𝐸𝐵 = 2𝑅𝑜
𝑓𝑐 + 𝑅0
𝑓𝑐 + 2𝑅0
68
BPSK (Cont.)
 +𝐴 cos 𝜔𝑐 𝑡 ⇒ "1" ,
𝜆𝑇𝑜 ≤ 𝑡 ≤ (𝜆 + 1)𝑇𝑜
 −𝐴 cos 𝜔𝑐 𝑡 ⇒ Acos(𝜔𝑐 𝑡 + 𝜋) ⇒ "0" , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜
 ⇒ 𝜑𝑃𝑆𝐾 𝑡 = 𝐴 𝑐𝑜𝑠 𝜔𝑐 𝑡 + 𝜃𝜆 , 𝜆𝑇𝑜 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑜
 ⇒ (PM)
 𝜃𝜆 constant with respect to t ⇒ instanteneous frequency is constant
 𝜑𝑃𝑆𝐾 𝑡 = 𝐴 𝑐𝑜𝑠𝜃𝜆 cos 𝜔𝑐 𝑡 − 𝐴 𝑠𝑖𝑛𝜃𝜆 sin(𝜔𝑐 𝑡)
 𝜑𝑃𝑆𝐾 𝑡 = 𝑎𝜆 cos 𝜔𝑐 𝑡 + 𝑏𝜆 sin 𝜔𝑐 𝑡 ⇒ 𝑄𝐴𝑀
 𝑎𝜆 = 𝐴 𝑐𝑜𝑠𝜃𝜆
 𝑏𝜆 = −𝐴 𝑠𝑖𝑛𝜃𝜆
 In binary 𝜃𝜆 = 0, π} ⇒ 𝑎𝜆 = ±𝐴 , 𝑏𝜆 = 0
 ⇒ 𝜑𝐵𝑃𝑆𝐾 𝑡 = ±𝐴 cos(𝜔𝑐 𝑡)
69
Constellation of BPSK
𝜓2 (𝑡)
𝑠2
𝑠1
𝜓1 (𝑡)
 We observe that 𝜓2 (𝑡) has nothing to do with
signals. Hence, only one basis function is
sufficient to represent the signals
70
BPSK Demodulation:
 Coherent:
𝑦(𝑡)
𝜑 𝑡 = y(t)cos(𝜔𝑐 𝑡)
LPF
cos(𝜔𝑐 𝑡)
Cannot use ED
71
Differential BPSK (DBPSK):
-
To use non-coherent detection.
Receiver detect the relative phase change between
successive modulation phases θk & θk-1
 𝜃𝑘 − 𝜃𝑘−1 = 0 ⇒ "0“
 𝜃𝑘 − 𝜃𝑘−1 = 𝜋 ⇒ "1“
72
DBPSK(Cont.)
𝐼𝑘 = "0", "1"

𝑎𝑘 = 2𝑑𝑘 − 1 = +𝐴, −𝐴
𝑑𝑘 = 𝐼𝑘 ⊕ 𝑑𝑘−1 = 0, 1
Delay To
Polar
generator
𝜑(𝑡) ± Acos(𝜔𝑐 𝑡)
cos(𝜔𝑐 𝑡)
⊕ 𝑋𝑂𝑅 𝑜𝑟 𝑀𝑜𝑑𝑢𝑙𝑜 2 𝑎𝑑𝑑𝑒𝑟
73
𝐼𝑘
1
0
0
1
1
0
𝑑𝑘 0
1
1
1
0
1
1
+𝐴
+𝐴
+𝐴
−𝐴
+𝐴
+𝐴
NRZI: Non return to
zero inverted
𝑎𝑘
+𝐴
−𝐴
74
DBPSK (Cont.)
 Sending "0" will keep the previous phase.
 Sending "1" change the previous phase by π.
 ⇒ when "0" is Tx : both the current & previous pulses
will be:
+A cos(ωc t)
or
−A cos(ωc t)
 ⇒ when "1" is Tx : the current pulse & the previous
pulse will differ by the sign
𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = ± cos 𝜔𝑐 𝑡 , 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 = ∓cos(𝜔𝑐 𝑡)
75
DBPSK Demodulator
𝑙(𝑡)
𝑟(𝑡)
𝜑(𝑡)
LPF
Decision
𝑙 𝑡 > 0 ⇒ "0"
𝑙 𝑡 < 0 ⇒ "1"
Delay To
𝐴2
(1
2
⇒"0": 𝑟 𝑡 = 𝜑 𝑡 . 𝜑 𝑡 − 𝑇𝑜 =
+ cos(2𝜔𝑐 𝑡))
𝐴2
𝑙 𝑡 =
2
−𝐴2
⇒“1": 𝑟 𝑡 = 𝜑 𝑡 . 𝜑 𝑡 − 𝑇𝑜 =
(1 + cos(2𝜔𝑐 𝑡))
2
−𝐴2
𝑙 𝑡 =
2
76
M-ary Digital Modulation
 To increase data rate or Reduce the Digital signal
Bandwidth.
 𝑀 = 2𝑘
, k: # of bits
, M: # of symbols (pulses)
 𝑘 = 𝑙𝑜𝑔2 𝑀
 k bits are assigned 1 of the M symbols.
 Send one symbol in 𝑇𝑀 .
 𝑇𝑀 is the symbol duration:
 # of symbols per sec is 𝑅𝑀 =
1
𝑇𝑀
Baud rate [Baud]
77
M-ary ASK:
 Like Binary ASK it is a special case of AM signal with Digital message signal.
 For AM with 𝒎 = 𝟏:
  𝜑𝑙 𝑡 = 𝑙 ∙ 𝑝 𝑡 cos(𝜔𝑐 𝑡) ,
𝑙 = 0,1,2, … . , (𝑀 − 1) , λ𝑇𝑀 ≤ 𝑡 ≤ (𝜆 + 1)𝑇𝑀
 Where 𝑝 𝑡 is a pulse shape.
e.g. M=4, k=2,
𝑡−𝑇𝑀
𝑝 𝑡 𝑖𝑠 𝑟𝑒𝑐𝑡
𝑇
𝑀
Bits
Amp.
00
0
01
+1
10
+3
11
+2
2
Demodulation:
Non-Coherent  ED
Coherent
Constellation
𝑠0 𝑠1
…..
𝑠𝑀−1
𝜓1 (𝑡)
78
M-ary ASK (Cont.)
 For AM with 𝒎 < 𝟏:
 𝜑𝑙 𝑡 = 𝑙 ∙ 𝑝 𝑡 cos(𝜔𝑐 𝑡) ,
𝑙 = 1,2, … . , 𝑀 , λ𝑇𝑀 ≤ 𝑡 ≤ (𝜆 + 1)𝑇𝑀
 Where 𝑝 𝑡 is a pulse shape.
e.g. M=4, k=2,
𝑡−𝑇𝑀
𝑝 𝑡 𝑖𝑠 𝑟𝑒𝑐𝑡
𝑇
𝑀
Bits
Amp.
00
+1
01
+2
10
+4
11
3
2
Demodulation:
Non-Coherent  ED
Coherent
Constellation
𝑠1 𝑠2
…..
𝑠𝑀
𝜓1 (𝑡)
79
M-ary ASK (Cont.)
 For AM with 𝒎 > 𝟏:
 𝜑𝑙 𝑡 = ±𝑙 ∙ 𝑝 𝑡 cos(𝜔𝑐 𝑡) ,
𝑙 = 1,3,5, … . , (𝑀 − 1) , λ ≤ 𝑡 ≤ (𝜆 + 1)𝑇𝑀
 Where 𝑝 𝑡 is a pulse shape.
e.g. M=4, k=2,
𝑡−𝑇𝑀
𝑝 𝑡 𝑖𝑠 𝑟𝑒𝑐𝑡
𝑇
2
Demodulation:
Coherent only
𝑀
Bits
Amp.
00
-3
01
-1
10
+3
11
+1
Constellation
…..
…..
𝜓1 (𝑡)
80
M-ary ASK (Cont.)
 PSD for 𝑚 > 1 and M=4:
𝑃(𝜔)
𝑆𝑌 𝜔 =
𝑇𝑀
5𝑃 𝜔
=
𝑇𝑀
∞
2
ℛ𝑛 𝑒 −𝑗𝑛𝜔𝑇𝑀
𝑛=−∞
2
𝑃 𝜔 =ℱ 𝑝 𝑡 ]
𝑝 𝑡 =
𝑡−𝑇𝑀 2
𝑟𝑒𝑐𝑡(
)
𝑇𝑀
(NRZ rect)
𝜔𝑇𝑀
)
2
𝑆𝑌 𝜔 = 5 ∙ 𝑇𝑀 𝑠𝑖𝑛𝑐 2 (
81
𝑆𝑀𝐴𝑆𝐾 𝜔
5𝑇𝑀 4
𝜔𝑐 − 2𝜋
−2𝑅M
−𝑅M
𝑇𝑀
𝜔𝑐
𝜔
𝜔𝑐 + 2𝜋 𝑇
𝑀
𝑓
𝐸𝐵 = 2𝑅𝑀
𝑅M
2𝑅M
𝜌 =?
82
M-ary FSK (Multi tone Signaling):
 Each pulse is transmitted as:
 𝜑𝑖 𝑡 = 𝐴 cos(𝜔𝑖 𝑡) , 𝜆𝑇𝑀 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑀 ,
𝑖 = 1,2, … , 𝑀
 𝜑𝑖 𝑡 = 𝐴 cos(2𝜋𝑓𝑖 𝑡) , 𝜆𝑇𝑀 ≤ 𝑡 ≤ (𝜆 + 1)𝑇𝑀
𝑖 = 1,2, … , 𝑀
 FSK is a special case of FM.
83
M-ary FSK (Cont.)
 One way to select 𝑓𝑖 is:
 𝑓𝑖 = 𝑓1 + 𝑖 − 1 𝛿𝑓 ,
𝑖 = 1,2,3, … , 𝑀
Lower
frequency
f1
Upper
frequency
f2
f3
fc
𝑓𝑀 = 𝑓1 + (𝑀 − 1)𝛿𝑓
2∆𝑓
 ∆𝑓: Frequency deviation
84
M-ary FSK (Cont.)
 2∆𝑓 = 𝑓𝑀 − 𝑓1 = (𝑀 − 1)𝛿𝑓
 ∆𝑓 =
(𝑀−1)𝛿𝑓
2
 As ∆𝑓 increase  Bandwidth increase
 As ∆𝑓 decrease  Bandwidth decrease
 For fixed value of M, ∆𝑓 is controlled through 𝛿𝑓.
𝑡𝑜𝑜 ℎ𝑖𝑔ℎ, 𝐵𝑊 𝑖𝑠 ℎ𝑖𝑔ℎ
 𝛿𝑓 ⇒
𝑡𝑜𝑜 𝑙𝑜𝑤, 𝑐𝑎𝑛 𝑛𝑜𝑡 𝑑𝑖𝑠𝑡𝑖𝑛𝑔𝑢𝑖𝑠ℎ 𝑝𝑢𝑙𝑠𝑒𝑠
85
M-ary FSK (Cont.)
 Then the question is:
 What is the minimum value of 𝛿𝑓 such that we can
distinguish a pulse from another?
 The answer is through (Orthogonality principle).
 The signal set 𝜑𝑖 (𝑡) is orthogonal set over the time
interval 𝑎, 𝑏 if the inner product of any two signals in
this set is zero.
86
M-ary FSK (Cont.)




𝐸𝑛 , 𝑖𝑓 𝑚 = 𝑛
𝜑𝑚 (𝑡), 𝜑𝑛 (𝑡) =
𝑡 𝜑𝑛 𝑡 𝑑𝑡 =
0, 𝑖𝑓 𝑚 ≠ 𝑛
𝐸𝑛 is the energy of 𝜑𝑛 (𝑡) over the interval 𝑎, 𝑏
If 𝐸𝑖 = 1 for every signal in the set then the set is called
an orthonormal signal set.
To transform an orthogonal set into an orthonormal set
we divide each signal by the square root of its energy
( 𝐸𝑖 ).
 The set
𝑏
𝜑
𝑎 𝑚
𝜑𝑖 (𝑡)
𝐸𝑖
∗
is orthonormal signal set.
87
M-ary FSK (Cont.)
∗

𝑏 𝜑𝑚 𝑡 𝜑𝑛 𝑡
𝑎
𝐸𝑚
𝐸𝑛
𝑑𝑡 =
𝐸𝑛
𝐸𝑛
= 1, 𝑖𝑓 𝑚 = 𝑛
0, 𝑖𝑓 𝑚 ≠ 𝑛
 Let 𝜑𝑖 𝑡 = 𝐴 cos(2𝜋𝑓𝑖 𝑡) for 𝑖 = 1,2, … , 𝑀 be a signal
set defined of the interval 0, 𝑇𝑀 .
 Then 𝜑𝑚 𝑡 = 𝐴 cos(2𝜋𝑓𝑚 𝑡) and 𝜑𝑛 𝑡 = 𝐴 cos(2𝜋𝑓𝑛 𝑡)
 To be orthogonal:
𝑇𝑀 2
𝐴
0
cos(2𝜋𝑓𝑚 𝑡) cos(2𝜋𝑓𝑛 𝑡) 𝑑𝑡 = 0, 𝑓𝑜𝑟 𝑚 ≠ 𝑛.
88
M-ary FSK (Cont.)
𝐴2
=
2
𝐴2
=
2
𝑇𝑀
cos(2𝜋 𝑓𝑚 + 𝑓𝑛 𝑡) + cos(2𝜋 𝑓𝑚 − 𝑓𝑛 𝑡) 𝑑𝑡
0
𝑇𝑀
𝑇𝑀
cos( 2𝜋 𝑓𝑚 + 𝑓𝑛 𝑡)𝑑𝑡 +
0
cos(2𝜋 𝑓𝑚 − 𝑓𝑛 𝑡)𝑑𝑡
0
Zero
𝐴2
sin(2𝜋 𝑓𝑚 + 𝑓𝑛 𝑇𝑀
sin(2𝜋 𝑓𝑚 − 𝑓𝑛 𝑇𝑀
=
𝑇 ∙
+ 𝑇𝑀
2 𝑀
2𝜋 𝑓𝑚 + 𝑓𝑛 𝑇𝑀
2𝜋(𝑓𝑚 − 𝑓𝑛 )𝑇𝑀
Since practically 𝑓𝑚 + 𝑓𝑛 𝑇𝑀 is very high.
89
M-ary FSK (Cont.)

sin 2𝜋 𝑓𝑚 −𝑓𝑛 𝑇𝑀
2𝜋 𝑓𝑚 −𝑓𝑛
=0
,
𝑚≠𝑛
 sin( 2𝜋 𝑓𝑚 − 𝑓𝑛 𝑇𝑀 ) = 0
𝑓𝑚 = 𝑓1 + (𝑚 − 1)𝛿𝑓
⇒ 𝑓𝑛 = 𝑓1 + (𝑛 − 1)𝛿𝑓
𝑓𝑚 − 𝑓𝑛 = (𝑚 − 𝑛)𝛿𝑓
⇒ sin(2𝜋 𝑚 − 𝑛 𝛿𝑓. 𝑇𝑀 = 0
⇒ 2𝜋 𝑚 − 𝑛 𝛿𝑓. 𝑇𝑀 = 𝜋 𝑜𝑟 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 𝑖𝑡
⇒ 𝛿𝑓 =
⇒ 𝛿𝑓 =
1
minimum when 𝑚
2𝑇𝑀 (𝑚−𝑛)
1
𝑅
= 𝑀
(min)
2𝑇𝑀
2
−𝑛 =1
e.g. Assume 3 orthogonal pulses
𝜓2 (𝑡)
𝑠2
𝑠1
𝜓1 (𝑡)
(𝑀 − 1)
⇒ ∆𝑓 =
4𝑇𝑀
& 𝑓𝑖 = 𝑓1 +
(𝑖−1)
2𝑇𝑀
𝑠3
𝜓3 (𝑡)
,
𝑖 = 1,2,3, … . , 𝑀
90
M-ary FSK (Cont.)
 𝜑𝑖 𝑡 = 𝐴 cos 2𝜋 𝑓1 +
𝑓4
𝑓3
𝑓2
𝑓3
(𝑖−1)
2𝑇𝑀
𝑓4
𝑡 , 𝑖 = 1,2, … … . . 𝑀
𝑓1
A
-A
Bits
Freq.
00
𝑓1
01
𝑓2
10
𝑓3
11
𝑓4
91
Each pulse can be visualized as an OOK
signal.
Σ
𝑆𝑀𝐹𝑆𝐾 𝜔
If 𝑧 𝑡 = 𝑥1 𝑡 + 𝑥2 𝑡 + ⋯ ⋯ ⋯ + 𝑥𝑀 𝑡
And 𝑥𝑖 𝑡 , 𝑥𝑗 𝑡 are mutually orthogonal for i,j=1,2,…,M
⇒ 𝑆 𝜔 = 𝑆𝑥1 𝜔 + 𝑆𝑥2 𝜔 + ⋯ ⋯ ⋯ 𝑆𝑥𝑀 (𝜔)
2𝑅𝑀
2𝑅𝑀
……………………………….
𝜔1
𝑓1 − 𝑅𝑜 𝑓
1
𝜔1 + 2π𝑅𝑀
𝜔𝑀 𝜔𝑀 + 2π𝑅𝑀
𝜔2
𝜔
𝑓
𝑓𝑀 + 𝑅𝑀
𝑓2
2∆𝑓 + 2𝑅𝑀
Orthogonal M-ary FSK
 𝑓𝑖 = 𝑓1 + 𝑖 − 1 𝛿𝑓 ,
𝑖 = 1,2,3, … , 𝑀
Lower
frequency
f1
Upper
frequency
f2
𝑅𝑀
2
f3
fc
𝑓𝑀 = 𝑓1 + (𝑀 − 1)𝛿𝑓
2∆𝑓
94
M-ary FSK (Cont.)
 Using Carson’s Rule:
𝐵𝑊 = 2∆𝑓 + 2𝐵
 Assume the base band signal y(t) is polar NRZ
 → Bandwidth of y t is 𝐵 =
1
𝑇𝑀
= 𝑅𝑀
  𝐵𝑊 𝑜𝑓 𝑀𝐹𝑆𝐾 = 2∆𝑓 + 2𝑅𝑀 =
𝑀−1
2𝑇𝑀
+
2
𝑇𝑀
=
𝑀+3
2𝑇𝑀
 Increasing M will increase the BW linearly but the
power will remains constant.
 For orthogonal BW = 2∆𝑓 + 𝑅𝑀 =
𝑀+1
2𝑇𝑀
≈
𝑀
2𝑇𝑀
95
M-ary FSK Demodulation:
-Coherent
cos(𝜔1 𝑡)
𝑡 = n𝑇𝑀
LPF
sampler
cos(𝜔2 𝑡)
𝑡 = n𝑇𝑀
LPF
sampler
𝜑𝑀𝐹𝑆𝐾 𝑡
……..
……..
……..
cos(𝜔𝑀 𝑡)
Comparator
Select
Largest
𝑡 = n𝑇𝑀
LPF
sampler
96
M-ary FSK Demodulation:
-Non-Coherent
𝐻𝑖 (𝜔) is a bandpass filter centered at 𝜔𝑖
𝑡 = n𝑇𝑀
𝐻1 (𝜔)
ED
sampler
𝑡 = n𝑇𝑀
𝐻2 (𝜔)
ED
sampler
𝜑𝑀𝐹𝑆𝐾 𝑡
Comparator
Select
Largest
……..
……..
……..
𝑡 = n𝑇𝑀
𝐻𝑀 (𝜔)
ED
sampler
97
M-ary PSK:
 Each pulse is transmitted as:
 𝜑𝑖 𝑡 = 𝐴 cos(𝜔𝑐 𝑡 + 𝜃𝑖 ) , 𝜆𝑇𝑀 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑀 ,
𝑖 = 1,2, … , 𝑀
 𝜑𝑖 𝑡 = 𝐴 cos(2𝜋𝑓𝑐 𝑡 + 𝜃𝑖 ) , 𝜆𝑇𝑀 ≤ 𝑡 ≤ (𝜆 + 1)𝑇𝑀
𝑖 = 1,2, … , 𝑀
 Where 𝑓𝑐 =
𝑛
𝑇𝑀
, n:integer
Tc
  𝑇𝑀 = 𝑛𝑇𝑐 , 𝑇𝑐 is the period of the carrier.
TM
98
M-ary PSK (Cont.)
 One way to choose 𝜃𝑖 is:
 𝜃𝑖 = 𝜃𝑜 +
2𝜋
(𝑖
𝑀
− 1),
𝑖 = 1,2, … , 𝑀
 𝜃𝑜 is an initial phase.
 𝜑𝑖 𝑡 = 𝐴 cos(2𝜋𝑓𝑐 𝑡 + 𝜃𝑖 )


= 𝐴 cos 𝜃𝑖 cos 𝜔𝑐 𝑡 − 𝐴 sin 𝜃𝑖 sin 𝜔𝑐 𝑡
= 𝑎𝑖 cos 𝜔𝑐 𝑡 − 𝑏𝑖 sin(𝜔𝑐 𝑡)
DSB-SCASK
DSB-SCASK
 𝑎𝑖 = 𝐴 cos(𝜃𝑖 ) : The In-phase component (I).
 𝑏𝑖 = 𝐴 sin(𝜃𝑖 ) : The Quadrature component (Q).
99
M-ary PSK (Cont.)
 𝜑𝑖 𝑡 = 𝑎𝑖 cos 𝜔𝑐 𝑡 − 𝑏𝑖 sin(𝜔𝑐 𝑡)
 The cos(𝜔𝑐 𝑡) and sin(𝜔𝑐 𝑡) are orthogonal.
 PSD of MPSK = PSD of 𝑎𝑖 cos 𝜔𝑐 𝑡 +PSD of 𝑏𝑖 sin(𝜔𝑐 𝑡)
 PSD of s(t) = 𝑚(𝑡) sin(𝜔𝑐 𝑡) is:
1
𝑆𝑠 𝜔 = 𝑆𝑚 𝜔 − 𝜔𝑐 + 𝑆𝑚 (𝜔 + 𝜔𝑐 )
4
  PSD of MPSK = 2 ∙ 𝑃𝑆𝐷 𝑜𝑓 𝐴𝑆𝐾
100
𝑆𝑀𝑃𝑆𝐾 𝜔
𝜔𝑐 − 2𝜋
𝑓𝑐 − 2𝑅𝑀
𝑓𝑐 − 𝑅𝑀
𝑇𝑀
𝜔𝑐
𝜔
𝜔𝑐 + 2𝜋 𝑇
𝑀
𝑓
𝐸𝐵 = 2𝑅𝑀
𝑓𝑐 + 𝑅𝑀
𝑓𝑐 + 2𝑅𝑀
101
M-ary PSK (Cont.)
𝑟=
 𝜃𝑖 =
𝑎𝑖 2 + 𝑏𝑖 2 =
𝐴2 𝑐𝑜𝑠 2 𝜃𝑖 + 𝐴2 𝑠𝑖𝑛2 𝜃𝑖 = 𝐴
−1 𝑏𝑖
𝑡𝑎𝑛
𝑎𝑖
 Let 𝑧 = 𝑎𝑖 + 𝑗𝑏𝑖 called the complex envelope (the
equivalent low-pass signal of 𝜑𝑖 (𝑡)).
𝜓2 (𝑡)
 𝜑𝑖 𝑡 = 𝑅𝑒 (𝑎𝑖 + 𝑗𝑏𝑖 )𝑒 𝑗𝜔𝑐 𝑡
= 𝑅𝑒 (𝑎𝑖 +𝑗𝑏𝑖 )(cos 𝜔𝑐 𝑡 + 𝑗 sin 𝜔𝑐 𝑡)
𝑏𝑖
𝑟
𝜃𝑖
𝑎𝑖
= 𝑅𝑒 𝑎𝑖 cos 𝜔𝑐 𝑡 + 𝑗𝑎𝑖 sin 𝜔𝑐 𝑡 + 𝑗𝑏𝑖 cos 𝜔𝑐 𝑡 − 𝑏𝑖 sin 𝜔𝑐 𝑡
= 𝑎𝑖 cos 𝜔𝑐 𝑡 − 𝑏𝑖 sin 𝜔𝑐 𝑡
𝜓1 (𝑡)
102
M-ary PSK (Cont.)
⇒ 𝑧 =
𝑎𝑖 + 𝑏𝑖 = 𝑟
 ⇒ ∠𝑧 = tan
𝑏𝑖
𝑎𝑖
= 𝜃𝑖
103
Constellation of 8-ary PSK
𝜓2 (𝑡)
All the points will stay on a circle
𝑀=8
𝑠3
𝑠2
𝑠4
𝑠1
𝑠5
𝜓1 (𝑡)
𝑠6
𝑠8
𝑠7
104
M-ary PSK (Cont.)
 Demodulation: Coherent only
𝑟 𝑡 =
𝐴
2
× 2 cos 𝜃𝑖 + cos(2𝜔𝑐 𝑡 + 𝜃𝑖 )
 𝑙 𝑡 = 𝐴𝑐𝑜𝑠(𝜃𝑖 )
𝜑𝑖 𝑡 = Acos(𝜔𝑐 𝑡 + 𝜃𝑖 )
𝑙(𝑡)
𝑟(𝑡)
LPF
2cos(𝜔𝑐 𝑡)
Why?
105
Quadriphase Shift keying (QPSK):
 Known also as Quaternary PSK.
 𝑀 = 4 ⇒ 𝑘 = 2.
 Each pulse is transmitted as:
 𝜑𝑖 𝑡 = 𝐴 cos(𝜔𝑐 𝑡 + 𝜃𝑖 ) , 𝜆𝑇𝑀 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑀 ,
𝑖 = 1,2,3,4


𝜋
4
2𝜋 𝑖−1
𝜃𝑖 = +
,
𝑖 = 1,2,3,4
4
𝜋
3𝜋
5𝜋
7𝜋
So 𝜃1 = , 𝜃2 = , 𝜃3 = , 𝜃4 =
4
4
4
4
106
Constellation of QPSK
𝜓2 (𝑡)
𝑀=4
dibit ≜ two bits
Bits
00
01
10
11
Phase
5𝜋
4
3𝜋
4
7𝜋
4
𝜋
4
dibit
𝑠2
01
𝑠1
11
𝜓1 (𝑡)
𝑠3
00
𝑠4
10
How to assign a code word for the neighbor pulse? Gray code
107
QPSK (Cont.)
𝜋
4
𝜋
4
5𝜋
4
3𝜋
4
7𝜋
4
108
QPSK (Cont.)
 Gray Code:
 Assume 𝑏1 𝑏2 𝑏3 … 𝑏𝑘 is a 𝑘 bit code word
𝑏𝑖 = "0", "1" , 𝑖 = 1,2, … , 𝑘
 And 𝑔1 𝑔2 𝑔3 … 𝑔𝑘 is the corresponding Gray code work.
𝑔𝑖 = "0", "1" , 𝑖 = 1,2, … , 𝑘
 Then:
𝑔1 = 𝑏1
𝑔𝑚 = 𝑏𝑚 ⊕ 𝑏𝑚−1 , 𝑚 ≥ 2
Where ⊕ is XOR
109
QPSK (cont.)
 ex:
𝑘=3
𝑘=2
b
g
b
g
000
000
00
00
001
001
01
01
010
011
10
11
011
010
11
10
100
110
101
111
110
101
111
100
110
QPSK (Cont.)
 𝑎1 = 𝐴 𝑐𝑜𝑠
𝜋
4
 𝑎2 = 𝐴 𝑐𝑜𝑠
3𝜋
4
=
𝐴
2
,
=−
𝐴
2
 𝑎3 = 𝐴 𝑐𝑜𝑠
5𝜋
4
=−
𝐴
2
 𝑎4 = 𝐴 𝑐𝑜𝑠
7𝜋
4
𝐴
2
=
,
,
,
𝑏1 = 𝐴 sin
𝜋
4
𝑏2 = 𝐴 sin
3𝜋
4
𝑏3 = 𝐴 sin
5𝜋
4
𝑏4 = 𝐴 sin
7𝜋
4
=
=
𝐴
2
𝐴
2
=−
𝐴
2
=−
𝐴
2
𝐴
𝑚𝑎𝑝𝑝𝑒𝑑 𝑡𝑜 "0"
2
𝐴
+ 𝑚𝑎𝑝𝑝𝑒𝑑 𝑡𝑜 "1”
2
 Let −
 And
111
QPSK (Cont.)
𝜓2 (𝑡)
b
g
00
00
01
01
10
11
11
10
𝒂𝒊
𝐴
−
2
𝐴
−
2
𝐴
+
2
𝐴
+
2
𝒃𝒊
𝐴
−
2
𝐴
+
2
𝐴
+
2
𝐴
−
2
𝑀=4
𝑠2
01
𝑠1
11
𝜓1 (𝑡)
𝑠3
00
𝑠4
10
112
QPSK (Cont.)
0 0
𝑎𝑖 cos(𝜔𝑐 𝑡)
1 1
1 1
0 1
1 0
BASK ≡BPSK
𝑏𝑖 sin(𝜔𝑐 𝑡)
BASK ≡BPSK
𝜑𝑖 𝑡 = 𝑎𝑖 cos 𝜔𝑐 𝑡 − 𝑏𝑖 sin(𝜔𝑐 𝑡)
113
QPSK (Cont.)
 PSD:
 𝑆𝑄𝑃𝑆𝐾 𝜔 = 𝑆𝐵𝑃𝑆𝐾 𝜔 + 𝑆𝐵𝑃𝑆𝐾 (𝜔)
=
𝐴2 𝑇𝑀
2
𝜔−𝜔𝑐 𝑇𝑀
2
𝑠𝑖𝑛𝑐 2
(𝜔+𝜔𝑐 )𝑇𝑀
2
+ 𝑠𝑖𝑛𝑐 (
)
2
𝑆𝑄𝑃𝑆𝐾 𝜔
𝜔𝑐 − 2𝜋
𝑓𝑐 − 2𝑅𝑀
𝑓𝑐 − 𝑅𝑀
𝑇𝑀
𝜔𝑐
𝜔
𝜔𝑐 + 2𝜋 𝑇
𝑀
𝑓
𝐸𝐵 = 2𝑅𝑀
𝑓𝑐 + 𝑅𝑀
𝑓𝑐 + 2𝑅𝑀
114
QPSK Generation:
01101
𝑑1 𝑑3 𝑑5 𝑑7 𝑑9
𝐴
2
cos 𝜔𝑐 𝑡
+
~
00 11 11 01 10
Polar NRZ
∑
DeMux
-
-π/2
𝑑1 𝑑2 𝑑3 𝑑4 𝑑5 𝑑6 𝑑7 𝑑8 𝑑9 𝑑10
0
0
1
𝑑2 𝑑4 𝑑6 𝑑8 𝑑10
1
1
01110
1
0
1
𝐴
2
1
𝜑𝑄𝑃𝑆𝐾 (𝑡)
sin 𝜔𝑐 𝑡
0
I
Q
𝑇𝑀
115
QPSK Demodulation
𝑥1 > 0 ⇒ 1
𝑥1 < 0 ⇒ 0
𝑥1 = 𝑎𝑖 /2
LPF
Decision
cos 𝜔𝑐 𝑡
𝜑𝑄𝑃𝑆𝐾 (𝑡)
~
Mux
-π/2
𝑥2 = −𝑏𝑖 /2
sin 𝜔𝑐 𝑡
LPF
Decision
𝑥2 < 0 ⇒ 1
𝑥2 > 0 ⇒ 0
116
Offset QPSK (OQPSK):
 In QPSK, the carrier phase changes only once every
𝑇𝑀 = 2𝑇𝑜 . If the two bits in a successive dibits change
simultaneously (i.e. 00 then come 11 or 10 then come
01 etc.), a 180𝑜 phase shift occurs.
180𝑜
𝑠2
01
𝜓2 (𝑡)
𝑠1
11
90𝑜
𝜓1 (𝑡)
𝑠3
00
𝑠4
10
117
OQPSK (Cont.)
 At 180𝑜 phase-shift, the amplitude of the transmitted
signal changes very rapidly costing amplitude
fluctuation.
 This signal may be distorted when is passed through
the filter or nonlinear amplifier.
118
OQPSK (Cont.)
2
1
0
-1
-2
0
1
2
3
4
5
6
7
8
Original Signal
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
0
1
2
3
4
5
6
7
8
Filtered signal
119
OQPSK (Cont.)
 To solve the amplitude fluctuation problem, the offset
QPSK was proposed.
 Offset QPSK delay the data in quadrature component
𝑇𝑀
by
= 𝑇𝑜 (half of symbol).
2
 Now, no way that both bits can change at the same
time, 𝜃𝑖 =
𝑏𝑖
−1
𝑡𝑎𝑛
𝑎𝑖
.
 In the offset QPSK, the phase of the signal can change
by 90 or 0 degree only while in the QPSK the phase of
the signal can change by 180 90 or 0 degree.
120
OQPSK (Cont.)
0
0
1
1
1
1
0
1
1
0
I
Q
121
121
OQPSK (Cont.)
Inphase
QPSK
1
0
0.5
0
1
-0.5
0
1
-1
0
1
2
3
4
5
6
7
8
1
Q phase
QPSK
0
0.5
0
-0.5
1
0
0
-1
0
1
2
3
4
5
4
5
6
7
8
2
1
QPSK
0
-1
-2
0
01
1
10
2
3
10
00
6
7
8
1
Inphase
Offset QPSK
0.5
0
0
1
0
1
-0.5
-1
0
1
2
3
4
5
6
7
8
7
8
1
Q phase
Offset QPSK
0.5
1
0
0
0
-0.5
-1
0
1
2
3
4
5
6
2
10
1
Offset QPSK
10
01
0
-1
00
-2
0
1
2
𝑇𝑀
3
4
5
6
7
8
122
OQPSK (Cont.)
 Possible paths for switching between the message
points in OQPSK.
𝜓2 (𝑡)
𝑠2
𝑠1
90𝑜
𝜓1 (𝑡)
𝑠3
𝑠4
123
Hybrid Amplitude-Phase
Modulation(QAM):
 A general form of PSK where the constraint that all
points in the constellation stay on a circle is broken,
this will result in QAM (Quadrature amplitude
modulation).
 Current protocols such as 802.11b wireless Ethernet
(Wi-Fi) and digital video broadcast (DVB), both utilize
64-QAM modulation. In addition, emerging wireless
technologies such as Worldwide Interoperability for
Microwave Access (WiMAX) and 802.11n.
124
QAM (Cont.)
 𝜑𝑖 𝑡 = 𝑎𝑖 cos 𝜔𝑐 𝑡 − 𝑏𝑖 sin(𝜔𝑐 𝑡)
M-ary ASK
M-ary ASK
𝜆𝑇𝑀 ≤ 𝑡 ≤ 𝜆 + 1 𝑇𝑀
𝑖 = 1,2, … , 𝑀
 𝜑𝑖 𝑡 = 𝑟𝑖 cos(2𝜋𝑓𝑐 𝑡 + 𝜃𝑖 ) , 𝜆𝑇𝑀 ≤ 𝑡 ≤ (𝜆 + 1)𝑇𝑀
𝑖 = 1,2, … , 𝑀
 Where 𝑓𝑐 =
𝑛
𝑇𝑀
, n:integer
  𝑇𝑀 = 𝑛𝑇𝑐 , 𝑇𝑐 is the period of the carrier.
125
QAM (Cont.)
 𝑎𝑖 : The In-phase component (I).
 𝑏𝑖 : The Quadrature component (Q).
 𝑟𝑖 =
 𝜃𝑖 =
𝑎𝑖 2 + 𝑏𝑖 2
𝑏𝑖
−1
𝑡𝑎𝑛
𝑎𝑖
 Let 𝑧 = 𝑎𝑖 + 𝑗𝑏𝑖 called the complex envelope (the
equivalent low-pass signal of 𝜑𝑖 (𝑡)).
126
M-ary PSK (Cont.)
 𝜑𝑖 𝑡 = 𝑅𝑒 (𝑎𝑖 + 𝑗𝑏𝑖 )𝑒 𝑗𝜔𝑐 𝑡
= 𝑅𝑒 (𝑎𝑖 +𝑗𝑏𝑖 )(cos 𝜔𝑐 𝑡 + 𝑗 sin 𝜔𝑐 𝑡)
= 𝑅𝑒 𝑎𝑖 cos 𝜔𝑐 𝑡 + 𝑗𝑎𝑖 sin 𝜔𝑐 𝑡 + 𝑗𝑏𝑖 cos 𝜔𝑐 𝑡 − 𝑏𝑖 sin 𝜔𝑐 𝑡
𝜓2 (𝑡)
= 𝑎𝑖 cos 𝜔𝑐 𝑡 − 𝑏𝑖 sin 𝜔𝑐 𝑡
⇒ 𝑧 =
𝑎𝑖 + 𝑏𝑖 = 𝑟𝑖
 ⇒ ∠𝑧 = tan
𝑏𝑖
𝑎𝑖
= 𝜃𝑖
𝑏𝑖
𝑟𝑖
𝜃𝑖
𝑎𝑖
𝜓1 (𝑡)
127
𝑆𝑄𝐴𝑀 𝜔
𝜔𝑐 − 2𝜋
𝑓𝑐 − 2𝑅𝑀
𝑓𝑐 − 𝑅𝑀
𝑇𝑀
𝜔𝑐
𝜔
𝜔𝑐 + 2𝜋 𝑇
𝑀
𝑓
𝐸𝐵 = 2𝑅𝑀
𝑓𝑐 + 𝑅𝑀
𝑓𝑐 + 2𝑅𝑀
The cos(𝜔𝑐 𝑡) and sin(𝜔𝑐 𝑡) are orthogonal.
PSD of QAM = PSD of 𝑎𝑖 cos 𝜔𝑐 𝑡 +PSD of 𝑏𝑖 sin(𝜔𝑐 𝑡)
 PSD of QAM = 2 ∙ (𝑃𝑆𝐷 𝑜𝑓 𝑀𝑎𝑟𝑦 𝐴𝑆𝐾)
 BW = Bandwidth of an M-ary ASK = 2𝑅𝑀 when the
pulse shape is rect NRZ.
128
QAM (Cont.)
 General Modulator:
𝑎𝑖
cos 𝜔𝑐 𝑡
+
~
∑
-π/2
𝑏𝑖
𝜑𝑖 (𝑡)
-
sin 𝜔𝑐 𝑡
129
QAM (Cont.)
 General Demodulator:
LPF
𝑎𝑖
LPF
𝑏𝑖
2 cos 𝜔𝑐 𝑡
𝜑𝑖 (𝑡)
~
π/2
−2 sin 𝜔𝑐 𝑡
130
QAM (Cont.)
 𝑘 = 𝑙𝑜𝑔2 𝑀
𝑒𝑣𝑒𝑛: 𝑠𝑞𝑢𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑒𝑙𝑙𝑎𝑡𝑖𝑜𝑛
𝑜𝑑𝑑: 𝑐𝑟𝑜𝑠𝑠 𝑐𝑜𝑛𝑠𝑡𝑒𝑙𝑙𝑎𝑡𝑖𝑜𝑛
 e.x. 𝑀 = 8 ⇒ 𝑘 = 3 ⇒ 𝑐𝑟𝑜𝑠𝑠 𝑐𝑜𝑛𝑠𝑡𝑒𝑙𝑙𝑎𝑡𝑖𝑜𝑛

𝑀 = 16 ⇒ 𝑘 = 4 ⇒ 𝑠𝑞𝑢𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑒𝑙𝑙𝑎𝑡𝑖𝑜𝑛
𝑘⇒
131
QAM Square constellation:
 𝑘: even
 Let 𝐿 = + 𝑀
 M-ary QAM constellation results from the Cartesian
product of one dimensional L-ary ASK constellation by
itself.
 1-dim X 1-dim = 2-dim
 2-dim X 1-dim = 3-dim
 2-dim X 2-dim = 4-dim
132
QAM (cont.)
 e.g.: 4-ary ASK
00 01
11
-3A
+A
-A
10
+3A
𝜓1 (𝑡)
 𝑀 = 16, 𝐿 = 4
 𝜓1 𝑡 𝑋 𝜓1 𝑡 ⇒ 2-dim
(-3A ,+3A),
(-3A ,+A),
(-3A ,-A),
(-3A ,-3A),
(-A ,+3A),
(-A , +A),
(-A , -A),
(-A , -3A),
(+A , +3A),
(+A , +A ),
( +A , -A ),
(+A , -3A),
(+3A , +3A)
(+3A , +A)
(+3A , -A)
(+3A , -3A)
133
QAM (Cont.)
−3,3
(−3,1)
 ⇒ 𝑎𝑖 , 𝑏𝑖 = 𝐴
(−3, −1)
(−3, −3)
 In general for any L:
1,3
(1,1)
(1, −1)
1, −3
(3,3)
(3,1)
(3, −1)
(3, −3)
(−𝐿 + 1, 𝐿 − 1)
(−𝐿 + 3, 𝐿 − 1)
……
(𝐿 − 1, 𝐿 − 1)
(−𝐿 + 1, 𝐿 − 3)
(−𝐿 + 3, 𝐿 − 3)
……
(𝐿 − 1, 𝐿 − 3)
⋮
⋮
⋮
⋮
(−𝐿 + 1, −𝐿 + 1)
(−𝐿 + 3, −𝐿 + 1)
……
(𝐿 − 1, −𝐿 + 1)
 𝑎𝑖 , 𝑏𝑖 =
𝐴
−1,3
(−1,1)
(−1, −1)
(−1, −3)
134
QAM (Cont.)
0010
0011
𝜓2 (𝑡)
+3A
0110
1010
1110
+A
0111
1111
𝑟𝑖
1011
𝑏𝑖
𝜃𝑖
-3A
-A
+A
+3A
𝑎𝑖
𝜓1 (𝑡)
-A
0001
0101
0000
0100
-3A
1101
1001
1100
1000
135
QAM (Cont.)
Odd –
numbered dibit
𝒂𝒊
Even –
numbered dibit
𝒃𝒊
00
−3𝐴
00
−3𝐴
01
−𝐴
01
−𝐴
11
+𝐴
11
+𝐴
10
+3𝐴
10
+3𝐴
4-ary ASK m> 1
4-ary ASK m> 1
136
+3𝐴cos(𝜔𝑐 𝑡)
−𝐴cos(𝜔𝑐 𝑡)
−3𝐴cos(𝜔𝑐 𝑡)
QAM (Cont.)
 Input sequence
101101010010
+𝐴sin(𝜔𝑐 𝑡)
𝑘 = 4, 𝐿 = 4
−𝐴sin(𝜔𝑐 𝑡)
+3𝐴sin(𝜔𝑐 𝑡)
𝐴 2cos(𝜔𝑐 𝑡 + 225𝑜 )
𝐴 10cos(𝜔𝑐 𝑡 + 18.4𝑜 )
𝐴 18cos(𝜔𝑐 𝑡 + 135𝑜 )
137
QAM Modulator:
100100
𝑑1 𝑑2 𝑑5 𝑑6 𝑑9 𝑑10
L-ary PAM
cos 𝜔𝑐 𝑡
+
~
𝑑1 𝑑2 𝑑3 𝑑4 𝑑5 𝑑6 𝑑7 𝑑8 𝑑9 𝑑10 𝑑11 𝑑12
∑
DeMux
101101010010
-π/2
-
𝜑𝑄𝐴𝑀 (𝑡)
sin 𝜔𝑐 𝑡
L-ary PAM
𝑑3 𝑑4 𝑑7 𝑑8 𝑑11 𝑑12
110110
138
QAM Demodulator:
𝑥1 = 𝑎𝑖 /2
LPF
Decision
cos 𝜔𝑐 𝑡
𝜑𝑄𝐴𝑀 (𝑡)
~
Mux
-π/2
𝑥2 = −𝑏𝑖 /2
sin 𝜔𝑐 𝑡
LPF
Decision
139
Cross Constellation:
 𝑘: odd
2 𝑘−1 + 4 ∗ 2𝑘−2 =
2𝑘−1 + 22 ∗ 2𝑘−3 =
2𝑘−3 𝑝𝑜𝑖𝑛𝑡𝑠
2𝑘−3
points
Square
2𝑘−1 𝑝𝑜𝑖𝑛𝑡𝑠
2𝑘−1 + 2𝑘−1 =
2 ∗ 2𝑘−1 = 2𝑘 = 𝑀 points
2𝑘−3
points
2𝑘−3 𝑝𝑜𝑖𝑛𝑡𝑠
140
Cross Constellation (Cont.)
𝑀=8⇒𝑘=3
𝜓2
𝜓1
141
Cross Constellation (Cont.)
𝜓2
𝜓1
𝑀 = 32 ⇒ 𝑘 = 5
25−1 = 24 = 16, 25−3
= 22 = 4
142