Lecture 11 - Analog Communication (I) · PDF fileColorado State University Dept of Electrical...
Transcript of Lecture 11 - Analog Communication (I) · PDF fileColorado State University Dept of Electrical...
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Colorado State University Dept of Electrical and Computer Engineering ECE423 – 1 / 22
Lecture 11 - Analog Communication (I)
James Barnes ([email protected])
Spring 2009
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Outline
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 2 / 22
● Communication Basics
● AM: amplitude modulation
● DSB-SC: double sideband - suppressed carrier
● QAM: quadrature amplitude modulation
● SSB: single sideband
Reference:
● Tretter Chapters 5-8 (On e-reserve),
● Haykin, Communication Systems, Third Edition
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Basics
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 3 / 22
The information we wish to send is in the form of a real signal (for example,voice). Some properties of real signals:Fourier Transform:
g(t) =
∫∞
−∞
G(f) e−j2πftdf. (1)
Then for real g(t), we have g(t) = g∗(t), and
g∗(t) =
∫∞
−∞
G∗(f) ej2πftdf =
∫∞
−∞
G∗(−f ′) e−j2πf ′tdf ′. (2)
So for real g(t), G∗(−f) = G(f), or
|G(f)| = |G(−f)|, Θ(−f) = Θ(f). (3)
Furthermore, for real, symmetric g(t) = g(−t), the above become
G∗(f) = G(f) = G(−f), Θ(f) = 0, (4)
i.e. G(f) is real and symmetric about f = 0.
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Modulation Theorem
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 4 / 22
If
m(t) ↔FT M(f), (5)
thenm(t)ej2πfct ↔FT M(f − fc) (6)
andm(t)e−j2πfct ↔FT M(f + fc). (7)
This is easily shown from the definition of Fourier Transform. Adding the two, weget
m(t)cos(2πfct) ↔FT 1
2[M(f − fc) + M(f + fc)]. (8)
So modulating m(t) by the cosine function produces two frequency-shiftedimages of the original baseband signal. This is the basis for DSB-SC.
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Hilbert Transform
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 5 / 22
Ref: Haykin.We use the Fourier Transform to separate out signals according to frequency:G(f). The Hilbert Transform plays a similar role for phase.
g(t) =1
π
∫∞
−∞
g(t)1
t − τdτ = g(t) ∗
1
πt. (9)
Convolution in time domain ≡ multiplication in frequency domain. We also knowthat 1
πt↔FT −jsgn(f), so can write above in frequency domain:
ˆG(f) = −jsgn(f) G(f). (10)
Importance of HT:
1. Used to realize phase selectivity (SSB Modulation/Demodulation);
2. Math basis for realizing bandpass signals.
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Hilbert Transform (2)
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 6 / 22
Example:
g(t) = cos(2πfct) ↔FT G(f) =
1
2[δ(f − fc) + δ(f + fc)]. (11)
Then using eqn(10) above,
ˆG(f) = −j
2[δ(f − fc) + δ(f + fc)] sgn(f) (12)
ˆG(f) =1
2j[δ(f − fc) − δ(f + fc) ↔
FT ˆg(t) = sin(2πfct) (13)
So the HT changes cosines into sines. Equivalent to 90◦ phase shift.
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Some Properties of Hilbert Transform
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 7 / 22
1. Signals g(t) and ˆg(t) have the same amplitude spectrum. HT operates onphase only
2.ˆ
g(t) = −g(t). Easily shown by noting that two HTs applied in series isequivalent to 180◦ degree phase shift, which is equivalent to multiplying by-1.
3. g(t) and ˆg(t) are othogonal:∫∞
−∞g(t) ˆg(t)dt = 0.
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Analytic Signals – Baseband
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 8 / 22
Let g(t) be a real baseband signal. Then it will have an amplitude spectrumsymmetric about f=0:
We can form two complex representations of g(t) using Hilbert Transforms:
g+(t) = g(t) + j ˆg(t), (14)
g−(t) = g(t) − j ˆg(t), (15)
g+(t) is called the positive pre-envelope of g(t) and g−(t) is the negativepre-envelope. Note that g−(t) = g∗+(t). We can express g(t) in terms of the ±envelopes:
g(t) =1
2[g+(t) + g−(t)]. (16)
These split the amplitude spectrum G(f) into two components G+(f) and G−(f)which are scaled pieces of the original spectrum having only positive andnegative frequencies, respectively.
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Analytic Signals – Baseband (2)
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 9 / 22
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Analytic Signals – Passband
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 10 / 22
We do the same for a real passband signal. This has a symmetric amplitudespectrum with two (typically narrowband) components centered about ±fc, thecarrier frequency.
We can writeg(t) = Re{ ˜g(t)ej2πfct}. (17)
˜g(t) is called the complex envelope of g(t). In terms of the pre-envelopes, we canwrite
g+(t) = ˜g(t)ej2πfct, g−(t) = ( ˜g(t))∗e−2πfct, g(t) =1
2[g+(t) + g−(t)]. (18)
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Analytic Signals – Passband (2)
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 11 / 22
The complex envelope can be split into two components: the in-phase andquadrature components:
˜g(t) = gI(t) + jgQ(t). (19)
Substituting (19) back into (17), we can write g(t) in terms of in-phase andquadrature components (canonical form):
g(t) = gI(t)cos(2πfct) − gQ(t)sin(2πfct) (20)
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Analytic Signals – Passband (3)
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 12 / 22
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Amplitude Modulation
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 13 / 22
First form of modulation (early radio). Simple, especially receiver (crystal set).Least efficient at using radiated power and spectrum.
Start with a carrier signal c(t) = Accos(2πfct). The real radiated signal s(t) givenby:
s(t) = Ac[1 + kam(t)]cos(2πfct), (21)
where m(t) is the message (baseband audio signal) and kais a constant calledthe amplitude sensitivity of the modulator. To avoid ”over-modulation”, a form ofdistortion, must have |kam(t)| < 1 ∀ t.Taking the Fourier Transform of (22), we have:
S(f) =Ac
2[δ(f − fc) + δ(f + fc)] +
kaAc
2[M(f − fc) + M(f + fc)]. (22)
The first bracketed term represents the carrier power (basically wasted) and thesecond term the two sideband power components, which is where the usefulinformation is.
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Amplitude Modulation (2)
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 14 / 22
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Amplitude Modulation (3)
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 15 / 22
Single-tone example:
Let m(t) = Amcos(2πfmt). Then s(t) is given by:
s(t) = Ac[1 + µcos(2πfct)cos(2πfct)], (23)
where µ = kaAmis called the modulation factor. Must have |µ| < 1.
By taking Fourier Transform (see Haykin), the average signal power is
● Carrier power = 1
2A2
c
● Upper side-frequency power = lower side-frequency power = 1
8µ2A2
c ,
⇒the sidebands can never get more that 33% of the total radiated power.
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Amplitude Modulation (4)
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 16 / 22
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Amplitude Modulation – Detection
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 17 / 22
Simple envelope detectors can be used, based on rectifying or squaring thesignal to eliminate the carrier. From (21),
s2(t) = A2c [1 + kam(t)]2[ej2πfct + e−j2πfct]2/4, (24)
ors2(t) = A2
c [1 + kam(t)]2[1 + cos(4πfct)]/2. (25)
The cosine term can be filtered out with a low pass filter, and the filter outputpassed through a square root circuit to recover m(t). Notice that the carrierproduces a DC offset which can be eliminated with capacitive coupling (highpass filtering).
[·]2
LPF ][ HPF
received
signal
K m(t)DAC
nts
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Double Side Band - Suppressed Carrier
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 18 / 22
A DSB-modulated signal has the following form:
s(t) = c(t)m(t) = Accos(2πfct)m(t). (26)
Every time m(t) crosses 0, s(t) undergoes a phase reversal. The spectrum of aDSB-SC signal has NO carrier component.
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Demodulating DSB-SC
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 19 / 22
All suppressed-carrier techniques require more elaboration detection than simpleenvelope detection – ”coherent” detection. The principle is shown in the figurebelow:
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Demodulating DSB-SC (2)
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 20 / 22
From the previous figure:
v(t) = A′
c cos(2πfct + φ)s(t) (27)
Substituting for s(t) using eqn(26) and multiplying out, we get
v(t) =1
2AcA
′
c[cos(4πfct + φ) + cos(φ)]m(t). (28)
The LPF eliminates the first cosine term in brackets and the output is:
vo(t) =1
2AcA
′
c cos(φ)m(t). (29)
We need a control loop to maintain φ = 0: a PLL.
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Costas Receiver Block Diagram
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 21 / 22
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Digital Costas Receiver – From Lab7 Notes
Colorado State University Dept of Electrical and Computer Engineering ECE423 – 22 / 22
-j sgn( )
received
signal
m(t)DAC
nto
x
11 z)(je
Hilbert
Transformer
BPFx
++1z
][nje
][1 nc
][2 nc
][n
][ns ][ˆ ns
][n
][nq
][nw
Voltage Controlled Oscillator (VCO)
Phase Discriminator
][nz
][nd