Lecture 2 Digital Modulation I-Hsiang Wang National Taiwan - - PowerPoint PPT Presentation

Lecture 2 Digital Modulation

I-Hsiang Wang

National Taiwan University ihwang@ntu.edu.tw

2018.09.13

Principle of Communications, Fall 2018

Outline

• Digital-to-analog & analog-to-digital: a signal space view
• Pulse amplitude modulation (PAM)
• Pulse shaping and the Nyquist criterion
• Quadrature amplitude modulation (QAM)
• The equivalent complex baseband representation
• Symbol mapping and constellation set

2

Architecture of Digital Modulation

• Three major components at Tx:
• Symbol mapping: bit sequence → symbol sequence
• Pulse shaping: symbol sequence → (baseband) waveform
• Up conversion: baseband waveform → passband waveform
• Corresponding components at Rx:
• Symbol demapping
• Filter + Sampling
• Down conversion

3

passband waveform

Symbol Mapper Pulse Shaper Filter + Sampler Symbol Demapper

discrete sequence

Up Converter Down Converter

baseband waveform

Noisy Channel coded bits

{ci} {ˆ ci}

{um} {ˆ um} xb(t) yb(t)

y(t) x(t)

Bit-to-Symbol Mapping

• To be designed in this layer:
• constellation set – collection of all possible symbols
• symbol mapping – how to map bits to symbols
• In this lecture, we introduce
• Constellation: standard PSK, standard PAM, standard QAM
• Mapping: Gray mapping
• Key system parameter: rate

4

passband waveform

Symbol Mapper Pulse Shaper Filter + Sampler Symbol Demapper

discrete sequence

Up Converter Down Converter

baseband waveform

Noisy Channel coded bits

{ci} {ˆ ci}

{um} {ˆ um} xb(t) yb(t)

y(t) x(t)

Sequence-to-Waveform

• Most pragmatic: Pulse Amplitude Modulation (PAM)
• To be designed in this layer:
• the modulating pulse + the receiver filter
• Nyquist criterion: a sufficient condition for the pulse to

satisfy in order to avoid aliasing effect

• Key system parameter: bandwidth

5

passband waveform

Symbol Mapper Pulse Shaper Filter + Sampler Symbol Demapper

discrete sequence

Up Converter Down Converter

baseband waveform

Noisy Channel coded bits

{ci} {ˆ ci}

{um} {ˆ um} xb(t) yb(t)

y(t) x(t)

Baseband-to-Passband

• Most pragmatic: Quadrature Amplitude Modulation (QAM)
• Essentially speaking, PAM with two branches:
• one mixed with cosine, the other with sine
• Equivalent complex baseband representation
• Key system parameter: center frequency

6

passband waveform

Symbol Mapper Pulse Shaper Filter + Sampler Symbol Demapper

discrete sequence

Up Converter Down Converter

baseband waveform

Noisy Channel coded bits

{ci} {ˆ ci}

{um} {ˆ um} xb(t) yb(t)

y(t) x(t)

Part I. Signal Space

7

A linear algebraic view for the conversion between sequences and waveforms

2018.09.13

Recap: Fourier Series

8

Analysis (waveform → sequence) Synthesis (sequence → waveform) x(t) → x[m]

x[m] → x(t)

x(t) =

∞

X

m=−∞

x[m]ej 2πm

T

t

x(t) =

∞

X

m=−∞

x[m] 1 √ T ej 2πm

T

t

x[m] = 1 T Z

T

x(t)e−j 2πm

T

t dt

x[m] = Z ∞

−∞

x(t) 1 √ T e−j 2πm

T

t dt

↓ rewrite ↓ rewrite φ∗

m(t)

φm(t)

Recap: Sampling Theorem

9

Analysis (waveform → sequence) Synthesis (sequence → waveform) x(t) → x[m]

x[m] → x(t)

↓ rewrite ↓ rewrite

x[m] = x(t)|t= m

2W = x

m

2W

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x(t) =

∞

X

m=−∞

x[m] sinc(2Wt − m)

x[m] =

1 √ 2W x( m 2W )

x(t) =

∞

X

m=−∞

x[m] √ 2Wsinc(2Wt − m)

= Z ∞

−∞

x(t) √ 2Wsinc(2Wt − m) dt

check! φ∗

m(t)

φm(t)

Signal Space Interpretation

10

Analysis (waveform → sequence) Synthesis (sequence → waveform) x(t) → x[m]

x[m] → x(t)

↓ rewrite into vectors ↓ rewrite into vectors

{x[m]} → {φm} → x x = P∞

m=−∞ x[m]φm

x ! φm ! x[m] x[m] = hx, φmi

x(t) → φm(t) → x[m] x[m] = R ∞

−∞ x(t)φ∗ m(t) dt

{x[m]} → {φm(t)} → x(t) x(t) = P∞

m=−∞ x[m]φm(t)

waveform ⟷ vector x(t)

x

integration ⟷ inner product ∞

−∞ u(t)v∗(t) dt

⟨u, v⟩

Fourier Basis

11

Analysis (waveform → sequence) Synthesis (sequence → waveform) x(t) → x[m]

x[m] → x(t)

x(t) =

∞

X

m=−∞

x[m]ej 2πm

T

t

x(t) =

∞

X

m=−∞

x[m] 1 √ T ej 2πm

T

t

x[m] = 1 T Z

T

x(t)e−j 2πm

T

t dt

x[m] = Z ∞

−∞

x(t) 1 √ T e−j 2πm

T

t dt

Fourier Basis: φm ≡ φm(t)

1 √ T exp(j 2π T mt),

m ∈ Z,

↓ rewrite ↓ rewrite

Orthonormal basis that spans all time-limited signals with duration T

φ∗

m(t)

φm(t)

Sinc Basis

12

Analysis (waveform → sequence) Synthesis (sequence → waveform) x(t) → x[m]

x[m] → x(t)

Sinc Basis:

↓ rewrite ↓ rewrite

Orthonormal basis that spans all band-limited signals with one-sided bandwidth W x[m] = x(t)|t= m

2W = x

m

2W

• <latexit 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x(t) =

∞

X

m=−∞

x[m] sinc(2Wt − m)

x[m] =

1 √ 2W x( m 2W )

x(t) =

∞

X

m=−∞

x[m] √ 2Wsinc(2Wt − m)

= Z ∞

−∞

x(t) √ 2Wsinc(2Wt − m) dt

φm ≡ φm(t) , √ 2Wsinc(2Wt − m), m ∈ Z

check! φ∗

m(t)

φm(t)

Signal Space View of Modulation

13

Analysis (waveform → sequence) Synthesis (sequence → waveform) x(t) → x[m]

x[m] → x(t)

{x[m]} → {φm} → x x = P∞

m=−∞ x[m]φm

x ! φm ! x[m] x[m] = hx, φmi

projection onto an

• rthonormal basis

expansion over an

• rthonormal basis

waveform ⟷ vector x(t)

x

integration ⟷ inner product ∞

−∞ u(t)v∗(t) dt

⟨u, v⟩

Part II. Pulse Amplitude Modulation

14

A pragmatic approach to convert symbols to baseband waveforms and back

2018.09.20

Illustration of Modulation in Time Domain

15

{um} → {φm(t)} → xb(t)

um

m

1

2

3

discrete seq. baseband waveform

t

xb(t)

φm(t)

basis waveform

t

m = 1

m = 2

m = 3

Even if the basis waveforms

• verlap, it still works.

Usually the modulator “buffers” the data symbols due to non- causality of basis waveforms.

Time-Shifted Pulses as Basis

• T = 1/2W : transmission interval
• W : operational bandwidth
• Much easier to design
• Desired properties of the pulse function p(t ):
• Time-limited (approximately): because of causality
• Band-limited: because of sharing with other systems

16

{um} → {φm(t)} → xb(t)

φm(t) = p(t − mT), T =

1 2W

xb(t) = ∞

m=−∞ um p(t − mT)

Pulse Shaper {um} xb(t)

φm(t)

t

m = 1

m = 2

m = 3

Demodulation

• Such a choice does not guarantee it to be orthonormal
• “Demodulation = projection”: no longer holds
• Forget about signal space and orthonormal basis for now
• Instead, mimic sampling theorem:
• Demodulation = filtering + sampling
• The filter q (t ) need to “match” the pulse p (t ) in some sense.

17

{um} → {φm(t)} → xb(t)

φm(t) = p(t − mT), T =

1 2W

xb(t) = ∞

m=−∞ um p(t − mT)

Pulse Shaper {um} xb(t)

PAM Modulation and Demodulation

• Key question: how to design pulse p (t ) and filter q (t )?
• A basic goal: able to reconstruct Tx symbols {um }

perfectly when there is no noise at all (yb(t ) = xb(t ))

18

yb(t) Filter q(t) T =

1 2W

ˆ um = ∞

−∞

yb(τ)q(mT − τ) dτ Pulse {um} xb(t) xb(t) =

∞

X

m=−∞

um p(t − mT)

p(t)

ISI-Free Condition for Reconstruction

19

yb(t) Filter q(t) T =

1 2W

ˆ um = ∞

−∞

yb(τ)q(mT − τ) dτ Pulse {um} xb(t) xb(t) =

∞

X

m=−∞

um p(t − mT)

p(t)

xb(t)

xb(τ)

What we want: A sufficient condition ˆ um = um ∀ m

ˆ um = (xb ∗ q)(mT) =

∞

X

k=−∞

uk g(mT − kT) =

∞

X

k=−∞

uk g((m − k)T))

g(t) , (p ∗ q)(t)

g(nT) = ( n 6= 0 1 n = 0

Ideal Nyquist

20

A sufficient condition (in time domain) An equivalent condition (in frequency domain) g(nT) = ( n 6= 0 1 n = 0

Need a condition in frequency domain due to band-limited requirement.

∞

X

m=−∞

˘ g ⇣ f − m T ⌘ = T, ∀ f

In the above condition, we can restrict the range of f to [–W , +W ]

21

Let’s give a simple proof. As in sampling, we multiply the waveform with : g(t)

ιT (t)

gs(t) = g(t)ιT (t) = ∞

n=−∞ g(nT)δ(t − nT)

Consider an impulse train ιT (t) ∞

n=−∞ δ(t − nT)

The waveform is ideal Nyquist iff g(t)

gs(t) = δ(t)

gs(t) = δ(t) ⇐ ⇒ ˘ gs(f) = 1

˘ gs(f) = (˘ g ∗ ˘ ιT )(f)

Key Lemma: the Fourier transform of a time-domain impulse train is also an impulse train in the frequency domain ˘ ιT (f) = 1

T

∞

m=−∞ δ(f − m T )

The proof is complete by observing ˘ gs(f) = (˘ g ∗ ˘ ιT )(f) = 1

T

∞

m=−∞ ˘

g(f − m

T )

Nyquist Criterion

22

Excessive bandwidth: Bb − W

← this should not be too large Typical choice: W ≤ Bb ≤ 2W

∞

X

m=−∞

˘ g ⇣ f − m T ⌘ = T, |f| ≤ W ≡ 1 2T

˘ g(f) f

1 2T ≡ W

− 1

2T ≡ −W

T ≡

1 2W

˘ g(f − m/T) ˘ g(f + m/T)

· · · · · · · · · · · ·

• When taking the typical choice , the Nyquist

criterion is simplified to band-edge symmetry:

Band-Edge Symmetry

23

ˆ g(f) T

✟ ✟ ✯

T − ˆ g(Wb−∆) f ˆ g(Wb+∆)

✟ ✟ ✙

Wb Bb

T − Re{˘ g(W − ∆)} Re{˘ g(W + ∆)} Re{˘ g(f)} W

Bb ∈ [W, 2W]

˘ g∗(W − ∆) + ˘ g(W + ∆) = T, ∀ ∆ ∈ [0, W]

⇐ ⇒ ( Re {˘ g(W − ∆)} + Re {˘ g(W + ∆)} = T Im {˘ g(W − ∆)} = Im {˘ g(W + ∆)} , ∀ ∆ ∈ [0, W]

• For the faster decay in time, one needs larger room for

smooth transition from T to 0 in frequency

• This room is 2×Excessive Bandwidth
• but needs to be normalized by the operational bandwidth
• Rolloff factor – normalized excessive bandwidth:

Excessive Bandwidth

24

ˆ g(f) T

✟ ✟ ✯

T − ˆ g(Wb−∆) f ˆ g(Wb+∆)

✟ ✟ ✙

Wb Bb

T − Re{˘ g(W − ∆)} Re{˘ g(W + ∆)} Re{˘ g(f)} W

2(Bb − W)

β , Bb − W W = Bb W − 1

Raised Cosine: a Widely Used Pulse

25

Time Domain

gβ(t) = 8 < :

π 4 sinc

⇣

1 2β

⌘ , |t| = T

2β

sinc t

T

cos( πβt

T )

1−4 β2t2

T 2

,

• <latexit 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β = 0.3

β = 0.5

β = 0

Decay to zero with speed as when ∼ 1

t3

t → ∞

β > 0

rolloff factor = β

Decay is faster when the rolloff factor is larger

Raised Cosine: a Widely Used Pulse

26

Frequency Domain

β = 0.3

β = 0.5

β = 0

˘ gβ(f) =      T |f| ≤ 1−β

2T

|f| > 1+β

2T

T cos2( πT

2β (|f| − 1−β 2T ))

1−β

2T < |f| ≤ 1+β 2T

The larger rolloff it has, the smoother it transits to zero in frequency, and hence converges to zero faster in the time domain.

rolloff factor = β

• A theorem:
• How to design pulse p(t ) and filter q(t )?
• For faster decay in the time-domain (less approximation error) in t

⟹ need “larger room” for smoother transition from T to 0 in freq. Choose ˘ q(f) = ˘ p∗(f), that is, q(t) = p∗(−t).

Choose ˘ p(f) such that |˘ p(f)|2 satisfies the Nyquist Criterion

Pulse Shaper and Receiver Filter

27

{p(t − mT) : m ∈ Z} form an orthonormal set

⇐ ⇒ |˘ p(f)|2 satisfies the Nyquist Criterion

Part III. Quadrature Amplitude Modulation

28

A pragmatic approach to convert baseband to passband waveforms and back

2018.09.27

Shift in Frequency, Naively

• Goal: shift baseband to passband with center freq. fc
• Recall: the frequency-shift property of Fourier Transform:
• A naive way: multiply the signal by a complex sinusoid
• But, imaginary signal does not exist in the real world
• So, let’s take the real part of it:

29

exp(j2πf0t)s(t)

F

← → ˘ s(f − f0)

x(t) = exp(j2πfct)xb(t)

x(t) = Re{exp(j2πfct)xb(t)} = xb(t) cos(2πfct), xb(t) ∈ R

Problem of the Naive Approach

30

For a real-valued waveform , its FT satisfies the following symmetry: s(t)

˘ s(f) = ˘ s∗(−f)

The baseband waveform is also real-valued, so its FT: xb(t)

Re{˘ xb(f)}

f

f

Im{˘ xb(f)}

After multiplying with , the occupied bandwidth is twice, but the symmetry remains ⟹ a waste of spectrum! cos(2πfct)

Mathematical solution: make the baseband waveform complex-valued Question: how to implement that?

• Let’s give the answer first, and explain why this is

implementing a complex baseband waveform

• Block diagram:

QAM Modulation: Real Domain

31

xb(t)

x(t) = x(I)

b (t)

√ 2 cos(2πfct) − x(Q)

b

(t) √ 2 sin(2πfct)

PAM p(t) √ 2 cos(2πfct) PAM p(t) − √ 2 sin(2πfct) {u(Q)

m }

{u(I)

m }

x(I)

b (t)

x(Q)

b

(t)

x(t)

in-phase component quadrature component

Equivalent Complex Baseband

• This is actually taking the real part of some complex-

valued signal:

• Block diagram:

32

x(t) = x(I)

b (t)

√ 2 cos(2πfct) − x(Q)

b

(t) √ 2 sin(2πfct)

= √ 2Re {xb(t) exp(j2πfct)}

PAM p(t)

x(t)

{um} xb(t) √ 2 exp(j2πfct) Re{·}

um u(I)

m + ju(Q) m

xb(t) x(I)

b (t) + jx(Q) b

(t)

Up-Conversion

33

PAM p(t) √ 2 cos(2πfct) PAM p(t) − √ 2 sin(2πfct) {u(Q)

m }

{u(I)

m }

x(I)

b (t)

x(Q)

b

(t)

x(t)

PAM p(t)

x(t)

{um} xb(t) √ 2 exp(j2πfct) Re{·}

34

PAM p(t) √ 2 cos(2πfct) PAM p(t) − √ 2 sin(2πfct) {u(Q)

m }

{u(I)

m }

x(I)

b (t)

x(Q)

b

(t)

x(t)

PAM p(t)

x(t)

{um} xb(t) √ 2 exp(j2πfct) Re{·}

• Re{˘

x(I)

b (f)}

1

• Im{˘

x(Q)

b

(f)}

1

Re{˘ xb(f)}

• 1

Only illustrate the real part of the passband frequency response. Imaginary part is similar.

• ≈

≈

• √

2 Real part

• ≈

≈

• 1

√ 2

Real part

• ≈

≈

• 1

√ 2

Real part

35

PAM p(t) √ 2 cos(2πfct) PAM p(t) − √ 2 sin(2πfct) {u(Q)

m }

{u(I)

m }

x(I)

b (t)

x(Q)

b

(t)

x(t)

PAM p(t)

x(t)

{um} xb(t) √ 2 exp(j2πfct) Re{·}

Only illustrate the real part of the passband frequency response. Imaginary part is similar.

√ 2 cos(2πfct) =

1 √ 2

• ej2πfct + e−j2πfct

− √ 2 sin(2πfct) =

j √ 2

• ej2πfct − e−j2πfct

• ≈

≈

• 1

√ 2

Real part

• ≈

≈

• 1

√ 2

Real part

• ≈

≈

• 1

√ 2

Real part

36

PAM p(t) √ 2 cos(2πfct) PAM p(t) − √ 2 sin(2πfct) {u(Q)

m }

{u(I)

m }

x(I)

b (t)

x(Q)

b

(t)

x(t)

PAM p(t)

x(t)

{um} xb(t) √ 2 exp(j2πfct) Re{·}

Only illustrate the real part of the passband frequency response. Imaginary part is similar.

Re{s(t)}

F

← → 1

2(˘

s(f) + ˘ s∗(−f))

Down-Conversion

37

y(t)

Step Filter 1 {f ≥ 0} yb(t) √ 2 exp(−j2πfct)

y(t)

√ 2 cos(2πfct) − √ 2 sin(2πfct) LPF 1 {|f| ≤ Bb} LPF 1 {|f| ≤ Bb} y(I)

b (t)

y(Q)

b

(t)

• ≈

≈

• 1

√ 2

Real part

• ≈

≈

• 1

√ 2

Real part

• ≈

≈

• 1

√ 2

Real part

38

y(t)

Step Filter 1 {f ≥ 0} yb(t) √ 2 exp(−j2πfct)

y(t)

√ 2 cos(2πfct) − √ 2 sin(2πfct) LPF 1 {|f| ≤ Bb} LPF 1 {|f| ≤ Bb} y(I)

b (t)

y(Q)

b

(t)

Only illustrate the real part of the passband frequency response. Imaginary part is similar.

• ≈

≈

• 1

√ 2

Real part Imaginary part

• ≈

≈

1

• ≈

≈

1 Real part

39

y(t)

Step Filter 1 {f ≥ 0} yb(t) √ 2 exp(−j2πfct)

y(t)

√ 2 cos(2πfct) − √ 2 sin(2πfct) LPF 1 {|f| ≤ Bb} LPF 1 {|f| ≤ Bb} y(I)

b (t)

y(Q)

b

(t)

Only illustrate the real part of the passband frequency response. Imaginary part is similar.

√ 2 cos(2πfct) =

1 √ 2

• ej2πfct + e−j2πfct

− √ 2 sin(2πfct) =

j √ 2

• ej2πfct − e−j2πfct

• ≈

≈

• 1

Imaginary part

• ≈

≈

• 1

Real part

40

y(t)

Step Filter 1 {f ≥ 0} yb(t) √ 2 exp(−j2πfct)

y(t)

√ 2 cos(2πfct) − √ 2 sin(2πfct) LPF 1 {|f| ≤ Bb} LPF 1 {|f| ≤ Bb} y(I)

b (t)

y(Q)

b

(t)

Only illustrate the real part of the passband frequency response. Imaginary part is similar.

√ 2 cos(2πfct) =

1 √ 2

• ej2πfct + e−j2πfct

− √ 2 sin(2πfct) =

j √ 2

• ej2πfct − e−j2πfct

• ≈

≈

• 1

√ 2

Real part

• ≈

≈

• 1

Im{˘ y(Q)

b

(f)}

• ≈

≈

• 1

Re{˘ y(I)

b (f)}

41

y(t)

Step Filter 1 {f ≥ 0} yb(t) √ 2 exp(−j2πfct)

y(t)

√ 2 cos(2πfct) − √ 2 sin(2πfct) LPF 1 {|f| ≤ Bb} LPF 1 {|f| ≤ Bb} y(I)

b (t)

y(Q)

b

(t)

Only illustrate the real part of the passband frequency response. Imaginary part is similar.

• ≈

≈

• 1

√ 2

Real part

• ≈

≈

• 1

Im{˘ y(Q)

b

(f)}

• ≈

≈

• 1

Re{˘ y(I)

b (f)}

42

y(t)

Step Filter 1 {f ≥ 0} yb(t) √ 2 exp(−j2πfct)

y(t)

√ 2 cos(2πfct) − √ 2 sin(2πfct) LPF 1 {|f| ≤ Bb} LPF 1 {|f| ≤ Bb} y(I)

b (t)

y(Q)

b

(t)

Only illustrate the real part of the passband frequency response. Imaginary part is similar.

• ≈

≈

• 1

√ 2

Real part

• ≈

≈

• 1

√ 2

Real part

• ≈

≈

• 1

Im{˘ y(Q)

b

(f)}

• ≈

≈

• 1

Re{˘ y(I)

b (f)}

43

y(t)

Step Filter 1 {f ≥ 0} yb(t) √ 2 exp(−j2πfct)

y(t)

√ 2 cos(2πfct) − √ 2 sin(2πfct) LPF 1 {|f| ≤ Bb} LPF 1 {|f| ≤ Bb} y(I)

b (t)

y(Q)

b

(t)

Only illustrate the real part of the passband frequency response. Imaginary part is similar.

Re{˘ yb(f)}

• 1
• ≈

≈

• 1

Im{˘ y(Q)

b

(f)}

• ≈

≈

• 1

Re{˘ y(I)

b (f)}

44

y(t)

Step Filter 1 {f ≥ 0} yb(t) √ 2 exp(−j2πfct)

y(t)

√ 2 cos(2πfct) − √ 2 sin(2πfct) LPF 1 {|f| ≤ Bb} LPF 1 {|f| ≤ Bb} y(I)

b (t)

y(Q)

b

(t)

Only illustrate the real part of the passband frequency response. Imaginary part is similar.

QAM Demodulation

45

y(t)

√ 2 cos(2πfct) − √ 2 sin(2πfct) Filter q(t) Filter q(t) T =

1 2W

T =

1 2W

{ˆ u(Q)

m }

{ˆ u(I)

m }

LPF 1 {|f| ≤ Bb} LPF 1 {|f| ≤ Bb} y(I)

b (t)

y(Q)

b

(t)

y(t)

Step Filter 1 {f ≥ 0} yb(t) √ 2 exp(−j2πfct) Filter q(t) T =

1 2W

{ˆ um}

merge

QAM Demodulation

46

y(t)

Step Filter 1 {f ≥ 0} yb(t) √ 2 exp(−j2πfct) Filter q(t) T =

1 2W

{ˆ um}

y(t)

√ 2 cos(2πfct) − √ 2 sin(2πfct) Filter q(t) Filter q(t) T =

1 2W

T =

1 2W

{ˆ u(Q)

m }

{ˆ u(I)

m }

Real-domain implementation Complex-domain equivalence

Passband Expansion

• In summary, under QAM, the transmitted waveform is
• Identify
• Rewrite

47

x(t) = x(I)

b (t)

√ 2 cos(2πfct) − x(Q)

b

(t) √ 2 sin(2πfct)

=

m

u(I)

m p(t − mT)

√ 2 cos(2πfct) −

m

u(Q)

m p(t − mT)

√ 2 sin(2πfct)

ψ(I)

m (t)

ψ(Q)

m (t)

p(t − mT) ← → φm(t) p(t − mT) √ 2 cos(2πfct) ← → ψ(I)

m (t)

−p(t − mT) √ 2 sin(2πfct) ← → ψ(Q)

m (t)

x(t) = X

m

u(I)

m ψ(I) m (t) + u(Q) m ψ(Q) m (t)

↑ is this an orthonormal expansion?

48

Theorem Consider an orthonormal set of waveforms {φm(t) : m ∈ Z}. Assume the Fourier transform exists for each φm(t) and is band-limited, that is, ˘ φm(f) = 0, ∀ |f| > Bb. Then for a center frequency fc > Bb, {ψ(I)

m (t), ψ(Q) m (t) | m ∈ Z} also

form an orthonormal set, where ψ(I)

m (t) , φm(t)

√ 2 cos(2πfct), ψ(Q)

m (t) , −φm(t)

√ 2 sin(2πfct).

In words, a baseband orthonormal basis remains orthonormal after up conversion