Wireless Communication Systems @CS.NCTU Lecture 1: Basics - - PowerPoint PPT Presentation

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Wireless Communication Systems @CS.NCTU Lecture 1: Basics - - PowerPoint PPT Presentation

Dec 09, 2023 314 likes •429 views

Wireless Communication Systems @CS.NCTU Lecture 1: Basics Instructor: Kate Ching-Ju Lin ( ) 1 Wireless Signal e ix = cos x + j sin x Sine wave noise phase change due to channel propagation delay y = hx + n h = e 2 j f

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SLIDE 1

Lecture 1: Basics

Instructor: Kate Ching-Ju Lin (林靖茹)

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Wireless Communication Systems

@CS.NCTU

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SLIDE 2

Wireless Signal

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  • Sine wave

y = hx + n

received signal transmitted signal channel noise

eix = cos x + j sin x

amplitude phase change due to propagation delay

h = αe−2jπfc(t+θ)

What is channel? Signal variation (amplitude and phase) over the air

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SLIDE 3

Constellation Diagram

  • Signal can be described as a sine wave
  • Rearranged as inphase and quadrature

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x(t) = A(t) cos(ωt + θ(t)) = A(t)ej(ωt+θ(t)) + e−j(ωt+θ(t)) 2 = Re[A(t)e−j(ωt+θ(t))] = Re[A(t)e−jθ(t)e−jωt] = Re[˜ x(t)e−jωt] = Re[(I(t) + jQ(t))e−jωt] = I(t) cos(ωt) + Q(t) sin(ωt)

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SLIDE 4

Constellation Diagram

  • Represent a wireless signal as a complex

number

  • Sine carrier: image part
  • Cosine carrier: real part
  • Why complex value?
  • Sine and Cosine are orthogonal with each other
  • Two carriers on the same frequency à rate⬆

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x(t) = A(t) cos(ωt + θ(t)) = I(t) cos(ωt) + Q(t) sin(ωt) = I(t) + jQ(t)

Q I I(t) (I(t), Q(t)) θ(t) Q(t) Constellation diagram

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SLIDE 5

Signal Power

  • Watt vs. Decibel (dBm)
  • dBm is usually used in radio
  • Able to express both very large and very small values

in a short form

  • dB: difference between two dBm values
  • ratio of two power = difference between two dBm

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PdBm = 10 log10(1000PW ) PW = 10PdBm/10 1000 P1toP2dB = 10 log10(P1 P2 ) = 10 log10(P1) − 10 log10(P2) = P1,dBm − P2,dBm

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SLIDE 6

SNR

  • Signal-to-Noise Ratio
  • In dB
  • From equation
  • Decoding SNR
  • Sent s=1+0i

but receive s’= a+bi

  • Signal power

= s2 =|1+0i|2

  • Noise power = |s-s’|2

= |(a+bi) – (1+0i)|2 = |(a-1)+bi|2

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S N 10 log10 S N

I Q

n s’ = a+bi

y = hx + n SNR = |h|2 E[|n|2]

SNR = |1 + 0i|2 |(a − 1) + bi|2

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SLIDE 7

Power vs. dB

  • Because of the log operation, double the

power produces 3dB gain

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P1 = 2 ∗ P2 ⇐ ⇒ P1,dB = P2,dB + 3dB SNRdB = 10 log10 SNR

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SLIDE 8

Path Loss

  • Attenuation reduction as the signal

propagates through the air

  • Friis Transmission Formula
  • λ: signal wavelength
  • Pt/Pr: transmitting/receiving power
  • Dt/Dr: directivity of transmitting/receiving antenna
  • Loss ∝ distance2

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Pr Pt = DtDr λ 4πd 2 Pr − Pt = Dt + Dr + 20 log10 λ 4πd

  • (in dB)

(in Watt)

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SLIDE 9

Shannon Capacity

  • The tight upper bound on the data rate

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C = B log2

  • 1 + S

N

  • = B log2 (1 + SNR)
  • B: bandwidth (Hz), e.g., WiFi with 20MHz
  • S and N is in Watt (SNR is power ratio, not in dB)
  • In low SNR regime, increasing SNR

can increase the rate significantly

  • In high SNR regime, the increase in

rate from SNR gain is relatively small

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SLIDE 10

Equalization

  • Reversal of distortion incurred by a signal

transmitted through a channel

  • Equalizer: recover the transmitted signal from

the received signal

  • a.k.a. decoding
  • Solution: MMSE, Zero-forcing, etc.
  • Example

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y = hx + n ⇒ x0 = y h = x + n h

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SLIDE 11

Coherence Time

  • The time over which a propagating wave may

be considered coherent (i.e., staying constant)

  • Why this is important?
  • To decode the signal, we need to estimate the

channel h

  • The time interval between consecutive channel

estimation should be shorter than coherence time

  • Otherwise, decoding can be erroneous due to

incorrect channel h

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