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Mobile Communications (KECE425) Lecture Note 14 4-16-2014 Prof.Young-Chai Ko 1

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Page 1: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Mobile Communications(KECE425)

Lecture Note 144-16-2014

Prof. Young-Chai Ko

1

Page 2: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

S(f)

f

Frequency Selective Fading Channels

• Frequency selective channel =) Inter-symbol interference (ISI)

s(t) r(t)H(f)

H(f)

f

R(f)

2

Page 3: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

Multi-Carrier Transmission

S(f)

fH(f)

• Each sub-block is transmitted so that it experiences the flat fading chan-

nels, that is, ISI-free channel.

3

Page 4: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

fc + f0

• Orthogonality condition for the receiver

- If we set f0 =

1NT , then we have the orthogonality among each branch of

the transmitter in the multi-carrier transmission.

X

cos(2⇡(fc + f0)t)

X

cos(2⇡(fc + 2f0)t)

X

cos(2⇡(fc +Nf0)t)

m1(t)

m2(t)

mN (t)

+ s(t)

total bandwidth

1/NT 1/NT

2

NT

...

fc +Nf0

2

NT

fc +Nf0

· · ·

fc + f0

fc + 2f0

fc + 3f0fc + 2f0

2

NT

(N � 2)f0

(N � 2)f0 +2

NT=

N � 1

NT

4

Page 5: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

• Transmit signal waveform

X

cos[2⇡(fc + kf0)t]

Passband filtermk(t)

s(t)

• Assume that there is no fading and no noise.

2

NT

X

cos[2⇡(fc + kf0)t]

Z NT

0(·) dts(t)

or equivalently

s(t) =NX

k=1

mk(t) cos[2⇡(fc + kf0)t]

0

1

2

Z NT

0mk(t) dt

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Page 6: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

X

cos[2⇡(fc + kf0)t]

Passband filtermk(t)

s(t) 2

NT

0

s(t) =NX

j=1

mj(t) cos(2⇡(fc + jf0)t)

s(t) cos(2⇡(fc + kf0)t) = cos(2⇡(fc + kf0)t)⇥NX

j=1

mj(t) cos(2⇡(fc + jf0)t)

=

NX

j=1

mj(t) cos(2⇡(fc + kf0)t)⇥ cos(2⇡(fc + jf0)t)

cos(2⇡(fc + kf0)t)⇥ cos(2⇡(fc + jf0)t) =

1

2

[cos (2⇡(2fc + (k + j)f0)t) + cos(2⇡(k � j)f0)t)]

6

Page 7: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

s(t) cos(2⇡(fc + kf0)t) =

1

2

NX

j=1

mj(t) [cos (2⇡(2fc + (k + j)f0)t) + cos(2⇡(k � j)f0)t)]

X

cos[2⇡(fc + kf0)t]

Passband filter

mk(t)s(t) 2

NT

0

=

mk(t)

2

+

mk(t)

2

cos(2⇡(2fc + 2kf0)t)

+

1

2

NX

j=1,j 6=k

mj(t) [cos (2⇡(2fc + (k + j)f0)t) + cos(2⇡(k � j)f0)t)]

7

Page 8: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

X

Passband filter

s(t)

2

NT

0

cos[2⇡(fc + f0)t]

m1(t)

X

Passband filter

2

NT

0

X

Passband filter

2

NT

0

cos[2⇡(fc + 2f0)t]

cos[2⇡(fc +Nf0)t]

m2(t)

mN (t)

......

8

Page 9: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

• In reality, the information bearing signal mk(t) is a signal with a certain

pulse which satisfies the Nyquist criterion, for example, rectangular pulse,

root-raised cosine pulse, and etc.

• In that case, the amplitude spectrum of mk(t) is sinc type.

· · ·· · ·

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Page 10: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

⇥s(t)

h(t)

r(t)+

n(t)

h(t) = ↵(t)ej�(t)

• Distribution of the amplitude ↵(t):

– Rayleigh fading channel

– Ricean fading channel

– Nakagami-m fading channel

• Flat fading channel model

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Page 11: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

• Received signal power

• Received signal

r(t) = ↵(t)ej�(t)s(t) + n(t)

0

Pr

PrdBm(d) = µdBm(d) + ✏dB

✏dB ⇠ N (0,�2✏ )

area mean

shadowing

local mean

↵2(t)Pr

tt

• Instantaneous power:

↵2(t)Pr or it is often written as ↵2(t)E[s2(t)]

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Page 12: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

• (Instantaneous) Signal-to-Noise ratio, �:

� =↵2(t)Es

N0

• Average Signal-to-Noise ratio, �:

E[�] = E

↵2(t)Es

N0

�=

E[↵2(t)]Es

N0=

⌦tEs

N0

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Page 13: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

CDF of Exponential RV

• For Rayleigh channel, ↵2(t) follows the exponential distribution:

p

2(x) =1

⌦p

e

�x/⌦p

• Then the CDF of ↵2(t) can e written as

P

2(x) = Pr[↵2< x] =

1

⌦p

Zx

0e

�t/⌦pdt

= 1� e�x/⌦p

x > 0

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Page 14: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

CDF of Instantaneous SNR

• (Instantaneous) Signal-to-Noise ratio, �:

� =↵2(t)Es

N0

p

(x) =1

e

�x/�

,

• PDF of �:

• CDF of �:

P

(x) = 1� e

�x/�

,

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Page 15: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

0 10 20 30 40 50 60 70 80 90 1000

1

2

3

4

5

6

7

8

� = 1

• Example of exponential random samples for � = 1.

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Page 16: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

0 10 20 30 40 50 60 70 80 90 100−25

−20

−15

−10

−5

0

5

10

� [dB]

� = 1 ! 0 [dB]

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Page 17: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

Outage Probability

• Outage probability, Pout

:

Pout

= Pr[� �T ]

= 1� e��T /�

−30 −25 −20 −15 −10 −5 0 5 1010−3

10−2

10−1

100

Pout

ex. � = 0 dB

�T dB

17

Page 18: Mobile Communications (KECE425)contents.kocw.net/KOCW/document/2014/korea/koyeongchae/... · 2016. 9. 9. · c +2kf 0)t) + 1 2 XN j=1,j6= k m j (t)[cos(2⇡(2f c +(k + j)f 0)t)+cos(2⇡(k

Prof. Y. -C. Ko

−30 −25 −20 −15 −10 −5 0 5 1010−4

10−3

10−2

10−1

100

• Outage probability for � = 0, 2, 4, 6, 8, 10 dB over Rayleigh channel.

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