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Noise

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It is a function of the carrier to noise power ratio (the average energy per bit ... menambah baik nisbah isyarat kepada hingar sebanyak dua kali bagi kes DSBSC. ... – PowerPoint PPT presentation

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Title: Noise


1
Noise
  • CHAPTER 5

2
Learning Outcomes
  • Student able to
  • Define and calculate bit error rate
  • Calculate noise in FM and AM

3
Noise - Bit Error Rate
  • It occurred in digital communication.
  • It is a function of the carrier to noise power
    ratio (the average energy per bit to noise power
    density ratio)
  • The carrier power is defined as below
  • CdBm 10log(Cwatts / 0.01).
  • The carrier to the noise power ratio is
  • C/N C/KTB

4
Noise - Bit Error Rate
  • The carrier to Noise power in dB is
  • C/N (dB) log (C/N)
  • CdBm - NdBm
  • The enery per bit is simply the energy of a
    single bit of information, Eb .
  • It is defined as below
  • Eb CTb (J/bit)
  • Eb energy of a single bit (joules per bit)
  • Tb time of a single bit (seconds)
  • C carrier power (watts)

5
Chapter 6 Noise - Bit Error Rate
  • in dBJ,
  • Eb(dBJ) 10 log Eb
  • Since Tb 1/fb,
  • Eb C/fb (J/bit)
  • Eb(dBJ) 10 log (C/fb )
  • 10 log c 10 log fb

6
Noise - Bit Error Rate
  • The Noise power density is
  • No N/B (W/Hz)
  • 10log(N/0.001) log B
  • Energy per bit to the noise power density is
    defined as
  • Eb / No C/N x B/fb
  • Energy per bit-to-noise power density ratio
    is simply the ratio of the energy of a single bit
    to the noise power present in 1 Hz of bandwidth.

7
Noise - Bit Error Rate
  • EXAMPLE 1
  • A ASK system has the following specification
  • C 10-12 W, fb 60 kbps, N 1.2x10-14 W
    B 120 kHz
  • Determine the following
  • (i) Carrier power in dBm
  • (ii) Noise power in dBm
  • (iii) Noise power density in dBm
  • (iv) Energy per bit in dBJ
  • (v) Carrier to noise power ratio in dB
  • (vi) Eb / No ratio

8
Noise - Bit Error Rate
  • SOLUTION TO EXAMPLE 1
  • (i) Carrier power
  • 10logC/0.001 10log(10-12/0.001
  • -90 dB
  • (ii) Noise power
  • 10logN/0.001 10log(1.2x10-14/0.001
  • -109.2 dB
  • (iii) Noise power density
  • No N/B 10logN/0.001 10logB
  • 10log(1.2x10-14/0.00
    1)-10log(120k)
  • -109.2 (50.79)
  • -160 dB

9
Noise - Bit Error Rate
  • SOLUTION TO EXAMPLE 1 (Continued)
  • (iv) Energy per bit in dBJ
  • Eb(dBJ) 10 log (C/fb )
  • 10log(10-12 /
    60 kbps)
  • -167.8 dB
  • (v) Carrier to Noise power ratio in dB
  • 10logC/N CdBm - NdBm -90
    (-109.2)

  • 19.2 dB
  • (vi) Eb / No 10logEb 10logNo
  • 19.2 10log(120kHz/60kbp
    s)
  • 19.2 3.0
  • 22.2 dB

10
Noise in AM DSBSC
Synchronous detector
Noise
Modulated signal
Hence
signal power Input
Noise power input
11
Noise in AM DSBSC
Hence
After the multiplier

After LPF
12
Noise in AM DSBSC
Output y(t)
signal power output

Noise power output
Hence
Persamaan menunjukkan bahawa pengesan telah
menambah baik nisbah isyarat kepada hingar
sebanyak dua kali bagi kes DSBSC.
13
Noise in Full AM
noise
Noise power input

Filtered out
14
Noise in Full AM
After LPF
DC value removed
It is shown that (SNR)o is always less than
(SNR)i as
15
Noise in FM
  • FM systems are far better at rejecting noise than
    AM systems. Noise generally is spread uniformly
    across the spectrum (the so-called white noise,
    meaning wide spectrum).
  • FM systems are inherently immune to random noise.
    In order for the noise to interfere, it would
    have to modulate the frequency somehow. But the
    noise is distributed uniformly in frequency and
    varies mostly in amplitude. As a result, there is
    virtually no interference picked up in the FM
    receiver.
  • FM is sometimes called "static free, " referring
    to its superior immunity to random noise.

16
Comparison Noise in AM Vs FM
  • Noise power output at IF AM receiver
  • Di mana di dalam perhubungan AM kebanyakkan
    isyarat yang dipancarkan adalah didominasi oleh
    isyarat pembawa maka kuasa isyarat yang
    dipancarkan adalah

17
  • for m 100
  • Therefore
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