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SPREAD SPECTRUM

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SPREAD SPECTRUM

Hiding Information in noise

- Military communication has always been concerned with the following two issues
- Security
- Jam resistance

- In civilian communications, above issues take on different interpretations
- privacy
- unintentional interference

- Spread spectrum is in effect a way to “hide” information
- Useful information is buried in noise. To an eavesdropper, the intercepted message looks juts like noise
- The intended receive however is able to recover the information from noise using a special “key”

- There are two main types of spread spectrum
- Direct Sequence(DS)
- Frequency Hopping(FH)

- in DS/SS, digital data is multiplied by another bitstream running several hundred times faster
- In FH/SS, carrier frequency, normally fixed, jumps around in a “random” manner known only to the intended receive

- Take the baseband digital data b(t) and modulate it by a “random” bit pattern c(t). The resulting bitstream is m(t)=c(t)b(t)

b(t)

Tb

Tc

c(t)

- There are a number of important parameters in SS
- b(t): data sequence
- c(t): spreading sequence
- Tb: bit length
- Tc: chip length
- N=Tb/Tc: number of chips per bit
- N=3 in this figure

b(t)

Tb

c(t)

Tc

- The classic jamming model is shown below. we will demonstrate that an SS signal provides superior protection against intentional jamming

m(t)

b(t)

r(t)

X

c(t)

i(t)

interference

- Clearly, randomness is at the heart of spread spectrum
- However, if truly random codes are used to spread the signal, receiver would never be able to recover the information
- Therefore, we need a “pseudo” random noise known as PN sequences. Pseudo because if you wait long enough, they will repeat

- To a casual observer, a PN sequence looks like a random alternations of +/-1.
- In truth, however, a PN sequence repeats. Can you spot the period here?
- The key to “cracking” the code is to find where the period ends

- It is said that spread spectrum signal looks like random noise to all others but why?
- Consider this

- PN sequences can be generated by a set of flip-flops with appropriate taps

+

1

0

0

So

S1

S2

output

Initial state: 100

1 0 0 1 1 0 1 1 1 0 1 1 1 0 1 0 1 0 0 0 1 1 0 0

output: 0 0 1 1 1 0 1 0

- The preceding sequence repeats itself with a period of 23-1=7
- In general, for an m-stage shift register, the period is at most
- If the period is equal to the above, we have maximal lengthor m-sequences

- # of 1’s are always one more than the number of 0’s
- Period: 2m-1
- Very desirable (tight) correlation

- Let c(t) be an m-sequence. Its autocorrelation function is given by

Tb

Shifted by <Tc

- The significant property of correlation here is that it can discriminate against the slightest shifts. In fact, shift of just a single chip drops the function by a factor of N

Rc()

1

-1/N

- Once you pick a length N, the question is how do we generate an m-sequence?
- N, fixes the number of shift register stages but you can connect them in many ways
- Only a few connections give you valid m-sequences(see Table 9.1 and Figure 9.4)

+

+

+

1

2

3

4

5

N=25-1=31, taps at [5,4,2,1]

- A PN sequence is generated using a feedback shift register of length 4. The chip rate is 107 pulses per second. Find
- a):PN sequence length
- b): Chip duration
- c):PN sequence period

- Answers
- a): if an m-sequence, period is 24-1=15. Less if not
- b): 1/107=10-7 sec
- c):T=NTc=15x10-7 sec

- Probably the single most important component of an SS system is a quantity called processing gain(PG)
- PG is defined by
PG=N=Tb/Tc

- In other words PGis given by the number of chips within a bit

- Bandwidth spreads by a factor equal to the processing gain
spread bandwidth Wss=(Tb/Tc)W=PGxW

- Want to know the bandwidth of a digital signal running at 28.8 Kb/secafter spreading
- Consider a m=19 stage shift register
- PN sequence period N=219-1~219
- There are 219 chips inside a bit, i.e. Tb=NTc
- Therefore, Rc=1/Tc=N/Tb=219x 28.8 Kb/sec

- Since bandwidth is proportional to bitrate, the new bandwidth is now 219 or 57 dB higher than the unspread signal

- The classic jamming model is shown below. we will demonstrate that an SS signal provides superior protection against intentional jamming

m(t)

b(t)

r(t)

X

c(t)

i(t)

interference

- A jammer or interference i(t) tries to interfere with a spread spectrum signal
- The corrupted spread spectrum signal at the receiver is put through a conventional correlation detector

r(t)

z(t)

c(t)

Data pn seq.

- Let’s walk the spread spectrum signal through the receiver

interference

desired data

- The jammer appears as c(t)*i(t). In other words we have created a spread spectrum signal out of the jammer!
- The bandwidth of a SS signal is very large making it look like white noise. Therefore, a lowpass filter integrator) will let the message b(t) through but will stop most of the jammer appearing as c(t)*i(t)

- So far we have looked at DS/SS in baseband.
- For the actual transmission we need to modulate the signal
- Spreading can be done either before or after carrier modulation. See Fig. 9.7, 9.8 and 9.9 while listening to this slide

- It can be shown that the SNR at the input and output of correlation detector is given by

- The improvement in SNR is caused by the processing gain, Tb/Tc. This ratio can be several hundreds or thousands
- SNR gain can be as high as 1000(30dB)

- A DS/BPSK in Gaussian noise had a BER of
- In the presence of jammer(but no noise)

- Comparison of the two BER expressions
- Equivalently, Eb=PTb where P is the average signal power. Then

- We just saw that processing gain helps counter jamming power
- The ratio of jammer power to signal power is called Jamming margin
J/P=PG/(Eb/No)

- In dB
jm=PG-Eb/No

- Digital data is running with bit-lengthTb=4.095 ms.This data is spread using a chip length of Tc=1 microsecond using DS/BPSK. What is the jamming margin if the required BER= 10-5.?
- In the presence of random noise alone we need Eb/No=10 to achieve BER= 10-5.

- The processing gain is Tb/Tc=4095. Plugging these numbers in the JM expression, we get
JM |db=10log4095-10log(10)=26.1 dB

- We can maintain BER at the desired level even in the presence of a jammer 26dB(400 times) higher than the desired signal

- Code Division Multiple Access is one of the two competing digital cellular standards (IS-54). The other is TDMA-based IS-136
- In this area, Comcast has adopted IS-136. Bell Atlantic and Sprint PCS have gone the way of CDMA.
- These digital services coincide with the AMPS infrastructure

- AMPS is an example of FDMA. Users are on all the time but on different frequency bands
- TDMA uses the same 30KHz band of AMPS but services 3 users. Users are on only during their time slot.
- In CDMA, there is neither frequency nor time sharing. Everyone is on simultaneously thus taking up the whole spectrum

- In CDMA, kth user’s signal is spread by a PN code ak unique to the subscriber
- M users can be on at the same time

- The familiar correlation receiver will do the job

X

b1

a1

b2

X

a2

b3

X

a3

- Transmitter and receiver always operate on a known frequency band. Once found, anyone can listen in
- Imagine a scenario where carrier frequency “hops” around in a random pattern
- This pattern is known only to the intended receiver thus nobody else can follow the hop

- One obvious way to implement FH is to use MFSK.
- In the conventional MFSK, carrier frequency jumps are controlled by the message
- In FH/MFSK, jumps are controlled by a PN sequence

- Slow frequency hopping
- Symbol rate Rs of the MFSK signal is an integer multiple of Rh, the hop rate; several symbols are transmitted on each frequency hop

three symbols,same carrier freq.

- Fast frequency hopping
- The hop rate Rh is an integer multiple of the MFSK symbol rate Rs; the carrier frequency will change several times even before the symbol ends.

one symbol

- k-bit segments of the PN code drive the synthesizer-->2^k frequencies

FH/MFSK

M-ary FSK

X

BPF

Freq

synthesizer

PN code

generator

- Chip: an individual FH/MFSK tone of shortest duration
- In general, Rc=max(Rh,Rs)
- For slow FH

1 FH chip

Rc=1 per sec

Rs=1 per sec

Rh=1/3 per sec

frequency

1/Rs

Rs

time

1/Rh

PN

001

110

011

001

4 FSK tones, 8 hops, PN period 16,

- Carrier frequency hops several times within one symbol

one symbol

frequency

time

symbol

4 MFSK tones, 2 hops per symbol(hop rate=bitrate), 8 possible hops