random processes and lsi systems what happedns when a random signal is processed by an lsi system?...

Random Processes and LSI Systems

What happedns when a random signal is processed by an LSI system? This is illustrated below, where x(n) and y(n) are random signals, and h(n) is a deterministic (i.e., nonrandom) LSI system.

The input, x(n) is a random signal, so y(n) is, too. Random in, random out.

x(n) y(n)h(n)


The LSI system, h(n) does exactly the same thing as it would if h(n) were deterministic, so:

x(n) y(n)h(n)

nxnhny or, in the frequency domain,

jjj eXeHeY


We can take the square of the magnitude,

222 jjj eXeHeY

Recall (from the previous lecture) that Sxx(), the spectral density of x(n), is related to by 2jeX

2

12

1lim j

Nxx eX

NES


so obviously

2

12

1lim j

Nyy eY

NES

and this can be rewritten as:

22

12

1lim jj

Nyy eXeH

NES

so obviously

2

12

1lim j

Nyy eY

NES

and this can be rewritten as:

22

12

1lim jj

Nyy eXeH

NES


Since the LSI system is deterministic, we can take it outside the expected value:

22

12

1lim j

N

jyy eX

NEeHS


What’s left inside the expected value is Sxx(), so

xx

jyy SeHS

2

isn’t that nice?


It can be shown that, if x(n) is a zero-mean sequence (which it is) and h(n) is an LSI system (which it is), then y(n) is also zero mean.

This means we can find the variance of y (the average output power by:

dSeHdS xxj

yyy

22

2

1

2

1

This means we can reduce the average power of a random signal (i.e., reduce the noise power) by attenuating parts of its spectral density function.


dSeHdS xxj

yyy

22

2

1

2

1

The above equation shows that we can attenuate parts of the output noise spectral density by making the system frequency response such that the system rejects those parts. The power of a wideband noise source can be reduced by lowpass (or bandpass) filtering the noise.


-

Sxx()

Suppose x(n) is a white noise sequence as shown. It’s average power is:

02

12xxxxy SdS


-

H()

Now, suppose h(n) is an ideal lowpass filter, c = /4

- -


-

So now the output spectral density is this:

- -

Syy()

Sxx()

And the average power of the output sequence is:

dSdS xxyyy

4

4

2 02

1

2

1

And the average power of the output sequence is:

04

1

02

1

2

1 4

4

2

xx

xxyyy

S

dSdS


So reducing the bandwidth by 75% also reduced the noise power by 75%.

Suppose the input to our LSI system h(n) is the sum of two random sequences, z(n) and g(n):


x(n)y(n)h(n)

z(n)

g(n)

If we know Szz(n) and Sgg(n), can we find Syy(n)? Yes!

We already know that


xx

jyy SeHS

2

So we need to find Sxx(n) in terms of Szz(n) and Sgg(n).

recall that: nrDTFTS xxxx

and

nkxkxEnrxx


nkgnkznkx

kgkzkx

so

nkgkzEnkzkgEnrnr

nkgkzEnkzkgE

nkgkgEnkzkzE

nkgkznkzkgnkgkgnkzkzE

nkgnkzkgkzEnr

ggzz

xx


Two random processes are independent if the outcome of one does not influence the other. For example, rolling two dice.

Two random processes are dependent if the outcome of one can influence the other. Example: drawing a 5-card poker hand.


If the two processes that produce z(n) and g(n) are independent, we can simplify the last expression:

nkgEkzEnkzEkgEnrnr

nkgkzEnkzkgEnrnrnr

ggzz

ggzzxx

and if they’re zero-mean processes, this further reduces to:

nrnrnr ggzzxx


and this means that the spectral density of x(n) is

nSnSnS ggzzxx

So, for two sequences z(n) and g(n), which are generated by independent, zero-mean random processes, summed together to form the input sequence x(n) to an LSI system h(n), the output noise spectral density is

nSeHnSeH

nSnSeHnS

ggj

zzj

ggzzj

yy

22

2


Finally, the average power of the output sequence y(n) is:

dnSeHdnSeHnS ggj

zzj

yy

22

2

1

2

1

White Noise

A zero-mean white noise sequence x(n) has the following proprerties:

E[x(n)] = 0

x(k) and x(k+n) are independent if n is nonzero.

White noise is independent with respect to any other sequence.

White Noise

If two random variables x and y are independent,

yExExyE

So, for white noise,

nkxEkxEnkxkxE

But E[x(n)] = 0, so

0 ,

0 ,022

nkxEnkxkxE

nnkxkxE

x

White Noise

We’ve previously seen that

nkxkxEnrxx

So we’ve just shown that for white noise,

nnr xxx 2

The power spectral density is the DTFT of the autocorrelation function, so for zero-mean white noise,

2xxxS

which is constant across the entire spectrum (white).

We can generate a zero-mean white noise sequence x(n) by randomly choosing, for each n, a real number between –/2 and /2. Each number must be independent of the choice of all others. This is a uniformly distributed, zero mean, white noise signal.

White Noise

Quantization Noise and Oversampling

Here’s a VERY practical example of exactly this type of signal: quantization noise.

Suppose we have an N-bit A/D converter with an input we designate xa(t). xa(t) can range from a minimum of -V volts to a maximum of +V volts. The A/D converter samples xa(t) every T seconds, and produces an N-bit binary number which approximates the quantized value of the sample xa(nT)


An N-bit binary number is used to represent the sampled input, xa(nT). This binary number has a finite number of possible values: 2N. Each of these values represents a range of possible input voltages, and there is one such range for each possible N-bit number. Thus, the total input range,

VtxV a

is divided into 2N smaller ranges.


Each of these smaller ranges can be expressed as

12

2

2212

2212

N

a

V

MVtxMV

where M is the actual N-bit number representing a particular sample.

The following figure illustrates these relationships for a 4-bit A/D converter (N = 4).


xa

M

M=0

M=1

V V V

V

4-bit A/D converter


If the output of the A/D converter is M, xa(nT) is somwhere in the range

22

1222

12

MVnTxMV a

So, if the output of the A/D converter is M, this says that the input is

22

12

MVnx


If we let the A/D converter output be the center of the range,

2

12ˆ

MVnx

we can rewrite the previous relationship as

2

ˆ

nxnx


or

2

ˆ

nxnx

where x(n) is the actual input signal to the A/D converter, and the A/D output is

We can express the quantization error as an error (or noise) sequence, e(n), yielding this:

nx

nenxnx ˆ


Written this way,

The A/D converter’s output can be thought of as the sum of two sequences: a signal sequence equal to the input signal sampled by an ideal, infinite-precision A/D converter, and the quantization noise sequence, e(n).

Note that the quantization noise can have any value between –/2 and /2, and its probability density function is uniform.

nenxnx ˆ


Consider the signal sequence, x(n). It’s average power is

nxEx22

The quantization noise sequence has this average power:

neEe22

so the A/D converter’s signal to noise ratio (SNR) is:

2

2

e

xSNR


We can write the SNR in terms of decibels:

22 log10log10log10 exdB SNRSNR

Naturally, we want the SNR to be as large as possible. This means we make the signal power as large as possible, by using the entire input range of the A/D converter.

We also minimize the noise power. One way to do this is to increase the number of bits, which may or may not be practical. We would like to have another way.


To see if there is another way, let’s investigate e(n).

We know that e(n) is a random variable with values in the range

22

ne

and it’s uniformly distributed in that range. Positive and negative values are equally likely, so it’s a zero-mean process.


It’s average power is:

12223

11

3

1 233

2

2

32

2

2

2

2

222

|

edee

deepeneEe


With this knowledge, we can write the SNR as

22

2

2

22

2

2

2

2

2

2

1234

1212

122

1212

NxN

x

N

xx

e

x

VV

VSNR

Or, in dB:

22

2

12log10log103log10

Nx

dB VSNR


If N > 8, we can use this approximation:

NV

VSNR

x

NxdB

02.6log1077.4

2log10log103log10

2

2

22

2

NN 212

substituting this in the previous expression for SNR gives us:


Which shows that each additional bit of precision improves the SNR by about 6 dB. This makes sense, since it cuts in half.

It’s not too much of a stretch to assume that quantization errors occurring at different times are independent. If this is assumed, then the quantization noise sequence, e(n), satisfies the conditions to be modeled as uniformly distributed white noise.


-

See()

e2

We’ve already seen that the power spectral density of a white noise sequence x(n) is

,2xxxS

so the power spectral density of e(n) is as shown below:


So the average power of the quantization noise sequence e(n) is spread over the discrete time frequency range from – to .

Remember that radians per sample in the discrete time domain maps to fs/2 in the continuous time domain, so the quantizing noise power is spread over the range – fs/2 to fs/2.

- fs/2 - fs/2

See(fT)

e2


If, for example, we quadruple the sampling frequency, but leave N alone, we take the quantizing noise power and spread it over a band four times as wide.

t,independen are and

Since ly.respective ,ˆ and , ,of' densities

spectralpower thebe and , ,Let ˆˆ

nenx

nxnenx

SSS xxeexx

eexxxx SSS ˆˆ

Suppose we have a bandlimited signal (bandwidth = B) with power spectral density shown in the next slide:

If we sample it at the minimum sample rate, fs = 2B, the spectral densities of the signal sequence and the quantization noise sequence are as shown next:

Sxx(f)


f

If we double the sample rate (fs = 4B), the signal and noise spectral densities are as shown:

Sxx(f)


f

e

If we double the sample rate again (fs = 8B), the signal and noise spectral densities are as shown:

Sxx(f)


f

e

Note that the true signal power and the true noise power are the same in all three, the apparent difference merely serves to show that the noise power is spread over a wider band.

Sxx(f)


f

e

This can be generalized for fs = 2MB, as shown below:

Sxx(f)


f

e

Notice that if M > 1 (i.e., if the signal is oversampled) some of the noise power is outside the signal bandwidth.

We can eliminate the portion of the noise power which is outside the signal bandwidth by using a digital lowpass filter to attenuate it.

Since this filter will not attenuate anything in the signal band, the signal (and the information it conveys) is unaffected. Unfortunately, the in-band noise is also not affected, but getting rid of the out-of-band noise is a good thing.


If M, the oversampling factor, is 1, we are sampling at the minimum rate, and the SNR is given by


NV

SNR xdB 02.6log1077.4

2

2

If we let M = 2 (oversampling by a factor of 2), and use an ideal lowpass filter to eliminate out-of-band noise, the result is as shown next:


Sxx(f)

f

e

This eliminates half the noise power, doubling the signal to noise ratio. If we let M = 4 (oversampling by 4), the SNR is doubled again:


Sxx(f)

f

e

In general, every time we double the sample rate, we can double the SNR. This is the same as increasing the SNR by 3 dB for every doubling of the sample rate.


The resolution of an A/D converter is the effective number of bits which, with M=1, would yield it’s signal to noise ratio. The SNR of an A/D converter (with M = 1) is given by:

NV


2

2

so oversampling by 4 effectively adds a bit to the converter’s resolution. If a digital filter eliminates the out-of-band quantization noise, that is.

Our oversampling system is as shown below:


xa(t) H(ej)A/D

fs = 2MB

nenxnx ˆ

Lowpassc = /M

nenxng ˆ

2

2

e

xSNR

2

2

e

xMSNR

The filter output, g(n), can be thought of as the sum of two components:

nenxng ˆ

The filter doesn’t change the signal, but it does affect the noise by eliminating the out-of-band portion. Notice that the noise at the filter input is denoted e(n), while the output noise is


Since the signal and noise components of the filter input were assumed to be independent, the spectral density of the filter output is given by:

eexx

eej

xxj

gg

SS

SeHSeHS

ˆˆ

22

ne

The signal power is still given by


nxEx22

The output noise power is

M

ddS

ee

M

M

e

M

M

eee

22ˆ

2ˆˆ

2ˆ

2

1

2

1


So the output SNR is:

inputfilter 22ˆ

2

outputfilter SNRM

MSNR

ee

x

So oversampling by a factor of M multiplies the output SNR by M.

In terms of dB, MSNRSNR dBdB log10

inputfilter outputfilter


This shows that if M = 2, we add 3 dB to the SNR, and if M = 4, we add 6 dB. 6 dB is also the SNR improvement we get by adding one bit of resolution. In other words, quadrupling the sample rate is equivalent to adding one bit to the A/D converter.

If we call the converter’s resolution in bits R, the relationship between R, N and M is:

MNR 2log2

1


If we have an N – bit converter, and need R bits of resolution, we can solve for M:

NRM 22

The dynamic range of an A/D converter is the SNR we get when the input is a full-scale sinusoid:

ftVtxa 2cos

The power of this signal is V2/2. by subsituting this for x

2 into

NV


2

2


we get

N

NDdB

02.676.1

02.62

1log1077.4

This shows that improving the resolution by 1 bit also improves the dynamic range by 6 dB.

The dynamic range of an oversampling A/D converter (and filter) is given by:


MNDdB log1002.676.1ngoversampli

random processes and lsi systems what happedns when a random signal is processed by an lsi system?...

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