6: Digital Representation of Analog Signals · 5 Summary 6: Digital Representation of Analog...

Sampling Theorem Pulse Code Modulation Differential Pulse Code Modulation Delta Modulation Summary

6: Digital Representation of Analog Signals

Y. Yoganandam, Runa Kumari, and S. R. Zinka

Department of Electrical & Electronics EngineeringBITS Pilani, Hyderbad Campus

Sep 16 – Sep 24, 2015

6: Digital Representation of Analog Signals Communication Systems, Dept. of EEE, BITS Hyderabad

Outline

1 Sampling Theorem

2 Pulse Code Modulation

3 Differential Pulse Code Modulation

4 Delta Modulation

5 Summary

Outline

1 Sampling Theorem

4 Delta Modulation

5 Summary

Sampling Theorem

1/TS-2/TS -1/TS 2/TS

Sampling Theorem

We obtain sampled signal g (t) by multiplying g (t) with impulse train δTs (t).Impulse train can be expressed as a Fourier series as shown below:

δTs (t) =1Ts

[1 + 2 cos ωst + 2 cos 2ωst + 2 cos 3ωst + · · · ] , ωs =2π

Therefore

g (t) = g (t) δTs (t)

[g (t) + 2g (t) cos ωst + 2g (t) cos 2ωst + 2g (t) cos 3ωst + · · · ] .

Applying Fourier transform on both sides of the above equation gives

G (ω) =1Ts

∑n=−∞

G (ω− nωs) . (1)

Sampling Theorem

δTs (t) =1Ts

Therefore

g (t) = g (t) δTs (t)

G (ω) =1Ts

∑n=−∞

G (ω− nωs) . (1)

Sampling Theorem

δTs (t) =1Ts

Therefore

g (t) = g (t) δTs (t)

G (ω) =1Ts

∑n=−∞

G (ω− nωs) . (1)

Sampling Theorem

δTs (t) =1Ts

Therefore

g (t) = g (t) δTs (t)

G (ω) =1Ts

∑n=−∞

G (ω− nωs) . (1)

Sampling Theorem

If we are to reconstruct g (t) from g (t), we should be able to recover G (ω)from G (ω) . It is possible only when

> 2B. (2)

The above equation can be rewritten as

fs > 2B, (3)

where, fs = 1/Ts, is the sampling frequency.

So, as long as fs is greater than twice the signal bandwidth B, we can recoverthe original signal g (t). The minimum sampling frequency fs = 2B requiredto recover g (t) is called Nyquist rate.

Sampling Theorem

> 2B. (2)

fs > 2B, (3)

Sampling Theorem

> 2B. (2)

fs > 2B, (3)

Sampling Theorem

> 2B. (2)

fs > 2B, (3)

Signal Reconstruction

Signal Reconstruction – Zero-Order Hold Circuit

Signal Reconstruction – Ideal Interpolation

1/TS-2/TS -1/TS 2/TS-1/2TS +1/2TS

Practical Issues with Sampling

Practical Sampling Rate

f0-2/TS -1/TS 2/TS1/TS

Smaller sampling rate

f1/TS-2/TS -1/TS 2/TS

Larger sampling rate

Lowpass Filtering – Ideal

Lowpass Filtering – Still Ideal

Lowpass Filtering – Practical

Aliasing (Spectral Folding)

Lost tailLost tail getsfolded back

Recoveredspectrum

The Antialiasing Filter

What happens if we implement antialiasing filter after sampling?

Practical Sampling

Applications of Sampling Theorem

Sampling Theorem – Applications

Time division multiplexing

CoderDigital

processor

Transmissionmedium

Digitalprocessor Decoder

Channel

Coder output(pulse code)

Ch. 3Ch. 6

Ch. 24

CoderDigital

processor

Transmissionmedium

Channel

Ch. 3Ch. 6

Ch. 24

CoderDigital

processor

Transmissionmedium

Channel

Ch. 3Ch. 6

Ch. 24

CoderDigital

processor

Transmissionmedium

Channel

Ch. 3Ch. 6

Ch. 24

CoderDigital

processor

Transmissionmedium

Channel

Ch. 24

CoderDigital

processor

Transmissionmedium

Channel

Ch. 3Ch. 6

Ch. 24

CoderDigital

processor

Transmissionmedium

Channel

Ch. 3Ch. 6

Ch. 24

CoderDigital

processor

Transmissionmedium

Channel

Ch. 3Ch. 6

Ch. 24

Outline

1 Sampling Theorem

4 Delta Modulation

5 Summary

Analog vs Digital & Continuous vs Discrete

PCM is a method of converting an analog signal into digital signal (A/Dconversion).

In order to understand PCM, understanding the difference among the termsanalog, digital, continuous, and discrete is very very important.

A Continuous Analog Signal

1234567

A Discrete Analog Signal

1234567

A Discrete (8-ary) Digital Signal

1234567

A Continuous (8-ary) Digital Signal

1234567

A Continuous Digital Signal – Binary PCM

1234567

t000 100 101 100 011 100 110 111 101 011 011 100 100

Pulse Code Modulation (PCM) – Trivia

Even though PCM is invented as early as 1926, It was the transistor thatmade PCM practicable in early 60s.

Average power of a PCM signal is less. However, the price paid is theincrease in the signal bandwidth.

PCM– Steps Involved

1 Sampling

2 Quantization

3 Encoding

4 Baseband transmission (we will discuss this topic in next chapter)

1 Sampling

2 Quantization

3 Encoding

1 Sampling

2 Quantization

3 Encoding

1 Sampling

2 Quantization

3 Encoding

1 Sampling

2 Quantization

3 Encoding

Quantization

Quantization Error

Quantization error is defined as

q (t) = m (t)−m (t). (4)

From the figure shown in the previous slide,

−∆v/2 ≤ q (t) ≤ ∆v/2, (5)

where∆v = 2mp/L.

Assuming that the error is equally likely to lie anywhere in the range(−∆v/2,+∆v/2), the mean square quantization error is

ˆ ∆v/2

−∆v/2q2dq =

(∆v)2

12. (6)

Quantization Error

q (t) = m (t)−m (t). (4)

−∆v/2 ≤ q (t) ≤ ∆v/2, (5)

where∆v = 2mp/L.

ˆ ∆v/2

−∆v/2q2dq =

(∆v)2

12. (6)

Quantization Error

q (t) = m (t)−m (t). (4)

−∆v/2 ≤ q (t) ≤ ∆v/2, (5)

where∆v = 2mp/L.

ˆ ∆v/2

−∆v/2q2dq =

(∆v)2

12. (6)

Quantization Error

q (t) = m (t)−m (t). (4)

−∆v/2 ≤ q (t) ≤ ∆v/2, (5)

where∆v = 2mp/L.

ˆ ∆v/2

−∆v/2q2dq =

(∆v)2

12. (6)

Signal to Quantization Noise Ratio (SQNR)

Now that we got mean square quantization error, we can obtain SQNR asshown below:

SQNR =So

m2 (t)

q2 (t)=

12m2 (t)

(∆v)2 . (7)

Since ∆v = 2mp/L,

SQNR = 3L2 m2 (t)m2

p. (8)

SQNR =So

m2 (t)

q2 (t)=

12m2 (t)

(∆v)2 . (7)

Since ∆v = 2mp/L,

SQNR = 3L2 m2 (t)m2

p. (8)

SQNR =So

m2 (t)

q2 (t)=

12m2 (t)

(∆v)2 . (7)

Since ∆v = 2mp/L,

SQNR = 3L2 m2 (t)m2

p. (8)

Non-Uniform Quantization

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

0.10.20.30.40.50.60.70.80.91.0

Speech signal magnitudes relativeto the rms of such magnitudes

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

0.10.20.30.40.50.60.70.80.91.0

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

0.10.20.30.40.50.60.70.80.91.0

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

0.10.20.30.40.50.60.70.80.91.0

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

0.10.20.30.40.50.60.70.80.91.0

Quantizationlevels

The Compressor

The Expandor

The Companding

µ Law

μ = 0

μ = 10μ = 100

μ = 1000

ln (1 + µ)ln(

1 +µmmp

), 0 ≤ m

mp≤ 1 (9)

µ Law

μ = 0

μ = 10μ = 100

μ = 1000

ln (1 + µ)ln(

1 +µmmp

), 0 ≤ m

mp≤ 1 (9)

µ Law

μ = 0

μ = 10μ = 100

μ = 1000

ln (1 + µ)ln(

1 +µmmp

), 0 ≤ m

mp≤ 1 (9)

A = 10A = 87.6

A = 1000

1+ln A

)0 ≤ m

mp≤ 1

1+ln A

(1 + ln Am

)1A ≤

mmp≤ 1

A = 10A = 87.6

A = 1000

1+ln A

)0 ≤ m

mp≤ 1

1+ln A

(1 + ln Am

)1A ≤

mmp≤ 1

A = 10A = 87.6

A = 1000

1+ln A

)0 ≤ m

mp≤ 1

1+ln A

(1 + ln Am

)1A ≤

mmp≤ 1

SQNR (µ Law)

0 10 20 30 40 50 60

Relative signal power m2(t), dB

L = 256μ = 0

μ = 255

No' 3L2

[ln (1 + µ)]2, µ2 �

m2 (t)

SQNR (µ Law)

0 10 20 30 40 50 60

L = 256μ = 0

μ = 255

No' 3L2

[ln (1 + µ)]2, µ2 �

m2 (t)

SQNR (µ Law)

0 10 20 30 40 50 60

L = 256μ = 0

μ = 255

No' 3L2

[ln (1 + µ)]2, µ2 �

m2 (t)

SQNR (A Law)

Search for the SQNR expression for A law case.

SQNR (A Law)

Search for the SQNR expression for A law case.

Binary PCM

1234567

t000 100 101 100 011 100 110 111 101 011 011 100 100

Binary PCM

1234567

t000 100 101 100 011 100 110 111 101 011 011 100 100

Pulse Shaping

0 Tb 2Tb

Tb = Ts / n

Pulse Shaping

0 Tb 2Tb

Tb = Ts / n

Pulse Shaping

0 Tb 2Tb

Tb = Ts / n

Pulse Shaping

0 Tb 2Tb

Tb = Ts / n

Pulse Shaping

0 Tb 2Tb

Tb = Ts / n

Pulse Shaping

0 Tb 2Tb

Tb = Ts / n

Pulse Shaping

0 Tb 2Tb

Tb = Ts / n = 1/2nB

Binary PCM – Transmission Bandwidth

For a binary PCM,L = 2n or n = log2 L. (11)

Since there should be at least 2B pulses per second, total number of bits persecond are 2nB.

So, minimum channel bandwidth is

BT = nB Hz. (12)

Binary PCM – SQNR

We have already derived the expression

SQNR = 3L2 m2 (t)m2

Since for binary PCM, L = 2n,

SQNR = 3 (2n)2 m2 (t)m2

p= c22n,

where c = 3 m2(t)m2

pfor uncompressed casea.

Since n = BT/B,SNR = c22BT/B. (13)

aFor µ law case, c = 3/ [ln (1 + µ)]2 .

Binary PCM – SQNR

SQNR = 3L2 m2 (t)m2

SQNR = 3 (2n)2 m2 (t)m2

p= c22n,

where c = 3 m2(t)m2

Binary PCM – SQNR

SQNR = 3L2 m2 (t)m2

SQNR = 3 (2n)2 m2 (t)m2

p= c22n,

where c = 3 m2(t)m2

Binary PCM – SQNR

SQNR = 3L2 m2 (t)m2

SQNR = 3 (2n)2 m2 (t)m2

p= c22n,

where c = 3 m2(t)m2

Binary PCM – SQNR

SQNR is given asSNR = c22n.

Applying dB scale on both sides of the above equation gives

(SQNR)dB = 10 log10 c + 20n log10 2 = (α + 6n) dB. (14)

An increase in the code word size by 1 bit, the SQNR increases by 6 dB and BTbecomes (n + 1)B.

Binary PCM – SQNR

Self Study

• Digital telephony

• T1 carriers

• Digital multiplexing

Self Study

• T1 carriers

Self Study

• T1 carriers

Self Study

• T1 carriers

Outline

1 Sampling Theorem

4 Delta Modulation

5 Summary

Why Differential Pulse Code Modulation (DPCM)?

PCM is not efficient as it requires bit-rate Rb = 2nB bits/sec (nB HzTransmission bandwidth).

There exist other methods that result in to less bit rates. They exploit theinherent characteristics of the underlying signal.

These techniques are also considered as compression techniques.

In analog messages, we can make a good guess about the sample value froma knowledge of the past sample values.

Proper exploitation of this redundancy leads to encoding a signal with alesser number of bits.

Instead of transmitting sample values, we can transmit the differencebetween the consecutive sample values.

Is there any other better estimation technique to reduce the amplitude of thetransmitted difference signal further?

m[k]m[k]>

We can reduce the amplitude of the difference signal by taking more numberof past samples into consideration.

m[k]m[k]>

Differential Estimation of m (t) and m [k]

For a given analog signal, using the Taylor series, we can predict (estimate)the future value as shown below:

m (t + ∆t) = m (t) + ∆tm (t) +(∆t)2

2!m (t) +

(∆t)3

3!...m (t) + · · · (15)

≈ m (t) + ∆tm (t) = m (t) (16)

A discrete version of the above equation is

m [k + 1] ≈ m [k] + Ts

[m [k]−m [k− 1]

]= 2m [k]−m [k− 1]. (17)

m [k] ≈ m [k] = 2m [k− 1]−m [k− 2]. (18)

m (t + ∆t) = m (t) + ∆tm (t) +(∆t)2

2!m (t) +

(∆t)3

3!...m (t) + · · · (15)

≈ m (t) + ∆tm (t) = m (t) (16)

m [k + 1] ≈ m [k] + Ts

[m [k]−m [k− 1]

]= 2m [k]−m [k− 1]. (17)

m [k] ≈ m [k] = 2m [k− 1]−m [k− 2]. (18)

m (t + ∆t) = m (t) + ∆tm (t) +(∆t)2

2!m (t) +

(∆t)3

3!...m (t) + · · · (15)

≈ m (t) + ∆tm (t) = m (t) (16)

m [k + 1] ≈ m [k] + Ts

[m [k]−m [k− 1]

]= 2m [k]−m [k− 1]. (17)

m [k] ≈ m [k] = 2m [k− 1]−m [k− 2]. (18)

m (t + ∆t) = m (t) + ∆tm (t) +(∆t)2

2!m (t) +

(∆t)3

3!...m (t) + · · · (15)

≈ m (t) + ∆tm (t) = m (t) (16)

m [k + 1] ≈ m [k] + Ts

[m [k]−m [k− 1]

]= 2m [k]−m [k− 1]. (17)

m [k] ≈ m [k] = 2m [k− 1]−m [k− 2]. (18)

The approximation becomes more accurate as we add more number of termsin the series (15). This requires more number of samples in the past.

So, a more general expression for estimation is

m [k] ≈ m [k] = a1m [k− 1] + a2m [k− 2] + · · ·+ aNm [k−N] . (19)

Transversal Filter used as a Line Predictor

DelayTs

Input m[k]

Output m[k]

DelayTs

aNaN-1a2

m [k] ≈ m [k] = a1m [k− 1] + a2m [k− 2] + · · ·+ aNm [k−N]

DelayTs

Input m[k]

Output m[k]

DelayTs

aNaN-1a2

DelayTs

Input m[k]

Output m[k]

DelayTs

aNaN-1a2

DPCM Transmitter and Receiver

m[k] d[k]

m[k] d[k] dq[k]

QuantizerTo channelΣ

m[k] d[k] dq[k]

QuantizerTo channel

m[k] d[k] dq[k]

Quantizer

Predictor

To channel

m[k] d[k] dq[k]

Quantizer

Predictor

To channel

m[k] d[k] dq[k]

Quantizer

Predictor

To channel

Input Output

dq[k] mq[k]Σ

mq[k]>

m[k] d[k] dq[k]

Quantizer

Predictor

To channel

Predictor

Input Output

dq[k] mq[k]Σ

mq[k]>

DPCM – SQNR

Compared to PCM, in DPCM, the quantization step size is reduced for agiven number of levels L which in turn reduces the quantization noise.

For the same number of bits/sample, the SQNR improves (over PCM) byabout 5.6 dB for a 2 step predictor.

For the same SQNR , we require 3-4 bits/ sample less than PCM

DPCM – SQNR

Adaptive DPCM (ADPCM)

In DPCM, the quantization step is fixed. However, the prediction error couldbe small or large, depending on the signal and the predictor accuracy.

So, in ADPCM, the quantization step is made adaptive, depending on theprediction error. Thus, compared to DPCM, ADPCM can further compress

the bit rates.

International Telecommunication Union, (ITU) specifies the standards andadapted ADPCM under G-726 Standard. G.726, for a 8 kHz sampled voice,

has bit rates as: 16, 24, 32 and 40 kbps.

This implies, 2, 3, 4 and 5 bits / sample. Standard PCM has 8 bits/ sample.

the bit rates.

Outline

1 Sampling Theorem

4 Delta Modulation

5 Summary

Delta Modulation (DM)

m[k] d[k] dq[k]

Quantizer

Predictor

To channel

Predictor

Input Output

dq[k] mq[k]Σ

mq[k]>

m[k] d[k] dq[k]

Quantizer

Predictor

To channel

Predictor

Input Output

dq[k] mq[k]Σ

mq[k]>

m[k] d[k] dq[k]

mq[k-1]

1-bitQuantizer

Delay Ts

To channel

Σ Input Output

dq[k] mq[k]Σ

mq[k-1]

Delay Ts

DPCM vs DM

Since we are using only first order predictor (delay Ts), there should be anincreased correlation between adjacent samples.

This can be achieved by oversampling (typically 4 times Nyquist rate) of thebaseband signal.

DPCM vs DM

m(t)mq(t)

Error d(t)

m(t)mq(t)

Error d(t)

m(t)mq(t)

Delta Modulation and Demodulation

m(t) d(t)Σ

Comparator

m(t) E

dq(t)d(t)Σ

Comparator

Samplerfrequency fs

m(t) E

dq(t)d(t)Σ

Comparator

Samplerfrequency fs

m(t) E

dq(t)d(t)

Integrator amplifier

Comparator

Samplerfrequency fs

m(t) E

dq(t)d(t)

Amplifier-Integrator

dq[k] m(t)~

Threshold and Overloading

StartupSlope

overload

m(t)mq(t)

Threshold effect

Slope overload noise can be decresed by increasing the step size. However,this increases the granular noise (which is similar to quantization noise).

No overload occurs if|m (t)| < σfs. (20)

So, for tone modulation (i. e., when m (t) = A cos ωt)

|m (t)| = ωA < σfs. (21)

StartupSlope

overload

m(t)mq(t)

Threshold effect

|m (t)| = ωA < σfs. (21)

StartupSlope

overload

m(t)mq(t)

Threshold effect

|m (t)| = ωA < σfs. (21)

StartupSlope

overload

m(t)mq(t)

Threshold effect

|m (t)| = ωA < σfs. (21)

StartupSlope

overload

m(t)mq(t)

Threshold effect

|m (t)| = ωA < σfs. (21)

Mitigating Slope Overload Effect (Double Integration)

Frequency, Hz

Relative amplitude, dB

100 200 500 1000 5000 10000

010203040

Single integrationin the

feedback loop

Frequency, Hz

100 200 500 1000 5000 10000

010203040

feedback loop

Frequency, Hz

100 200 500 1000 5000 10000

010203040

feedback loop

Frequency, Hz

100 200 500 1000 5000 10000

010203040

feedback loop

Double integrationin the

feedback loop

Frequency, Hz

100 200 500 1000 5000 10000

010203040

feedback loop

Voicespectrum

Double integrationin the

feedback loop

Mitigating Slope Overload Effect (ADM)

Slopeoverload

m(t)mq(t)

Threshold effect

Slopeoverload

m(t)mq(t)

>Threshold effect

m(t)mq(t)

Σ− ∆ Modulator

Comparator

Samplerfrequency fs

m(t) E

dq(t)d(t)

Σ− ∆ Modulator

Comparator

Samplerfrequency fs

m(t) E

dq(t)d(t)

Σ− ∆ Modulator

Comparator

Samplerfrequency fs

m(t) E

dq(t)d(t)

dq[k] m(t)~

Σ− ∆ Modulator

Comparator

Samplerfrequency fs

m(t) E

dq(t)d(t)

dq[k] m(t)~

Comparator

Samplerfrequency fs

m(t) E

Σ− ∆ Modulator

Comparator

Samplerfrequency fs

m(t) E

dq(t)d(t)

dq[k] m(t)~

Comparator

Samplerfrequency fs

m(t) E

Σ LPFm(t)~

Σ− ∆ Modulator

Comparator

Samplerfrequency fs

m(t) E

dq(t)d(t)

dq[k] m(t)~

Comparator

Samplerfrequency fs

m(t) E

Outline

1 Sampling Theorem

4 Delta Modulation

5 Summary

Summary

• ∆v = 2mp/L, where L = 2n

• q2 = (∆v)2

• SQNRuniform = m2

12m2(t)(∆v)2 = 3L2 m2(t)

(uniform quantization)

• SQNRµ ' 3L2

[ln(1+µ)]2, µ2 � m2

• Minimum transmission bandwidth required for PCM is Rb2

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