# Intro to Digital Signal Processing by Robert Nowak, et al - HTML preview

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Now we can plot this result (Figure 1.66):

Figure 1.66.

The sequence from 0 to 4 (the underlined part of the sequence) is the regular convolution result. From this illustration we can see that it is 5-periodic!

General Result

We can compute the regular convolution result of a convolution of an M-point signal x[ n] with an N-point signal h[ n] by padding each signal with zeros to obtain two M+ N−1 length sequences and computing the circular convolution (or equivalently computing the IDFT of

H[ k] X[ k] , the product of the DFTs of the zero-padded signals) (Figure 1.67).

Figure 1.67.

Note that the lower two images are simply the top images that have been zero-padded.

DSP System

Figure 1.68.

The system has a length N impulse response, h[ n]

1. Sample finite duration continuous-time input x( t) to get x[ n] where n={0, , M−1} .

2. Zero-pad x[ n] and h[ n] to length M+ N−1 .

3. Compute DFTs X[ k] and H[ k]

4. Compute IDFTs of X[ k] H[ k]

where n={0, , M+ N−1} .

5. Reconstruct y( t)

1.12. Ideal Reconstruction of Sampled Signals*

Reconstruction of Sampled Signals

How do we go from x[ n] to CT (Figure 1.69)?

Figure 1.69.

Step 1

Place x[ n] into CT on an impulse train s( t) (Figure 1.70).

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Figure 1.70.

Step 2

Pass xs( t) through an idea lowpass filter H LP( Ω) (Figure 1.71).

Figure 1.71.

If we had no aliasing then xr[ t]= xc( t) , where x[ n]= xc( nT) .

Ideal Reconstruction System

Figure 1.72.

In Frequency Domain:

1. Xs( Ω)= X( ΩT) where X( ΩT) is the DTFT of x[ n] at digital frequency ω= ΩT .

2. Xr( Ω)= H LP( Ω) Xs( Ω)

Result

Xr( Ω)= H LP( Ω) X( ΩT)

In Time Domain:

1.

2.

Result

Figure 1.73.

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h LP( t) "interpolates" the values of x[ n] to generate xr( t) (Figure 1.74).

Figure 1.74.

(1.5)

Sinc Interpolator

1.13. Amplitude Quantization*

The Sampling Theorem says that if we sample a bandlimited signal s( t) fast enough, it can be

recovered without error from its samples s( nTs) , n∈{ , -1, 0, 1, } . Sampling is only the first phase of acquiring data into a computer: Computational processing further requires that the

samples be quantized: analog values are converted into digital form. In short, we will have performed analog-to-digital (A/D) conversion.

(a)

(b)

Figure 1.75.

A three-bit A/D converter assigns voltage in the range [-1, 1] to one of eight integers between 0 and 7. For example, all inputs having values lying between 0.5 and 0.75 are assigned the integer value six and, upon conversion back to an analog value, they all become 0.625. The width of a single quantization interval Δ equals

. The bottom panel shows a signal going through the analog-to-digital,

where B is the number of bits used in the A/D conversion process (3 in the case depicted here). First it is sampled, then amplitude-quantized to three bits. Note how the sampled signal waveform becomes distorted after amplitude quantization. For example the two signal values between 0.5 and 0.75 become 0.625. This distortion is irreversible; it can be reduced (but not eliminated) by using more bits in the A/D converter.

A phenomenon reminiscent of the errors incurred in representing numbers on a computer prevents

signal amplitudes from being converted with no error into a binary number representation. In

analog-to-digital conversion, the signal is assumed to lie within a predefined range. Assuming we

can scale the signal without affecting the information it expresses, we'll define this range to be

[–1, 1] . Furthermore, the A/D converter assigns amplitude values in this range to a set of integers.

A B -bit converter produces one of the integers {0, 1, , 2 B−1} for each sampled input.

Figure 1.75 shows how a three-bit A/D converter assigns input values to the integers. We define a quantization interval to be the range of values assigned to the same integer. Thus, for our

example three-bit A/D converter, the quantization interval Δ is 0.25 ; in general, it is

.

Exercise 6.

Recalling the plot of average daily highs in this frequency domain problem, why is this plot so jagged? Interpret this effect in terms of analog-to-digital conversion.

The plotted temperatures were quantized to the nearest degree. Thus, the high temperature's

amplitude was quantized as a form of A/D conversion.

Because values lying anywhere within a quantization interval are assigned the same value for

computer processing, the original amplitude value cannot be recovered without error.

Typically, the D/A converter, the device that converts integers to amplitudes, assigns an amplitude

equal to the value lying halfway in the quantization interval. The integer 6 would be assigned to

the amplitude 0.625 in this scheme. The error introduced by converting a signal from analog to

digital form by sampling and amplitude quantization then back again would be half the

quantization interval for each amplitude value. Thus, the so-called A/D error equals half the width

of a quantization interval:

. As we have fixed the input-amplitude range, the more bits available

in the A/D converter, the smaller the quantization error.

To analyze the amplitude quantization error more deeply, we need to compute the signal-to-noise

ratio, which equals the ratio of the signal power and the quantization error power. Assuming the

signal is a sinusoid, the signal power is the square of the rms amplitude:

. The

illustration details a single quantization interval.

Figure 1.76.

A single quantization interval is shown, along with a typical signal's value before amplitude quantization s( nTs) and after Q( s( nTs)) .

ε denotes the error thus incurred.

Its width is Δ and the quantization error is denoted by ε. To find the power in the quantization

error, we note that no matter into which quantization interval the signal's value falls, the error will

have the same characteristics. To calculate the rms value, we must square the error and average it

over the interval.

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Since the quantization interval width for a B-bit converter equals

, we find that the

signal-to-noise ratio for the analog-to-digital conversion process equals

()

Thus, every bit increase in the A/D converter yields a 6 dB increase in the signal-to-noise ratio.

The constant term 10log101.5 equals 1.76.

Exercise 7.

This derivation assumed the signal's amplitude lay in the range [-1, 1] . What would the amplitude

quantization signal-to-noise ratio be if it lay in the range [– A, A] ?

The signal-to-noise ratio does not depend on the signal amplitude. With an A/D range of [– A, A] ,

the quantization interval

and the signal's rms value (again assuming it is a sinusoid) is .

Exercise 8.

How many bits would be required in the A/D converter to ensure that the maximum amplitude

quantization error was less than 60 db smaller than the signal's peak value?

Solving 2– B=.001 results in B=10 bits.

Exercise 9.

Music on a CD is stored to 16-bit accuracy. To what signal-to-noise ratio does this correspond?

A 16-bit A/D converter yields a SNR of 6×16+10log101.5=97.8 dB.

Once we have acquired signals with an A/D converter, we can process them using digital hardware

or software. It can be shown that if the computer processing is linear, the result of sampling,

computer processing, and unsampling is equivalent to some analog linear system. Why go to all

the bother if the same function can be accomplished using analog techniques? Knowing when

digital processing excels and when it does not is an important issue.

1.14. Classic Fourier Series*

The classic Fourier series as derived originally expressed a periodic signal (period T) in terms of

harmonically related sines and cosines.

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The complex Fourier series and the sine-cosine series are identical, each representing a signal's

spectrum. The Fourier coefficients, ak and bk , express the real and imaginary parts respectively of the spectrum while the coefficients ck of the complex Fourier series express the spectrum as a

magnitude and phase. Equating the classic Fourier series to the complex Fourier series, an extra factor of two and complex conjugate become necessary to relate the Fourier coefficients in each.

Exercise 10.

Derive this relationship between the coefficients of the two Fourier series.

Write the coefficients of the complex Fourier series in Cartesian form as ck= Ak+ ⅈBk and

substitute into the expression for the complex Fourier series.

Simplifying each term in the sum using Euler's formula,

We now combine terms that have the same

frequency index in magnitude. Because the signal is real-valued, the coefficients of the complex

Fourier series have conjugate symmetry:

or Ak= Ak and Bk=– Bk . After we add the

positive-indexed and negative-indexed terms, each term in the Fourier series becomes

. To obtain the classic Fourier series, we must have 2 Ak= ak and 2 Bk=– bk .

Just as with the complex Fourier series, we can find the Fourier coefficients using the

orthogonality properties of sinusoids. Note that the cosine and sine of harmonically related

frequencies, even the same frequency, are orthogonal.

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These

orthogonality relations follow from the following important trigonometric identities.

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These identities allow you to substitute a sum of sines and/or cosines for a product of them. Each

term in the sum can be integrating by noticing one of two important properties of sinusoids.

The integral of a sinusoid over an integer number of periods equals zero.

The integral of the square of a unit-amplitude sinusoid over a period T equals .

To use these, let's, for example, multiply the Fourier series for a signal by the cosine of the

l th harmonic

and integrate. The idea is that, because integration is linear, the integration

will sift out all but the term involving al .

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The first and third terms are zero; in the second, the only non-zero term in the sum results when

the indices k and l are equal (but not zero), in which case we obtain

. If k=0= l , we obtain a 0 T .

Consequently,

All of the Fourier coefficients can be found similarly.

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Exercise 11.

The expression for a 0 is referred to as the average value of s( t) . Why?

The average of a set of numbers is the sum divided by the number of terms. Viewing signal

integration as the limit of a Riemann sum, the integral corresponds to the average.

Exercise 12.

What is the Fourier series for a unit-amplitude square wave?

We found that the complex Fourier series coefficients are given by

. The coefficients are

pure imaginary, which means ak=0 . The coefficients of the sine terms are given by

bk=–(2Im( ck)) so that

Thus, the Fourier series for the square wave is

()

Example 1.13.

Let's find the Fourier series representation for the half-wave rectified sinusoid.

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Begin with the sine terms in the series; to find bk we must calculate the integral

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Using our trigonometric identities turns our integral of a product of sinusoids into a sum of

integrals of individual sinusoids, which are much easier to evaluate.

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Thus,

b 2= b 3= =0

On to the cosine terms. The average value, which corresponds to a 0 , equals . The remainder

of the cosine coefficients are easy to find, but yield the complicated result

()

Thus, the Fourier series for the half-wave rectified sinusoid has non-zero terms for the

average, the fundamental, and the even harmonics.

Glossary

Definition: FFT

(Fast Fourier Transform) An efficient computational algorithm for computing the

m10249m102493DFT.

Solutions

Chapter 2. Random Signals

2.1. Introduction to Random Signals and Processes*

Before now, you have probably dealt strictly with the theory behind signals and systems, as well

as look at some the basic characteristics of signals and systems. In doing so you have developed an important foundation; however, most electrical engineers do not get to work in this type of

fantasy world. In many cases the signals of interest are very complex due to the randomness of the

world around them, which leaves them noisy and often corrupted. This often causes the

information contained in the signal to be hidden and distorted. For this reason, it is important to

understand these random signals and how to recover the necessary information.

Signals: Deterministic vs. Stochastic

For this study of signals and systems, we will divide signals into two groups: those that have a

fixed behavior and those that change randomly. As most of you have probably already dealt with

the first type, we will focus on introducing you to random signals. Also, note that we will be

dealing strictly with discrete-time signals since they are the signals we deal with in DSP and most

real-world computations, but these same ideas apply to continuous-time signals.

Deterministic Signals

Most introductions to signals and systems deal strictly with deterministic signals. Each value of

these signals are fixed and can be determined by a mathematical expression, rule, or table.

Because of this, future values of any deterministic signal can be calculated from past values. For

this reason, these signals are relatively easy to analyze as they do not change, and we can make

accurate assumptions about their past and future behavior.

Figure 2.1. Deterministic Signal

An example of a deterministic signal, the sine wave.

Stochastic Signals

Unlike deterministic signals, stochastic signals, or random signals, are not so nice. Random

signals cannot be characterized by a simple, well-defined mathematical equation and their future

values cannot be predicted. Rather, we must use probability and statistics to analyze their

behavior. Also, because of their randomness, average values from a collection of signals are usually studied rather than analyzing one individual signal.

Figure 2.2. Random Signal

We have taken the above sine wave and added random noise to it to come up with a noisy, or random, signal. These are the types of signals that we wish to learn how to deal with so that we can recover the original sine wave.

Random Process

As mentioned above, in order to study random signals, we want to look at a collection of these

signals rather than just one instance of that signal. This collection of signals is called a random

process