Patrick F. Dunn, Measurement, Data Analysis, and Sensor Fundamentals for Engineering and Science, 2nd Edition

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326 Measurement and Data Analysis for Engineering and Science

The numerator in Equation 8.58 can be expanded to yield

(τ) =

E[x(t) · x(t + τ)] − x

E[x(t + τ)] − x

E[x(t)] + x

. (8.59)

Substitution of Equations 8.55 and 8.56 into Equation 8.59 results in an

expression that relates the autocorrelation function to its coeﬃcient

(τ) =

(τ) − x

(8.60)

(τ) = ρ

(τ)σ

+ x

. (8.61)

Some limits can be placed on the value of R

(τ). Because −1 ≤ ρ

(τ) ≤

1, R

(τ) is bounded as

−σ

+ x

≤ R

(τ) ≤ σ

+ x

. (8.62)

Now it can be shown (see Chapter 5) that

E[x

] = σ

+ x

(8.63)

by expanding E[(x − x

)

]. So, the maximum value that R

(τ) can have is

E[x

]. It follows from Equation 8.55 that

(0) = E[x

]. (8.64)

That is, the maximum value of R

(τ) occurs at τ = 0. Using Equation 8.63

and the deﬁnition of the autocorrelation coeﬃcient (Equation 8.60),

(0) = 1. (8.65)

Further, as τ → ∞, there is no correlation between x(t) and x(t+τ) because

x(t) is the signal of a random variable. That is,

(τ → ∞) = 0, (8.66)

which implies that

(τ → ∞) = x

. (8.67)

Finally, R

(τ) is an even function because

(−τ) = E[x(t)x(t − τ)] = E[x(t − τ)x(t)] = E[x(t

∗

)x(t

∗

+ τ)] = R

(τ),

(8.68)

where t

∗

= t − τ , noting x(t) is stationary. So, R

(τ) is symmetric about

the τ = 0 axis.

A generic autocorrelation function and its corresponding autocorrela-

tion coeﬃcient having these properties is displayed in Figure 8.16. Values

of R

(τ) and ρ

(τ) that are greater than their respective limiting values

Regression and Correlation 327

FIGURE 8.16

Typical autocorrelation function and coeﬃcient.

as τ → ∞ indicate a positive correlation at that particular value of τ. Con-

versely, a negative correlation is indicated for values less than that limiting

value. Values equal to the limiting value signify no correlation. Note that

(τ) experiences decreasing oscillations about a value of 0 as τ → ∞.

This always will be the case for a stationary signal provided there are no

deterministic components in the signal other than a nonzero mean.

Example Problem 8.6

Statement: Determine the autocorrelation coeﬃcient for the signal x(t) = A sin(ωt).

Then plot the coeﬃcient for values of ω = 1 rad/s and A = 1.

Solution: The autocorrelation function, only for the range 0 ≥ τ ≥ T /2π, needs to

be examined, because x(t) in this example is a periodic function of period T = 2π/ω.

Equation 8.55 for this periodic function becomes

(τ) =

2π/ω

sin(ωt) sin(ωt + φ)dt,

where φ = ωτ . Performing and evaluating the integral,

(τ) =

[sin

(ωt) cos(φ) + sin(ωt) cos(ωt) sin(φ)]

2π/ω

cos(ωτ).

Now σ

A/2, so, according to Equation 8.58,

= cos(ωτ ).

The plot of ρ

(τ) for values of ω = 1 rad/s and A = 1 is presented in Figure 8.17.

It shows that the sine function has a positive, perfect autocorrelation at values of τ =

328 Measurement and Data Analysis for Engineering and Science

FIGURE 8.17

Autocorrelation of a sine function.

T = 2π when ω = 1 and a negative, perfect autocorrelation at values of τ = T /2 = π.

Further, when τ = T/4 = π/2 or τ = 3T/4 = 3π/2 there is no correlation of the sine

function with itself.

8.10.2 *Cross-Correlation

Expressions for the cross-correlation function and coeﬃcient can be devel-

oped in the same manner as that done for the case of autocorrelation.

For the stationary signals x(t) and y(t) there are two cross-correlation

functions deﬁned as

(τ) ≡ E[x(t) · y(t + τ)] (8.69)

and

(τ) ≡ E[y(t) · x(t + τ)]. (8.70)

(τ) denotes the cross-correlation of x with y and R

(τ) that of y with

x. Further, because the signals are stationary

(τ) = E[x(t − τ )y(t)] = R

(−τ) (8.71)

and

(τ) = E[y(t − τ )x(t)] = R

(−τ). (8.72)

Regression and Correlation 329

FIGURE 8.18

Typical cross-correlation of two signals.

So, in general, R

(τ) 6= R

(τ) and both are not even with respect to τ.

The corresponding two cross-correlation coeﬃcients are deﬁned as

(τ) ≡

(τ) − x

(8.73)

and

(τ) ≡

(τ) − x

, (8.74)

where both coeﬃcients are bounded between values of −1 and 1. Thus, both

functions are bounded between values of −σ

+ x

and σ

+ x

Finally, as τ → ∞ both functions tend to the value of x

because no

correlation between the random signals x(t) and y(t) would be expected at

that limit.

Typically, two signals will experience a maximum cross-correlation at

some value of τ = τ

, which corresponds to a phase lag between the two

signals, where φ = ωτ

. This is shown for a typical cross-correlation in

Figure 8.18.

Example Problem 8.7

Statement: Determine the cross-correlation coeﬃcient ρ

(τ) for the the signals

x(t) = A sin(ωt) and y(t) = B cos(ωt). Then plot the coeﬃcient for the value of ω = 1

rad/s.

330 Measurement and Data Analysis for Engineering and Science

FIGURE 8.19

Cross-correlation of sine and cosine functions.

Solution: The cross-correlation function only for the range 0 ≤ τ ≤ T /2π needs

to be examined because x(t) and y(t) are periodic functions of period T = 2π/ω.

Equation 8.69 for these periodic functions becomes

(τ) =

2π/ω

sin(ωt) cos(ωt + φ)dt,

where φ = ωτ . Performing and evaluating the integral,

(τ) =

[−

ω cos(2ωt + φ) −

t sin(φ)]

2π/ω

−AB

sin(φ).

Now σ

= A/

√

2 and σ

= B/

√

2, so, according to Equation 8.73,

(τ) = −sin(ωτ).

Note the minus sign in this expression. The plot of ρ

(τ) for the value of ω = 1 rad/s

is given in Figure 8.19. For a delay time value of τ = π/2, ρ

(τ) = 1. This is because

the value of cos(t + π/2) exactly equals that of sin(t). Similar reasoning can be used to

explain the value of ρ

(τ) = −1 when τ = 3π/2, where the cosine and sine values are

equal but opposite in sign.

The M-ﬁle sigcor.m determines and plots the autocorrelations and

cross-correlation of discrete data that is user-speciﬁed. This M-ﬁle normal-

izes the correlations such that the autocorrelations at zero time lag are

identically 1.0. An example plot generated using sigcor.m for a ﬁle con-

taining eight sequential measurements of x and y data is shown in Figure

8.20. Note that both autocorrelations have a value of 1.0 at zero time lag.

Regression and Correlation 331

FIGURE 8.20

Autocorrelations and cross-correlation of discrete data.

8.11 *Higher-Order Analysis

Higher-order (m > 2) regression analysis can be performed in a manner

similar to that developed for linear least-squares regression analysis. This

will result in m + 1 algebraic normal equations. These can be solved most

easily using methods of linear algebra to obtain the expressions for the m+1

regression coeﬃcients.

For higher-order regression analysis, the coeﬃcients a

through a

the expression

+ a

+ ... + a

= y

(8.75)

are found using the method of minimizing D as described in the previous

section. The resulting equations are

332 Measurement and Data Analysis for Engineering and Science

N + a

i=1

+ a

i=1

+ ··· +a

i=1

+ a

i=1

+ a

i=1

+ ··· +a

i=1

m+1

i=1

+ a

i=1

+ a

i=1

+ ··· +a

i=1

m+2

i=1

. , and (8.76)

i=1

+ a

i=1

m+1

+ a

i=1

m+2

+ ··· +a

i=1

In expanded matrix notation, the set of Equations 8.76 becomes







···

m+1

···

m+2

m+1

m+2

···































where the summations are from i = 1 to N. Or, in matrix notation, this

becomes

[E][a] = [F ]. (8.77)

[E], [a], and [F ] represent the matrices shown in the expanded form. The

solution to Equation 8.77 for the regression coeﬃcients is

[a] = [E]

−1

[F ]. (8.78)

[E]

−1

is the inverse of the coeﬃcient matrix.

Example Problem 8.8

Statement: An experiment is conducted in which a ball is dropped with an initial

downward velocity, v

, from a tall tower. The distance fallen, y(m), is recorded at

time, t, intervals of one second for a period of six seconds. The resulting time, distance

data are 0, 0; 1, 7; 2, 21; 3, 48; 4, 81; 5, 131; and 6, 185. The distance equation is

y = y

+ v

t + 0.5gt

, where g is the local gravitational acceleration. This equation

is of the form y = a

+ a

t + a

. Using higher-order regression analysis, determine

the values of the regression coeﬃcients a

, a

, and a

. From their values determine the

values of v

and g. Finally, plot the data with the regression ﬁt.

Regression and Correlation 333

FIGURE 8.21

Higher-order regression ﬁt example.

Solution: The solution is obtained using MATLAB’s left-division method by typing

t\y, where [t] is







1 0 0

1 1 1

1 2 4

1 3 9

1 4 16

1 5 25

1 6 36







and [y] is







131

185







The resulting regression coeﬃcient matrix is





0.5238

0.3214

5.0833





Thus, the best-ﬁt expression is

0.5238 + 0.3214t + 5.0833t

= y.

The data and Equation 8.11 are shown in Figure 8.21. Also, v

= 0.3214 and g =

10.1666.

334 Measurement and Data Analysis for Engineering and Science

8.12 *Multi-Variable Linear Analysis

Linear least-squares regression analysis can be extended to situations in-

volving more than one independent variable. This is known as multi-variable

linear regression analysis and results in m+1 algebraic equations with m+1

regression coeﬃcient unknowns. This system of equations can be solved us-

ing methods of linear algebra.

Multi-variable linear regression analysis for a system of three indepen-

dent variables [15] with the regression coeﬃcients a

through a

in the ex-

pression

+ a

= R

(8.79)

yields a system of four equations, which is

N + a

i=1

+ a

i=1

+ a

i=1

+ a

i=1

+ a

i=1

+ a

i=1

+ a

i=1

+ a

i=1

+ a

i=1

, and

i=1

+ a

i=1

+ a

i=1

+ a

i=1

. (8.80)

In expanded matrix notation, the set of Equations 8.80 becomes





































where the summations are from i = 1 to N. Or, in matrix notation, this

becomes

[G][a] = [R]. (8.81)

[G], [a], and [R] represent the matrices shown in the expanded form. The

solution to Equation 8.81 for the regression coeﬃcients is

[a] = [G]

−1

[R]. (8.82)

Regression and Correlation 335

[G]

−1

is the inverse of the coeﬃcient matrix.

Example Problem 8.9

Statement: An experiment is conducted in which the values of three independent

variables, x, y, and z, are selected and then the resulting value of the dependent variable

R is measured. This procedure is repeated six times for diﬀerent combinations of x, y,

and z values. The [x,y,z,R] data values are [1,3,1,17; 2,4,2,24; 3,5,1,25; 4,4,2,30; 5,3,1,24;

6,3,2,31]. Determine the regression coeﬃcients for the multi-variable regression ﬁt of

the data. Then, using the resulting best-ﬁt expression, determine the calculated values

of R in comparison to their respective measured values.

Solution: The solution is obtained using MATLAB’s left-division method by typing

G\R, where [G] for this example is







1 1 3 1

1 2 4 2

1 3 5 1

1 4 4 2

1 5 3 1

1 6 3 2







and [R] is













The resulting regression coeﬃcient matrix is







3.4865

2.0270

2.2162

4.3063







Thus, the best-ﬁt expression is

3.4865 + 2.0270x + 2.2162y + 4.3063z = R. (8.83)

The calculated values of R are obtained by typing G*ans after the regression coeﬃcient

solution is obtained. The values are 16.4685, 25.0180, 24.9550, 29.0721, 24.5766, and

30.9099. All calculated values agree with their respective values to within a diﬀerence

of less than 1.0.