Craig Chapt7

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7

Response of SDOF Systems

to Periodic Excitation:

Frequency-Domain Analysis

In Chapter 4 you studied the response of SDOF systems to harmonic excitation and

became familiar with important concepts such as resonance. You also learned how to

simplify the analysis of viscous-damped systems by using complex frequency response.

These concepts from Chapter 4 are now extended to determine the response of SDOF

systems to periodic excitation.

Upon completion of this chapter you should be able to:

Determine the Fourier series representation of a periodic function using the real

form of the Fourier series.

Determine the Fourier series representation of a periodic function using the com-

plex form of the Fourier series.

Determine the steady-state response of an SDOF system to periodic excitation

using either the real form or the complex form.

Use the basic definition to determine the Fourier transform of a transient functionp(t).

Show that the impulse-response function for a linear system and the frequency-

response function for the system form a Fourier transform pair.

Apply the FFT algorithm to compute the Fourier transform of a periodic functionp(t), and plot the magnitude and phase of the Fourier transform computed.

7.1 RESPONSE TO PERIODIC EXCITATION: REAL

FOURIER SERIES

Forces acting on structures are frequently periodic, or can be approximated closely by

periodic forces. For example, the forces exerted on an automobile traveling at constant

speed over certain roadway surfaces can be considered to be periodic. Figure 7.1 showsa periodic function with period T1, that is,

p(t+ T1) = p(t) (7.1)

A periodic function can be separated into its harmonic components by means of a

Fourier series expansion. In this section we consider real Fourier series. In Section 7.2

184


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7.1 Response to Periodic Excitation: Real Fourier Series 185

p(t)

T1

t

Figure 7.1 Periodic function with period T1.

complex Fourier series are introduced. The complex form is very useful when combinedwith the complex frequency-response function of Chapter 4 to study the steady-state

response of damped systems.

7.1.1 Real Fourier Series

The periodic function p(t) may be separated into its harmonic components by means of

a Fourier series expansion. The real Fourier series expansion ofp(t) may be defined as

p(t) = a0 +

n=1

an cos n1t+

n=1

bn sin n1t (7.2)

where

1 =2

T1(7.3)

is the fundamental frequency (in rad/s), and an and bn are the coefficients of the nth

harmonic. The coefficients a0, an, and bn are related to p(t) by the equations

a0 =1

T1

+T1

p(t)dt= average value of p(t)

an =2

T1

+T1

p(t) cos n1t dt, n = 1, 2, . . .

bn = 2T1

+T1

p(t) sin n1t dt, n = 1, 2, . . .

(7.4)

where is an arbitrary time.

Although theoretically, a Fourier series representation of p(t) may require an infi-

nite number of terms, in actual practice p(t) can generally be approximated with


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186 Response of SDOF Systems to Periodic Excitation: Frequency-Domain Analysis

sufficient accuracy by a relatively small number of terms. Example 7.1 illustrates the

Fourier series representation of a square wave.

Example 7.1 (a) Determine expressions for the coefficients of a real Fourier series

representation of the square wave shown in Fig. 1. Write the Fourier series representa-

tion ofp(t). (b) Plot truncated series employing, respectively one, two, and three terms

of the Fourier series.

p(t)

p0

p0

T1

t

Figure 1 Square wave.

SOLUTION (a) The integrals in Eqs. 7.4 can be evaluated over the period T1/2 T

(a) Determine the Fourier transform of this rectangular pulse. Express the transform as

a function of the frequency variable . (b) Plot the Fourier transform.

SOLUTION (a) From Eq. 7.21,

P () = p(t)e

it

dt =TT p0e

it

dt (1)

Therefore,

P () =p0

i (eiT eiT) (2)

which can also be written in terms of the sinc function (sin )/.

P () = 2 p0TsinT

TAns. (a) (3)

(b) The Fourier transform P () is therefore a real function. It can be plotted

versus the frequency variable (Fig. 1) and compared with the discrete Fourier series

of Example 7.4.

2p0T

0

4p/T 2p/T 2p/T 4p/T

P()

Figure 1 Fourier transform of a symmetric rectangular pulse.

Reference [7.1] contains a table of a number of Fourier transform pairs.


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7.3.2 Frequency-Response Functions

In Eq. 7.14 we found that the response of an SDOF system to periodic excitation can

be expressed in the form

u(t) =

n=

Unei(n1t) (7.14)

where, from Eq. 7.15,Un = HnPn (7.15)

Following the procedure of Eqs. 7.17 through 7.24, we obtain the following Fourier

transform pair for the response of an SDOF system:

U( f ) =

u(t)ei(2f t) dt (7.25)

u(t) =

U(f)ei(2f t) df(7.26)

where

U( f ) = H( f ) P ( f ) (7.27)

which is the product of the system frequency-response function, H ( f ), and the Fourier

transform of the excitation, P ( f ). Therefore, the response can be expressed by the

following inverse Fourier transform:

u(t) =

H(f)P(f)ei(2f t) df (7.28)

In some cases a table of Fourier transform pairs can be used to evaluate this inverse

transform.[7.1,7.2] However, evaluation of this definite integral generally involves contour

integration in the complex plane, which is beyond the scope of this book.

7.3.3 Parameter Identification

Of great importance (as demonstrated in Chapter 18) is the fact that Eq. 7.27 can be

written symbolically in the form

H ( f ) =U ( f )

P ( f )(7.29)

The system frequency-response function, H ( f ), can thus be obtained from Fourier

transforms of the measured time histories of excitation p(t) and response u(t). Important

system parameters (e.g., the undamped natural frequency n and damping factor of

the system) can then be extracted from this system frequency-response function. For

example, for a viscous-damped SDOF system, H ( f ) is given by (Eq. 7.13)

H ( f ) =1/k

[1 (f/fn)2] + i(2f/fn)(7.30)

Such parameter identification procedures are discussed in Chapter 18 and in modal

analysis references such as Refs. [7.4] and [7.5].


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7.4 Relationship Between Complex Frequency Response and Unit Impulse Response 199

In Section 7.5 we describe how the Fourier integral can be approximated by the

discrete Fourier transform (DFT), which, in turn, can be evaluated numerically by use

of the fast Fourier transform (FFT) algorithm.

7.4 RELATIONSHIP BETWEEN COMPLEX FREQUENCY

RESPONSE AND UNIT IMPULSE RESPONSE

The complex frequency response, H () or H ( f ), of a linear system describes its

response characteristics in the frequency domain, and the unit impulse response h(t)

describes the systems response in the time domain. We now show that the unit impulse

function and the system frequency-response function form a Fourier transform pair.

From Eq. 7.23, the Fourier transform of the unit impulse excitation is

P ( f ) =

p(t)ei(2f t) dt = 1 (7.31)

Therefore, the response to this unit impulse is, from Eq. 7.28,

h(t) =

H(f)ei(2f t) df (7.32)

But from Eq. 7.24, the equation that defines the inverse Fourier transform, this unit

impulse response function is just the expression for the inverse Fourier transform of

the system frequency-response function H ( f ). Conversely, it follows that the system

frequency-response function H ( f ) is the Fourier transform of the unit impulse response

function h(t), that is,

H ( f ) F[h(t)] =

h(t)ei(2f t) dt (7.33)

This Fourier transform pair relationship between a systems impulse-response function

in the time domain and its frequency-response function in the frequency domain is

illustrated in Fig. 7.4. For example, for a viscous-damped SDOF system, H ( f ) is

given by Eq. 7.30 and h(t) by Eq. 5.30, both repeated here.

H ( f ) =1/k

[1 (f/fn)2] + i(2f/fn)(7.30)

h(t) =1

mdent sindt, t > 0 (5.30)

Time Domain

h(t)

Impulse Response

F F1

Frequency DomainH ( f )

Frequency -Response Function

Figure 7.4 Relationship between impulse response and frequency response.


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It is left as an exercise for the reader to show that these SDOF system functions satisfy

Eq. 7.33.

In Chapter 18 it will be noted that some parameter-estimation algorithms work

directly with the frequency-response function H ( f ); these are called frequency-domainalgorithms. Other parameter-estimation algorithms make use of the impulse-response

function h(t); these are called time-domain algorithms.

7.5 DISCRETE FOURIER TRANSFORM AND FAST

FOURIER TRANSFORM

Although the Fourier integral techniques discussed in Section 7.3 provide a means for

determining the transient response of a system, numerical implementation of the Fourier

integral became a practical reality only with the publication of the Cooley Tukey

algorithm for the fast Fourier transform in 1965.[7.3] Since that date, the FFT has virtually

led to a revolution in many areas of technology, including the area of vibration testing

(e.g., see Chapter 18 and Refs. [7.4] and [7.5]).Two steps are involved in the numerical evaluation of Fourier transforms.[7.1] First,

discrete Fourier transforms (DFTs), which correspond to Eqs. 7.23 and 7.24, are derived.

Then an efficient numerical algorithm, the fast Fourier transform (FFT), is used to the

compute the DFTs.

7.5.1 Discrete Fourier Transform Pair

For numerical treatment of the Fourier transform, it is necessary first to define a dis-

crete Fourier transform pair corresponding to the continuous Fourier transform pair of

Eqs. 7.23 and 7.24. To be transformed, a continuous function must first be sampled at

discrete time intervals t. Second, due to computer memory and execution-time lim-

itations, only a finite number N of these sampled values can be utilized. Figure 7.5a

illustrates a sampled waveform with N= 16 samples taken at t= 0.25-sec intervals.

1 0 1 2 3 4 5

0.5

0

0.5

1

1.5

5

0

5

t(sec)

(a)

n

(b)

p(t)

0 4 8 12 16

Mag(P)

Figure 7.5 Sampled waveform (et with t= 0.25 sec, N= 16): (a) discrete-time representation; (b) discrete-frequency representation.


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7.5 Discrete Fourier Transform and Fast Fourier Transform 201

The effects of this sampling and truncation are to approximate the continuous sig-

nal by a periodic signal of period T1 = Nt, sampled at times tm = mt, m =

0, 1, . . . , ( N 1).The total sample time is T1, so the fundamental-frequency sinusoid that fits within

this sample time has a period T1. Therefore, the frequency interval of the discrete Fourier

transform is

f =1

T1=

1

Nt(7.34)

The DFT consists of N samples extending up to a maximum frequency of Nf.

Specific features of sampling and truncation that are illustrated in Fig. 7.5 are dis-

cussed following Example 7.7. Sampling and truncation are discussed in greater detail

in Section 18.4.

Since a period T1 consists of N samples, the integral of Eq. 7.23 is replaced by the

following finite sum:

P (fn) =N1m=0

p(tm)ei2(mt)(nf ) t (7.35)

Finally, the discrete Fourier transform (DFT) can be written

P (fn) = t

N1m=0

p(tm)ei(2m(n/N)), n = 0, 1, . . . , N 1 (7.36)

The inverse DFT can be obtained from Eq. 7.24 in a similar manner. Thus,

p(tm) =

N1

n=0 P (fn)ei2(mt)(nf )f (7.37)

This expression for the inverse Fourier transform (IFT) can be written as

p(tm) =1

Nt

N1n=0

P (fn)ei(2m(n/N)), m = 0, 1, . . . , N 1 (7.38)

Equations 7.36 and 7.38 define a discrete Fourier transform pair that is consistent with

the continuous Fourier transform.

The discrete Fourier transform approximates the continuous Fourier transform at

discrete frequencies fn. The accuracy of a DFT representation depends on the sampling

interval t and the number of samples, or block size, N. In Section 18.4 we discuss

these effects in much greater detail.

Reference [7.1] presents a graphical derivation and a theoretical derivation of theDFT pair. The resulting discrete Fourier transform (DFT), written in present notation, is

P (fn) =

N1m=0

p(tm)ei(2m(n/N)), n = 0, 1, . . . , N 1 (7.39)


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where fn = nf. The corresponding inverse DFT is

p(tm

) =1

N

N1

n=0

P (fn

)ei(2m(n/N)), m = 0, 1, . . . , N

1 (7.40)

This form is the one that is utilized in computing FFTs.[7.6] However, a scale factor

t is required to produce an equivalence between this DFT form and the continuous

Fourier transform.

7.5.2 Fast Fourier Transform Algorithm

The fast Fourier transform (FFT) is not a new type of transform but rather, an efficient

numerical algorithm for evaluating the DFT. Its importance lies in the fact that by

eliminating most of the repetition in the calculation of a DFT, it permits much more

rapid computation of the DFT.

Either Eq. 7.39 or 7.40 can be cast in the form

Am =

N1n=0

BnWmn

N , m = 0, 1, . . . , N 1 (7.41)

where

WN = ei(2/N) (7.42)

A measure of the amount of computation involved in Eq. 7.41 is the number of complex

products implied by the form of the equation and the range of m. It is clear that there are

N sums, each of which requires N complex products, or there are N2 products required

for computing all of the Ams. By taking advantage of the cyclical nature of powers

of WN, the total computational effort can be drastically reduced. Figure 7.6 shows the

repetition cycle for W

mn

8 . The number of complex products for the FFT algorithm isgiven by (N/2) log2 N. For example, if N= 512, the number of FFT operations is less

than 1% of the corresponding number of DFT operations. Signal-processing software

products invariably provide for computation of the FFT.[7.6]

W68 =W814=

W78 =W815=

W28 =W810=

W38 =W811=

W48 =W812=

W58 =W813=

W08 =W88=

W18 =W89=

Figure 7.6 Cyclical nature of WmnN for N = 8.


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7.5 Discrete Fourier Transform and Fast Fourier Transform 203

7.5.3 FFT Computations

In Example 7.7 the FFT command in the Matlab computer program is used to com-

pute the Fourier transform of a square wave, similar to the one treated analytically in

Example 7.4 in Section 7.2. Computation of FFTs of nonperiodic signals is discussed

in Section 18.4.

Example 7.7 Let p(t) be the square wave defined over one period by the function

p(t) =

2

0

0.0 t < 0.5 sec0.5 sec t < 1.0 sec

represented by 16 samples over the finite interval 0 sec t < 1 sec. Use the FFTcommand in Matlab to determine the Fourier transform of this square wave, and plot

the real and imaginary parts of the resulting Fourier transform.

SOLUTION Relative to an offset average value of 1.0, the square wave is antisym-metric in the time window 0.5 sec t 0.5 sec. The 16 discrete-time sample pointsare shown in Fig. 1a. These 16 data points constitute the input to the FFT algorithm.

(See the Comments following this example.)

0 0.5 1

0

1

2

t(sec)

(a)

(b)

p(t)

0 2 4 6 80.5

0

0.5

1

1.5

f(Hz)

(P)

0 2 4 6 81

0.5

0

0.5

1

f(Hz)

(P)

Figure 1 (a) Time- and (b) frequency-domain representations of a square wave.


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In Fig. 1b the results of a 16-point FFT are plotted versus the discrete frequency

fn for 0 fn fN/2. The real part (top right) and the imaginary part(lower right) canbe compared with the plots in Fig. 1 of Example 7.4, noting that the square wave in

Example 7.4 is an antisymmetric function with an average value of zero, whereas thesquare wave in this example is an antisymmetric function relative to an average value

of 1.0. Other differences are explained in the Comments that follow.

Comments Regarding Matlab FFT Input/Output

1. The FFT algorithm is executed with the command

y = fft(x,N)

where N is the block size and x is the input sequence, that is, the vector of

discrete-time samples of the input function. For structural dynamics applications,

x will be a vector of real numbers. It is desirable to let N be a power of 2.

2. The FFT treats the first N numbers in the input sequence as one period of aperiodic function. For example, the rectangular pulse in Fig. 1a of Example 7.7

will correspond to a square wave that is represented by a Fourier series whose

fundamental frequency is 1.0 sec, with the sampled value at t= 1 sec repeat-

ing the sampled value at t= 0 sec. This topic is treated in greater detail in

Section 18.4.

3. Where there is a step discontinuity in the input function, the sampled value is

taken as the average of the value prior to the jump and the value after the jump,

as shown in Fig. 1a of Example 7.7 and in Fig. 7.5a.

4. Because P (fn) and P (fn) are complex conjugates of each other, the secondhalf of the Fourier coefficients are usually not plotted (e.g., as in Fig. 1b of

Example 7.7). Thus, the complex values in the output vector are sequenced by

Matlab as follows:

y(1) = P (f0) = average value of input (real)

y(2) = P (f+1)

...

y(N/2 + 1) = P (f+N/2)

y(N/2 + 2) = P (fN/2+1)

...

y(N) = P (f1)

5. To treat the FFT output values as approximations to the complex Fourier coef-

ficients given by Eq. 7.8, it is necessary to divide the output values y(n) by the

block size N.5

5This scaling is necessitated by the scaling factor T1 introduced in Eq. 7.18, together with the additionalscaling factor oft between Eqs. 7.36 and 7.39.


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Problems 205

6. To convert the input sequence number m to input sample time tm in seconds,

it is necessary to use tm = mt. Similarly, to convert the frequency scale of

the output from coefficient number n to frequency fn in hertz, one must use

fn = nf = n/Nt.

REFERENCES

[7.1] E. O. Brigham, The Fast Fourier Transform, Prentice-Hall, Englewood Cliffs, NJ, 1974.

[7.2] R. A. Gabel and R. A. Roberts, Signals and Systems, 3rd ed., Wiley, New York, 1987.

[7.3] J. W. Cooley and J. W. Tukey, An Algorithm for Machine Calculation of Complex

Fourier Series, Math Computation, Vol. 19, 1965, pp. 297301.

[7.4] D. J. Ewins, Modal Testing: Theory, Practice and Application, 2nd ed., Research Studies

Press, Baldock, Hertfordshire, England, 2000.

[7.5] N. M. M. Maia and J. M. M. Silva, ed., Theoretical and Experimental Modal Analysis,

Wiley, New York, 1997.

[7.6] Using MATLAB, Version 6, The MathWorks, Natick, MA, 2002, pp. 12-41 to 12-48.

PROBLEMS

For problems whose number is preceded by a C,

you are to write a computer program and use it to

produce the plot(s) requested. Note: Matlab .m-files

for many of the plots in this book may be found on

the books website.

Problem Set 7.1

7.1 Determine the real Fourier series for the periodic

excitation function in Fig. P7.1.

T1/4

p0

p0

p(t)

t

3T1/4

Figure P7.1

7.2 Determine the real Fourier series for the periodic

excitation function in Fig. P7.2.

p(t)

Half-sine wavesp0

T1/2 T1

t

Figure P7.2

C 7.3 (a) Determine the real Fourier series for the

square wave shown in Fig. P7.3. (b) By comparing

your result from part (a) with the answer in Eq. 10 of

Example 7.1, what do you observe to be the effect(s) of

the phase shift of the square wave? (c) Run the Mat-

lab .m-file sd2hw7 3.m, and print out the plots pro-

duced. Modify this .m-file to produce the input signal of

Example 7.1, and print out the plots produced.

p(t)

p0

T1/4 5T1/4 t3T1/4

p0

Figure P7.3


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7.4 (a) Determine the real Fourier series for the periodic

excitation function in Fig. P7.4. (b) By comparing your

result from part (a) with the answer in Eq. 10 of Example

7.1, what do you observe to be the effect(s) of the upward

shift of the square wave?

2p0

p(t)

T0/2 T0/2

t

Figure P7.4

7.5 The undamped SDOF system shown in Fig. P7.5

is subjected to the excitation p(t) given in Problem 7.1.

(a) Determine a Fourier series expression for the steady-

state response of the system ifn = 41. (b) Sketch the

spectra for Pn and for Un, similar to those shown in

Fig. 7.2.

km

p(t)

Figure P7.5

7.6 The undamped SDOF system shown in Fig. P7.5

is subjected to the excitation p(t) given in Fig. P7.4.

(a) Determine a Fourier series expression for the steady-

state response of the system if n = 61. (b) Sketch

the spectra for Pn and for Un similar to those shown

in Fig. 7.2.

Problem Set 7.27.7 (a) Determine the coefficients of the complex

Fourier series that corresponds to the excitation p(t)

given in Fig. P7.1. (b) Sketch the real and imaginary

parts and the magnitude of the complex coefficients as

in Example 7.4.

7.8 (a) Determine the coefficients of the complex


in Fig. P7.3. (b) Sketch the real and imaginary parts

and the magnitude of the complex coefficients as in

Example 7.4.

7.9 (1) Determine the coefficients of the complex


in Fig. P7.4. (2) Sketch the real and imaginary parts

and the magnitude of the complex coefficients as in

Example 7.4.

7.10 The undamped SDOF system of Fig. P7.10a is

subjected to the sawtooth base motion z(t) illustrated

in Fig. 7.10b. Let

z(t) =

n=Zne

i(n1t) , w(t) =

n=Wne

i(n1t)

where Wn = HnZn. (a) Determine Hn. (b) Determine

Zn. (These will have the same form as Pn of Problem

7.1.) (c) Determine Wn, and sketch |Wn|/Z versus

frequency order, as in Example 7.4.

z(t)u(t)

m w= u zk

z(t)

(a)

(b)

Z

T1Z

t

Figure P7.10

Problem Set 7.3

7.11 Determine the continuous Fourier transform for the

rectangular pulse in Fig. P7.11. (Note that this involves

a time shifting of the pulse in Example 7.6.)


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Problems 207

p(t)

p0

t

2T

Figure P7.11

7.12 Determine the real and imaginary parts of the

continuous Fourier transform of each of the following

functions:

(a) p(t) = p0e|t| Note: Absolute value of time t.

(b) p(t) =

0, t < 0

p0

sin(2f0t), 0

t

1

2f0

0, t >1

2f0

Problem Set 7.4

7.13 Using the unit impulse response function h(t) of

Eq. 5.30 and the definition in Eq. 7.33 of the continuous

Fourier transform H( f ), show that the Fourier transform

of the unit impulse response function is the frequency

response function H( f ) given by Eq. 7.30. That is,

using Eq. 7.33, show that h(t) and H( f ) form a Fourier

transform pair, as illustrated in Fig. 7.4.

Problem Set 7.5

C 7.14 (a) Using the input sequence given in Fig. 1a

of Example 7.7, use the Matlab FFT command to

compute a 16-point Fourier transform to verify the (real

and imaginary) results shown in Figs. 1b. (Note: You can

find the Matlab .m-file for Example 7.7 as sd2ex7 7.m

on the books website.) (b) What is the resulting value

of the coefficient (P1)16pt? Compare this answer with

the value given in Eq. 5 of Example 7.4. (c) Repeat

part (a) to calculate a 32-point transform and a 64-

point transform. Are the values of (P1)16pt, (P1)32pt,

and (P1)64pt converging to the value given in Eq. 5 of

Example 7.4?

C 7.15 Let p(t) be the periodic function defined by

the 16-point discrete-time sawtooth wave shown in

Fig. P7.15. (a) Use the FFT command in Matlab to

determine the 16-point discrete Fourier transform (DFT)

of this sawtooth wave. (b) Plot the real and imaginary

parts of this DFT transform.

0 0.25 0.5 0.75 11

0.5

0

0.5

1

Time

p(t)

Figure P7.15

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Craig Chapt7