Frequency Response by Analysis of Data Records

In a frequency response measurement that uses a sinusoidal test signal, the input is В sin cot and the output is M sin (cot + ф). The amplitude of the frequency response is M/B, and the phase is ф. If two sinusoidal signals with different frequencies were used simultaneously, the input SI would be

SI = B, sin со^ + B2 sin co2t

Because of the principle of superposition for linear systems, the output would be

SO = M1 sin(aj! Г + Фі) + M2 sin(a>2t + ф2)


80 = S! IG(jcoі)| sin^t + фк) + B2G(jco2) sin(a>2t + ф2)

The same idea may be used with general, nonsinusoidal, periodic signals. The input may be expressed (exactly) as a Fourier series


8l(t) = (aj2) + £ (ak cos cokt + bk sin cokt) (2.9.1)

k= і

where <ok = 2knjT and T is the period. The output is


80(t) = (aJ2)G(0) cos ф0 + £ akG(jcok) cos(cokt + фк)

k= 1

+ bkG(jcok) sin(o)kt + фк) (2.9.2)

The complex form of these expansions is


<5J(t) = X ck exp()wkt) (2.9.3)

к = — oo


50(t) = £ CkG(jcok) exp(jcokt) (2.9.4)

к = — oo

image089 Подпись: (2.9.5)

If the input and output signals are Fourier analyzed at a harmonic frequency, the results are

Подпись: (2.9.6)j»nT

(1 /nT) 80(t) exp(~jcokt) dt = Ck G(jcok

J 0

where n is the number of periods of data analyzed. The frequency response is obtained as follows:

C(. , (1 /nT) j"r 50(f) exp( — jcokt) dt }C°k (1/nT) jjr dl(t) exp( —j(okt) dt


The results are obtained from the following expressions:

G(M) = (R + jS)/( V + jW)




R = (1/nT) SO(t) cos cokt dt

J 0



S = —(1/nT) SO(t) sin cokt dt

J 0


/* nT

V = (1/nT) dl(t) cos cokt dt

J 0


W= -(1/nT) SI(t) sin coktdt Jo


These lead to the following results:

Re{G(jcut)} = (RF+ SW)/(V2 + W2)


Im{G(M)} = (SF — RW)/(V2 + W2)


|G(M)| = [(R2 + S2)/(V2 + IF2)]1/2


^{G(M)} = arctan[(SF — RW)/(RV + S1F)]


The frequency response also may be obtained from the power spectra. If the input signal is given by


dl{t) = X Ck exp{jcokt)

к = — со


then the output is


dO(t) = X °k exp(jwtt)

k = — со



Подпись: (2.9.19)Dk = CkG(j(ak)

The cross-power spectrum PI0 and the power spectrum of the input P,, are given by

PI0 = DkC. k = G(jcok)CkC_k (2.9.20)

P„ = CkC. k (2.9.21)

Then the frequency response is simply

G(M) = P, o/P„ (2-9-22)

One way to obtain these power spectra is by Fourier analysis of the cross­correlation function and the autocorrelation function of the input (see Eqs. (2.7.3) and (2.7.4)). Thus a prc :edure that may be used to give the fre­quency response is

G(jw) = F{C12}/F{Cn} (2.9.23)

The analyst would compute the correlation functions and then Fourier analyze them. As will be seen later, this requires greater computational effort than direct Fourier analysis of the input and output signals. But the correla­tion functions are sometimes worth the effort because they can be interpreted directly to provide information on system dynamics (see Section 3.6).

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