Как выбрать гостиницу для кошек
14 декабря, 2021
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)
or
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
00
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
00
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
00
<5J(t) = X ck exp()wkt) (2.9.3)
к = — oo
00
50(t) = £ CkG(jcok) exp(jcokt) (2.9.4)
к = — oo
![]() |
![]() |
If the input and output signals are Fourier analyzed at a harmonic frequency, the results are
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 |
(2.9.7) |
The results are obtained from the following expressions: |
|
G(M) = (R + jS)/( V + jW) |
(2.9.8) |
where |
|
лпГ R = (1/nT) SO(t) cos cokt dt J 0 |
(2.9.9) |
лпГ S = —(1/nT) SO(t) sin cokt dt J 0 |
(2.9.10) |
/* nT V = (1/nT) dl(t) cos cokt dt J 0 |
(2.9.11) |
W= -(1/nT) SI(t) sin coktdt Jo |
(2.9.12) |
These lead to the following results: |
|
Re{G(jcut)} = (RF+ SW)/(V2 + W2) |
(2.9.13) |
Im{G(M)} = (SF — RW)/(V2 + W2) |
(2.9.14) |
|G(M)| = [(R2 + S2)/(V2 + IF2)]1/2 |
(2.9.15) |
^{G(M)} = arctan[(SF — RW)/(RV + S1F)] |
(2.9.16) |
The frequency response also may be obtained from the power spectra. If the input signal is given by |
|
CO dl{t) = X Ck exp{jcokt) к = — со |
(2.9.17) |
then the output is |
|
oo dO(t) = X °k exp(jwtt) k = — со |
(2.9.18) |
where |
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).