VI. Power Spectra
Amplitude and Phase We start with the Fourier series representation of X(t) in real notation: A() X(t) = + [ A(n) cos(nωt) + B(n) sin(nωt)] 2 n=1
he waveform of the observed segment exactly determines the values of the Fourier coefficients: 2 A(n) = X(t) cos(nω t)dt 2 B(n) = X(t) sin(nω t)dt and
As long as X(t) is continuous, the complete Fourier series obtained in this manner will reconstruct the original waveform without error. Once A(n) and B(n) have been determined in this way, they can be plotted as functions of frequency (nω ). hese spectra are defined only for integer values of n. For this reason, they are called discrete spectra or line spectra.
A(n) and B(n) both refer to the same frequency; together they are equivalent to the complex spectrum: Z(n) = A(n) - jb(n) 2
Z(n) is a complex number. he absolute value or amplitude of Z(n) is defined as: Z(n) = Z(n)Z(n) * = 2 2 A(n) + B(n) 2 he amplitude of Z(n) displayed as a function of n (or frequency) is referred to as the amplitude spectrum of x(t).
Consider the following trigonometric expression: F HG 2 2 A B Acos x + Bsin x = A + B cos x + sin x 2 2 2 2 A + B A + B A Letting: cosθ = and sinθ = 2 2 2 2 A + B A + B and letting x = nω, and using the following trigonometric expression: cos x cosθ + sin x sinθ = cos( x θ) B I KJ
We find that: cos( ω ) + sin( ω ) = 2 + 2 cos ω θ A n n t B n n t A n B n n t n Notice that the amplitude is (A 2 + B 2 ) 1/2 rather than (A 2 + B 2 ) 1/2 / 2. his is because the expression is written in real notation, and the amplitude at n has equal contributions from the complex components at n and n.
Notice that tan(θ) = (B/A). From this, it follows that the angle θ, called the phase, is: θ(n) = arctan F H G B(n) A(n) I K J When displayed as a function of n (or frequency), θ yields the phase spectrum of X(t).
Power Z(n) 2 is referred to as power: Z n 2 = 2 2 + B( n) A n When displayed as a function of n or frequency, it becomes the power spectrum of X(t). he term power comes from the field of electrical engineering, where power dissipated in an electrical circuit is proportional to the mean square voltage applied. 4
he total power of a time series can be determined either directly from the time series or from its spectral components. In the time domain, power is defined as the mean square (ms) value: 1 2 ms = X t dt or its root : 1 2 rms = X t dt
Parseval s heorem Parseval's theorem relates the mean square value and the sum of spectral powers as: 2 n= - 1 X t dt = Z(n) 2 hus the total power of X(t) can be determined in the time domain as the mean square value, or in the frequency domain as the sum of spectral powers.
he complex spectrum is symmetric around ω =, so the complex components at n and n are equal. he total power at any frequency (in real notation) is equally divided between complex components at nω and -nω. 2 2 Z(n) = Z(-n) for n and 2 2 2 otal Power at n = Z(n) + Z(-n) = 2 Z(n) for 1 n Nb: n is both negative and positive in complex notation; it is only non-negative in real notation.
An Example Now we consider a simple example in which we determine the total power for a cosine function with amplitude of 1 at an arbitrary frequency kω. For X(t)=cos(kω t): (1) Derive the values for A(n) and B(n) for all values of n. (2) Using A(n) and B(n), compute the total power of X(t).
First, solve for A(n): 2 2 A( n ) = X ( t ) cos( nω t ) dt = cos( kω t ) cos( nω t ) dt In the case where k n, we can use the following integral form: z cosax cos bx dx = b g sin a b x sin a+ b x b g + 2b g 2 a b a+ b b g with a = kω, b = nω, x = t (provided a b)
cos( ω ) cos( ω ) It follows that: ( k n) ( k+ n) ( k n) ( k+ n) 2 sin k n ωt sin k+ n ωt 2 k n ω 2 k+ n ω A n = k t n t dt = + sin k n 2π sin k+ n 2π sin k n sin k+ n = + 2 k n ω 2 k+ n ω 2 k n ω 2 k+ n ω = + = 2 ω 2 ω 2 ω 2 ω Conclusion: A(n) = for all n where k n.
Now consider the case where k = n. We can use the following integral form: 2 x sin 2ax cos ax dx = + 2 4a Again, we let a = kω, b = nω, x = t. hen: Conclusion: A(n) = 1 when k = n. ( 2kω t) 2 2 t sin A( k ) = cos( kωt) = + 2 4kω 2 2 sin( 2k 2π ) sin( 2k ) = + 2 4k 2 4k ω ω 2 sin( 2k 2π ) sin 2 = + 1 2 4kω 4k = = ω 2
Next, solve for B(n): 2 2 B ( n ) = X( t) sin( nω t) dt = cos( kω t) sin( nω t) dt In the case where k n, we can use the following integral form: cos ax sin bx dx = ( + ) cos ( ) 2( a+ b) 2( b a) cos a b x b a x with a = kω, b = nω, x = t (provided a b)
cos( ω ) sin( ω ) It follows that: ( k+ n) ( n k) ( k+ n) ( n k) 2 cos k+ n ω t cos n k ω t B n = k t n t dt = 2 k+ n ω 2 n k ω cos k+ n 2π cos n k 2π cos k+ n cos n k = + 2 k+ n ω 2 n k ω 2 k+ n ω 2 n k ω 1 1 1 1 = + = 2 ω 2 ω 2 ω 2 ω Conclusion: B(n) = for all n where k n.
Now consider the case where k = ± n. We can use the following integral form: cos 2ax sin ax cos ax dx = - 4a Again, we let a = kω, b = nω, x = t. hen: ( 2 2π ) cos( 2 ) Conclusion: B(n) = when k = n. ( 2kω t) 4kω 2 B ( k ) = cos( kω t ) sin( kω t ) dt = cos cos k k 1 1 = + = + = 4kω 4kω 4kω 4kω
Summary k A(n) B(n) n n 1
o solve for total power: bg 2 otal Power = Z n Z(k) + Z( k) = 2 [ 2 2 2 2 A + B ] = n= 4 = 1 2
More Examples (1) wo time series may be very different and yet have identical power spectra. hey would necessarily have different phase spectra.
(2) Reconstruction of a time series using only the components having largest power yields a smoothed version of the time series.
(3) he power spectrum is useful for distinguishing between time series that may look alike bet are generated by very different processes, e.g. beat (2 close frequency components) vs amplitude modulation (1 component that changes amplitude over time).
(4) he power spectrum of the EEG is useful for distinguishing among different brain states, e.g. delta for deep sleep, alpha for resting awake, and broad-band for focused awake.
(5) he power spectrum of a time series only displays the average power over the length of the time series. wo very different time functions, e.g. low amplitude with persistent rhythmic activity vs irregularly occurring bursts of high amplitude rhythmic activity, will have the same power spectrum if their average power is the same.
Relation Between Power Spectrum and Autocovariance Remember that for a signal of length, the estimate of the autocovariance function for an aperiodic function is defined as: ˆ XX C ( τ ) = 1 τ X(t)X(t + τ )dt
We can now express X(t) and X(t+τ) as their Fourier series representations. X(t) = n= - m= - Z(n)e X(t + τ ) = Z(m)e jnω t jm ω (t+ τ )
We now substitute these Fourier series representations in the autocovariance equation: τ ˆ 1 jn t jm (t+ ) XX = ( n) e ω e ω τ τ n,m= - C ( Z )( Z(m) )dt 1 = Z(n)Z(m) n= - m= - τ jnω t jmω t jmωτ e e e Since τ is constant with respect to the integral: dt Cˆ XX τ 1 t = Z(n)Z(m) e dt n,m= - jmωτ j(n+m) ω ( τ ) e
Next, consider the expression within the [] in this equation. First of all, if (n+m), then the integral of the exponential in the [] is approximately equal to. It is exactly equal to when τ = since (n+m) is an integer multiplier of ω, and each sine and cosine component thus has an integer number of cycles in interval. For this reason, the integral over one period of a sine or cosine wave has positive and negative values that exactly cancel. When the interval is reduced to length -τ, the positive and negative values of the cosine terms will not exactly cancel, but the residual integral will still be close to. he result is that we need not consider values of the integral in the [] for which (n+m) since the contribution of these terms to the autocovariance function is negligible. Next, consider the situation where (n+m) =. In this case, the exponential inside the integral in [] is equal to 1, and the integral will equal the length over which the integration is performed. When τ =, this length is, and multiplication of the integral by 1/ causes the term in [] to exactly equal 1. When the length is -τ, division by 1/ yields terms that are negligibly less than 1. he result is that value of the integral in the [] needs only be considered in the case where (n+m) =, and the term in [] is approximately 1 in that case.
herefore, we conclude that: $C τ = Z(n)Z(m)e XX b g n,m= - jmω τ Since n + m =, m = n. We thus substitute n for m, giving: -jn C XX = Z(n)Z(-n)e n= - $ τ ω τ
We have already seen that Z(n) and Z(-n) are complex conjugates: Z(n) = Z(-n) = A(n) - jb(n) 2 A(n) + jb(n) 2
herefore, the product of Z(n) and Z(-n) is the power. hus: 2 2 Z(n)Z(-n) = A(n ) + B(n) 4 b g 2 -jnω τ C $ XX τ = Z(n) e n= - = Z(n) 2
Note that: < n< and b g b g 2 2 Z n = Z n b g herefore: 2 jn C XX = Z(n) e n= - $ τ ω τ his relation shows that the autocovariance function can be expressed as a complex Fourier series whose coefficients are the power values.
his means that the autocovariance and power spectrum form a Fourier transform pair, so that: Z(n) Cˆ XX 2 1 -jn Cˆ τ XX e ωτ = d and ( τ ) = Z(n) e n= - 2 jnω τ τ
Note that the Fourier transform of the autocovariance function is a real-valued function. his is because the products of all the sine terms of the exponential and C xx (τ) are zero. his results from the fact that C xx (τ) is symmetric around τ =, and the sine functions are all anti-symmetric. his confirms what we already know, namely that the power spectrum is a real function. It is also why the Fourier transform of C xx (τ) is called Fourier cosine transformation. Note also that, since real and imaginary terms have been combined in the power, it is not possible to determine phase from the Fourier representation of the autocovariance. Since the phase information is not available, it is not possible to deduce the original waveform of the time series from the autocovariance. Any number of time series with different phase structure could produce the same autocovariance function.
Finally, remember that C XX () is the variance of X(t). Assuming that the mean of X(t) is, the variance is equivalent to the mean square value. hen: 2 1 CXX = X t dt = Z(n) e = Z(n) 2 2 n= - n= - his proves Parseval s theorem.