Signals and Systems
Problem Set:Power Spectral Density and White Noise
Problem Set
For all questions, time signals x [n] are real-valued and wide sense stationary.
Problem 1
Show that the auto-correlation function is even, that is
Rxx [k] = Rxx [−k] .
Problem 2
Show that the power spectral density function is even, that is
Sxx (Ω) = Sxx (−Ω) .
Problem 3
Show that the power spectral density is real.
Problem 4
Show that the power spectral density is nonnegative, that is
Sxx (Ω) ≥ 0 .
Note: This is not an easy question!
Problem 5
Let
y [n] =x [n] + x [n− 1]
2,
where x [n] is white noise. Calculate the auto-correlation function Ryy [k] and the power spectraldensity Syy (Ω).
Problem 6
Let
y [n] = x1 [n] + x2 [n] ,
where x1 [n] and x2 [n] are independent and zero mean, implying that
E (x1 [n1]x2 [n2]) = 0 ∀n1, n2 .
Show that
Syy (Ω) = Sx1x1(Ω) + Sx2x2
(Ω) .
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Problem 7
Show that the cross-correlation function satisfies
Rxy [k] = Ryx [−k]
and that
Sxy (Ω) = Syx (−Ω) .
Problem 8
Come up with a simple example where the power spectral density Sxy (Ω) is complex.
Problem 9
Using Matlab, calculate the auto-correlation function and the power spectral density of thesignal
x [n] = sin
(
10πn
N
)
+N (0, 1) ,
where N = 1024 is the number of elements of x, and N (0, 1) is Gaussian noise with mean zeroand variance one.
Note: When numerically calculating the auto-correlation function of an N -sample signal, we canuse the following definition:
1
N
N−1∑
n=0
x [n]x [n− k]
which assumes that x[n] is periodic. An example of code that performs this summation appearsin the file white noise.m, available on the course website. When calculating the power spectraldensity, use the Discrete Fourier Transform. (i.e. The Matlab command fft)
Plot the signal, its auto-correlation, and its power spectral density. Confirm that the powerspectral density is real and positive. (Hint: numerical errors may result in small imaginary
values, so you may wish to use the command real.) Also calculate the signal’s DFT andcompare the square of the absolute value of the Fourier coefficients with the power spectraldensity. By what factor do they differ? Can you explain this?
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Sample Solutions
Problem 1 (Solution)
The auto-correlation function is defined as
Rxx [k] = E (x [n]x [n− k])
which gives us
Rxx [−k] = E (x [n]x [n+ k]) .
Substituting m = n+ k we get
Rxx [−k] = E (x [m− k]x [m]) ,
which is exactly Rxx [k].
Problem 2 (Solution)
The power spectral density is the Fourier Transform of the auto-correlation function:
Sxx (Ω) =+∞∑
k=−∞
Rxx [k] e−jΩk
which gives us
Sxx (−Ω) =
+∞∑
k=−∞
Rxx [k] e+jΩk
Substituting l = −k we get
Sxx (−Ω) =+∞∑
l=−∞
Rxx [−l] e−jΩl
Using the result of Problem 1 (Rxx [k] = Rxx [−k]) this is identical to
Sxx (−Ω) =
+∞∑
l=−∞
Rxx [l] e−jΩl ,
which is Sxx (Ω).
Problem 3 (Solution)
We can rewrite the power spectral density as
Sxx (Ω) =
−1∑
k=−∞
Rxx [k] e−jΩk +Rxx [0] +
∞∑
k=1
Rxx [k] e−jΩk .
Again using the result that Rxx [k] = Rxx [−k], this is identical to
Sxx (Ω) = Rxx [0] +
∞∑
k=1
Rxx [k](
e−jΩk + e+jΩk)
,
where the complex conjugates can be rewritten as
Sxx (Ω) = Rxx [0] +
∞∑
k=1
Rxx [k] (2 cos (Ωk)) .
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Problem 4 (Solution)
First proof
We may consider the autocorrelation as an infinite sum
Rxx [k] = E (x [n]x [n− k])
= limN→∞
1
2N + 1
N∑
n=−N
x [n]x [n− k] .
Using the definition of power spectral density we may then write
Sxx (Ω) =∞∑
k=−∞
Rxx [k] e−jΩk
=
∞∑
k=−∞
limN→∞
1
2N + 1
N∑
n=−N
x [n]x [n− k] e−jΩk
= limN→∞
1
2N + 1
N∑
n=−N
x [n]
∞∑
k=−∞
x [n− k] e−jΩk
= limN→∞
1
2N + 1
N∑
n=−N
x [n]∞∑
m=−∞
x [m] e−jΩ(n−m)
= limN→∞
1
2N + 1
N∑
n=−N
x [n] e−jΩn
∞∑
m=−∞
x [m] ejΩm.
Now consider that since for N ≥ 0,
1
2N + 1> 0.
Using this, and the fact that x[n] is a real-valued sequence, we find that
sgn (Sxx (Ω)) = sgn
(
∞∑
n=−∞
x [n] e−jΩn
∞∑
m=−∞
x [m] ejΩm
)
= sgn (X (Ω)X (−Ω))
= sgn(
X (Ω)X (Ω))
= sgn(
|X (Ω) |2)
= 1.
Second proof
Assume that the signal x [n] is the input signal into a system. Now suppose our system is anideal bandpass filter H (Ω) of the following form:
5
1
ΩΩ0
w
−Ω0
w
|H (Ω)|
When applying an linear time-invariant system to our input signal, the power spectral densityof the output is
Syy (Ω) = |H (Ω)|2 Sxx (Ω) .
We now apply the inverse Fourier Transform to calculate the auto-correlation of the outputsignal Ryy [0] of the bandpass filter:
Ryy [0] =1
2π
∫ π
−π
|H (Ω)|2 Sxx (Ω) dΩ
=1
π
∫ Ω0+w
2
Ω0−w
2
Sxx (Ω) dΩ .
For the output signal to be real, Ryy [0] must be nonnegative. If we let w go to zero, the limitcase can be expressed as
Ryy [0] ≈ limw→0
w
πSxx (Ω0) .
For Ryy [0] to be nonnegative, Sxx (Ω0) must be nonnegative. Since Ω0 was arbitrarily chosen,this must hold for all Ω. Therefore,
Sxx (Ω) ≥ 0 ∀Ω .
Problem 5 (Solution)
The auto-correlation function is
Ryy [k] = E (y [n] y [n− k])
= E
(
x [n] + x [n− 1]
2
x [n− k] + x [n− k − 1]
2
)
=1
4E (x [n]x [n− k] + x [n]x [n− k − 1] + x [n− 1] x [n− k] + x [n− 1] x [n− k − 1]) .
Since Ryy [k] only depends on the difference of the indices, this can be rewritten as
Ryy [k] =1
4(2E (x [n]x [n− k]) +E (x [n]x [n− k − 1]) + E (x [n]x [n− k + 1])) ,
which, assuming x to be white noise, gives
Ryy [k] =
0 for |k| ≥ 2 ,14 for |k| = 1 ,12 for k = 0 .
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The Fourier Transform is then easy to apply:
Syy (Ω) =+∞∑
k=−∞
Ryy [k] e−jΩk
Syy (Ω) =1
4ejΩ +
1
2+
1
4e−jΩ
=1
2+
1
2cos (Ω)
Problem 6 (Solution)
The auto-correlation function is
Ryy [k] = E (y [n] y [n− k])
= E ((x1[n] + x2[n]) (x1[n− k] + x2[n− k]))
= E (x1[n]x1[n− k] + x1[n]x2[n− k] + x2[n]x1[n− k] + x2[n]x2[n − k]) .
Because of the independance of x1 and x2, the middle two terms are zero and we can rewritethe correlation function to
Ryy [k] = E (x1[n]x1[n− k]) + E (x2[n]x2[n− k])
= Rx1x1[k] +Rx2x2
[k]
Due to the linearity of the Fourier Transform,
Syy (Ω) = Sx1x1(Ω) + Sx2x2
(Ω)
Problem 7 (Solution)
We write down the definition of cross-correlation, then use the fact that reindexing does notchange it, and see that this immediately produces the desired result:
Rxy [−k] = E (x [n] y [n+ k])
= E (x [n− k] y [n])
= Ryx [k]
We use the above result to prove the cross-power spectral densities relationship:
Sxy (Ω) =∞∑
k=−∞
Rxy [k] e−jΩk
=
∞∑
k=−∞
Ryx [−k] e−jΩk
=
∞∑
m=−∞
Ryx [m] ejΩm
= Syx (−Ω)
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Problem 8 (Solution)
Assuming real signals x and y, Rxy [k] is always real. Therefore, signals resulting in a complexpower spectral density must fulfill
Rxy [k] 6= Rxy [−k] ,
meaning that the decomposition used in Problem 3 does not eliminate the complex components.A simple example of this can be constructed assuming x [n] to be white noise and
y [n] = x [n− 1] ,
resulting in
Rxy [1] = E (x[n]x[n− 2]) = 0 ,
Rxy [−1] = E (x[n]x[n]) = 1 .
It is easy to show that Rxy [k] is zero for all k 6= −1. The power spectral density is then
Sxy (Ω) =+∞∑
k=−∞
Rxy [k] e−jΩk
Sxy (Ω) = 1ejΩ
= cos (Ω) + j sin (Ω)
Problem 9 (Solution)
The following are Matlab commands. We begin by creating the signal and plotting it:
>> N = 1024;
>> n = 1:N; n = n(:);
>> x = sin(10*pi*n/N) + randn(N,1);
>> figure; plot(x);
Now we use some code from white noise.m (see announcement of 12 Nov on the class website)to estimate the signal’s auto-correlation function. We then plot it:
>> xx = [x;x];
>> r = zeros(N,1);
>> for k = 1:N r(k) = sum(x.*xx(k:N+k-1) ); end
>> r = r/N;
>> figure; plot(r);
We now calculate its power spectral density, and check that it is real positive:
>> s = fft(r);
>> max(abs(imag(s)))
>> min(real(s))
The second command above returned 2.1839e-11. We assume that this is small enough to beexplained by numerical error, and that the power spectral density is therefore real. The lastcommand returned 6.2887, which confirms that the power spectral density is positive.To compare to square of the absolute value of the Fourier coefficients with the power spectraldensity, we do:
8
>> plot(abs(fft(x)).^2./real(s))
We see from this that the square of the absolute value of the Fourier coefficients are exactly N
times larger than the power spectral density. When Matlab calculates the DFT with fft, itdoes not use the divisor N . (This may be seen by typing help fft at the Matlab commandprompt.) An inspection of the first proof of Problem 4 shows that using this definition of theDFT, the only difference between the power spectral density and the square of the absolutevalue of the Fourier coefficients, is the factor of N calculated above.
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