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This example shows how to measure the power of deterministic periodic signals. Although continuous in time, periodic deterministic signals produce discrete power spectra.

We provide two examples of how to measure a signal's average power. The examples use sine waves and assume a load impedance of 1 ohm.

In general signals can be classified into three broad categories, power signals, energy signals, or neither. Deterministic signals which are made up of sinusoids is an example of power signals which have infinite energy but finite average power. Random signals also have finite average power and fall into the category of power signals. A transient signal is an example of energy signals which start and end with zero amplitude. There are still other signals which cannot be characterized as either power or energy signals.

In our first example we estimate the average power of a sinusoidal signal with a peak amplitude of 1 volt and a frequency component at 128 Hz.

Fs = 1024; t = 0:1/Fs:1-(1/Fs); A = 1; F1 = 128; x = A*sin(2*pi*t*F1);

Let us look at a portion of the signal in the time domain.

idx = 1:128; plot(t(idx),x(idx)) ylabel('Amplitude') xlabel('Time (seconds)') axis tight grid

The theoretical average power (mean-square) of each complex sinusoid is ${\mathit{A}}^{2}/4$, which in our example is 0.25 or –6.02 dB. So, accounting for the power in the positive and negative frequencies results in an average power of $2\times {\mathit{A}}^{2}/4$.

power_theoretical = (A^2/4)*2

power_theoretical = 0.5000

We can also compute in dB the power contained in the positive frequencies:

pow2db(power_theoretical/2)

ans = -6.0206

To measure the signal's average power we call `periodogram`

and specify the `'power'`

option.

periodogram(x,hamming(length(x)),[],Fs,'centered','power') ylim([-10 -5.5])

As we can see from the zoomed-in portion of the plot, each complex sinusoid has an average power of roughly –6 dB.

Another way to calculate a signal's average power is by "integrating" the area under the PSD curve.

periodogram(x,hamming(length(x)),[],Fs,'centered','psd')

One thing to notice in this plot is that the peaks of the spectrum plot do not have the same height as when we plotted the power spectrum. The reason is because when taking Power Spectral Density (PSD) measurements it is the area under the curve (which is the measure of the average power) that matters. We can verify that by calling the `bandpower`

function, which uses the rectangle approximation to integrate under the curve to calculate the average power.

[Pxx_hamming,F] = periodogram(x,hamming(length(x)),[],Fs,'psd'); power_freqdomain = bandpower(Pxx_hamming,F,'psd')

power_freqdomain = 0.5000

Since according to Parseval's theorem the total average power in a sinusoid must be equal whether it is computed in the time domain or the frequency domain, we can verify our signal's estimated total average power by summing up the signal in the time domain.

power_timedomain = sum(abs(x).^2)/length(x)

power_timedomain = 0.5000

For the second example we estimate the total average power of a signal containing energy at multiple frequency components: one at DC, one at 100 Hz, and another at 200 Hz.

Fs = 1024; t = 0:1/Fs:1-(1/Fs); Ao = 1.5; A1 = 4; A2 = 3; F1 = 100; F2 = 200; x = Ao + A1*sin(2*pi*t*F1) + A2*sin(2*pi*t*F2);

Let us look at a portion of the signal.

idx = 1:128; plot(t(idx),x(idx)) grid ylabel('Amplitude') xlabel('Time (seconds)')

Like in the previous example, the theoretical average power of each complex sinusoid is ${\mathit{A}}^{2}/4$. The signal's DC average power is equal to its peak power since it is constant and therefore is given by ${\mathit{A}}_{0}^{2}$. Accounting for the power in the positive and negative frequencies results in a signal's total average power (sum of the average power of each harmonic component) of ${\mathit{A}}_{0}^{2}+2\times {\mathit{A}}_{1}^{2}/4+2\times {\mathit{A}}_{2}^{2}/4$.

power_theoretical = Ao^2 + (A1^2/4)*2 + (A2^2/4)*2

power_theoretical = 14.7500

By calculating the average power of each unique frequency component in dB we see that the theoretical results match the mean-square spectrum plot below.

pow2db([Ao^2 A1^2/4 A2^2/4])

`ans = `*1×3*
3.5218 6.0206 3.5218

To measure the signal's average power we once again use the `periodogram`

function to calculate and plot the power spectrum of the signal.

periodogram(x,hamming(length(x)),[],Fs,'centered','power') ylim([0 7])

As in the first example, we can estimate the signal's total average power by "integrating" under the PSD curve.

periodogram(x,hamming(length(x)),[],Fs,'centered','psd')

Once again the height of the peaks of the spectral density plot at a specific frequency component may not match the ones of the plot of the power spectrum for the reasons noted in the first example.

[Pxx, F] = periodogram(x, hamming(length(x)),[],Fs,'centered','psd'); power_freqdomain = bandpower(Pxx,F,'psd')

power_freqdomain = 14.7500

Again, we can verify the signal's estimated average power by invoking Parseval's theorem and summing up the signal in the time domain.

power_timedomain = sum(abs(x).^2)/length(x)

power_timedomain = 14.7500

You may have noticed that while the height of the peaks of the power and power spectral density plots are different, they differ by a constant ratio.

Pxx = periodogram(x,hamming(length(x)),[],Fs,'centered','psd'); Sxx = periodogram(x,hamming(length(x)),[],Fs,'centered','power'); plot(F,Sxx./Pxx) grid axis tight xlabel('Frequency (Hz)') title('Ratio of Power Spectrum to Power Spectral Density')

ratio = mean(Sxx./Pxx)

ratio = 1.3638

This ratio is related to the two-sided equivalent noise bandwidth (ENBW) of the window. You can compute this ratio directly by calling ENBW with the window and its corresponding sample rate.

bw = enbw(hamming(length(x)),Fs)

bw = 1.3638

In the previous sections, power was measured from one or multiple sinusoids having a frequency that coincided with a bin. Peak power estimates are usually less accurate when the signal frequency is out of bin. To see this effect, create a sinusoid with a non-integer number of cycles over a one-second period.

Fs = 1024; t = 0:1/Fs:1-(1/Fs); A = 1; F = 20.4; x = A*sin(2*pi*F*t); nfft = length(x); power_theoretical = pow2db(A^2/4*2);

Create a Hamming window and a flat top window.

w1 = hamming(length(x)); w2 = flattopwin(length(x));

Compute the periodogram of `x`

using the Hamming window. Zoom in on the peak.

h1 = figure; stem(F,power_theoretical,'BaseValue',-50); [Pxx1,f1] = periodogram(x,w1,nfft,Fs,'power'); hold on plot(f1,pow2db(Pxx1),'+-') axis([0 40 -45 0]) legend('Theoretical','Periodogram') xlabel('Frequency (Hz)') ylabel('Power (dB)') title('Periodogram Power Spectrum Estimate') grid

The peak power estimate is below the theoretical peak, and the frequency of the peak estimate differs from the true frequency.

[Pmax,imax] = max(Pxx1); dPmax_w1 = pow2db(Pmax) - power_theoretical

dPmax_w1 = -1.1046

dFreq = f1(imax) - F

dFreq = -0.4000

**Reduce Amplitude Error with Zero-Padding**

To see why this is happening, compute the periodogram using a larger number of FFT bins.

[Pxx2,f2] = periodogram(x,w1,100*nfft,Fs,'power'); figure stem(F,power_theoretical,'BaseValue',-50) hold on plot(f1,pow2db(Pxx1),'+') plot(f2,pow2db(Pxx2)) hold off axis([0 40 -40 0]) legend('Theoretical Peak','nfft = 1024','nfft = 102400') xlabel('Frequency (Hz)') ylabel('Power (dB)') title('Periodogram Power Spectrum Estimate') grid

In the original periodogram, the spectral peak is located between two bins, and for that reason the estimated peak is below the theoretical peak. Increasing the number of FFT bins gives a better picture of the spectrum, although this may be a computationally expensive way to improve peak measurements.

**Reduce Amplitude Error with a Flat Top Window**

Another way to produce a better estimate for the peak amplitude is to use a different window. Compute the periodogram of `x`

using the flat top window.

[Pxx,F1] = periodogram(x,w2,nfft,Fs,'power'); figure(h1) plot(F1,pow2db(Pxx)) legend('Theoretical','Hamming','Flat Top') hold off

The flat top window is broad and flat. It produces a peak estimate closer to the theoretical value when `x`

does not contain an integer number of cycles, and hence the spectral peak does not fall exactly on a bin.

dPmax_w2 = pow2db(max(Pxx)) - power_theoretical

dPmax_w2 = -6.2007e-04

The broader peak that the flat top window produces could be a disadvantage when trying to resolve closely spaced peaks, and the frequency of the measured peak is again different from the frequency of the theoretical peak.

**Reduce Amplitude Error with Reassigned Periodogram**

Now add the `'reassigned'`

flag to `periodogram`

. Periodogram reassignment uses phase information, which is normally discarded, to reassign the signal to its center of energy. The procedure can result in sharper spectral estimates. Plot the reassigned periodogram of `x`

and zoom in on the peak. Use the Hamming window and the flat top window.

[RPxx1,~,~,Fc1] = periodogram(x,w1,nfft,Fs,'power','reassigned'); [RPxx2,~,~,Fc2] = periodogram(x,w2,nfft,Fs,'power','reassigned'); stem(F,power_theoretical,'*','BaseValue',-40) hold on stem(Fc1,pow2db(RPxx1),'BaseValue',-50) stem(Fc2,pow2db(RPxx2),'BaseValue',-50) hold off legend('Theoretical','Hamming Reassignment','Flattop Reassignment') xlabel('Frequency (Hz)') ylabel('Power (dB)') title('Periodogram Power Spectrum Estimate') axis([19.5 21 -4 -2]) grid

The reassigned estimates of power are closer to the theoretical value for both windows, with the flat top window producing the best peak measurement.

[RPxx1max,imax1] = max(RPxx1); [RPxx2max,imax2] = max(RPxx2); dPmax_reassign_w1 = pow2db(RPxx1max) - power_theoretical

dPmax_reassign_w1 = -0.0845

dPmax_reassign_w2 = pow2db(RPxx2max) - power_theoretical

dPmax_reassign_w2 = -1.1131e-05

The frequency estimates are also improved using the reassigned periodogram, with the flat top window again giving the best results.

Fc1(imax1)-F

ans = -0.0512

Fc2(imax2)-F

ans = 5.6552e-04