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

We will provide two examples of how to measure a signal's average power. The examples will 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 can't be characterized as either a power or energy signal.

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

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

Let's look at a portion of our signal in the time domain.

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

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

power_theoretical = (A^2/4)*2

power_theoretical = 0.5000

in dB the power contained in the positive frequencies only is:

10*log10(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'); v = axis; axis([v(1) v(2) -10 -5.5]); % Zoom in Y. hgcf = gcf; hgcf.Color = [1,1,1];

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

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

figure periodogram(x, hamming(length(x)), [], Fs, 'centered', 'psd'); hgcf = gcf; hgcf.Color = [1,1,1];

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's 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 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's 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'll estimate the total average power of a signal containing energy at multiple frequency components: one at DC, one at 100Hz, and another at 200Hz.

Fs = 1024; t = 0:1/Fs:1-(1/Fs); Ao = 1.5; % Vpeak @ DC A1 = 4; % Vpeak A2 = 3; % Vpeak F1 = 100; % Hz F2 = 200; % Hz x = Ao + A1*sin(2*pi*t*F1) + A2*sin(2*pi*t*F2); % Let's look at a portion of our signal. idx = 1:128; plot(t(idx),x(idx)); grid; ylabel('Amplitude'); xlabel('Time (sec)'); hgcf = gcf; hgcf.Color = [1,1,1];

Like the previous example, the theoretical average power of each complex sinusoid is A^2/4. The signal's DC average power is equal to its peak power since it's constant and therefore is given by Ao^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 Ao^2 + (A1^2/4)*2 + (A2^2/4)*2.

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'll see that the theoretical results match the mean-square spectrum plot below.

[10*log10(Ao^2) 10*log10(A1^2/4) 10*log10(A2^2/4)]

ans = 3.5218 6.0206 3.5218

To measure the signal's average power we will once again use the periodogram function to calculate and plot the power spectrum of the signal.

periodogram(x, hamming(length(x)), [], Fs, 'centered', 'power'); v = axis; axis([v(1) v(2) 0 7]); % Zoom in Y hgcf = gcf; hgcf.Color = [1,1,1];

As in the first example, estimating the signal's total average power by "integrating" under the PSD curve we get:

periodogram(x, hamming(length(x)), [], Fs, 'centered', 'psd'); hgcf = gcf; hgcf.Color = [1,1,1];

Note that 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 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 on axis tight; xlabel('Frequency (Hz)'); title('Ratio between Power Spectrum and 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

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