Periodogram of sinusoid: why power is -6 dB instead of -3 dB?

Hi all,

I'm trying to understand the periodogram function. If I take the periodogram of a sinusoid, the power of the positive and negative frequency components are -6 dB, whereas I would expect them to be -3 dB (the power is evenly split between the positive and negative frequency).

Similarly, if I perform the multiplication of two sinusoids, the power of the resulting frequency components are -12 dB, whereas I would expect them to be -6 dB.

Is there something wrong with my code or with my understanding?

Thanks for the help.

clear all; close all; clc

fs = 10e3;   %samples
fc = 500;    %sinusoid freq
t = linspace(0,1,fs);
y = sin(2*pi*fc.*t);

[P,F] = periodogram(y,[],length(y),fs,'power','centered');
plot(F,10*log10(P)) %plot 10*log10() to convert to dBW
xlabel('Freq. in Hz')
ylabel('PSD (dBW)')

[dpb]

Well, what they're doing is explained here...I'll have to think some more about the "why" part...

https://www.mathworks.com/help/signal/examples/measuring-the-power-of-deterministic-periodic-signals.html?s_tid=answers_rc2-1_p4_BOTH

[J D]

Thanks dpb, it sounds like I just need to change the amplitude of the sine wave to sqrt(2). Doing so, the power is then 0 dB and the power of the individual frequency components is then -3 dB.

Answer 1

[Honglei Chen]

The periodogram shows the power density. The magnitude at those frequency for a sinusoid is 1/2, so the power is 1/4, which corresponds to -6 dB.

HTH