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

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J D on 2 May 2019
Answered: Honglei Chen on 9 May 2019
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)')


J D on 3 May 2019
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.

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Answers (1)

Honglei Chen
Honglei Chen on 9 May 2019
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.


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