# Confusing THD value for a Signal without Harmonics

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weinan on 5 Mar 2024
Commented: weinan on 6 Mar 2024
When I use the offical example 'Determine THD for a Signal with Two Harmonics', I change the reference signal without harmonics and change the fundamental frequency from 100Hz to 8Hz, shown as follows:
t = 0:0.001:1-0.001;
x = 2*cos(2*pi*8*t);
Next, obtain the total harmonic distortion explicitly and using thd
r = thd(x,1000,10)
which yields r = -58.9845 dB, almost 11% THD rate!
Why a simple sinusoidal wave can cause such a THD rate using thd function?

David Goodmanson on 5 Mar 2024
Edited: David Goodmanson on 5 Mar 2024
Hi weinan,
MODIFIED
The thd process widens the peaks, as shown by the plot produced when you invoke thd. When the frequency is as small as 8 Hz, the left hand side of the peak is cut off a 0 frequency (first plot) and the code picks the second peak at the nonpeak spot that you see. If you raise the frequency to, say, 20 Hz (second plot), all the peak is included in the plot and and the resulting thd is -296.41 dB which is small by any standard.
##### 3 CommentsShow 1 older commentHide 1 older comment
David Goodmanson on 5 Mar 2024
Hi weinan, after I understood the issue I went back and modified the answer.
weinan on 6 Mar 2024
Thx, David. From your point of view, it is due to the fact that the thd func somehow cuts off the fundamental and picks the wrong second harmonics when the fundamental frequency is selected small, which, in result, causing a relatively big thd rate.
Thx again!

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