Eduardo, thanks a lot.
May I see the deriving procedure of your equation? If you have any documents to describe it, that would be great for me.
If you derived this equation on your own, could you please show your deriving process of this equation?
My work is to apply more complicated sinusoidal signal to an equation (e.g., your equation) describing a clipped pure sine or cosine.
The deriving procedure should be so much helpful for my work.
If you solve by hand the integrals, the best way is splitting the interval (albbg message). Unfortunately, the resulting expressions are quit long.
By software, you must write the functions in a way that the software interprets. And, if possible, of fast execution.
The synthesis of the clip may be written as:
(a) Fclip(x) = f(x)*U(x) + fc1*V(x)
where
U(x) = 0 in the range of clipping and 1 otherwise.
V(x) = 1 in the range of clipping and 0 otherwise.
fc1: Value at clip.
While it would be most natural write U and V
as a function of the amplitude
U(clip-f(x)) = 1 if clip-f(x)> 0, 0 otherwise
V(clip-f(x)) = 0 if clip-f(x)> 0, 1 otherwise
But at least Derive6 is
not efficient handling these expressions.
Returning to (a), writing in terms of the Heaviside step (because the software does not have a built-in pulse function).
Considering the clipping angle at x=pi/2-w
V(x) = H(x-pi/2-w) - H(x-pi/2+w)
U(x) = 1 - V(x)
fc1 = sin(pi/2-w) = cos(w)
Note: H(x): Heaviside step, STEP(x) in Derive6 notation.
After some algebraic steps arrive to the expression that I showed in the previous message.
Of course, this method is only of interest if you are going to evaluate the Fourier coefficients by software
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The method is also applicable with a double clipping (positive and negative) and not necessarily symmetrical.
(b) Fclip(x) = f(x)*U(x) + fc1*V1(x) + fc2*V2(x)
with
V1(x) = 1 in clipping interval1 , 0 otherwise.
fc1 = Value at clip1
V2(x) = 1 in clipping interval2 , 0 otherwise.
fc2 = Value at clip2
U(x) = 1 - V1(x) - V2(x)