Showing that a square wave of unity amplitude can be bound by the subtraction of its fundamental component...
Consider a normal square wave that has a switching angle of $alpha = cos^{-1}left(frac{pi}{4} m right)$. (Note that these values are found from Fourier Analysis of a square wave with varying (symmetric) switching angles). Here, $m$ is considered to be the amplitude of the fundamental component of the square wave. Let's call the square wave $f_{sqr}(t)$.
At $alpha = cos^{-1}left(frac{pi}{4} m right)$, $f_{sqr}(t)$ switch from $0 rightarrow1$.
At $pi - alpha$, $f_{sqr}(t)$ switches from $1rightarrow0$.
At $pi + alpha$, $f_{sqr}(t)$ switches from $0rightarrow -1$.
At $2pi - alpha$, $f_{sqr}(t)$ switches from $-1rightarrow 0$.
Knowing this, the amplitude of the fundamental component of the sine wave contained within the square wave is $mcdot sin(theta) = frac{4}{pi}cos(alpha) sin(theta)$.
A third order harmonic of amplitude $m_3 =frac{0.5-m sin(alpha)}{sin(3alpha)}$ can be added such that:
$$-0.5leq f_{sqr}(t)-mcdot sin(theta) - m_3 sin(3theta) leq 0.5$$
Clearly, if $alpha = 0$ (corresponding to $m = 4/pi$), the bound above doesn't hold. However, if $m = 1.26$ the bound does hold.
My question is this: which value of $m$, analytically, provides the bound at which the above inequality doesn't hold?
Thank you.
trigonometry inequality fourier-analysis boundary-value-problem
asked Nov 27 at 17:24
Lerbi
838
add a comment |
Consider a normal square wave that has a switching angle of $alpha = cos^{-1}left(frac{pi}{4} m right)$. (Note that these values are found from Fourier Analysis of a square wave with varying (symmetric) switching angles). Here, $m$ is considered to be the amplitude of the fundamental component of the square wave. Let's call the square wave $f_{sqr}(t)$.
At $alpha = cos^{-1}left(frac{pi}{4} m right)$, $f_{sqr}(t)$ switch from $0 rightarrow1$.
At $pi - alpha$, $f_{sqr}(t)$ switches from $1rightarrow0$.
At $pi + alpha$, $f_{sqr}(t)$ switches from $0rightarrow -1$.
At $2pi - alpha$, $f_{sqr}(t)$ switches from $-1rightarrow 0$.
Knowing this, the amplitude of the fundamental component of the sine wave contained within the square wave is $mcdot sin(theta) = frac{4}{pi}cos(alpha) sin(theta)$.
A third order harmonic of amplitude $m_3 =frac{0.5-m sin(alpha)}{sin(3alpha)}$ can be added such that:
$$-0.5leq f_{sqr}(t)-mcdot sin(theta) - m_3 sin(3theta) leq 0.5$$
Clearly, if $alpha = 0$ (corresponding to $m = 4/pi$), the bound above doesn't hold. However, if $m = 1.26$ the bound does hold.
My question is this: which value of $m$, analytically, provides the bound at which the above inequality doesn't hold?
Thank you.
trigonometry inequality fourier-analysis boundary-value-problem
asked Nov 27 at 17:24
Lerbi
838
add a comment |
Consider a normal square wave that has a switching angle of $alpha = cos^{-1}left(frac{pi}{4} m right)$. (Note that these values are found from Fourier Analysis of a square wave with varying (symmetric) switching angles). Here, $m$ is considered to be the amplitude of the fundamental component of the square wave. Let's call the square wave $f_{sqr}(t)$.
At $alpha = cos^{-1}left(frac{pi}{4} m right)$, $f_{sqr}(t)$ switch from $0 rightarrow1$.
At $pi - alpha$, $f_{sqr}(t)$ switches from $1rightarrow0$.
At $pi + alpha$, $f_{sqr}(t)$ switches from $0rightarrow -1$.
At $2pi - alpha$, $f_{sqr}(t)$ switches from $-1rightarrow 0$.
Knowing this, the amplitude of the fundamental component of the sine wave contained within the square wave is $mcdot sin(theta) = frac{4}{pi}cos(alpha) sin(theta)$.
A third order harmonic of amplitude $m_3 =frac{0.5-m sin(alpha)}{sin(3alpha)}$ can be added such that:
$$-0.5leq f_{sqr}(t)-mcdot sin(theta) - m_3 sin(3theta) leq 0.5$$
Clearly, if $alpha = 0$ (corresponding to $m = 4/pi$), the bound above doesn't hold. However, if $m = 1.26$ the bound does hold.
My question is this: which value of $m$, analytically, provides the bound at which the above inequality doesn't hold?
Thank you.
trigonometry inequality fourier-analysis boundary-value-problem
asked Nov 27 at 17:24
Lerbi
838
Consider a normal square wave that has a switching angle of $alpha = cos^{-1}left(frac{pi}{4} m right)$. (Note that these values are found from Fourier Analysis of a square wave with varying (symmetric) switching angles). Here, $m$ is considered to be the amplitude of the fundamental component of the square wave. Let's call the square wave $f_{sqr}(t)$.
At $alpha = cos^{-1}left(frac{pi}{4} m right)$, $f_{sqr}(t)$ switch from $0 rightarrow1$.
At $pi - alpha$, $f_{sqr}(t)$ switches from $1rightarrow0$.
At $pi + alpha$, $f_{sqr}(t)$ switches from $0rightarrow -1$.
At $2pi - alpha$, $f_{sqr}(t)$ switches from $-1rightarrow 0$.
Knowing this, the amplitude of the fundamental component of the sine wave contained within the square wave is $mcdot sin(theta) = frac{4}{pi}cos(alpha) sin(theta)$.
A third order harmonic of amplitude $m_3 =frac{0.5-m sin(alpha)}{sin(3alpha)}$ can be added such that:
$$-0.5leq f_{sqr}(t)-mcdot sin(theta) - m_3 sin(3theta) leq 0.5$$
Clearly, if $alpha = 0$ (corresponding to $m = 4/pi$), the bound above doesn't hold. However, if $m = 1.26$ the bound does hold.
My question is this: which value of $m$, analytically, provides the bound at which the above inequality doesn't hold?
Thank you.
trigonometry inequality fourier-analysis boundary-value-problem
trigonometry inequality fourier-analysis boundary-value-problem
asked Nov 27 at 17:24
Lerbi
838
asked Nov 27 at 17:24
Lerbi
838
asked Nov 27 at 17:24
Lerbi
838
asked Nov 27 at 17:24
Lerbi
838
asked Nov 27 at 17:24
Lerbi
838
838
add a comment |
add a comment |
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