Fourier Analysis of Common Switch Waveform

Rectangular Waveform with Arbitrary Duty Cycle

Before deriving the Fourier series, consider a simple periodic rectangular waveform. This waveform is one of the most important examples in power electronics because many switching waveforms can be approximated or directly represented by this form.

Let the waveform have:

Figure 1. Switch Waveform

amplitude $V$,
switching period $T_s$,
duty cycle $D$, where $0\le D\le1$.

The duty cycle is defined as the fraction of each switching period during which the waveform remains at $V$:

D=\frac{T_{\mathrm{on}}}{T_s}

Therefore, the ON-time is

T_{\mathrm{on}}=DT_s

and the remaining OFF-time is

T_{\mathrm{off}}=(1-D)T_s.

We define the rectangular waveform over one period as follows:

v(t)= \begin{cases} V, & 0\leq t<DT_s,\\ 0, & DT_s\leq t<T_s. \end{cases}

and the waveform repeats every switching period:

v(t+T_s)=v(t).

Thus, the waveform has a value of $V$ for a fraction $D$ of every period and a value of zero for the remaining fraction $1-D$.

Fundamental Period and Frequency

The fundamental period of the waveform is $ T_S $. Therefore, the fundamental frequency is $f_s=1/T_s$, and the corresponding fundamental angular frequency is $\omega_s=2\pi f_s=2\pi/T_s$.

The Fourier-series components occur at integer multiples of the fundamental frequency. Thus, the angular frequencies are $\omega_s,,2\omega_s,,3\omega_s,\ldots$, or equivalently, the frequencies are $f_s,,2f_s,,3f_s,\ldots$.

We can now derive the Fourier series from the general Fourier-series equations while keeping the duty cycle $D$ arbitrary throughout the derivation.

Note that the rectangular waveform is periodic and discontinuous (piecewise-constant). During each switching period, the voltage remains at $V$ for the ON-time $DT_s$ and at zero for the remaining time $ T_S $.

At $t=DT_s$, the voltage changes instantaneously from $V$ to zero. At the beginning of the next switching period, it returns to $V$.

Thus, although the waveform is continuous between switching instants, it is discontinuous at the switching transitions.

Fourier Series Derivation

For the rectangular waveform, start from the general Fourier series:

f(t)\equal\frac{a_0}{2} +\sum_{n=1}^{\infty} \left[ a_n\cos(n\omega_s t) + b_n\sin(n\omega_s t) \right]

with

\omega_s=\frac{2\pi}{T_s}

The Fourier coefficients are

a_0=\frac{2}{T_s}\int_0^{T_s}v(t),dt

a_n=\frac{2}{T_s}\int_0^{T_s}v(t)\cos(n\omega_s t),dt

b_n=\frac{2}{T_s}\int_0^{T_s}v(t)\sin(n\omega_s t),dt

Since the waveform is $V$ only during $0\leq t<DT_s$ and zero for the remainder of the period, the coefficient calculations reduce to the ON interval.

DC Coefficient

a_0=\frac{2}{T_s}\int_0^{DT_s}V,dt

a_0=\frac{2V}{T_s}\left[t\right]_0^{DT_s}

a_0=\frac{2V}{T_s}(DT_s)

\boxed{a_0=2VD}

Therefore,

\boxed{\frac{a_0}{2}=VD}

Cosine Coefficient

a_n=\frac{2V}{T_s}\int_0^{DT_s}\cos(n\omega_s t),dt

a_n= \frac{2V}{T_s} \left[ \frac{\sin(n\omega_s t)}{n\omega_s} \right]_0^{DT_s}

a_n= \frac{2V}{T_s} \frac{\sin(n\omega_s DT_s)}{n\omega_s}

Using $\omega_s=2\pi/T_s$,

a_n= \frac{2V}{T_s} \frac{ \sin\left( n\frac{2\pi}{T_s}DT_s \right) }{ n\frac{2\pi}{T_s} }

a_n= \frac{V}{n\pi}\sin(2\pi nD)

Therefore,

\boxed{ a_n=\frac{V}{n\pi}\sin(2\pi nD) }

Sine Coefficient

b_n=\frac{2V}{T_s}\int_0^{DT_s}\sin(n\omega_s t),dt

b_n= \frac{2V}{T_s} \left[ -\frac{\cos(n\omega_s t)}{n\omega_s} \right]_0^{DT_s}

b_n= \frac{2V}{T_s} \left[ -\frac{\cos(n\omega_s DT_s)}{n\omega_s} + \frac{1}{n\omega_s} \right]

b_n= \frac{2V}{T_s} \frac{ 1-\cos(n\omega_s DT_s) }{ n\omega_s }

Using $\omega_s=2\pi/T_s$,

b_n= \frac{V}{n\pi} \left[ 1-\cos(2\pi nD) \right]

Therefore,

\boxed{ b_n= \frac{V}{n\pi} \left[ 1-\cos(2\pi nD) \right] }

Substituting $a_0$, $a_n$, and $b_n$ into the Fourier-series expression gives

\boxed{ v(t)\equal VD + \sum_{n=1}^{\infty} \left[ \frac{V}{n\pi}\sin(2\pi nD)\cos(n\omega_s t) + \frac{V}{n\pi} \left(1-\cos(2\pi nD)\right) \sin(n\omega_s t) \right] }

Starting from the general trigonometric form for arbitrary duty cycle $D$ ,

we identify

a_n=\frac{V}{n\pi}\sin(2\pi nD)

and

b_n=\frac{V}{n\pi}\left[1-\cos(2\pi nD)\right]

The magnitude-phase form is

v(t)=VD+\sum_{n=1}^{\infty}C_n\cos(n\omega_s t-\phi_n)

where

C_n=\sqrt{a_n^2+b_n^2}

Substituting $a_n$ and $b_n$,

C_n= \sqrt{ \left[ \frac{V}{n\pi}\sin(2\pi nD) \right]^2 + \left[ \frac{V}{n\pi}(1-\cos(2\pi nD)) \right]^2 }

Factor out $V/(n\pi)$:

C_n= \frac{V}{n\pi} \sqrt{ \sin^2(2\pi nD) + \left[1-\cos(2\pi nD)\right]^2 }

Expand the square:

C_n= \frac{V}{n\pi} \sqrt{ \sin^2(2\pi nD) + 1 -2\cos(2\pi nD) + \cos^2(2\pi nD) }

Using $\sin^2\theta+\cos^2\theta=1$,

C_n= \frac{V}{n\pi} \sqrt{ 2-2\cos(2\pi nD) }

Using the identity $1-\cos(2x)=2\sin^2x$,

2-2\cos(2\pi nD)=4\sin^2(\pi nD)

Therefore,

\boxed{ C_n= \frac{2V}{n\pi} \left|\sin(\pi nD)\right| }

The absolute value is required because a magnitude is always non-negative.

The phase angle is

\phi_n=\operatorname{atan2}(b_n,a_n)

\boxed{ \phi_n= \operatorname{atan2} \left( 1-\cos(2\pi nD), \sin(2\pi nD) \right) }

An equivalent and more compact phase expression can be obtained from the harmonic term itself:

a_n\cos(n\omega_s t)+b_n\sin(n\omega_s t)

Using the identities $\sin(2x)=2\sin x\cos x$ and $1-\cos(2x)=2\sin^2x$,

= \frac{2V}{n\pi}\sin(\pi nD) \left[ \cos(\pi nD)\cos(n\omega_s t) + \sin(\pi nD)\sin(n\omega_s t) \right]

Using $\cos(A-B)=\cos A\cos B+\sin A\sin B$,

= \frac{2V}{n\pi}\sin(\pi nD) \cos\left(n\omega_s t-\pi nD\right)

Therefore, a compact magnitude-phase representation is

\boxed{ v(t)= VD + \sum_{n=1}^{\infty} \frac{2V}{n\pi} \sin(\pi nD) \cos\left(n\omega_s t-\pi nD\right) }

If you want the magnitude strictly non-negative, then write it as

\boxed{ v(t)= VD + \sum_{n=1}^{\infty} C_n \cos(n\omega_s t-\phi_n) }

with

\boxed{ C_n= \frac{2V}{n\pi} \left|\sin(\pi nD)\right| }

and the corresponding phase chosen according to the sign of $\sin(\pi nD)$.

Special Case: $D=0.5$

For a 50% duty cycle, substitute $D=0.5$ directly into the general Fourier-series result.

The DC component becomes

VD=\frac{V}{2}

For the cosine coefficient,

a_n=\frac{V}{n\pi}\sin(n\pi)=0

For the sine coefficient,

b_n=\frac{V}{n\pi}\left[1-\cos(n\pi)\right]

Since $\cos(n\pi)=(-1)^n$,

b_n=\frac{V}{n\pi}\left[1-(-1)^n\right]

Therefore,

\boxed{ b_n= \begin{cases} \dfrac{2V}{n\pi}, & n=1,3,5,\ldots\ 0, & n=2,4,6,\ldots \end{cases} }

Thus, the Fourier series becomes

\boxed{ v(t)= \frac{V}{2} + \frac{2V}{\pi} \left[ \sin(\omega_s t) + \frac{1}{3}\sin(3\omega_s t) + \frac{1}{5}\sin(5\omega_s t) + \frac{1}{7}\sin(7\omega_s t) +\cdots \right] }

So for $D=0.5$, the waveform contains a DC component of $V/2$ and only odd harmonics.

Special Case $D=0.5$ in Magnitude-Phase Form

For $D=0.5$, the general harmonic magnitude

C_n=\frac{2V}{n\pi}\left|\sin(\pi nD)\right|

becomes

C_n=\frac{2V}{n\pi}\left|\sin\left(\frac{n\pi}{2}\right)\right|

For even $n$,

\sin\left(\frac{n\pi}{2}\right)=0

so all even harmonics vanish.

For odd $n=1,3,5,\ldots$,

\left|\sin\left(\frac{n\pi}{2}\right)\right|=1

therefore

\boxed{ C_n= \frac{2V}{n\pi}, \qquad n=1,3,5,\ldots }

and

\boxed{ C_n=0, \qquad n=2,4,6,\ldots }

For the 50% duty-cycle waveform, the non-zero harmonics can be written with a phase of $90^\circ$ in cosine form:

\boxed{ v(t)= \frac{V}{2} + \frac{2V}{\pi}\cos(\omega_s t-90^\circ) + \frac{2V}{3\pi}\cos(3\omega_s t-90^\circ) + \frac{2V}{5\pi}\cos(5\omega_s t-90^\circ) +\cdots }

Equivalently,

\boxed{ v(t)= \frac{V}{2} + \sum_{\substack{n=1\ n\ \mathrm{odd}}}^{\infty} \frac{2V}{n\pi} \cos\left(n\omega_s t-90^\circ\right) }

This is the magnitude-phase form of the 50% duty-cycle rectangular waveform.

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Rectangular Waveform with Arbitrary Duty Cycle

Fundamental Period and Frequency

Fourier Series Derivation

DC Coefficient

Cosine Coefficient

Sine Coefficient

Special Case: D=0.5D=0.5

Special Case D=0.5D=0.5 in Magnitude-Phase Form

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Special Case: $D=0.5$

Special Case $D=0.5$ in Magnitude-Phase Form