PWM: exact and approximate time and frequency domain solutions

Exact Solution
Approximate Time Domain Solution
Approximate Frequency Domain Solution

Exact Derivation of PWM output

The figure below shows a PWM signal, x(t), with period T=1S and a duty cycle of 0.6 (or 60%) as a solid black line. Also shown is the ouput, y(t), of a system with time constant τ=0.5S as a blue line.

We will find

• an expression for the voltage during T₁, while the input is high, and the output is increasing,
• an expression for the output, y(t), during T₁, while the input is low and the output is decreasing,
• the values of y_max and y_min (the maximum and minimum values of the output,
• the "ripple" in the output (i.e., y_max-y_min) which depends on T₁,
• the maximum, or worst case, ripple,
• the average value of y(t).

PWM output increasing (during T₁)

First consider the output during the time interval T₁ which we will call y_high(t). The voltage starts at y₁ and is asymptotic to 1 (as shown by the dotted line)

\[{y_{high}}\left( t \right) = {y_{high}}\left( \infty \right) + \left( {{y_{high}}\left( {{0^ + }} \right) - {y_{high}}\left( \infty \right)} \right){e^{ - t/\tau }} = 1 + \left( {{y_{min}} - 1} \right){e^{ - t/\tau }}\]

PWM output decreasing (during T₂)

To make the math during this section easier, we will redefine the origin of the time axis so that t=0 when the PWM input falls (in previous graphs, t=0 when PWM input rises). Now consider the output during the time interval T₂ which we will call y_low(t). With this modified time origin, the voltage starts at y₂ and is asymptotic to 0 (as shown by the dotted line)

\[{y_{low}}\left( t \right) = {y_{low}}\left( {0^ + } \right){e^{ - t/\tau }} = y_{max}{e^{ - t/\tau }}\]

Finding y_min and y_max.

We have expressions for y_low(t) and y_high(t) in terms of y_min and y_max, but we don't yet know y_min and y_max. However, if you examine the graphs you can see that

\[y_{max} = y_{high} \left( T_1 \right) = 1 + \left( {{y_{min}} - 1} \right){e^{ - T_1/\tau }}\]

and

\[y_{min} = y_{low} \left( T_2 \right) =y_{max} {e^{ - T_2/\tau }}\]

We can solve these equations (using the fact the T=T₁+T₂) to get

\[{y_{\max }} = \frac{{1 - {e^{ - {T_1}/\tau }}}}{{1 - {e^{ - T/\tau }}}},\quad \quad {y_{\min }} = {y_{\max }}{e^{ - {T_2}/\tau }} = {e^{ - {T_2}/\tau }}\frac{{1 - {e^{ - {T_1}/\tau }}}}{{1 - {e^{ - T/\tau }}}} = \frac{{{e^{ - {T_2}/\tau }} - {e^{ - T/\tau }}}}{{1 - {e^{ - T/\tau }}}}\]

Expression for y_high(t) and y_low(t)

\[\begin{align}
{y_{high}}\left( t \right) &= 1 + \left( {{y_{\min }} - 1} \right){e^{ - t/\tau }} = 1 + \left( {{e^{ - T/\tau }}\frac{{{e^{{T_1}/\tau }} - 1}}{{1 - {e^{ - T/\tau }}}} - 1} \right){e^{ - t/\tau }} = 1 + \left( {{e^{ - T/\tau }}\frac{{{e^{{T_1}/\tau }} - 1}}{{1 - {e^{ - T/\tau }}}} - \frac{{1 - {e^{ - T/\tau }}}}{{1 - {e^{ - T/\tau }}}}} \right){e^{ - t/\tau }} \\
&= 1 + \left( {\frac{{{e^{ - T/\tau }}{e^{{T_1}/\tau }} - 1}}{{1 - {e^{ - T/\tau }}}}} \right){e^{ - t/\tau }} \\
{y_{low}}\left( t \right) &= {y_{max}}{e^{ - t/\tau }} = \frac{{1 - {e^{ - {T_1}/\tau }}}}{{1 - {e^{ - T/\tau }}}}{e^{ - t/\tau }} \\
\end{align} \]

Expression for output ripple (i.e., y_ripple = y_max-y_min)

The ripple is dependent on the value of T₁, so we will write it as a function of T₁.

\[ y_{ripple}\left( {{T_1}} \right) = {y_{\max }} - {y_{\min }} = \frac{{1 - {e^{ - {T_1}/\tau }}}}{{1 - {e^{ - T/\tau }}}} - \frac{{{e^{ - T/\tau }}{e^{{T_1}/\tau }} - {e^{ - T/\tau }}}}{{1 - {e^{ - T/\tau }}}} = \frac{{1 - {e^{ - {T_1}/\tau }} - {e^{ - T/\tau }}{e^{{T_1}/\tau }} + {e^{ - T/\tau }}}}{{1 - {e^{ - T/\tau }}}}\]

Maximum (worst case ripple)

The ripple depends on the value of T₁, so we need to find the value of T₁ that maximizes the ripple. There is no ripple when T₁=0, or T₁=T. The ripple is symmetric about T₁/T=0.5 (if the output is high 40% of the time and low 60%, the values of y_min and y_max are just switched from the case when the output is high 60% of the time), it is reasonable to assume the peak occurs at T₁/T=0.5.

Just to make sure this is correct, we can prove it by differentiating our expression for y_ripple(T₁) with respect to T₁ and setting it to zero

\[\begin{align}
\frac{{\partial y_{ripple}\left( {{T_1}} \right)}}{{\partial {T_1}}} &= \frac{{{e^{{T_1}/\tau }}{\mkern 1mu} {e^{ - T/\tau }} - {e^{ - {T_1}/\tau }}}}{{\tau {\mkern 1mu} \left( {{e^{ - T/\tau }} - 1} \right)}} = 0{\text{,}}\quad {\text{so}} \\
{e^{{T_1}/\tau }}{\mkern 1mu} {e^{ - T/\tau }} - {e^{ - {T_1}/\tau }} &= 0 \\
{e^{{T_1}/\tau }}{\mkern 1mu} {e^{ - T/\tau }} &={e^{ - {T_1}/\tau }},\quad {\text{take the logarithm}} \\
\frac{{{T_1}}}{\tau } - \frac{T}{\tau } &= - \frac{{{T_1}}}{\tau }{\text{,}}\quad {\text{or}} \\
{T_1} &= \frac{T}{2} \\
\end{align} \]

which shows that the peak is at T₁/T=0.5, as we expected, and the worst case ripple is

\[{y_{ripple,\max }} = \frac{{1 - {e^{ - T/(2\tau )}} - {e^{ - T/\tau }}{e^{T/(2\tau )}} + {e^{ - T/\tau }}}}{{1 - {e^{ - T/\tau }}}} = \frac{{1 - 2{e^{ - T/(2\tau )}} + {e^{ - T/\tau }}}}{{1 - {e^{ - T/\tau }}}} = \frac{{{{\left( {1 - {e^{ - T/(2\tau )}}} \right)}^2}}}{{1 - {e^{ - T/\tau }}}}\]

Average value of PWM output

There are several ways to find the average value of the PWM output. Here we will find the average value through integration. This is conceptually easy, but computationally difficult. In the section describing the frequency domain approximation we will use another techniques that is conceptually very simple, but is conceptually more difficult (we will also find it in the discussion of the time domain approximation, which is somewhere between the two both conceptually and computationally). We start with the definition of the average:

\[PWM\;output\;average = \frac{1}{T}\left( {\int\limits_0^T {{y_{PWMout}}\left( t \right)dt} } \right)\]

The notation on the next line is a little tricky. We split the two integral into two parts, one where the function is rising, and one where it is falling. Recall that when we defined the latter function, we reset the time origin; this means when we integrate y_low(t), we integrate from 0 to T₂, instead of T₁ to 1. The rest of the derivation is some straightforward calculus and algebra.

\[\begin{aligned} PWM\;output\;average &= \frac{1}{T}\left( {\int\limits_0^{{T_1}} {{y_{high}}\left( t \right)dt} + \int\limits_0^{{T_2}} {{y_{low}}\left( t \right)dt} } \right) = \frac{1}{T}\left( {\int\limits_0^{{T_1}} {\left( {1 + \left( {\frac{{{e^{ - T/\tau }}{e^{{T_1}/\tau }} - 1}}{{1 - {e^{ - T/\tau }}}}} \right){e^{ - t/\tau }}} \right)dt} + \int\limits_0^{{T_2}} {\frac{{1 - {e^{ - {T_1}/\tau }}}}{{1 - {e^{ - T/\tau }}}}{e^{ - t/\tau }}dt} } \right) \\ &= \frac{1}{T}\left( {\frac{{\tau {\mkern 1mu} \left( {{e^{{T_1}/\tau }} + {e^{\frac{{T - {T_1}}}{\tau }}} - {e^{T/\tau }} - 1} \right)}}{{{\text{1}} - {e^{T/\tau }}}} + \frac{{{T_1} + \tau - {T_1}{\mkern 1mu} {e^{T/\tau }} + \tau {\mkern 1mu} {e^{T/\tau }} - \tau {\mkern 1mu} {e^{{T_1}/\tau }} - \tau {\mkern 1mu} {e^{\frac{{T - {T_1}}}{\tau }}}}}{{{\text{1}} - {e^{T/\tau }}}}} \right) \\ &= \frac{{\tau {e^{{T_1}/\tau }} + \tau {e^{\frac{{T - {T_1}}}{\tau }}} - \tau {e^{T/\tau }} - \tau + {T_1} + \tau - {T_1}{\mkern 1mu} {e^{T/\tau }} + \tau {\mkern 1mu} {e^{T/\tau }} - \tau {\mkern 1mu} {e^{{T_1}/\tau }} - \tau {\mkern 1mu} {e^{\frac{{T - {T_1}}}{\tau }}}}}{{T\left( {{\text{1}} - {e^{T/\tau }}} \right)}} \\ &= \frac{{{T_1}}}{T} \\ &= PWM\;input\;average \\ \end{aligned} \]

This is a power result! It states that the output of the PWM output is the same as that of the PWM input. So, for the case described here, since the duty cycle is 0.6, the input average is 0.6, as is the output average.

1^st Order Time Domain Derivation of PWM output

The exact analysis is very laborious, but we can make an accurate linear model if the pulse width modulated square wave is much faster than the system being driven; in other words if T<<τ. The figure below shows a PWM signal, x(t), with period T=1S and a duty cycle of 0.6 (or 60%) as a solid black line. Also shown is the exact ouput, y(t), of a system with time constant τ=0.5S as a blue line, and a the linear approximation in red. You can see they are quite close:

Average value

We want to start by finding a relationship between T₁ and y_avg.

• First consider the plot during the interval labeled T₁, as the PWM output is rising (see image at right)
  Note that the graph at the right has the time axis shifted so t=0 is set to the time when the linear approximation (red) goes through the average PWM output (y_avg, shown in magenta).
  • We want to match the slope of the exponential (shown as dotted red) with the slope of the linear approximation (solid red) as it passes through the average.
  • The slope of the linear approximation is (y_max-y_min)/T₁.
  • The slope of the exponential (see above) is Δy_T1/τ = (1-y_avg)/τ. (Recall Δy_T1 is the difference between the asymptotic value (1) of the exponential and the initial value (y_avg).
  • The two slopes are equal so $\frac{{{y_{\max }} - {y_{\min }}}}{{{T_1}}} = \frac{{\Delta {y_{T1}}}}{\tau } = \frac{{1 - {y_{avg}}}}{\tau }$.
    
• Now consider the plot during the interval labeled T₂, as the PWM output is false (see image at right)
  Note that the graph at the right has the time axis shifted so t=0 is set to the time when the linear approximation (red) goes through the average PWM output (y_avg, shown in magenta).
  • We want to match the slope of the exponential (shown as dotted red) with the slope of the linear approximation (solid red) as it passes through the average.
  • The slope of the linear approximation is (y_min-y_max)/T₂.
  • The slope of the exponential (see above) is Δy_T2/τ = -y_avg/τ. (Recall Δy_T2 is the difference between the asymptotic value (0) of the exponential and the initial value (y_avg).
  • The two slopes are equal so $\frac{{{y_{\min }} - {y_{\max }}}}{{{T_2}}} = \frac{{\Delta {y_{T2}}}}{\tau } = - \frac{{{y_{avg}}}}{\tau }$.
• We know that the change in PWM output during T₁ is just the slope multiplied by T₁ or ${\Delta _{rising}} = \left( {\frac{{1 - {y_{avg}}}}{\tau }} \right){T_1}$
• Likewise, the change in output during T₂ is just the slope multiplied by T₂ or ${\Delta _{falling}} = \left( { - \frac{{{y_{avg}}}}{\tau }} \right){T_2}$
• During one cycle of the input, the sum of the changes during T₁ and T₂ must add to zero (so the cycle can repeat). We also use the fact that T₂=T-T₁
  \[\begin{align} {\Delta _{rising}} + {\Delta _{falling}} &= 0 = \left( {1 - {y_{avg}}} \right)\frac{{{T_1}}}{\tau } - {y_{avg}}\frac{{{T_2}}}{\tau },\quad \tau {\text{ cancels from equation}} \\ 0 &= \left( {1 - {y_{avg}}} \right){T_1} - {y_{avg}}\left( {T - {T_1}} \right) \\ {y_{avg}} \cdot T &= {T_1} \\ {y_{avg}} &= \frac{{{T_1}}}{T} \\ \end{align} \]

so the average of the PWM output is equal to the average of the PWM input and is equal to T₁/T.

Ripple

We can easily find the ripple by solving for Δy_rising or -Δy_falling (since they are equal).

\[{y_{\max }} - {y_{\min }} = - \Delta_{falling} = {y_{ripple}} = {y_{mid}}\frac{{{T_2}}}{\tau } = \frac{{{T_1}}}{T}\frac{{{T_2}}}{\tau } = \frac{{{T_1}\left( {T - {T_1}} \right)}}{{T \cdot \tau }}\]

Worst case ripple

The ripple depends on the value of T₁, so we need to find the value of T₁ that maximizes the ripple. There is no ripple when T₁=0, or T₁=T. The ripple is symmetric about T₁/T=0.5 (if the output is high 40% of the time and low 60%, the values of y_min and y_max are just switched from the case when the output is high 60% of the time), it is reasonable to assume the peak occurs at T₁/T=0.5. Just to make sure this is correct, we can prove it by differentiating our expression for y_ripple(T₁) with respect to T₁ and setting it to zero

\[y_{ripple}\left( T_1 \right) = \frac{{{T_1}\left( {T - {T_1}} \right)}}{{T \cdot \tau }}\]\[\frac{{\partial {y_{ripple}}\left( {{T_1}} \right)}}{{\partial {T_1}}} = \frac{{T - 2{T_1}}}{{T \cdot \tau }} = 0\]

which shows that the peak is at T₁/T=0.5, as we expected, and the worst case ripple is

\[{y_{ripple,\max }} = \frac{{\frac{T}{2}\left( {T - \frac{T}{2}} \right)}}{{T \cdot \tau }} = \frac{T}{{4\tau }}\]

 

Final words

Because this derivation is so much simpler than the exact derivation, and because the results are similarly less complex, this is usually a much more fruitful way of thinking about pulse width modulation.

 

1^st Order Frequency Domain Derivation of PWM output

Understanding Pulse Width Modulation in the context of a frequency domain analysis is in some ways the simplest (the mathematics is quite simple), and in some ways the most complex (it requires a knowledge of transfer functions, sinusoidal steady state, and Fourier Series. We assume we have a low pass system, H_sys(s), with and PWM input, x(t), and we want to find the system output, y(t).

To start this analysis consider a PWM input (in this case we have the pulse is symmetric around t=0 because it makes the math easier, but this has no impact on our calculation of the average value of the output, nor on our calculation of the ripple.

The duty cycle is given by T₁/T, in this case 0.6. The average is also 0.6 (since the input has a value of one 60% of the time, and is otherwise equal to 0). The Fourier Series representation for this waveform is derived here, and is given by

\[x(t) = {a_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos \left( {n{\omega _{pwm}}t} \right)} \]

where

\[a_0=average=\frac{T_1}{T}\]

and otherwise,

\[{a_n} = {2 \over {n\pi }}\sin \left( {n\pi {{T_1} \over T}} \right)\]

In these equations ω_pwm is the frequency of the PWM signal so $\omega_{pwm}=\frac{2\pi}{T}$.

From a frequency domain perspective PWM works by attenuating the time varying parts of the signal (all of the cosine terms), but not the constant (i.e., average) term, a₀. Let's examine how that works.

Consider the case when T₁/T=0.6. The first five values of a_n are given below, along with the frequency of oscillation, and (to the right) a graph showing the amplitudes (i.e., a_n) at these frequencies.

n	an	ω
0	0.600	0
1	0.606	ωpwm
2	-0.187	2·ωpwm
3	-0.125	3·ωpwm
4	0.151	4·ωpwm

We have already stated that the system being driven is a low-pass system, so let's assume a first order system, with a break frequency of ω_low, so

\[{H_{sys}}(s) = \frac{1}{{1 + \frac{s}{{{\omega _{low}}}}}}\]

Note that ω_low=1/τ, where τ is the time constant we used in the time domain analyses on this page.

Recall from sinusoidal steady state analysis that if the input to a system, H_sys(s), is

\[A \cdot \cos \left( {\omega t} \right)\]

that the output is

\[A \cdot M \cdot \cos \left( {\omega t + \theta } \right)\]

where

\[M = \left| {{H_{sys}}(j\omega )} \right| = \left| {\frac{1}{{1 + \frac{{j\omega }}{{{\omega _{kiw}}}}}}} \right| = \frac{1}{{\sqrt {\left( {1 + {{\left( {\frac{\omega }{{{\omega _{kiw}}}}} \right)}^2}} \right)} }},\quad \quad \theta = \angle {H_{sys}}(j\omega ) = - \arctan \left( {\frac{\omega }{{{\omega _{kiw}}}}} \right)\]

We will not need to know the angle, θ, if we are only interested in the magnitude of the variations (ripple) in the output. It is value is included here for completeness.

Such a system with sinusoidal input and output is shown below.

Each of the component cosines from the input is at a frequency of n·ω_pwm (with a frequency of 0 (i.e., a constant) associated with the a₀ term).

\[x(t) = {a_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos \left( {n{\omega _{pwm}}t} \right)} \]

These individual component will have their magnitude and phase changed by the system in accordance with the magnitude and phase of the transfer function so that the output is given by:

\[y(t) = {a_0} \cdot {H_{sys}}(j \cdot 0) + \sum\limits_{n = 1}^\infty {{a_n}\left| {{H_{sys}}(jn{\omega _{pwm}})} \right|\cos \left( {n{\omega _{pwm}}t + \angle {H_{sys}}(jn{\omega _{pwm}})} \right)} \]

If we call a_n·|H_sys(j·n·ω_pwm)|=α_n (and ∠H(j·n·ω_pwm) we can rewrite the series as

\[y(t) = {\alpha _0} + \sum\limits_{n = 1}^\infty {{\alpha _n}\cos \left( {n{\omega _{pwm}}t + {\theta _n}} \right)} \]

Note: from here on out weLet's add |H_sys(j·n·ω_pwm) and α_n to our previous table (for the case where ω_low=ω_pwm/3). Immediately you see that all of the coefficients of the cosine terms in the output (the α_n terms) are smaller than the corresponding terms in the input (the a_n terms) by a factor of |H_sys(j·n·ω_pwm|, except for a₀ (which represents the constant,or average, part of the PWM input). Since

 

 

n	an	ω	|Hsys(j·n·ωpwm)|	αn
0	0.600	0	1.000	0.600
1	0.606	ωpwm	0.316	0.191
2	-0.187	2·ωpwm	0.164	-0.031
3	-0.125	3·ωpwm	0.110	-0.014
4	0.151	4·ωpwm	0.083	0.013

We can also show this visually. There are three graphs below. The first shows the values of a_n at various frequencies (this was shown previously, above). The next graph shows the magnitude of the transfer function (|H_sys(j·n·ω)|) over the same frequency range. The last graph shows α_n, and is simply the product of the first two graphs.

By decreasing ω_low, we can attenuate the higher frequency parts of the signal even more. Here ω_low=ω_pwm/10.

Average value

Clearly, as long as H(j0)=1, the average value of the input is equal to the average value of the output. In other words, the constant (a₀) term of the input is equal to the constant (α₀) term of the output because α₀=|H(j0)|·a_n.

Ripple

Since the coefficients of the Fourier Series drop off rapidly as n increases, we create a first order approximation to the output with just the first two terms of the series. In other words we can estimate the output as a constant, plus the lowest frequency (n=1) cosine:

\[y(t) = {\alpha _0} + {\alpha _1}\cos \left( {{\omega _{pwm}}t + {\theta _{_1}}} \right)\] where

We assume all of the ripple is from this cosine, so the peak to peak ripple is 2·α1.

\[\begin{align} {\alpha _1} &= {a_1}\left| {{H_{sys}}(j{\omega _{pwm}})} \right| \\ &= {a_1}\left| {\frac{1}{{1 + \frac{{j{\omega _{pwm}}}}{{{\omega _{low}}}}}}} \right| = \frac{{{a_1}}}{{\sqrt {1 + {{\left( {\frac{{{\omega _{pwm}}}}{{{\omega _{low}}}}} \right)}^2}} }} = \frac{{\frac{2}{\pi }\sin \left( {\pi \frac{T}{{{T_1}}}} \right)}}{{\sqrt {1 + {{\left( {\frac{{{\omega _{pwm}}}}{{{\omega _{low}}}}} \right)}^2}} }} \\ y_{ripple} &= 2 \cdot {\alpha _1} = \frac{{\frac{4}{\pi }\sin \left( {\pi \frac{T}{{{T_1}}}} \right)}}{{\sqrt {1 + {{\left( {\frac{{{\omega _{pwm}}}}{{{\omega _{low}}}}} \right)}^2}} }} \\ \end{align} \]

We can put this in terms of time domain quantities by noting that the time constant of the system is τ=1/ω_low, and the period of the pwm signal is given by T=2·π/ω_pwm.

\[y_{ripple} = \frac{{\frac{4}{\pi }\sin \left( {\pi \frac{T}{{{T_1}}}} \right)}}{{\sqrt {1 + {{\left( {2\pi \frac{\tau }{T}} \right)}^2}} }}\]

Worst case ripple

The worst case ripple happens when the sinusoidal term in the expression for the ripple is equal to 1, or when T₁=T/2, in which case

\[{y_{ripple,\max }} = \frac{{\frac{4}{\pi }}}{{\sqrt {1 + {{\left( {\frac{{{\omega _{pwm}}}}{{{\omega _{low}}}}} \right)}^2}} }} = \frac{{\frac{4}{\pi }}}{{\sqrt {1 + {{\left( {2\pi \frac{\tau }{T}} \right)}^2}} }}\]

An Example

If we have a PWM input with T=1S (so ω_pwm=2·π/T=6.28 rad/S) with a duty cycle of 0.6 (or 60%), pass it through a system with time constant τ=0.5S (so ω_low =2rad/S), we get the output shown below. The average is a magenta line, the ijnput is black, the exact output is blue, and the approximate output (with just the a₀ and a₁ terms of the Fourier Series) is green.

Note that the Fourier Series approximation is slightly shifted relative to the pulses. This is because of the angle, θ, of the transfer function (which we have been ignoring because it does not affect the size of the ripple).

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