Introduction
The bandwidth of a square wave depends on the rise time of the signal. That is, as the rise time of the signal decreases, the bandwidth increases. A common factor is ${0.35}/{t_{rise}}$ which is a rule of thumb approximation for the bandwidth of a signal given the $10-90\%$ risetime. The derivation and calculation of this factor is the subject of this article.
First, the bandwidth of a signal refers to the spectral content of the signal. Typically defined as the spectral content that lies within some threshold, often 3dB below a reference point.
-3dB Point
Recall that -3dB corresponds to half power, not amplitude. The amplitude of half power is then: $10^{-3/20} \approx 0.707$. This can be confirmed using Ohm’s law. $V^2 \div R$.
For example: $1V^2 \div 50 \Omega=20mW$. $0.707V^2 \div 50 \Omega = 10mW$.
The calculation for bandwidth given rise time of a signal can be performed using the cutoff frequency equation for a single pole RC circuit. Specifically, the bandwidth is described as the cutoff frequency of the circuit or where the output of the system is at half power which is $-3dB$.
\[\Large f_c = \frac{1}{\omega RC} = \frac{1}{2\pi RC}\]Terms
- $R$ - Resistance, in Ohms ($\Omega$).
- $C$ - Capacitance, in Farads.
- $2 \pi$ - Conversion factor from angular frequency.
- $f_c$ - Cutoff frequency ($-3dB$).
Dimensional Analysis of $RC$
Dimensional analysis involves the evaluation of dimensions, or units, of an equation to gain additional understanding of an equation. For example, in electronics the product of $R$ and $C$ is called $\tau$ (tau), or the time constant of a circuit. A dimensional analysis reveals that the product of $R$ and $C$ has units of seconds.
\[\Large [\Omega\cdot F] \equiv [\frac{s}{\textcolor{WildStrawberry}{\cancel{\textcolor{black}{F}}}} \cdot \textcolor{WildStrawberry}{\cancel{\textcolor{black}{F}}}] \equiv [s]\]Additionally, the inverse of a period is frequency. A dimensional analysis reveals where the frequency component comes from.
\[\Large \frac{1}{RC} \rightarrow s^{-1} \equiv f\]Rise Time
The rise time quantifies the time a signal takes to transition between two reference points, typically specified as a percentage of maximum amplitude. For example $10\%$ to $90\%$ and $20\%$ to $80\%$ are the most common. The following calculations use $10\%$ to $90\%$.
Calculations
The factors for the final bandwidth equation come from the charging equation.
\[\Large V_{out} = V_{in} \left ( 1 - e^{\frac{-t}{RC}} \right)\]This equation is rearranged to solve for $RC$ given the measured rise time. These factors can then be used to determine the $-3dB$ frequency content, or bandwidth of the signal.
10% Calculation ($t_1$)
Step by step calculations for $t_1$ are as follows:
| Step | Calculation | Explanation |
|---|---|---|
| 1 | $\Large V_{out} = V_{in} \left( 1-e^{\frac{-t_{1}}{RC}} \right)$ | Starting Equation |
| 2 | $\Large \frac{V_{out}}{V_{in}} = 1 - e^{\frac{-t_{1}}{RC}}$ | Move $V_{in}$ to left side by dividing both sides by $V_{in}$. |
| 3 | $\Large 0.1 = 1-e^{\frac{-t_{1}}{RC}}$ | Replace $V_{out}/V_{in}$ with the reference point (10%). |
| 4 | $\Large 0.1 - 1 = -e^{\frac{-t_1}{RC}}$ | Subtract $1$ from both sides. |
| 5 | $\Large 0.9 = -e^{\frac{-t_1}{RC}}$ | Simplify. |
| 6 | $\Large -0.9 = e^{\frac{-t_1}{RC}}$ | Multiply both sides by $-1$. |
| 7 | $\Large - \ln(0.9) = \ln \left( e^{\frac{-t_1}{RC}} \right)$ | Isolate the exponent by taking the natural log. |
| 8 | $\Large - \ln(0.9) = \frac{-t_1}{RC}$ | Simplify. |
| 9 | $\Large - \ln(0.9)\cdot RC = -t_1$ | Multiply both sides by $RC$ to isolate time. |
| 10 | $\Large -1\cdot -0.10536051\cdot RC = t_1$ | Multiply both sides by $-1$ to isolate positive time. |
| 12 | $\Large t_1 = 0.10536051 \cdot RC$ | Simplify. |
$V_{out}/V_{in}$ Term
The term $V_{out}/V_{in}$ comes from the ratio of the risetime. $10\%$ would be $10\%$ of the maximum value for instance. The decimal value is then $0.1$.
90% Calculation ($t_2$)
Following the same steps as used for $t_1$, $t_2$ can be calculated. Note that for brevity some steps have been combined.
| Step | Calculation | Explanation |
|---|---|---|
| 1 | $\Large V_{out} = V_{in} \left( 1-e^{\frac{-t_{1}}{RC}} \right)$ | Starting Equation. |
| 2 | $\Large \frac{V_{out}}{V_{in}} = 1 - e^{\frac{-t_{1}}{RC}}$ | Move $V_{in}$ to left side by dividing both sides by $V_{in}$. |
| 3 | $\Large 0.9 - 1 = -e^{\frac{-t_2}{RC}}$ | Replace $V_{out}/V_{in}$ with the reference point ($90\%$) and subtract $1$ from both sides. |
| 4 | $\Large -0.1 = e^{\frac{-t_2}{RC}}$ | Multiply both sides by $-1$. |
| 5 | $\Large - \ln(0.1) = \ln \left( e^{\frac{-t_2}{RC}} \right)$ | Isolate the exponent by taking the natural log of both sides. |
| 6 | $\Large - \ln(0.1) = \frac{-t_2}{RC}$ | Simplify. |
| 7 | $\Large - \ln(0.1)\cdot RC = -t_2$ | Multiply both sides by RC to cancel the denominator term. |
| 8 | $\Large -1\cdot -2.30258509 \cdot RC = t_2$ | Multiply by $-1$ to isolate $t_2$. Simplify logarithm. |
| 9 | $\Large t_2 = 2.30258509 \cdot RC$ | Simplify. |
Solve for $RC$
Now that the factors have been determined, solving the difference results in an equation in terms of $RC$ that can be used in the bandwidth equation.
\[\Large t_{rise} = \textcolor{WildStrawberry}{t_2} - \textcolor{BurntOrange}{t_1}\] \[\Large t_{rise} = \textcolor{WildStrawberry}{2.30258509}\textcolor{Cerulean}{RC} - \textcolor{BurntOrange}{0.10536051}\textcolor{Cerulean}{RC}\] \[\Large t_{rise} = 2.19722458\textcolor{Cerulean}{RC}\]Rearrange and solve for RC:
\[\Large \frac{t_{rise}}{2.19722458} = \textcolor{Cerulean}{RC}\]Bandwidth Factor
Now that all equations have been solved, they can be input into the cutoff frequency equation to determine the bandwidth.
\[\Large BW = \frac{1}{2\pi \textcolor{Cerulean}{RC}} = \frac{1}{2\pi \textcolor{Cerulean}{\left( \frac{t_{rise}}{2.19722458} \right)}}\]Distribute denominator:
\[\Large 2\pi \textcolor{Cerulean}{\left( \frac{t_{rise}}{2.19722458} \right)} = \frac{2\pi t_{rise}}{2.19722458}\]Take the inverse:
\[\Large BW= \frac{2.19722458}{2\pi t_{rise}}\]Simplify:
\[\Large BW = \frac{2.19622458}{2\pi} \cdot \frac{1}{t_{rise}} = \frac{0.34969915299}{t_{rise}}\] \[\Large BW_{(10-90\%)} \approx \frac{0.35}{t_{rise}}\]Summary
The bandwidth equation can be used to determine the frequency content of a signal provided the measured rise time and calculated constants. Bandwidth is inversely proportional to rise time: as rise time decreases, bandwidth increases. Further, the constants depend on what point the risetime has been measured from. These constants are calculated by solving the time constant equation.
For the standard $10\%-90\%$ risetime $0.35$ can be used. Similar derivations can determine the factor for a $20\%-80\%$ risetime as well as an arbitrary risetime based upon application needs.