# Pulse Repetition Rate and Duty Cycle

A series of pulses such as those appearing in Fig. 1 is called a pulse train. The varying widths and heights may contain information that can be decoded at the receiving end.
Fig. 1: Pulse train.
If the pattern repeats itself in a periodic manner as shown in Fig. 2(a) and (b), the result is called a periodic pulse train.
The period (T) of the pulse train is defined as the time differential between any two similar points on the pulse train, as shown in Figs. 2(a) and (b).
The pulse repetition frequency (prf), or pulse repetition rate (prr), is defined by
$$\bbox[10px,border:1px solid grey]{\text{prf (or prr)} =\frac{1}{T} \, (\mathrm{Hz}} \, \text{or pulses/s}) \tag{1}$$
Applying Eq. (1) to each waveform of Fig. $2$ will result in the same pulse repetition frequency since the periods are the same. The result clearly reveals that the shape of the periodic pulse does not affect the determination of the pulse repetition frequency.
Fig. 2: Periodic pulse trains.
The pulse repetition frequency is determined solely by the period of the repeating pulse. The factor that will reveal how much of the period is encompassed by the pulse is called the duty cycle, defined as follows:
$$\text{Duty cycle} =\frac{\text { pulse width }}{\text { period }} \times 100 \%$$
or
$$\text{Duty cycle} =\frac{t_{p}}{T} \times 100 \%$$
For Fig. 2(a) (a square-wave pattern),
$$\text{Duty cycle} =\frac{0.5 T}{T} \times 100 \%=\mathbf{5 0} \%$$
and for Fig. 2(b),
$$\text{Duty cycle} =\frac{0.2 T}{T} \times 100 \%=\mathbf{2 0} \%$$
The above results clearly reveal that
the duty cycle provides a percentage indication of the portion of the total period encompassed by the pulse waveform.
Example 1: Determine the pulse repetition frequency and the duty cycle for the periodic pulse waveform of Fig. 3.
Fig. 3: Example 1.
Solution:
\begin{aligned} T &=(15-6) \mu \mathrm{s}=9 \mu \mathrm{s} \\ \text{prf} &=\frac{1}{T}=\frac{1}{9 \mu \mathrm{s}} \cong \mathbf{1 1 1 . 1 1} \mathbf{k H z} \\ \text { Duty cycle } &=\frac{t_{p}}{T} \times 100 \%=\frac{(8-6) \mu \mathrm{s}}{9 \mu \mathrm{s}} \times 100 \% \\ &=\frac{2}{9} \times 100 \% \cong \mathbf{2 2 . 2 2 \%} \end{aligned}
Example 2: Determine the pulse repetition frequency and the duty cycle for the oscilloscope pattern of Fig. $4$ having the indicated sensitivities.
Fig. 4: Example 2.
Solution:
\begin{aligned} T &=(3.2 \text { div. })(1 \mathrm{~ms} / \text { div. })=3.2 \mathrm{~ms} \\t_{p}\\ &=(0.8 \text { div. })(1 \mathrm{~ms} / \text { div. })=0.8 \mathrm{~ms} \\ \text{prf} &=\frac{1}{T}=\frac{1}{3.2 \mathrm{~ms}}=\mathbf{3 1 2 . 5} \mathbf{~ H z} \end{aligned}
$$\text{Duty cycle} =\frac{t_{p}}{T} \times 100 \%=\frac{0.8 \mathrm{~ms}}{3.2 \mathrm{~ms}} \times 100 \%=\mathbf{2 5 \%}$$