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.
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
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.
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:
or
For Fig. 2(a) (a square-wave pattern),
and for Fig. 2(b),
The above results clearly reveal that
Fig. 1: 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
$$ \bbox[10px,border:1px solid grey]{\text{prf (or prr)} =\frac{1}{T} \, (\mathrm{Hz}} \, \text{or pulses/s}) \tag{1}$$

Fig. 2: Periodic pulse trains.
$$ \text{Duty cycle} =\frac{\text { pulse width }}{\text { period }} \times 100 \% $$
$$ \text{Duty cycle} =\frac{t_{p}}{T} \times 100 \% $$
$$\text{Duty cycle} =\frac{0.5 T}{T} \times 100 \%=\mathbf{5 0} \% $$
$$\text{Duty cycle} =\frac{0.2 T}{T} \times 100 \%=\mathbf{2 0} \% $$
Example 1: Determine the pulse repetition frequency and the duty cycle for the periodic pulse waveform of Fig. 3.
Solution:

Fig. 3: Example 1.
$$\begin{aligned}
T &=(15-6) \mu \mathrm{s}=9 \mu \mathrm{s} \\
\text{prf} &=\frac{1}{T}=\frac{1}{9 \mu \mathrm{s}} \cong \mathbf{1 1 1 . 1 1} \mathbf{k H z} \\
\text { Duty cycle } &=\frac{t_{p}}{T} \times 100 \%=\frac{(8-6) \mu \mathrm{s}}{9 \mu \mathrm{s}} \times 100 \% \\
&=\frac{2}{9} \times 100 \% \cong \mathbf{2 2 . 2 2 \%}
\end{aligned}$$
Example 2: Determine the pulse repetition frequency and the duty cycle for the oscilloscope pattern of Fig. $ 4 $ having the indicated sensitivities.
Solution:

Fig. 4: Example 2.
$$\begin{aligned}
T &=(3.2 \text { div. })(1 \mathrm{~ms} / \text { div. })=3.2 \mathrm{~ms} \\t_{p}\\
&=(0.8 \text { div. })(1 \mathrm{~ms} / \text { div. })=0.8 \mathrm{~ms} \\
\text{prf} &=\frac{1}{T}=\frac{1}{3.2 \mathrm{~ms}}=\mathbf{3 1 2 . 5} \mathbf{~ H z}
\end{aligned}$$
$$\text{Duty cycle} =\frac{t_{p}}{T} \times 100 \%=\frac{0.8 \mathrm{~ms}}{3.2 \mathrm{~ms}} \times 100 \%=\mathbf{2 5 \%} $$
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