Properties of a Pulse Waveform
Amplitude
For most applications, the amplitude of a pulse waveform is defined as the peak-to-peak value. Of course, if the waveforms all start and return to the zero-volt level, then the peak and peak-to-peak values are synonymous. For the purposes of this text, the amplitude of a pulse waveform is the peak-to-peak value, as illustrated in Figs. 1 and 2.
Fig. 1: Ideal pulse waveform.
Fig. 2: Actual pulse waveform.
Pulse Width
For the ideal pulse of Fig. 1, the pulse width is the same at any level, whereas $t_p$ for the waveform of Fig. 2 is a very specific value.Base-Line Voltage
The waveforms of Figs. 1 and 2 both have a 0-V base-line voltage. In Fig. 3(a) the base-line voltage is 1 V, whereas in Fig. 3(b) the base-line voltage is $-4 V$
Fig. 3: Defining the base-line voltage.
Positive-Going and Negative-Going Pulses
The waveform of Fig. 3(a) is a positive-going pulse, whereas the waveform of Fig. 3(b) is a negative-going pulse. Even though the base-line voltage of Fig. 4 is negative, the waveform is positive-going (with an amplitude of 10 V) since the voltage increased in the positive direction from the base-line voltage.
Fig. 4: Positive-going pulse.
Rise Time ($t_r$) and Fall Time ($t_f $)
Of particular importance is the time required for the pulse to shift from one level to another. The rounding (defined in Fig. 5) that occurs at the beginning and end of each transition makes it difficult to define the exact point at which the rise time should be initiated and terminated. For this reason, Note that there is no requirement that $t_r$ equal $t_f $.
Fig. 5: Defining $t_r$ and $t_f$
Tilt
An undesirable but common distortion normally occurring due to a poor low-frequency response characteristic of the system through which a pulse has passed appears in Fig. 6. The drop in peak value is called tilt, droop, or sag. The percentage tilt is defined by $$\bbox[10px,border:1px solid grey]{ \% \text{tilt} = \frac{V_1 - V_2}{V} \times 100\%} \tag{1}$$ where V is the average value of the peak amplitude as determined by $$ \bbox[10px,border:1px solid grey]{V = \frac{V_1 - V_2}{V}} \tag{2}$$
Fig. 6: Defining tilt.
Fig. 7: Defining preshoot, overshoot, and ringing.
Example 1: Determine the following for the pulse waveform of Fig. 8:
a. positive- or negative-going?
b. base-line voltage
c. pulse width
d. maximum amplitude
e. tilt
Solution:
a. positive-going
b. $ V_{b}=-4 \mathrm{~V} $
c. $ t_{p}=(12-7) \mathrm{ms}=\mathbf{5} \mathbf{~ m s} $
d. $ V_{\max }=8 \mathrm{~V}+4 \mathrm{~V}=\mathbf{1 2} \mathbf{V} $
e.
(Remember, $ V $ is defined by the average value of the peak amplitude.
a. positive- or negative-going?
b. base-line voltage
c. pulse width
d. maximum amplitude
e. tilt
Fig. 8: Example 1.
a. positive-going
b. $ V_{b}=-4 \mathrm{~V} $
c. $ t_{p}=(12-7) \mathrm{ms}=\mathbf{5} \mathbf{~ m s} $
d. $ V_{\max }=8 \mathrm{~V}+4 \mathrm{~V}=\mathbf{1 2} \mathbf{V} $
e.
$$ V=\frac{V_{1}+V_{2}}{2}=\frac{12 \mathrm{~V}+11 \mathrm{~V}}{2}=\frac{23 \mathrm{~V}}{2}=11.5 \mathrm{~V} $$
$$\% tilt =\frac{V_{1}-V_{2}}{V} \times 100 \%=\frac{12 \mathrm{~V}-11 \mathrm{~V}}{11.5 \mathrm{~V}} \times 100 \%=\mathbf{8 . 6 9 6 \%} $$

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