Question

Algebraicallycalculate the rise time and fall timevalues for a sine wave at an arbitrary frequency f0(expressed in Hz). Assume that the sine wave has zeroaverage, thus Vmin= −Vmax.Express the rise time in units of the period T= 1/ f0.Using your algebraic results, calculate the rise time inμs for the sinusoidal signal at 50 kHz. Keep 4 significantdigits.

Answer 1

185.95 μsec

Step-by-step explanation:

Given that:

Sine wave =
`Amsinw_(o)t`

`V_(min)=-V_(max)`

Sinusodial signal
`f_(o)` = 50 kHz =
`5*10^4Ht`

Rise time is said to be defined as the time needed for a pulse to rise from 10% - 90% of maximum rate.

∴

`t_1` = 10% = 0.1 Am

`t_2` =90% = 0.9 Am

Using sine wave for
`t_1`; we have:

`Amsinw_(o)t_1` = 0.1 Am

`sinw_(o)t_1` = 0.1

`w_(o)t_1 = sin^(-1)(0.1)`

`w_(o)t_1` = 5.7392

`t_1=(5.7392)/(2 \pi f_0)`

`t_1=(0.9134)/( f_0)`

Using sine wave for
`t_2` ; we have:

`Amsinw_(o)t_2` = 0.9 Am

`sinw_(o)t_2 = 0.9`

`w_ot_2` =
`sin^(-1)(0.9)`

`w_ot_2` = 64.158

`t_2` =
`(64.1581)/(2 \pi f_o)`

`t_2` =
`(10.211)/(f_o)`

Change in rise time
`t_r` =
`t_2` -
`t_1`

`t_r` =
`(10.211)/(f_o)-(0.9134)/( f_0)`

`t_r` =
`(10.211-0.9134)/(f_o)`

`t_r` =
`(9.2976)/(f_o)`

since;
`f_(o)` = 50 kHz =
`5*10^4Ht`

`t_r` =
`(9.2976)/(5*10^4)`

`t_r` = 1.85952 × 10⁻⁴

`t_r` = 185.952 × 10⁻⁶ sec

`t_r` = 185.95 μ sec

∴ The rise in time in (μ sec) for the sinosodial signal at 50 kHz = 185.95 μ sec