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

Answer:

185.95 μsec

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

$t_1=\frac{5.7392}{2 \pi f_0}$

$t_1=\frac{0.9134}{ f_0}$

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

$t_2$ = $\frac{64.1581}{2 \pi f_o}$

$t_2$ = $\frac{10.211}{f_o}$

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

$t_r$ = $\frac{10.211}{f_o}-\frac{0.9134}{ f_0}$

$t_r$ = $\frac{10.211-0.9134}{f_o}$

$t_r$ = $\frac{9.2976}{f_o}$

since;   $f_{o}$  = 50 kHz = $5*10^4Ht$

$t_r$ = $\frac{9.2976}{5*10^4}$

$t_r$ = 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