Answer :
Answer:
185.95 μsec
Explanation:
Given that:
Sine wave = [tex]Amsinw_{o}t[/tex]
[tex]V_{min}=-V_{max}[/tex]
Sinusodial signal [tex]f_{o}[/tex] = 50 kHz = [tex]5*10^4Ht[/tex]
Rise time is said to be defined as the time needed for a pulse to rise from 10% - 90% of maximum rate.
∴
[tex]t_1[/tex] = 10% = 0.1 Am
[tex]t_2[/tex] =90% = 0.9 Am
Using sine wave for [tex]t_1[/tex]; we have:
[tex]Amsinw_{o}t_1[/tex] = 0.1 Am
[tex]sinw_{o}t_1[/tex] = 0.1
[tex]w_{o}t_1 = sin^{-1}(0.1)[/tex]
[tex]w_{o}t_1[/tex] = 5.7392
[tex]t_1=\frac{5.7392}{2 \pi f_0}[/tex]
[tex]t_1=\frac{0.9134}{ f_0}[/tex]
Using sine wave for [tex]t_2[/tex] ; we have:
[tex]Amsinw_{o}t_2[/tex] = 0.9 Am
[tex]sinw_{o}t_2 = 0.9[/tex]
[tex]w_ot_2[/tex] = [tex]sin^{-1}(0.9)[/tex]
[tex]w_ot_2[/tex] = 64.158
[tex]t_2[/tex] = [tex]\frac{64.1581}{2 \pi f_o}[/tex]
[tex]t_2[/tex] = [tex]\frac{10.211}{f_o}[/tex]
Change in rise time [tex]t_r[/tex] = [tex]t_2[/tex] - [tex]t_1[/tex]
[tex]t_r[/tex] = [tex]\frac{10.211}{f_o}-\frac{0.9134}{ f_0}[/tex]
[tex]t_r[/tex] = [tex]\frac{10.211-0.9134}{f_o}[/tex]
[tex]t_r[/tex] = [tex]\frac{9.2976}{f_o}[/tex]
since; [tex]f_{o}[/tex] = 50 kHz = [tex]5*10^4Ht[/tex]
[tex]t_r[/tex] = [tex]\frac{9.2976}{5*10^4}[/tex]
[tex]t_r[/tex] = 1.85952 × 10⁻⁴
[tex]t_r[/tex] = 185.952 × 10⁻⁶ sec
[tex]t_r[/tex] = 185.95 μ sec
∴ The rise in time in (μ sec) for the sinosodial signal at 50 kHz = 185.95 μ sec