Question

The transfer function of a system is given by,

\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{1 - s}}{{1 + s}}\)

Let the output of the system be \({v_0}\left( t \right) = {V_m}\sin \left( {\omega t + \phi } \right)\) for the input, \({v_i}\left( t \right) = {V_m}\sin \left( {\omega t} \right)\). Then the minimum and maximum values of ϕ (in radians) are respectively
1. \(- \frac{\pi }{2}\;and\frac{\pi }{2}\)
2. \(- \frac{\pi }{2}\;and\;0\)
3. \(0\;and\frac{\pi }{2}\)
4. \(- \pi \;and\;0\)

Answer 1

Correct Answer - Option 4 : \(- \pi \;and\;0\)

Transfer function,

\(\frac{{{V_o}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{1 - s}}{{1 + s}}\)

V_i (t) = V_m sin(ωt)

V_o (t) = V_m sin(ωt + ϕ)

Here, ϕ = tan^-1 (-ω) - tan^-1(ω)

= - tan¹(ω) - tan^-1 (ω)

= - 2 tan^-1(ω)

At ω = 0, 

⇒ ϕ = - 2 tan-1(0) = 0 (since tan^-1(0) = 0 )

At ω = ∞,

⇒ - 2 tan-1(∞) = - 2 × π/2 = - π (since tan-1(∞) = π/2)

Range of ϕ = (-π, 0)