Answer:
a) 262 Hz
b) 18.2°
Explanation:
a) The maximal sinusoidal input frequency is given by:
[tex]|H(jw)|=\frac{K}{\sqrt{1+w^2\tau^2} } \\\\where\ |H(jw)|=magnitude, w=angular\ frequency, \tau=time\ constant\\\\[/tex]
The magnitude that will keep the output error less than 5% is 95% of its DC value = 0.95K. hence |H(jw)| = 0.95K, τ = 20 ms = 0.02 s
[tex]|H(jw)|=\frac{K}{\sqrt{1+w^2\tau^2} }\\\\0.95K=\frac{K}{\sqrt{1+w^2(0.02)^2} } \\\\0.95=\frac{1}{\sqrt{1+0.0004w^2} }\\\\squaring\ both\ sides:\\\\0.9025=\frac{1}{1+0.0004w^2}\\ \\0.9025(1+0.0004w^2)=1\\\\0.9025+0.000361w^2=1\\\\0.000361w^2=0.0975\\\\w^2=270.083\\\\w=16.4\ rad/s\\\\f=\frac{w}{2\pi}=\frac{16.4}{2\pi} =2.62\ Hz[/tex]
b)
[tex]The\ phase\ angle(\phi)=tan^{-1}(\frac{-w\tau}{1} )\\\\\phi=tan^{-1}(\frac{-16.4*0.02}{1} )=-18.2^o[/tex]
