Question

A first-order low-pass instrument has a time constant of 20 ms. Find the maximal sinusoidal input frequency that will keep output error due to frequency response less than 5%. Find the phase angle at this frequency

Answer 1

a) 262 Hz

b) 18.2°

Step-by-step explanation:

a) The maximal sinusoidal input frequency is given by:

`|H(jw)|=(K)/(√(1+w^2\tau^2) ) \\\\where\ |H(jw)|=magnitude, w=angular\ frequency, \tau=time\ constant\\\\`

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

`|H(jw)|=(K)/(√(1+w^2\tau^2) )\\\\0.95K=(K)/(√(1+w^2(0.02)^2) ) \\\\0.95=(1)/(√(1+0.0004w^2) )\\\\squaring\ both\ sides:\\\\0.9025=(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=(w)/(2\pi)=(16.4)/(2\pi) =2.62\ Hz`

b)

`The\ phase\ angle(\phi)=tan^(-1)((-w\tau)/(1) )\\\\\phi=tan^(-1)((-16.4*0.02)/(1) )=-18.2^o`