Explanation:
The ratio of amplitude of two successive maxima is [tex]e^{-YT_{d} }[/tex] .
[tex]e^{-YT_{d} } = \frac{1}{2}[/tex]
by taking log we get [tex]Y = \frac{1}{T_{d} } ln2 = f_{d} ln2[/tex] ∵ [tex]f_{d} = \frac{1}{T_{d} }[/tex]
(a) we know that
[tex]W_{o} = (W_{d}^{2} + Y^{2})^{\frac{1}{2} }[/tex]
or [tex]f_{o} = [f_{d}^{2} + (\frac{Y}{2\pi })^{2} ]^{\frac{1}{2} }[/tex] ∴ W= 2πf
[tex]f_{o} = f_{d}[ 1+ (\frac{ln2}{2\pi })^{2}]^{\frac{1}{2} }[/tex] ∵ [tex]Y = f_{d}ln2[/tex]
we got [tex]f_{0} = 100.6 Hz[/tex]
(b)
we know
[tex](W_{d}^{2} - Y^{2})^{\frac{1}{2} } = W_{r}[/tex]
or [tex]f_{r} ^{2} = [ f_{d} ^{2} -(\frac{Y}{2\pi } )^{2}]} ^{\frac{1}{2} }[/tex] ∵ W =2πf
[tex]= f_{d}[ 1- (\frac{ln2}{2\pi })^{2}]^{\frac{1}{2} }[/tex]
[tex]f_{r} = 99.4 Hz[/tex]
