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Question

A signal x(t)=e−tu(t) is passed through an ideal low pass filter with cut-off frequency ofωcrad/sec, then the value of ωc for the output energy of low pass filter is 90% of theinput energy will be ________rad/sec6.314

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Solution

The correct option is A 6.314Given input signal x(t)=e−tu(t)Energy of the input signal Ei=∫∞−∞|x(t)|2dt=∫∞−∞|e−tu(t)|2dt=∫∞0e−2tdt=[e−2t−2]∞0=12=0.5JNow, ESD [Energy spectral density] of the output y(t) is given byEy(ω)=|H(ω)|2Ex(ω)∴Input ESD,Ex(ω)=|X(ω)|2X(ω)=FT[x(t)]=FT[e−tu(t)]=11+jωEx(ω)=|X(ω)|2=11+ω2For a LPF,the square of the transfer function is given as|H(ω)|2={1,|ω<ωc|0,otherwiseEy(ω)=|H(ω)|2Ex(ω)=⎧⎪⎨⎪⎩11+ω2|ω|<ωc0otherwise∴The total energy of the output signal is E0E0=12π∫∞−∞Ey(ω)dω[∴Ey(ω)is an even function of(ω)]=1π∫∞0Ey(ω)dω=1π∫ωc011+ω2dω0.90[12]=1πtan−1(ωc)tan−1ωc=0.90×π2ωc=tan[0.90×π2]ωc=6.314rad/sec

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