Inverse fourier transform?

Question

system has the frequency response

H(w)=10/(jw+10)

find output y(t) if x(t)=2+2cos(50t + pi/2)

Y(w)=X(w)H(w)

y(t) is the inv fourier transform of Y(w).

I'm having problems with the inverse Fourier. The claim is that you use partial fraction expansion..i have no idea.

Answer 1

x(t)=2+2cos(50t + π/2)  Ã¢Â†Â”  x(jω)=4π δ(ω) + [δ(ω-50)+ δ(ω+50)] e^ -jωπ/2

Y(ω)=X(ω)H(ω) = [40π δ(ω) + [δ(ω-50)+ δ(ω+50)] 10 e ^-jωπ/2] ÷ [jω + 10]

Y(ω) is the sum of three terms (no need for partial fraction expansion here):

40π δ(ω) ÷ [jω + 10]   and  [10 δ(ω-50) e^ -jωπ/2] ÷ [jω + 10] and [10 δ(ω+50) e^ -jωπ/2] ÷ [jω + 10].

First term:

y(t) = 1/ 2π ∫ y(jω) e jωt dω

y(t) = 1/ 2π ∫ [40π δ(ω) ÷ [jω + 10]]   e^jωt dω

y(t) = 2

Second term:

y(t) = 1/ 2π ∫ y(jω) e^jωt dω

y(t) = 1/ 2π ∫ [[10 δ(ω-50) e^-jωπ/2] ÷ [jω + 10]] e^jωt dω

y(t) = (1/ 2π) [( e^-j25π) ÷ (j5 + 1)] e^j50t

Third term:

y(t) = 1/ 2π ∫ y(jω) e^jωt dω

y(t) = 1/ 2π ∫ [[10 δ(ω+50) e^-jωπ/2] ÷ [jω + 10]] e^jωt dω

y(t) = (1/ 2π) [( e^j25π) ÷ (1-j5)] e^-j50t

The denominators of the 2nd and 3rd terms are made equal by multiplying by the complex conjugate:

2nd term:

y(t) = (1/ 2π) [( e^-j25π) ÷ (j5 + 1)] e^j50t x [ (1-j5) / (1-j5) ]

y(t) = (1/ 52π) [ e^j(50t-25) + (5/j) e^j(50t-25)]

3rd term:

y(t) = (1/ 2π) [( e^j25π) ÷ (j5 - 1)] e^j50t x [ (1+j5) / (1+j5) ]

y(t) = (1/ 52π) [e^-j(50t-25) - (5/j) e^-j(50t-25)]

Summing all 3 terms produces the final answer (with an Euler conversion):

y(t) = 2 + (1/ 52π) [2cos25π + 10sin(50t-25)]

y(t) = 2 + (1/ 52π) [10sin(50t-25)-2]