Prove the properly of, g_1(t)g_2(t) doubleheadarrow G_1(f) G_2(f),where g_1(t) d

ID: 1716542 • Letter: P

Question

Prove the properly of, g_1(t)g_2(t) doubleheadarrow G_1(f) G_2(f),where g_1(t) doubleheadarrow G_1(f) and g_2(t) doubleheadarrow(f) use the above property to prove, A(t) doubleheadarrow sin c^2(f) where Pi(t) Pi(t) = A(t) use the above property to figure out the signals, of which Fourier transformations are sin c^3(f) and sin c^3(f) use the above property to calculate the Fourier transformation of summation_n = -infinity^infinity g(t)delta (t - nT_s) where n is an integer and g(t) doubleheadarrow G(f).

Explanation / Answer

1)multiplication property:

Let g1(t)<->G1(f) and g2(t)<->G2(f) then we have

g1(t)g2(t)<-> G1(s)G2(f-s)ds=G1(f)*G2(f)

where* denotes convolution operation.

F[g1(t)g2(t)]= g1(t)g2(t)e-j2ftdt=

1)multiplication property:

Let g1(t)<->G1(f) and g2(t)<->G2(f) then we have

g1(t)g2(t)<-> G1(s)G2(f-s)ds=G1(f)*G2(f)

where* denotes convolution operation.

F[g1(t)g2(t)]= g1(t)g2(t)e-j2ftdt= g1(t)( G2(p)ej2td)e-j2ftdt

g1(t)G()e-j2(f-p)tddt= G2(f-s)ds g1(t)e-j2ftdt

= G2(f-s)G1(s)ds=G1(f)*G2(f)

2)g(t)dt=G(f)2df

This is known as rayleighs energy theorem.

Sinc(t)<->(u(f+1/2)-u(f-1/2) by using the rayleighs energy theorem.

Sinc2(t)dt= (u(f+1/2)-u(f-1/2))2df=1

3)Sinc3(t)dt=u(f)(f+1/2) (u(f+1/2)-u(f-1/2))2df=1

4)g(t)(t-nTs)

Noting (t)<->1 and using the timeshifting property,we have(t-t0)<->e-jfts

F[ (t-nt)]= e-j2ftnx(jw)

[ (t-nt]=g(t) 1/T e-j2ftn

g1(t)( G2(p)ej2td)e-j2ftdt