Digital Communications_Question

[jeanpierre_dsp]

Hi ,

My major is mechanical eng but I started master in electrical eng. I have problems with signal processing and digital communications.

Here I wrote the questions, could you tell me how to solve it or the way that where i need to start ? I need to solve them until wednesday.

 

[jeanpierre_dsp]

Please, I really need help.

Thanks

JP

 

[mahdithdn]

Hi,

1 - sinc(w*t) = sin(pi*w*t)/(pi*w*t) is a rectangular signal between [-w/2,w/2] in frequency domain.
the question is ambiguous but if we assume s(t) a passband signal and we want the quadrature component to be zero the positive-frequency part of the signal should be symmetric around f0 => f0 = w/4 = 1024/4 = 256 is the solution.

2 - A passband signal around can be represented around f0 as: x(t) = xi(t)*cos(2*pi*f0*t) - xq(t)*sin(2*pi*f0*t), in which xi(t) is the inphase component and xq(t) is the quadrature component.
comparing this with the first part of question xi(t) = sqrt(2)*(sin(pi*t)/(pi*t)), and xq(t) = 0.
in the sqcond part we now cos(500*pi*t) = cos(501*pi*t)*cos(pi*t) + sin(501*pi*t)*sin(pi*t)
substituting this to x(t) => xi(t) = sqrt(2)*(sin(pi*t)/(pi*t))*cos(pi*t), and xq(t) = -sqrt(2)*(sin(pi*t)/(pi*t))*sin(pi*t)

3 - Pe = ∫ min(p1(x),p2(x))dx, where p1(x) = N(√Eb,N0/2), p2 = -N(√Eb,N0/2)
Pe = ∫ min(p1(x),p2(x))dx ≤ ∫ (p1(x)^.5)(p2(x)^.5)dx = ∫ (1/√pi*N0) exp(-x^2/N0)*exp(-Eb/N0)dx = exp(-Eb/N0)*∫ (1/√pi*N0) exp(-x^2/N0)dx = exp(-Eb/N0)
Pe = Q(√2*Eb/N0) ≤ exp(-Eb/N0) => Q(x) ≤ e^((-x^2)/2)

regards

 

[jeanpierre_dsp]

thank you very much...