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## Homework Statement

creating a nulling filter that filters the normalised radian frequency of (0.6pi )

Find the difference equation?

## Homework Equations

**Do we use the exponential form or cosine form.**

Ae^j(wn + ∅)

A=amplitude

w=normalised frequency

∅ = phase angle

## The Attempt at a Solution

Therefore

x[n]=e^j(0.6*pi*n)

y[n]=e^j(0.6*pi*n) xH( e^j(0.6*pi*n))

H( e^j(0.6∏n))=1/2(1+e^-j(0.6*pi)

H( e^j(0.6∏n))=Ʃbke^-j(0.6*pi*k)

b0=1/2 and b1=1/2

therefore is the difference equation

y[n]= 1/2x[n]+1/2x[n-1]