# Discrete Time question

1. Dec 2, 2005

### perryben

w[n] = exp(j*pi*n)

w[n] = (−1)^n

Hi, can anyone off hand verify why these two exressions are equal? Thanks

2. Dec 2, 2005

### Muzza

exp(i * pi * n) = cos(pi * n) + isin(pi * n).

You figure out the rest.

3. Dec 2, 2005

### BerkMath

I am assuming n takes on integer values. In which case and expansion of exp(i*pi*n)=cos(pi*n)+i*sin(pi*n). Now for n in Z, sin(pi*n)=0, since: sin(0)=sin(pi)=sin(2pi)=sin(3pi)=...=0. Similarily, cos(pi*n)=(-1)^n, since cos(0)=cos(2pi)=cos(4pi)=..cos(2kpi)=1, and cos(pi)=cos(3*pi)=...cos((2k+1)pi)= -1. Therefore exp(i*pi*n)=(-1)^n for any n in Z.

4. Dec 2, 2005

### uart

The easiest way is just to write $$e^{j \pi n} = (e^{j \pi})^n$$ and note that $$e^{j \pi} = -1$$