Question

Descrit fourier transformation of unit step function

Answer 1

u[n]=f[n]+g[n]

Where:

f[n]=12 for −∞<n<∞

and

g[n]={12 for n≥0−12 for n<0

do:

δ[n]=g[n]−g[n−1]

U Know DTFT of δ[n] is 1 and DTFT of g[n]−g[n−1]→G(ejω)−e−jωG(ejω) so: 1=G(ejω)−e−jωG(ejω)

therefore

G(ejω)=11−e−jω

and we know that the DTFT of f[n]→F(ejω)=π∑∞k=−∞δ(ω−2πk)

finally :

u[n]=f[n]+g[n]→U(ejω)=F(ejω)+G(ejω)

U(ejω)=11−e−jω+π∑∞k=−∞δ(ω−2πk)

share improve this answer

Firstly, you need to be aware that

u[n]=∑n=0∞δ[n]

And according to Accumulation property of DTFT which implies that:

y[n]=∑m=−∞nx[m]

∑m=−∞nx[m]⟺11−e−jωX(ejω)+πX(ej0)∑k=−∞∞δ(ω−2πk) ∗

Now assume that:

x[n]=∑m=−∞ng[m]

g[n]=δ[n]

G(ejω)=1

Substitute in *:

11−e−jωG(ejω)+πG(ej0)∑k=−∞∞δ(ω−2πk)

Now refine the last formula:

HINT: G(ejω)=1G(ej0)=1

11−e−jω+π∑k=−∞∞δ(ω−2πk