Signals and Systems - Electrical Engineering

590 C H A P T E R 10: Fourier Analysis of Discrete-Time Signals and Systems

nExample 10.8

For the signalx[n]=αnu[n], 0< α <1, find the magnitude and the phase of its DTFTX(ejω).

Solution

The DTFT ofx[n] is

X(ejω)=

1

1 −αz−^1

∣

∣z=ejω=^1

1 −αe−jω

since the Z-transform has a region of convergence|z|> α that includes the unit circle. Its

magnitude is

|X(ejω)|=

1

√

( 1 −αcos(ω))^2 +α^2 sin^2 (ω)

which is an even function ofωgiven that cos(ω)=cos(−ω) and sin^2 (−ω)=(−sin(ω))^2 =

sin^2 (ω). The phase is given by

θ(ω)=−tan−^1

[

αsin(ω)

1 −αcos(ω)

]

which is an odd function ofω. As functions ofω, the numerator is odd and the denominator is

even, so that the argument of the inverse tangent is odd, which is in turn odd. n

nExample 10.9

For a discrete-time signal

x[n]=cos(ω 0 n+φ) −π≤φ < π

determine how the magnitude and the phase responses of the DTFTX(ejω)change withφ.

Solution

The signalx[n] has a DTFT

X(ejω)=π

[

e−jφδ(ω−ω 0 )+ejφδ(ω+ω 0 )

]

Its magnitude is

|X(ejω)|=|X(e−jω)|=π[δ(ω−ω 0 )+δ(ω+ω 0 )]

for all values ofφ. The phase ofX(ejω)is

θ(ω)=φδ(ω+ω 0 )−φδ(ω−ω 0 )

=







φ ω=−ω 0

−φ ω=ω 0

0 otherwise