Sinusoidal Steady-State Response of Digital Filters - Digital Signal Processing: Fundamentals and Applications - page 814

Substituting z ¼ e j U , we get the residue as
R ¼ Ve jf x Hðe j U Þ
Then substituting R ¼ Ve jf x Hðe j U Þ back into Equation (D.4) results in
YðzÞ
z

¼ Ve jf x Hðe j U Þ

z e j U þ sum of the rest of partial fractions

(D.6)

and multiplying z on both sides of Equation (D.6) leads to
YðzÞ¼ Ve jf x Hðe j U Þz

z e j U þ z sum of the rest of partial fractions

(D.7)

Taking the inverse z-transform leads to two parts of the solution:

yðnÞ¼Ve jf x Hðe j U Þe j U n þ Z 1 ð z sum of the rest of partial fractions Þ (D.8)

From Equation (D.8) , we have the steady-state response
y ss ðnÞ¼Ve jf x Hðe j U Þe j U n (D.9)

and the transient response
y tr ðnÞ¼Z 1

ð z sum of the rest of partial fractions Þ (D.10)

Note that since the digital filter is a stable system, and the locations of its poles must be inside the unit
circle on the z-plane, the transient response will settle to zero eventually. To develop the filter
magnitude and phase responses, we write the digital steady-state response as
y ss ðnÞ¼V Hðe j U Þ e j U þjf x þ : Hðe j U Þ (D.11)

Comparing Equation (D.11) and Equation (D.1) , it follows that

Amplitude of the steady-state output
Amplitude of the sinusoidal input

Magnitude response ¼

(D.12)

Hðe j U Þ

¼ Hðe j U Þ ¼ V

V

¼ e jf x þj : Hðe j U Þ
e jf x

¼ e j : Hðe j U Þ ¼ : Hðe j U Þ

Phase response (D.13)

Thus we conclude that

Frequency response: Hðe j U Þ¼HðzÞj z¼e j U

(D.14)

Since Hðe j U Þ¼jHðe j U Þj : Hðe j U Þ

Magnitude response: Hðe j U Þ

(D.15)

Phase response: : Hðe j U Þ (D.16)