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Sinusoidal Steady-State Response of Digital Filters
APPENDIX
SINUSOIDAL STEADY-STATE RESPONSE OF DIGITAL FILTERS
D
D.1 SINUSOIDAL STEADY-STATE RESPONSE Analysis of the sinusoidal steady-state response of the digital filters will lead to the development of the magnitude and phase responses of digital filters. Let us look at the following digital filter with a digital transfer function H(z) and a complex sinusoidal input xðnÞ ¼ VejðΩn + φx Þ ,
(D.1)
where Ω ¼ ωT is the normalized digital frequency, while T is the sampling period and y(n) denotes the digital output, as shown in Fig. D.1. The z-transform output from the digital filter is then given by Y ðzÞ ¼ HðzÞXðzÞ:
(D.2)
jφx
z Since XðzÞ ¼ Ve zejΩ , we have
Y ðz Þ ¼
Vejφx z H ðzÞ: z ejΩ
(D.3)
Based on the partial fraction expansion, Y(z)/z can be expanded as the following form: Y ðzÞ Vejφx R H ðzÞ ¼ + sum of the rest of partial fractions: ¼ z ejΩ z z ejΩ
(D.4)
Multiplying the factor (z ejΩ) on both sides of Eq. (D.4) yields Vejϕx H ðzÞ ¼ R + z ejΩ ðsum of the rest of partial fractionsÞ:
Substituting z ¼ ejΩ, we get the residue as
x (n) Ve j
n j
X ( z)
(D.5)
R ¼ Vejϕx H ejΩ : y ( n)
x
H ( z)
yss (n)
ytr (n)
Y ( z)
FIG. D.1 Steady-state response of the digital filter.
833
834
APPENDIX D SINUSOIDAL STEADY-STATE RESPONSE OF DIGITAL FILTERS
Then substituting R ¼ VejϕxH(ejΩ) back into Eq. (D.4) results in Y ðzÞ Vejφx H ðejΩ Þ + sum of the rest of partial fractions, ¼ z z ejΩ
(D.6)
and multiplying z on both sides of Eq. (D.6) leads to Y ðz Þ ¼
Vejφx HðejΩ Þz + z sum of the rest of partial fractions: z ejΩ
(D.7)
Taking the inverse z-transform leads to two parts of the solution:
yðnÞ ¼ Vejφx H ejΩ ejΩn + Z 1 ðz sum of the rest of partial fractionsÞ:
(D.8)
From Eq. (D.8), we have the steady-state response
yss ðnÞ ¼ Vejφx H ejΩ ejΩn
(D.9)
ytr ðnÞ ¼ Z 1 ðz sum of the rest of partial fractionsÞ:
(D.10)
and the transient response
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 be settled to zero eventually. To develop filter magnitude and phase responses, we write the digital steady-state response as jΩ yss ðnÞ ¼ V H ejΩ ejΩ + jφx + ∠Hðe Þ :
(D.11)
Comparing Eq. (D.11) and Eq. (D.1), it follows that Amplitude of the steady state output Amplitude of the sinusoidal input V HðejΩ Þ jΩ ¼ H e ¼ V
Magnitude response ¼
Phase response ¼
Thus we conclude that
jΩ jΩ ejφx + j∠Hðe Þ ¼ ej∠Hðe Þ ¼ ∠H ejΩ : ejφx
Frequency response ¼ H ejΩ ¼ H ðzÞjz¼ejΩ :
Since H(ejΩ) ¼ jH(ejΩ)j ∠ H(ejΩ)
Magnitude response ¼ H ejΩ Phase response ¼ ∠H ejΩ :
(D.12)
(D.13)
(D.14)
(D.15) (D.16)
D.2 PROPERTIES OF THE SINUSOIDAL STEADY-STATE RESPONSE From Euler’s identity and trigonometric identity, we know that ejðΩ + k2πÞ ¼ cos ðΩ + k2π Þ + jsin ðΩ + k2π Þ ¼ cosΩ + jsinΩ ¼ ejΩ ,
(D.17)
D.2 PROPERTIES OF THE SINUSOIDAL STEADY-STATE RESPONSE
835
where k is an integer taking values of k ¼ 0, 1, 2, ⋯. Then:
Frequency response : H ejΩ ¼ H ejðΩ + k2πÞ
(D.18)
Magnitude frequency response : H ejΩ ¼ H ejðΩ + k2πÞ
(D.19)
Phase response : ∠H ejΩ ¼ ∠H ejΩ + 2kπ :
(D.20)
Clearly, the frequency response is periodical, with a period of 2π. Next, let us develop the symmetric properties. Since the transfer function is written as H ð zÞ ¼
Y ðzÞ b0 + b1 z1 + ⋯ + bM zM , ¼ XðzÞ 1 + a1 z1 + ⋯ + aN zN
(D.21)
Substituting z ¼ ejΩ into Eq. (D.21) yields b0 + b1 ejΩ + ⋯ + bM ejMΩ H ejΩ ¼ : 1 + a1 ejΩ + ⋯ + aN ejNΩ
(D.22)
Using Euler’s identity, e jΩ ¼ cos Ω j sin Ω, we have ðb0 + b1 cosΩ + ⋯ + bM cosMΩÞ jðb1 sin Ω + ⋯ + bM sinMΩÞ H ejΩ ¼ : ð1 + a1 cosΩ + ⋯ + aN cosNΩÞ jða1 sin Ω + ⋯ + aN sinNΩÞ
(D.23)
ðb0 + b1 cos Ω + ⋯ + bM cos MΩÞ + jðb1 sinΩ + ⋯ + bM sinMΩÞ H ejΩ ¼ : ð1 + a1 cosΩ + ⋯ + aN cos NΩÞ + jða1 sinΩ + ⋯ + aN sin NΩÞ
(D.24)
Similarly,
Then the magnitude response and phase response can be expressed as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðb0 + b1 cos Ω + ⋯ + bM cos MΩÞ2 + ðb1 sin Ω + ⋯ + bM sinMΩÞ2 jΩ H e ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 + a1 cos Ω + ⋯ + aN cos NΩÞ2 + ða1 sin Ω + ⋯ + aN sinNΩÞ2 ðb1 sinΩ + ⋯ + bM sin MΩÞ ∠H ejΩ ¼ tan 1 b0 + b1 cosΩ + ⋯ + bM cosMΩ ða1 sinΩ + ⋯ + aN sin NΩÞ tan 1 1 + a1 cos Ω + ⋯ + aN cos NΩ
(D.25)
(D.26)
Based in Eq. (D.24), we also have the magnitude and phase response for H(e jΩ) as
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðb0 + b1 cos Ω + ⋯ + bM cos MΩÞ2 + ðb1 sinΩ + ⋯ + bM sinMΩÞ2 jΩ H e ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 + a1 cos Ω + ⋯ + aN cos NΩÞ2 + ða1 sinΩ + ⋯ + aN sinNΩÞ2 b1 sin Ω + ⋯ + bM sinMΩ ∠H ejΩ ¼ tan 1 b0 + b1 cos Ω + ⋯ + bM cos MΩ a1 sinΩ + ⋯ + aN sinNΩ : tan 1 1 + a1 cosΩ + ⋯ + aN cos NΩ
(D.27)
(D.28)
836
APPENDIX D SINUSOIDAL STEADY-STATE RESPONSE OF DIGITAL FILTERS
Comparing (D.25) with (D.27) and (D.26) with (D.28), respectively, we conclude the symmetric properties as jΩ jΩ H e ¼ H e
jΩ
∠H e
¼ ∠H e
jΩ
(D.29) (D.30)
SINUSOIDAL STEADY-STATE RESPONSE OF DIGITAL FILTERS
D
D.1 SINUSOIDAL STEADY-STATE RESPONSE Analysis of the sinusoidal steady-state response of the digital filters will lead to the development of the magnitude and phase responses of digital filters. Let us look at the following digital filter with a digital transfer function H(z) and a complex sinusoidal input xðnÞ ¼ VejðΩn + φx Þ ,
(D.1)
where Ω ¼ ωT is the normalized digital frequency, while T is the sampling period and y(n) denotes the digital output, as shown in Fig. D.1. The z-transform output from the digital filter is then given by Y ðzÞ ¼ HðzÞXðzÞ:
(D.2)
jφx
z Since XðzÞ ¼ Ve zejΩ , we have
Y ðz Þ ¼
Vejφx z H ðzÞ: z ejΩ
(D.3)
Based on the partial fraction expansion, Y(z)/z can be expanded as the following form: Y ðzÞ Vejφx R H ðzÞ ¼ + sum of the rest of partial fractions: ¼ z ejΩ z z ejΩ
(D.4)
Multiplying the factor (z ejΩ) on both sides of Eq. (D.4) yields Vejϕx H ðzÞ ¼ R + z ejΩ ðsum of the rest of partial fractionsÞ:
Substituting z ¼ ejΩ, we get the residue as
x (n) Ve j
n j
X ( z)
(D.5)
R ¼ Vejϕx H ejΩ : y ( n)
x
H ( z)
yss (n)
ytr (n)
Y ( z)
FIG. D.1 Steady-state response of the digital filter.
833
834
APPENDIX D SINUSOIDAL STEADY-STATE RESPONSE OF DIGITAL FILTERS
Then substituting R ¼ VejϕxH(ejΩ) back into Eq. (D.4) results in Y ðzÞ Vejφx H ðejΩ Þ + sum of the rest of partial fractions, ¼ z z ejΩ
(D.6)
and multiplying z on both sides of Eq. (D.6) leads to Y ðz Þ ¼
Vejφx HðejΩ Þz + z sum of the rest of partial fractions: z ejΩ
(D.7)
Taking the inverse z-transform leads to two parts of the solution:
yðnÞ ¼ Vejφx H ejΩ ejΩn + Z 1 ðz sum of the rest of partial fractionsÞ:
(D.8)
From Eq. (D.8), we have the steady-state response
yss ðnÞ ¼ Vejφx H ejΩ ejΩn
(D.9)
ytr ðnÞ ¼ Z 1 ðz sum of the rest of partial fractionsÞ:
(D.10)
and the transient response
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 be settled to zero eventually. To develop filter magnitude and phase responses, we write the digital steady-state response as jΩ yss ðnÞ ¼ V H ejΩ ejΩ + jφx + ∠Hðe Þ :
(D.11)
Comparing Eq. (D.11) and Eq. (D.1), it follows that Amplitude of the steady state output Amplitude of the sinusoidal input V HðejΩ Þ jΩ ¼ H e ¼ V
Magnitude response ¼
Phase response ¼
Thus we conclude that
jΩ jΩ ejφx + j∠Hðe Þ ¼ ej∠Hðe Þ ¼ ∠H ejΩ : ejφx
Frequency response ¼ H ejΩ ¼ H ðzÞjz¼ejΩ :
Since H(ejΩ) ¼ jH(ejΩ)j ∠ H(ejΩ)
Magnitude response ¼ H ejΩ Phase response ¼ ∠H ejΩ :
(D.12)
(D.13)
(D.14)
(D.15) (D.16)
D.2 PROPERTIES OF THE SINUSOIDAL STEADY-STATE RESPONSE From Euler’s identity and trigonometric identity, we know that ejðΩ + k2πÞ ¼ cos ðΩ + k2π Þ + jsin ðΩ + k2π Þ ¼ cosΩ + jsinΩ ¼ ejΩ ,
(D.17)
D.2 PROPERTIES OF THE SINUSOIDAL STEADY-STATE RESPONSE
835
where k is an integer taking values of k ¼ 0, 1, 2, ⋯. Then:
Frequency response : H ejΩ ¼ H ejðΩ + k2πÞ
(D.18)
Magnitude frequency response : H ejΩ ¼ H ejðΩ + k2πÞ
(D.19)
Phase response : ∠H ejΩ ¼ ∠H ejΩ + 2kπ :
(D.20)
Clearly, the frequency response is periodical, with a period of 2π. Next, let us develop the symmetric properties. Since the transfer function is written as H ð zÞ ¼
Y ðzÞ b0 + b1 z1 + ⋯ + bM zM , ¼ XðzÞ 1 + a1 z1 + ⋯ + aN zN
(D.21)
Substituting z ¼ ejΩ into Eq. (D.21) yields b0 + b1 ejΩ + ⋯ + bM ejMΩ H ejΩ ¼ : 1 + a1 ejΩ + ⋯ + aN ejNΩ
(D.22)
Using Euler’s identity, e jΩ ¼ cos Ω j sin Ω, we have ðb0 + b1 cosΩ + ⋯ + bM cosMΩÞ jðb1 sin Ω + ⋯ + bM sinMΩÞ H ejΩ ¼ : ð1 + a1 cosΩ + ⋯ + aN cosNΩÞ jða1 sin Ω + ⋯ + aN sinNΩÞ
(D.23)
ðb0 + b1 cos Ω + ⋯ + bM cos MΩÞ + jðb1 sinΩ + ⋯ + bM sinMΩÞ H ejΩ ¼ : ð1 + a1 cosΩ + ⋯ + aN cos NΩÞ + jða1 sinΩ + ⋯ + aN sin NΩÞ
(D.24)
Similarly,
Then the magnitude response and phase response can be expressed as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðb0 + b1 cos Ω + ⋯ + bM cos MΩÞ2 + ðb1 sin Ω + ⋯ + bM sinMΩÞ2 jΩ H e ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 + a1 cos Ω + ⋯ + aN cos NΩÞ2 + ða1 sin Ω + ⋯ + aN sinNΩÞ2 ðb1 sinΩ + ⋯ + bM sin MΩÞ ∠H ejΩ ¼ tan 1 b0 + b1 cosΩ + ⋯ + bM cosMΩ ða1 sinΩ + ⋯ + aN sin NΩÞ tan 1 1 + a1 cos Ω + ⋯ + aN cos NΩ
(D.25)
(D.26)
Based in Eq. (D.24), we also have the magnitude and phase response for H(e jΩ) as
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðb0 + b1 cos Ω + ⋯ + bM cos MΩÞ2 + ðb1 sinΩ + ⋯ + bM sinMΩÞ2 jΩ H e ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 + a1 cos Ω + ⋯ + aN cos NΩÞ2 + ða1 sinΩ + ⋯ + aN sinNΩÞ2 b1 sin Ω + ⋯ + bM sinMΩ ∠H ejΩ ¼ tan 1 b0 + b1 cos Ω + ⋯ + bM cos MΩ a1 sinΩ + ⋯ + aN sinNΩ : tan 1 1 + a1 cosΩ + ⋯ + aN cos NΩ
(D.27)
(D.28)
836
APPENDIX D SINUSOIDAL STEADY-STATE RESPONSE OF DIGITAL FILTERS
Comparing (D.25) with (D.27) and (D.26) with (D.28), respectively, we conclude the symmetric properties as jΩ jΩ H e ¼ H e
jΩ
∠H e
¼ ∠H e
jΩ
(D.29) (D.30)