Lecture #3:

SINUSOIDAL STEADY-STATE RESPONSE AND DT FILTERS

Motivation: DT filtering has a number of advantages over CT filtering

Outline:

Acknowledgment: Don MacLellan of the Bose Corporation provided

the data on the Bose music system.

I. EXAMPLES OF DISCRETE-TIME SYSTEMS

Filtering DT time functions

Demo showing filtering of daily DJIA from 1930-1992.

Filtering a DT image

An image can be considered as a two-dimensional signal or as a set of one-dimensional signals.

Filtering a DT image, cont’d

This shows the same image before and after application of DT,third-order lowpass filter to each row.

This shows the same image before and after application of DT,third-order lowpass filter to each column.

Design of a high-performance automobile audio system

Description of system

If you can afford a car like a Cadillac, you will want an excellent music reproduction system. The incremental cost is small!

Bose music systems are available for a number of different automobiles including the Cadilllac Eldorado and Seville. The system contains 4 audio channels consisting of driver amplifiers and speakers — two located in the front and two in the back.

Each channel has a block diagram as shown below.

The equalizer filter is used to compensate for the undesirable acoustic characteristics of the speaker and the enclosure (the car interior).

Without the equalizer, the frequency response of the system linking the sound source to the sound at the location of the driver’s head is complicated. This frequency response depends upon such factors as the speaker electromechanics, the dimensions of the car interior, the acoustic characteristics of the interior surface of the car, etc.

Without the equalizer, the frequency response of the system linking the sound source to the sound at the location of the driver’s head differs for the front and rear speaker pair.

Without the equalizer, the frequency response of the system linking the sound source to the sound at the location of the driver’s head is different for the Cadillac Eldorado and the Seville.

Equalization

Equalizers must be designed to remove the undesirable characteristics and generate a frequency response that is pleasant for listening to music. The equalized frequency response for the Bose music system is proprietary.

The equalizer must have a complex frequency response. Equalizers for different speakers in the same car differ and equalizers for different car models differ. Bose music systems are available for over 100 different models of cars and all are equalized individually.

The problem

Equalization filters can be designed using active filters such as the one shown below.

For this second-order lowpass filter, the values of 4 resistors and 2 capacitors need to be set. Typically, resistors for such an application have tolerance of 1%; the capacitances have tolerances of 5-10%.

For an equalization filter with 5 poles, the values of the order of 5 resistors and 5 capacitors need to be chosen to match the desired frequency response. Since there are 4 different equalization filters, the values of about 40 components must be chosen to match the equalizer characteristics for each model of car. This has been done in the past with analog filters. It is difficult to design complex analog filters with multiple components that:

(e.g., to accommodate variability in models, interiors, etc.).

The digital solution

There are advantages to using digital rather than analog filters and Bose is now installing digital filter systems for car music systems. The digital equalization filter consists of an analog-to-digital converter, a DT filter, and a digital-to-analog converter. Both the input and output of the DT filter are sequences of numbers, i.e., DT signals.

This is an example of DT processing of a CT signal. The filter characteristics can be modified by changes in software rather than in changes in hardware.

II. REVIEW STEADY-STATE RESPONSE

1/ Definition

Suppose we have a DT LTI system with an input that is an eternal complex geometric sequence

x [ n ] = Xz n x [ n ] = Xz n size 12{x \[ n \] = ital "Xz" rSup { size 8{n} } } {}

As with CT systems, steady state occurs if z is in the ROC of H~(z)H~(z) size 12{ {H} cSup { size 8{ "~" } } \( z \) } {} . Under those conditions,

y [ n ] = X H ~ ( z ) z n y [ n ] = X H ~ ( z ) z n size 12{y \[ n \] =X {H} cSup { size 8{ "~" } } \( z \) z rSup { size 8{n} } } {}

2/ The z-plane

Consider a causal, stable system with pole-zero diagram shown below

For z in the shaded region, the response to x[n]=Xznx[n]=Xzn size 12{x \[ n \] = ital "Xz" rSup { size 8{n} } } {} is the steady-state response .

y [ n ] = Yz n = X H ~ ( z ) z n y [ n ] = Yz n = X H ~ ( z ) z n size 12{y \[ n \] = ital "Yz" rSup { size 8{n} } =X {H} cSup { size 8{ "~" } } \( z \) z rSup { size 8{n} } } {}

3/ Sinusoidal steady-state response and DT frequency response

a/ DT sinusoids

x [ n ] = cos ( 2 πϕ n ) x [ n ] = cos ( 2 πϕ n ) size 12{x \[ n \] ="cos" \( 2 ital "πϕ"n \) } {}

Listen to sinusoidal sounds with MATLAB.

b/ Relation of frequency response to system function

c/ Vectorial interpretation

The frequency response is H~(z)H~(z) size 12{ {H} cSup { size 8{ "~" } } \( z \) } {} for z=ejΩz=ejΩ size 12{z=e rSup { size 8{j %OMEGA } } } {} or H~(ejΩ)H~(ejΩ) size 12{ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) } {} where ΩΩ size 12{ %OMEGA } {} is the DT radian frequency, i.e., evaluated on the unit circle.

The radian frequency is the angle that the unit vector makes with the real axis. H~(ejΩ)H~(ejΩ) size 12{ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) } {} can be evaluated vectorially by computing pole and zero vectors.

d/ DT radian frequency and DT frequency

e/ Properties of H~(ejΩ)H~(ejΩ) size 12{ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) } {} — periodicity

Note

H ~ ( e j ( Ω + 2π ) ) = H ~ ( e j Ω e j2π ) ) = H ~ ( e j Ω ) H ~ ( e j ( Ω + 2π ) ) = H ~ ( e j Ω e j2π ) ) = H ~ ( e j Ω ) size 12{ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j \( %OMEGA +2π \) } } \) = {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } e rSup { size 8{j2π \) } } \) = {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) } {}

which shows that the DT frequency response is periodic with period 2π. This is evident from the vector diagram.

Therefore, it is only necessary to plot the DT frequency response over 2π2π size 12{2π} {} radians.

f/ Properties of H~(ejΩ)H~(ejΩ) size 12{ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) } {} — symmetry

For a DT LTI system, if the input x[n]=δ[n]x[n]=δ[n] size 12{x \[ n \] =δ \[ n \] } {} then the output is the unit sample response which we designate h[n]. In the Z-transform domain, X~(z)=1X~(z)=1 size 12{ {X} cSup { size 8{ "~" } } \( z \) =1} {} , and

Thus,

which evaluated on the unit circle is

H ~ ( e j Ω ) = ∑ n = − ∞ ∞ h [ n ] e − j Ω n H ~ ( e j Ω ) = ∑ n = − ∞ ∞ h [ n ] e − j Ω n size 12{ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h \[ n \] e rSup { size 8{ - j %OMEGA n} } } } {}

We assume that h[n] is a real sequence. Then we expand the exponential

H ~ ( e j Ω ) = ∑ n = − ∞ ∞ h [ n ] ( cos ( Ω n ) − j sin ( Ω n ) ) H ~ ( e j Ω ) = ∑ n = − ∞ ∞ h [ n ] ( cos ( Ω n ) − j sin ( Ω n ) ) size 12{ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h \[ n \] \( "cos" \( %OMEGA n \) - j"sin" \( %OMEGA n \) \) } } {}

H ~ ( e j Ω ) = H ~ r ( e j Ω ) + j H ~ i ( e j Ω ) H ~ ( e j Ω ) = H ~ r ( e j Ω ) + j H ~ i ( e j Ω ) size 12{ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) = {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) +j {H} cSup { size 8{ "~" } } rSub { size 8{i} } \( e rSup { size 8{j %OMEGA } } \) } {}

Therefore, H~(ejΩ)H~(ejΩ) size 12{ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) } {} can be expanded into its real and imaginary parts and the magnitude computed as follows.

H~r(ejΩ)=∑n=−∞∞h[n]cos(Ωn)H~r(ejΩ)=∑n=−∞∞h[n]cos(Ωn) size 12{ {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h \[ n \] "cos" \( %OMEGA n \) } } {} even function of ΩΩ size 12{ %OMEGA } {},

H~i(ejΩ)=−∑n=−∞∞h[n]sin(Ωn)H~i(ejΩ)=−∑n=−∞∞h[n]sin(Ωn) size 12{ {H} cSup { size 8{ "~" } } rSub { size 8{i} } \( e rSup { size 8{j %OMEGA } } \) = - Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h \[ n \] "sin" \( %OMEGA n \) } } {} odd function of ΩΩ size 12{ %OMEGA } {},

∣H~(ejΩ)=Hr2~(ejΩ)+jHi2~(ejΩ)∣∣H~(ejΩ)=Hr2~(ejΩ)+jHi2~(ejΩ)∣ size 12{ \lline {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) = sqrt { {H rSub { size 8{r} rSup { size 8{2} } } } cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) +j {H rSub { size 8{i} rSup { size 8{2} } } } cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) } \lline } {} even function of ΩΩ size 12{ %OMEGA } {},

The angle can be computed as follows,

∠ H ~ ( e j Ω ) = { tan − 1 H ~ i ( e j Ω ) H ~ r ( e j Ω ) for H ~ r ( e j Ω ) > 0 π + tan − 1 H ~ i ( e j Ω ) H ~ r ( e j Ω ) for H ~ r ( e j Ω ) < 0 ∠ H ~ ( e j Ω ) = { tan − 1 H ~ i ( e j Ω ) H ~ r ( e j Ω ) for H ~ r ( e j Ω ) > 0 π + tan − 1 H ~ i ( e j Ω ) H ~ r ( e j Ω ) for H ~ r ( e j Ω ) < 0 size 12{∠ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) = left lbrace matrix { "tan" rSup { size 8{ - 1} } left [ { { {H} cSup { size 8{ "~" } } rSub { size 8{i} } \( e rSup { size 8{j %OMEGA } } \) } over { {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) } } right ] matrix { {} # {} } ital "for" matrix { {} # {} } {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) >0 {} ## π+"tan" rSup { size 8{ - 1} } left [ { { {H} cSup { size 8{ "~" } } rSub { size 8{i} } \( e rSup { size 8{j %OMEGA } } \) } over { {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) } } right ] matrix { {} # {} } ital "for" matrix { {} # {} } {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) <0{} } right none } {}

But, since ±n2π±n2π size 12{ +- n2π} {} can always be added to the angle, and since tan−1H~i(ejΩ)H~r(ejΩ)tan−1H~i(ejΩ)H~r(ejΩ) size 12{"tan" rSup { size 8{ - 1} } left [ { { {H} cSup { size 8{ "~" } } rSub { size 8{i} } \( e rSup { size 8{j %OMEGA } } \) } over { {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) } } right ]} {} is an odd function of ΩΩ size 12{ %OMEGA } {},

∠ H ~ ( e j Ω ) = { − tan − 1 H ~ i ( e j Ω ) H ~ r ( e j Ω ) for H ~ r ( e j Ω ) > 0 − π − tan − 1 H ~ i ( e j Ω ) H ~ r ( e j Ω ) for H ~ r ( e j Ω ) < 0 ∠ H ~ ( e j Ω ) = { − tan − 1 H ~ i ( e j Ω ) H ~ r ( e j Ω ) for H ~ r ( e j Ω ) > 0 − π − tan − 1 H ~ i ( e j Ω ) H ~ r ( e j Ω ) for H ~ r ( e j Ω ) < 0 size 12{∠ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) = left lbrace matrix { - "tan" rSup { size 8{ - 1} } left [ { { {H} cSup { size 8{ "~" } } rSub { size 8{i} } \( e rSup { size 8{j %OMEGA } } \) } over { {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) } } right ]` matrix { {} # {} } ` ital "for" matrix { {} # {} } ` {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) `>0 {} ## - π - "tan" rSup { size 8{ - 1} } left [ { { {H} cSup { size 8{ "~" } } rSub { size 8{i} } \( e rSup { size 8{j %OMEGA } } \) } over { {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) } } right ] matrix { {} # {} } ital "for"` matrix { {} # {} } {H} cSup { size 8{ "~" } } rSub { size 8{r} } \( e rSup { size 8{j %OMEGA } } \) `<0{} } right none } {}

Therefore, ∠H~(ejΩ)∠H~(ejΩ) size 12{∠ {H} cSup { size 8{ "~" } } \( e rSup { size 8{j %OMEGA } } \) } {} is an odd function of ΩΩ size 12{ %OMEGA } {}.

g/ Plotting the frequency response

Plots of the frequency response of a DT lowpass filter show the period in ΩΩ size 12{ %OMEGA } {} is 2π2π size 12{2π} {}, the magnitude is an even function, and the angle is an odd function of ΩΩ size 12{ %OMEGA } {}. All information about the frequency response is contained in the range 0 ≤ ΩΩ size 12{ %OMEGA } {} ≤ ππ size 12{π} {}.

H ~ ( z ) = ∑ n = − ∞ ∞ h [ n ] z − n H ~ ( z ) = ∑ n = − ∞ ∞ h [ n ] z − n size 12{ {H} cSup { size 8{ "~" } } \( z \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h \[ n \] z rSup { size 8{ - n} } } } {}

h/ Plotting scales for the frequency response

Three frequency scales are commonly used. Radian frequency ΩΩ size 12{ %OMEGA } {} is related to frequency ϕϕ size 12{ϕ} {} by

Ω = 2 πϕ Ω = 2 πϕ size 12{ %OMEGA =2 ital "πϕ"} {}

MATLAB uses a normalized frequency defined as

normalized frequency = Ω π = 2ϕ normalized frequency = Ω π = 2ϕ size 12{ ital "normalized"` ital "frequency"`=` { { %OMEGA } over {π} } =2ϕ} {}

The different frequencies are displayed in the complex z plane below.

Three frequency scales are displayed along the unit circle. Ω is the radian frequency and φ is the frequency.

III. DT FILTERS

A variety of DT filters can be synthesized.

∑ k = 0 K a k y [ n + k ] = ∑ l = 0 L b 1 x [ n + 1 ] ∑ k = 0 K a k y [ n + k ] = ∑ l = 0 L b 1 x [ n + 1 ] size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } y \[ n+k \] = Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{1} } x \[ n+1 \] } } } {}

have poles (in addition to those at z = 0 or z=∞z=∞ size 12{z= infinity } {}) and infinite impulse responses and are called infinite impulse response or IIR filters.

y [ n ] = ∑ l = 0 L b 1 x [ n + 1 ] y [ n ] = ∑ l = 0 L b 1 x [ n + 1 ] size 12{y \[ n \] = Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{1} } x \[ n+1 \] } } {}

consist of all zeros (except at z = 0 or z=∞z=∞ size 12{z= infinity } {}), have finite impulse responses, and are called finite impulse response or FIR filters.

Demo of relation of pole-zero diagram, time waveforms, vector diagrams, and frequency response.

1/ Causal, first-order, IIR, lowpass filter

h [ n ] = α n u [ n ] for 0 < α < 1 ⇔ Z H ~ ( z ) = 1 1 − αz − 1 for ∣ z ∣ > ∣ a ∣ h [ n ] = α n u [ n ] for 0 < α < 1 ⇔ Z H ~ ( z ) = 1 1 − αz − 1 for ∣ z ∣ > ∣ a ∣ size 12{h \[ n \] =α rSup { size 8{n} } u \[ n \] `` matrix { {} # {} } ital "for" matrix { {} # {} } ```0<α<1`` { dlrarrow } cSup { size 8{Z} } `` {H} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - αz rSup { size 8{ - 1} } } } `` matrix { {} # {} } ` ital "for" matrix { {} # {} } ``` \lline z \lline > \lline a \lline } {}

A pole at 0 < α < 1 yields a lowpass frequency response. As |α| decreases, the unit sample response approaches a unit sample, the frequency response magnitude approaches 1 (0 dB) and the angle approaches 0.

2/ Causal, first-order, IIR, highpass filter

h [ n ] = α n u [ n ] for − 1 < α < 0 ⇔ Z H ~ ( z ) = 1 1 − αz − 1 for ∣ z ∣ > ∣ α ∣ h [ n ] = α n u [ n ] for − 1 < α < 0 ⇔ Z H ~ ( z ) = 1 1 − αz − 1 for ∣ z ∣ > ∣ α ∣ size 12{h \[ n \] =α rSup { size 8{n} } u \[ n \] `` matrix { {} # {} } ital "for"`` matrix { {} # {} } - 1<α<0``` { dlrarrow } cSup { size 8{Z} } ``` {H} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - αz rSup { size 8{ - 1} } } } `` matrix { {} # {} } ` ital "for"` matrix { {} # {} } ` \lline z \lline `>` \lline α \lline } {}

A pole at −1 < αα size 12{α} {} < 0 yields a highpass frequency response. As ∣α∣∣α∣ size 12{ \lline α \lline } {} decreases, the unit sample response approaches a unit sample, the frequency response magnitude approaches 1 (0 dB) and the angle approaches 0.

3/ Causal, first-order, FIR, lowpass filter

h [ n ] = δ [ n ] − αδ [ n − 1 ] for − 1 < α < 0 ⇔ Z H ~ ( z ) = 1 − az − 1 for z ≠ 0 h [ n ] = δ [ n ] − αδ [ n − 1 ] for − 1 < α < 0 ⇔ Z H ~ ( z ) = 1 − az − 1 for z ≠ 0 size 12{h \[ n \] =δ \[ n \] - ital "αδ" \[ n - 1 \] ``` matrix { {} # {} } ital "for"`` matrix { {} # {} } ` - 1<α<0 { dlrarrow } cSup { size 8{Z} } {H} cSup { size 8{ "~" } } \( z \) =1 - ital "az" rSup { size 8{ - 1} } matrix { {} # {} } ital "for"` matrix { {} # {} } `z <> 0} {}

A filter with a zero located at −1< αα size 12{α} {}<0 has a lowpass frequency response and a finite impulse response.

4/ Causal, first-order, FIR, highpass filter

h [ n ] = δ [ n ] − αδ [ n − 1 ] for − 1 < α < 0 ⇔ Z H ~ ( z ) = 1 − az − 1 for z ≠ 0 h [ n ] = δ [ n ] − αδ [ n − 1 ] for − 1 < α < 0 ⇔ Z H ~ ( z ) = 1 − az − 1 for z ≠ 0 size 12{h \[ n \] =δ \[ n \] - ital "αδ" \[ n - 1 \] ` matrix { {} # {} } ital "for" matrix { {} # {} } ` - 1<α<0 { dlrarrow } cSup { size 8{Z} } {H} cSup { size 8{ "~" } } \( z \) =1 - ital "az" rSup { size 8{ - 1} } matrix { {} # {} } ital "for"`` matrix { {} # {} } z <> 0} {}

A filter with a zero located at 0 < αα size 12{α} {} < 1 has a highpass frequency response and a finite impulse response.

5/ Causal, second-order, IIR, bandpass filter

H ~ ( z ) = 1 ( 1 − αe j Ω 0 z − 1 ) ( 1 − αe − j Ω 0 z − 1 ) ∣ z ∣ > ∣ α ∣ H ~ ( z ) = 1 ( 1 − αe j Ω 0 z − 1 ) ( 1 − αe − j Ω 0 z − 1 ) ∣ z ∣ > ∣ α ∣ size 12{ {H} cSup { size 8{ "~" } } \( z \) = { {1} over { \( 1 - αe rSup { size 8{j %OMEGA rSub { size 6{0} } } } z rSup { - 1} size 12{ \) \( 1 - αe rSup { - j %OMEGA rSub { size 6{0} } } } size 12{z rSup { - 1} } size 12{ \) }} } ``` \lline z \lline `>` \lline α \lline } {}

A filter with a pair of complex conjugate poles has a bandpass frequency response. The center frequency of the bandpass filter is determined by the radian frequency of the pole location.

H ~ ( z ) = 1 ( 1 − αe j Ω 0 z − 1 ) ( 1 − αe − j Ω 0 z − 1 ) ∣ z ∣ > ∣ α ∣ H ~ ( z ) = 1 ( 1 − αe j Ω 0 z − 1 ) ( 1 − αe − j Ω 0 z − 1 ) ∣ z ∣ > ∣ α ∣ size 12{ {H} cSup { size 8{ "~" } } \( z \) = { {1} over { \( 1 - αe rSup { size 8{j %OMEGA rSub { size 6{0} } } } z rSup { - 1} size 12{ \) \( 1 - αe rSup { - j %OMEGA rSub { size 6{0} } } } size 12{z rSup { - 1} } size 12{ \) }} } `` matrix { {} # {} } ` \lline z \lline `>` \lline α \lline } {}

A filter with a pair of complex conjugate poles has a bandpass frequency response. The bandwidth of the bandpass filter decreases as the pole location approaches the unit circle.

6/ Causal, second-order, FIR, bandreject filter

H ~ ( z ) = ( 1 − αe j Ω 0 z − 1 ) ( 1 − αe − j Ω 0 z − 1 ) , ∣ z ∣ ≠ 0 H ~ ( z ) = ( 1 − αe j Ω 0 z − 1 ) ( 1 − αe − j Ω 0 z − 1 ) , ∣ z ∣ ≠ 0 size 12{ {H} cSup { size 8{ "~" } } \( z \) = \( 1 - αe rSup { size 8{j %OMEGA rSub { size 6{0} } } } z rSup { - 1} size 12{ \) \( 1 - αe rSup { - j %OMEGA rSub { size 6{0} } } } size 12{z rSup { - 1} } size 12{ \) , matrix { {} # {} } ` \lline z \lline <> 0}} {}

{}

A filter with a pair of complex conjugate zeros has a bandreject frequency response. The bandwidth of the bandreject filter decreases as the zero location approaches the unit circle.

7/ Causal, second-order, notch filter

H ~ ( z ) = ( 1 − e jπ / 4 z − 1 ) ( 1 − e − jπ / 4 z − 1 ) ( 1 − 0 . 9 e jπ / 4 z − 1 ) ( 1 − 0 . 9 e − jπ / 4 z − 1 ) ∣ z ∣ > 0 . 9 H ~ ( z ) = ( 1 − e jπ / 4 z − 1 ) ( 1 − e − jπ / 4 z − 1 ) ( 1 − 0 . 9 e jπ / 4 z − 1 ) ( 1 − 0 . 9 e − jπ / 4 z − 1 ) ∣ z ∣ > 0 . 9 size 12{ {H} cSup { size 8{ "~" } } \( z \) = { { \( 1 - e rSup { size 8{jπ/4} } z rSup { size 8{ - 1} } \) \( 1 - e rSup { size 8{ - jπ/4} } z rSup { size 8{ - 1} } \) } over { \( 1 - 0 "." 9e rSup { size 8{jπ/4} } z rSup { size 8{ - 1} } \) \( 1 - 0 "." 9e rSup { size 8{ - jπ/4} } z rSup { size 8{ - 1} } \) } } matrix { {} # {} } \lline z \lline >0 "." 9} {}

IV. CONCLUSIONS ON DT FILTERS

Exercises for this lecture.

Solutions of Exercises.

Content actions

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This work is licensed by Vu Dinh Thanh, Truc Pham-Dinh, Anh Tuan Hoang, and Tam Huynh-Ngoc under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.

Last edited by Vietnam OpenCourseWare on Jul 3, 2009 6:22 am -0500.