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In addition to the lowpass frequency response, other basic ideal responses are often needed in practice. The ideal highpass filterrejects signals with frequencies below a certain value and passes those with frequencies above that value. The ideal bandpass filter passes only aband of frequencies, and the ideal band reject filter completely rejects a band of frequencies. These ideal frequency responses are illustrated in [link] .

Figure one contains four cartesian graphs. Each graph's horizontal axis ranges in value from -5 to 5 in increments of 5, and their vertical axes are labeled Ideal response, ranging in value from -1 to 1.5 in increments of 0.5. The first graph, labeled Ideal lowpass, contains a horizontal line from (-5, 0) to (5, 0) and a vertical line from (0, 0) to (0, 1.5). The graph also contains line segments from approximately (-1, 0) to (-1, 1), (1, 0) to (1, 1) and (-1, 1) to (1, 1). The second graph, labeled Ideal highpass, contains a horizontal line from (-5, 0) to (5, 0) and a vertical line from (0, 0) to (0, 1.5). The graph also contains line segments from approximately (-5, 1) to (-1, 1), (-1, 1) to (-1, 0) (1, 0) to (1, 1), and (1, 1) to (5, 1). The third graph, labeled ideal bandpass, contains a horizontal line from (-5, 0) to (5, 0) and a vertical line from (0, 0) to (0, 1.5). The graph also contains line segments from approximately (-2, 0) to (-2, 1), (-2, 1) to (-1, 1), (-1, 1) to (-1, 0), (1, 0) to (1, 1), (1, 1) to (2, 1) and (2, 1) to (2, 0). The fourth graph, labeled ideal bandreject, contains a horizontal line from (-5, 0) to (5, 0) and a vertical line from (0, 0) to (0, 1.5). The graph also contains line segments from approximately (-5, 1) to (-2, 1), (-2, 1) to (-2, 0), (-1, 0) to (-1, 1), (1, 0) to (1, 1), (-1, 1) to (1, 1), (2, 0) to (2, 1), and (2, 1) to (5, 1).
The Basic Four Ideal Frequency Responses

This section presents a method for designing the three new filters by using a frequency transformation on the basic lowpass design. When used onthe four classical IIR approximations (e.g. Butterworth, Chebyshev, inverse-Chebyshev, and Elliptic Function), the optimality is preserved.This procedure is used in the FREQXFM() subroutine of Program 8 in the appendix.

Change the bandedge

The classical filters have all been developed for a bandedge of ω 0 = 1 . That is where the Butterworth filter has a magnitude squared of one half: | F | = 0 . 5 or the Chebyshev filter has its passband edge or the Inverse Chebyshev has its stopband edge or the Elliptic filterhas its passband edge. To scale the bandedge, simply replace s by K s or: s K s where K is reciprocal of the new desired bandedge. What happened to the prototype filter at ω = 1 will now happen at ω = 1 / K . It is simply a linear scaling of the ω axis. This change can be done before the conversions below or after.

The highpass filter

The frequency response illustrated in [link] b can be obtained from that in [link] a by replacing the complex frequency variable s in the transfer function by 1 / s . This change of variable maps zero frequency to infinity, maps unity into unity, andmaps infinity to zero. It turns the complex s plane inside out and leaves the unit circle alone.

In the design procedure, the desired bandedge ω 0 for the highpass filter is mapped by 1 / ω 0 to give the bandedge for the prototype lowpass filter. This lowpass filter is next designed byone of the optimal procedures already covered and then converted to a highpass transfer function by replacing s by 1 / s . If an elliptic-function filter approximation is used, both the passbandedge ω p and the stopbandedge ω s are transformed. Because most optimal lowpass design procedures give the designed transferfunction in factored form from explicit formulas for the poles and zeros, the transformation can be performed on each pole andzero to give the highpass transfer function in factored form.

The bandpass filter

In order to convert the lowpass filter of [link] a into that of [link] c, a more complicated frequency transformation is required. In order to reduce confusion, the complexfrequency variable for the prototype analog filter transfer function will be denoted by p and that for the transformed analog filter by s . The transformation is given by

p = s 2 + ω 0 2 s

This change of variables doubles the order of the filter, maps the origin of the s -plane to both plus and minus j ω 0 , and maps minus and plus infinity to zero and infinity. The entire ω axis of the prototype response is mapped between zero and plus infinity on thetransformed responses. It is also mapped onto the left-half axis between minus infinity and zero. This is illustrated in [link] .

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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Source:  OpenStax, Digital signal processing and digital filter design (draft). OpenStax CNX. Nov 17, 2012 Download for free at http://cnx.org/content/col10598/1.6
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