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| | You have searched the classifieds for a utilized hot tub and discovered what looks like a wonderful deal. Or possibly a friend has supplied to let you get theirs, or possibly you have even been given a utilized tub as a freebie. Is it truly a excellent deal? There is no way to tell for confident until you have it installed and running.<br><br>The initial and most crucial issue is to by no means acquire and pay for any utilized spa or hot tub that you have not noticed in operation and tested entirely. Purchasing a "dry" tub is a disappointment waiting to occur. You may possibly get lucky and almost everything will perform fine, but you also stand the opportunity of obtaining a price range-busting disaster on your hands.<br><br>You have to also maintain in mind that merely moving a tub puts a entire new set of stresses on the plumbing, tub shell, and frame. An unnoticable crack in the spa sitting at its original place will most likely be produced worse by loading it on a truck and moving it. This is particularly accurate if the tub has been repaired sometime in the previous. So just simply because every little thing was fine when you checked it out, does not imply it will be specifically the exact same when you finish your move and installation.<br><br>Even with a cost-free hot tub you will probably will have some hidden and possibly unexpected expenses:<br><br>* You have to uncover a way to move it. A lot of "transportable" hot tubs will not match in the back of a pickup truck.<br><br>* You have to hook up the electrical (and unless you are a licensed electrician this is something that ought to in no way be [http://www.Reddit.com/r/howto/search?q=attempted attempted] by a homeowner). This may well involve trenching for the electrical line and operating a 220 volt drop, in addition to the electrical supplies.<br><br>* When you get it all set up you have to fix something you broke moving it, and repair harm or defects that were not apparent when you inspected it.<br><br>* You will also most likely require to acquire other supplies (chemicals for the water, a testing kit, a new cartridge filter, and possibly a hot tub cover).<br><br>* Check the cartridge filter to make certain it is in place and that the filter itself is not torn or just completely worn out.<br><br>* Does it come with a hot tub cover? If the cover is cracked, torn, discolored, waterlogged, or a spa cover is not included in the deal, then you may well be hunting at a substantial additional expense.<br><br>When inspecting the hot tub, be certain and verify the following:<br><br>* Use a thermometer to check the accuracy of the hot tub thermostat. If it is out of calibration that is not necessarily a problem, but is data you want to know to operate the tub safely. If it will not heat the water to normal operating temperature (normally 104 degrees) that is another issue entirely. There may possibly be a heater dilemma.<br><br>* Open up any of the doors that give you access to the gear or places underneath the tub. Look for any leaks or signs of water. I discovered [http://www.nexopia.com/users/fruitcoach3/blog/2192-preventing-breakins-throughout-holiday-season Nexopia | Blog] by searching webpages. Check around the pump for any indication that the pump seals are leaking and will need to be replaced.<br><br>* Listen to the sound of the pump operating. It ought to be a low steady hum. [http://www.agdfh.com/showthread.php?tid=28382 Restoring Flat Tires On A Scooter] includes additional info concerning the inner workings of it. Any other sound need to tell you that you may well have a pump replacement in your near future.<br><br>* Turn on the blower and once more, listen to the motor. It must sound a lot like a vacuum cleaner and if you hear any knocking, pinging, or scraping this is not typical and means you might be whipping out a credit card soon.<br><br>Any issues that you discover can be utilised as points for negotiating a reduced purchase cost. If you point out these concerns to the seller, you may get a cost concession which will make the deal far more eye-catching for you.<br><br>With all this mentioned, there are some great values in utilised hot tubs. Just do not fail to verify items out meticulously so you will not regret your decision. Budget for some unexpected expenses, and cross your fingers..<br><br>If you liked this post and you would like to receive additional info about [http://evasivedairy4709.soup.io free health insurance] kindly visit the web-site. |
| The '''bilinear transform''' (also known as '''[[Arnold Tustin|Tustin]]'s method''') is used in [[digital signal processing]] and discrete-time [[control theory]] to transform continuous-time system representations to discrete-time and vice versa.
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| The bilinear transform is a special case of a [[conformal map]]ping (namely, the [[Möbius transformation]]), often used to convert a [[transfer function]] <math> H_a(s) \ </math> of a [[linear]], [[time-invariant]] ([[LTI system theory|LTI]]) filter in the [[continuous function|continuous]]-time domain (often called an [[analog filter]]) to a transfer function <math> H_d(z) \ </math> of a linear, shift-invariant filter in the [[discrete signal|discrete]]-time domain (often called a [[digital filter]] although there are analog filters constructed with [[switched capacitor]]s that are discrete-time filters). It maps positions on the <math> j \omega \ </math> axis, <math> Re[s]=0 \ </math>, in the [[s-plane]] to the [[unit circle]], <math> |z| = 1 \ </math>, in the [[complex plane|z-plane]]. Other bilinear transforms can be used to warp the [[frequency response]] of any discrete-time linear system (for example to approximate the non-linear frequency resolution of the human auditory system) and are implementable in the discrete domain by replacing a system's unit delays <math> \left( z^{-1} \right) \ </math> with first order [[all-pass filter]]s.
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| The transform preserves [[BIBO stability|stability]] and maps every point of the [[frequency response]] of the continuous-time filter, <math> H_a(j \omega_a) \ </math> to a corresponding point in the frequency response of the discrete-time filter, <math> H_d(e^{j \omega_d T}) \ </math> although to a somewhat different frequency, as shown in the [[#Frequency warping|Frequency warping]] section below. This means that for every feature that one sees in the frequency response of the analog filter, there is a corresponding feature, with identical gain and phase shift, in the frequency response of the digital filter but, perhaps, at a somewhat different frequency. This is barely noticeable at low frequencies but is quite evident at frequencies close to the [[Nyquist frequency]].
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| == Discrete-time approximation ==
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| The bilinear transform is a first-order approximation of the natural logarithm function that is an exact mapping of the z-plane to the s-plane. When the [[Laplace transform]] is performed on a discrete-time signal (with each element of the discrete-time sequence attached to a correspondingly delayed [[Dirac delta function|unit impulse]]), the result is precisely the [[Z transform]] of the discrete-time sequence with the substitution of
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| :<math>
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| \begin{align}
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| z &= e^{sT} \\
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| &= \frac{e^{sT/2}}{e^{-sT/2}} \\
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| &\approx \frac{1 + s T / 2}{1 - s T / 2}
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| \end{align}
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| </math>
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| where <math> T \ </math> is the [[numerical integration]] step size of the [[trapezoidal rule]] used in the bilinear transform derivation.<ref>{{cite book |title=Discrete Time Signal Processing Third Edition |last=Oppenheim |first=Alan |year=2010 |publisher=Pearson Higher Education, Inc. |location=Upper Saddle River, NJ |isbn=978-0-13-198842-2 |page=504}}</ref> The above bilinear approximation can be solved for <math> s \ </math> or a similar approximation for <math> s = (1/T) \ln(z) \ \ </math> can be performed.
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| The inverse of this mapping (and its first-order bilinear approximation) is
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| :<math>
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| \begin{align}
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| s &= \frac{1}{T} \ln(z) \\
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| &= \frac{2}{T} \left[\frac{z-1}{z+1} + \frac{1}{3} \left( \frac{z-1}{z+1} \right)^3 + \frac{1}{5} \left( \frac{z-1}{z+1} \right)^5 + \frac{1}{7} \left( \frac{z-1}{z+1} \right)^7 + \cdots \right] \\
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| &\approx \frac{2}{T} \frac{z - 1}{z + 1} \\
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| &= \frac{2}{T} \frac{1 - z^{-1}}{1 + z^{-1}}
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| \end{align}
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| </math>
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| The bilinear transform essentially uses this first order approximation and substitutes into the continuous-time transfer function, <math> H_a(s) \ </math>
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| :<math>s \leftarrow \frac{2}{T} \frac{z - 1}{z + 1}.</math> | |
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| That is
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| :<math>H_d(z) = H_a(s) \bigg|_{s = \frac{2}{T} \frac{z - 1}{z + 1}}= H_a \left( \frac{2}{T} \frac{z-1}{z+1} \right). \ </math>
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| == Stability and minimum-phase property preserved ==
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| A continuous-time causal filter is [[BIBO stability|stable]] if the [[Pole (complex analysis)|poles]] of its transfer function fall in the left half of the [[complex number|complex]] [[s-plane]]. A discrete-time causal filter is stable if the poles of its transfer function fall inside the [[unit circle]] in the [[complex plane|complex z-plane]]. The bilinear transform maps the left half of the complex s-plane to the interior of the unit circle in the z-plane. Thus filters designed in the continuous-time domain that are stable are converted to filters in the discrete-time domain that preserve that stability. | |
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| Likewise, a continuous-time filter is [[minimum-phase]] if the [[Zero (complex analysis)|zeros]] of its transfer function fall in the left half of the complex s-plane. A discrete-time filter is minimum-phase if the zeros of its transfer function fall inside the unit circle in the complex z-plane. Then the same mapping property assures that continuous-time filters that are minimum-phase are converted to discrete-time filters that preserve that property of being minimum-phase.
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| == Example ==
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| As an example take a simple [[low-pass]] [[RC filter]]. This continuous-time filter has a transfer function
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| :<math>\begin{align} | |
| H_a(s) &= \frac{1/sC}{R+1/sC} \\
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| &= \frac{1}{1 + RC s}.
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| \end{align}</math>
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| If we wish to implement this filter as a digital filter, we can apply the bilinear transform by substituting for <math>s</math> the formula above; after some reworking, we get the following filter representation:
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| :{|
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| |-
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| |<math>H_d(z) \ </math>
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| |<math> =H_a \left( \frac{2}{T} \frac{z-1}{z+1}\right) \ </math>
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| |-
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| |<math>= \frac{1}{1 + RC \left( \frac{2}{T} \frac{z-1}{z+1}\right)} \ </math>
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| |-
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| |<math>= \frac{1 + z}{(1 - 2 RC / T) + (1 + 2RC / T) z} \ </math>
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| |-
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| |<math>= \frac{1 + z^{-1}}{(1 + 2RC / T) + (1 - 2RC / T) z^{-1}}. \ </math>
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| |}
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| The coefficients of the denominator are the 'feed-backward' coefficients and the coefficients of the numerator are the 'feed-forward' coefficients used to implement a real-time [[digital filter]].
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| == Frequency warping ==
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| To determine the frequency response of a continuous-time filter, the [[transfer function]] <math> H_a(s) \ </math> is evaluated at <math>s = j \omega \ </math> which is on the <math> j \omega \ </math> axis. Likewise, to determine the frequency response of a discrete-time filter, the transfer function <math> H_d(z) \ </math> is evaluated at <span style="vertical-align:+30%;"><math>z = e^{ j \omega T} \ </math></span> which is on the unit circle, <math> |z| = 1 \ </math>. When the actual frequency of <math> \omega \ </math> is input to the discrete-time filter designed by use of the bilinear transform, it is desired to know at what frequency, <math> \omega_a \ </math>, for the continuous-time filter that this <math> \omega \ </math> is mapped to.
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| :<math>H_d(z) = H_a \left( \frac{2}{T} \frac{z-1}{z+1}\right) \ </math> | |
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| :{|
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| |-
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| |<math>H_d(e^{ j \omega T}) \ </math>
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| |<math>= H_a \left( \frac{2}{T} \frac{e^{ j \omega T} - 1}{e^{ j \omega T} + 1}\right) \ </math>
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| |-
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| |<math>= H_a \left( \frac{2}{T} \cdot \frac{e^{j \omega T/2} \left(e^{j \omega T/2} - e^{-j \omega T/2}\right)}{e^{j \omega T/2} \left(e^{j \omega T/2} + e^{-j \omega T/2 }\right)}\right) \ </math>
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| |-
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| |<math>= H_a \left( \frac{2}{T} \cdot \frac{\left(e^{j \omega T/2} - e^{-j \omega T/2}\right)}{\left(e^{j \omega T/2} + e^{-j \omega T/2 }\right)}\right) \ </math>
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| |- | |
| |
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| |<math>= H_a \left(j \frac{2}{T} \cdot \frac{ \left(e^{j \omega T/2} - e^{-j \omega T/2}\right) /(2j)}{\left(e^{j \omega T/2} + e^{-j \omega T/2 }\right) / 2}\right) \ </math>
| |
| |-
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| |
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| |<math>= H_a \left(j \frac{2}{T} \cdot \frac{ \sin(\omega T/2) }{ \cos(\omega T/2) }\right) \ </math>
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| |-
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| |
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| |<math>= H_a \left(j \frac{2}{T} \cdot \tan \left( \omega T/2 \right) \right) \ </math>
| |
| |}
| |
| | |
| This shows that every point on the unit circle in the discrete-time filter z-plane, <span style="vertical-align:+30%;"><math>z = e^{ j \omega T} \ </math></span> is mapped to a point on the <math>j \omega \ </math> axis on the continuous-time filter s-plane, <math>s = j \omega_a \ </math>. That is, the discrete-time to continuous-time frequency mapping of the bilinear transform is
| |
| | |
| :<math> \omega_a = \frac{2}{T} \tan \left( \omega \frac{T}{2} \right) </math> | |
| | |
| and the inverse mapping is
| |
| | |
| :<math> \omega = \frac{2}{T} \arctan \left( \omega_a \frac{T}{2} \right). </math>
| |
| | |
| The discrete-time filter behaves at frequency <math>\omega \ </math> the same way that the continuous-time filter behaves at frequency <math> (2/T) \tan(\omega T/2) \ </math>. Specifically, the gain and phase shift that the discrete-time filter has at frequency <math>\omega \ </math> is the same gain and phase shift that the continuous-time filter has at frequency <math> (2/T) \tan(\omega T/2) \ </math>. This means that every feature, every "bump" that is visible in the frequency response of the continuous-time filter is also visible in the discrete-time filter, but at a different frequency. For low frequencies (that is, when <math>\omega \ll 2/T</math> or <math>\omega_a \ll 2/T</math>), <math>\omega \approx \omega_a \ </math>.
| |
| | |
| One can see that the entire continuous frequency range
| |
| | |
| : <math> -\infty < \omega_a < +\infty \ </math>
| |
| | |
| is mapped onto the fundamental frequency interval
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| : <math> -\frac{\pi}{T} < \omega < +\frac{\pi}{T}. \ </math>
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| | |
| The continuous-time filter frequency <math> \omega_a = 0 \ </math> corresponds to the discrete-time filter frequency <math> \omega = 0 \ </math> and the continuous-time filter frequency <math> \omega_a = \pm \infty \ </math> correspond to the discrete-time filter frequency <math> \omega = \pm \pi / T. \ </math>
| |
| | |
| One can also see that there is a nonlinear relationship between <math> \omega_a \ </math> and <math> \omega. \ </math> This effect of the bilinear transform is called '''''frequency warping'''''. The continuous-time filter can be designed to compensate for this frequency warping by setting <math> \omega_a = \frac{2}{T} \tan \left( \omega \frac{T}{2} \right) \ </math> for every frequency specification that the designer has control over (such as corner frequency or center frequency). This is called '''''pre-warping''''' the filter design.
| |
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| When designing a digital filter as an approximation of a continuous time filter, the frequency response (both amplitude and phase) of the digital filter can be made to match the frequency response of the continuous filter at frequency <math> \omega_0 </math> if the following transform is substituted into the continuous filter transfer function.<ref>Astrom, Karl J. ''Computer Controlled Systems, Theory and Design'' Second Edition. ISBN 0131686003. Prentice-Hall, 1990, pp 212</ref> This is a modified version of Tustin's transform shown above. However, note that this transform becomes the above transform as <math> \omega_0 \to 0 </math>. That is to say, the above transform causes the digital filter response to match the analog filter response at DC.
| |
| | |
| :<math>s \leftarrow \frac{\omega_0}{\tan(\frac{\omega_0 T}{2})} \frac{z - 1}{z + 1}.</math> | |
| | |
| The main advantage of the warping phenomenon is the absence of aliasing distortion of the frequency response characteristic, such as observed with [[Impulse invariance]]. It is necessary, however, to compensate for the frequency warping by pre-warping the given frequency specifications of the continuous-time system. These pre-warped specifications may then be used in the bilinear transform to obtain the desired discrete-time system.
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| ==See also==
| |
| | |
| * [[Impulse invariance]]
| |
| * [[Matched Z-transform method]]
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| | |
| ==References==
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| | |
| {{refimprove|date=February 2011}}
| |
| {{reflist}}
| |
| | |
| {{DSP}}
| |
| | |
| {{DEFAULTSORT:Bilinear Transform}}
| |
| [[Category:Digital signal processing]]
| |
| [[Category:Transforms]]
| |
| [[Category:Control theory]]
| |
You have searched the classifieds for a utilized hot tub and discovered what looks like a wonderful deal. Or possibly a friend has supplied to let you get theirs, or possibly you have even been given a utilized tub as a freebie. Is it truly a excellent deal? There is no way to tell for confident until you have it installed and running.
The initial and most crucial issue is to by no means acquire and pay for any utilized spa or hot tub that you have not noticed in operation and tested entirely. Purchasing a "dry" tub is a disappointment waiting to occur. You may possibly get lucky and almost everything will perform fine, but you also stand the opportunity of obtaining a price range-busting disaster on your hands.
You have to also maintain in mind that merely moving a tub puts a entire new set of stresses on the plumbing, tub shell, and frame. An unnoticable crack in the spa sitting at its original place will most likely be produced worse by loading it on a truck and moving it. This is particularly accurate if the tub has been repaired sometime in the previous. So just simply because every little thing was fine when you checked it out, does not imply it will be specifically the exact same when you finish your move and installation.
Even with a cost-free hot tub you will probably will have some hidden and possibly unexpected expenses:
* You have to uncover a way to move it. A lot of "transportable" hot tubs will not match in the back of a pickup truck.
* You have to hook up the electrical (and unless you are a licensed electrician this is something that ought to in no way be attempted by a homeowner). This may well involve trenching for the electrical line and operating a 220 volt drop, in addition to the electrical supplies.
* When you get it all set up you have to fix something you broke moving it, and repair harm or defects that were not apparent when you inspected it.
* You will also most likely require to acquire other supplies (chemicals for the water, a testing kit, a new cartridge filter, and possibly a hot tub cover).
* Check the cartridge filter to make certain it is in place and that the filter itself is not torn or just completely worn out.
* Does it come with a hot tub cover? If the cover is cracked, torn, discolored, waterlogged, or a spa cover is not included in the deal, then you may well be hunting at a substantial additional expense.
When inspecting the hot tub, be certain and verify the following:
* Use a thermometer to check the accuracy of the hot tub thermostat. If it is out of calibration that is not necessarily a problem, but is data you want to know to operate the tub safely. If it will not heat the water to normal operating temperature (normally 104 degrees) that is another issue entirely. There may possibly be a heater dilemma.
* Open up any of the doors that give you access to the gear or places underneath the tub. Look for any leaks or signs of water. I discovered Nexopia | Blog by searching webpages. Check around the pump for any indication that the pump seals are leaking and will need to be replaced.
* Listen to the sound of the pump operating. It ought to be a low steady hum. Restoring Flat Tires On A Scooter includes additional info concerning the inner workings of it. Any other sound need to tell you that you may well have a pump replacement in your near future.
* Turn on the blower and once more, listen to the motor. It must sound a lot like a vacuum cleaner and if you hear any knocking, pinging, or scraping this is not typical and means you might be whipping out a credit card soon.
Any issues that you discover can be utilised as points for negotiating a reduced purchase cost. If you point out these concerns to the seller, you may get a cost concession which will make the deal far more eye-catching for you.
With all this mentioned, there are some great values in utilised hot tubs. Just do not fail to verify items out meticulously so you will not regret your decision. Budget for some unexpected expenses, and cross your fingers..
If you liked this post and you would like to receive additional info about free health insurance kindly visit the web-site.