I need an expression that represents attenuation/insertion loss in filters as a function of frequency?
930095
This question cannot be answered without knowledge of the order of the filter.\n\nEach order attentuates 20 dB/decade for frequencies above its pole.
978796
Study of data with respect to different spectral responses
464204
A frequency oscillator.
124381
You can use the equation c = fλ.\nNow f = c/λ = 3 x 10*8/600 x 10-9 = 5 x 10*14 Hz
774319
frequency translation phase-locked loop (PLL) is provided for detecting the audio information from a television signal. The PLL includes a mixer and a phase detector, responsive to the intermediate picture and sound carriers respectively, which carriers are amplified by a single IF amplifier. A first filter is coupled between the output of the television receiver tuner and the input of the IF amplifier and has a frequency response for passing the picture and sound carriers. The input of the mixer is coupled to the output of the IF amplifier and is substantially responsive to the amplified picture carrier. The phase detector has a first input also coupled to the output of the IF amplifier and is substantially responsive to the amplified sound carrier and a second input coupled to the output of the mixer. A second filter coupled to one of the first and second inputs of the phase detector has an amplitude versus frequency response which provides a predetermined attenuation at the frequency of the picture carrier and substantially less attenuation at the frequency of the sound carrier. A lowpass filter coupled to the output of the phase detector provides the detected audio information.
436592
You'll need something which measures the amplitude and frequency of the sound waves to 'see' a difference in the noises.  You can probably calculate a trigonometric function based on one particular wave within the measurement, but it won't hold true as the sounds fluctuate at random.  Possibly, you could even program a computer to calculate a random value between set parameters to account for this fluctuation.\nI'm not a math person.  Can't help you much more than that.
1013560
The answer depends upon what sort of wave you are talking about and the medium it is travelling. Say for example if it is a light wave through vacuum, then \n\nWavelength = (2.99875 X 10^8) / Frequency.\n\nIf the same is sound through air at 60% humidity and at sea level, then\n\nWavelength = (340) / Frequency.\n\nIf it is a sound wave through water (4 degrees celsius), then\n\nWavelength = (1440) / Frequency.\n\nThe value inside the bracket is the speed of the wave in the respective medium under respective conditions.
751077
Yes...\n\nThe refraction indecces\n\nn =c/v, but c = wave lenght * frequency
471384
In no particular order ..... \n\nAt low frequencies the necessarily high values of L and C can result in an inconveniently large size.  \n\nAn active filter, which can be designed to exhibit and work with higher impedance values means that smaller values of C can be used.\n\nAdditionally the use of a gyrator configuration can eliminate the need for L.\n\nA passive filter is inherently lossy, at least to some extent.\nAn active filter can have gain.\n\nAn active filter is usually smaller than a passive one.\n\nA passive filter doesn't require a power source.
292420
Yes.  Convert the discrete time signal to a series of impulses.  Send the impulses through a digital to analog converter.  Send the output of the D/A through a low pass filter.  \n\nOppenheim describes the characteristics of an ideal reconstruction filter.  See pages 197-201.
45332
An attenuator is used to reduce the level of a signal.  A single attenuator will not work the same with any load, because the input and output impedances of the attenuator should be matched to the characteristic impedances of the system in which it is used.\n\nOne of the simplest forms of attenuator is a resistive T or pi network, which is basically just an electronic "divider" circuit. If you know the attenuator resistor and output impedance values, you just apply Ohm's Law to calculate the input/output ratio.  There's a good example, with calculations required, at http://www.radio-electronics.com/info/circuits/resistor_attenuator/resistor_attenuator.php
880092
A filter with 2 xtals in it is called a Half-Lattice filter. You can cascade two of these, or use a Full-Lattice filter to employ 4 xtals.\nI hope this helps you out.
254300
http://en.wikipedia.org/wiki/Time-division_multiplexing
522069
well.. refractive index is a function of frequency
866698
Check out this website. They discuss both direct and reverb loudness. It seems the steady-state reverb level depends on surface area and absorption coefficient while direct is 1/R^2 dependent. Unfortunately the symbols in their equations are not always defined right below the equation but they will be somewhere on the page. If you see a symbol that looks like 2 vertical sides and the top of a square, it's pi. Also they're not too careful about parentheses. One eq. you'll notice says\nWreverb = Wsource * (1 - a / Sa) but I'm betting it should be (1 - a) / Sa. (Otherwise they're using the fraction a/Sa which reduces to 1/S with 'a' irrelevant). Good luck. Maybe you'll find a more readable site.\nP.S. Of course one major effect you'll notice if your program material extends into the deep bass is a falloff of low-frequency response below about 22 Hz. From the ref., Fcutoff = c / 2 * longest dimension, should be c / (2 * longest dimension).
394827
yes you can. For real low freq. you can actually do a DC measurement. Just slow the sweep rate way down and you will see it.
946504
Intensity varies as the square of the distance.\nThat should help