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In many applications, one never has to deal directly with the scaling functions or wavelets. Only the coefficients h ( n ) and h 1 ( n ) in the defining equations [link] and [link] and c ( k ) and d j ( k ) in the expansions [link] , [link] , and [link] need be considered, and they can be viewed as digital filters and digital signals respectively [link] , [link] . While it is possible to develop most of the results of wavelet theory using only filter banks, we feel that both the signalexpansion point of view and the filter bank point of view are necessary for a real understanding of this new tool.

Analysis – from fine scale to coarse scale

In order to work directly with the wavelet transform coefficients, we will derive the relationship between the expansion coefficients at a lowerscale level in terms of those at a higher scale. Starting with the basic recursion equation from [link]

φ ( t ) = n h ( n ) 2 φ ( 2 t - n )

and assuming a unique solution exists, we scale and translate the time variable to give

φ ( 2 j t - k ) = n h ( n ) 2 φ ( 2 ( 2 j t - k ) - n ) = n h ( n ) 2 φ ( 2 j + 1 t - 2 k - n )

which, after changing variables m = 2 k + n , becomes

φ ( 2 j t - k ) = m h ( m - 2 k ) 2 φ ( 2 j + 1 t - m ) .

If we denote V j as

V j = Span k { 2 j / 2 φ ( 2 j t - k ) }

then

f ( t ) V j + 1 f ( t ) = k c j + 1 ( k ) 2 ( j + 1 ) / 2 φ ( 2 j + 1 t - k )

is expressible at a scale of j + 1 with scaling functions only and no wavelets. At one scale lower resolution, wavelets are necessary forthe “detail" not available at a scale of j . We have

f ( t ) = k c j ( k ) 2 j / 2 φ ( 2 j t - k ) + k d j ( k ) 2 j / 2 ψ ( 2 j t - k )

where the 2 j / 2 terms maintain the unity norm of the basis functions at various scales. If φ j , k ( t ) and ψ j , k ( t ) are orthonormal or a tight frame, the j level scaling coefficients are found by taking the inner product

c j ( k ) = f ( t ) , φ j , k ( t ) = f ( t ) 2 j / 2 φ ( 2 j t - k ) d t

which, by using [link] and interchanging the sum and integral, can be written as

c j ( k ) = m h ( m - 2 k ) f ( t ) 2 ( j + 1 ) / 2 φ ( 2 j + 1 t - m ) d t

but the integral is the inner product with the scaling function at a scale of j + 1 giving

c j ( k ) = m h ( m - 2 k ) c j + 1 ( m ) .

The corresponding relationship for the wavelet coefficients is

d j ( k ) = m h 1 ( m - 2 k ) c j + 1 ( m ) .

Filtering and down-sampling or decimating

In the discipline of digital signal processing, the “filtering" of a sequence of numbers (the input signal) is achieved by convolvingthe sequence with another set of numbers called the filter coefficients,taps, weights, or impulse response. This makes intuitive sense if you think of a moving average with the coefficients being the weights.For an input sequence x ( n ) and filter coefficients h ( n ) , the output sequence y ( n ) is given by

y ( n ) = k = 0 N - 1 h ( k ) x ( n - k )

There is a large literature on digital filters and how to design them [link] , [link] . If the number of filter coefficients N is finite, the filter is called a Finite Impulse Response (FIR) filter. If the number is infinite,it is called an Infinite Impulse (IIR) filter. The design problem is the choice of the h ( n ) to obtain some desired effect, often to remove noise or separate signals [link] , [link] .

The Down Sampler of Decimator
The Down Sampler of Decimator

In multirate digital filters, there is an assumed relation between the integer index n in the signal x ( n ) and time. Often the sequence of numbers are simply evenly spaced samples of a function of time. Twobasic operations in multirate filters are the down-sampler and the up-sampler. The down-sampler (sometimes simply called a sampler ora decimator) takes a signal x ( n ) as an input and produces an output of y ( n ) = x ( 2 n ) . This is symbolicallyshown in [link] . In some cases, the down-sampling is by a factor other than two andin some cases, the output is the odd index terms y ( n ) = x ( 2 n + 1 ) , but this will be explicitly stated if it is important.

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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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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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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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Wavelets and wavelet transforms. OpenStax CNX. Aug 06, 2015 Download for free at https://legacy.cnx.org/content/col11454/1.6
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