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Two intuitively reasonable approaches to developing the equations for the FDM-TDM transmultiplexer are presented in this section. The first emulates [link] . We first develop the equations for a digital counterpart of the analog tuners used in the filter bank and then observe that significant computational improvements can be obtained when the tuning frequencies are linked together in a simple way. The second subsection starts from a different point, that of using the discrete Fourier transform as a spectral channelizer. We ultimately find out that these two approaches yield essentially the same analytical results.

Figure one is a flow chart. From left to right, a title begins the chart. It reads, Baseband Input. An arrow pointing to the right follows. Above the arrow is the description x_c(t). The arrow points at a box containing the label A/D. Below this box is an unlabeled arrow pointing up at the box. To the right of the A/D box is another arrow pointing to the right. Above this arrow is the description x(k). The arrow points at a circle containing a large x. Below this circle is a large arrow pointing up at the circle, with the description e^(-j2πf_0kT) beside it. To the right of the circle is another arrow pointing to the right, and above it is the description ρ(k). The arrow points at a box containing the label Digital Filter h(k). To the right of this box is a bigger arrow pointing to the right. Above this arrow is the label y-bar(k). The arrow points at another box, labeled Decimation M. To the right of this box is a final arrow not pointing at anything, with the description y(r).
Using a digital Tuner to Extract One FDM Channel

The transmux as a bank of single channel digital tuners

Fundamental equations for a single-channel digital tuner

The input FDM signal is assumed to be the continuous-time waveform x c ( t ) . The analog-to-digital converter shown in [link] samples this waveform at the uniform rate of f s samples per second, producing the discrete-time sequence x ( k ) , where x ( k ) x c ( t = k T ) , the integer k is the time index, and T is the sampling interval given by T = 1 f s . The spectrum of this sequence is shifted down in frequency by multiplying it by a complex exponential of the form e - j 2 π f 0 k T , where f 0 is the desired amount of the frequency downconversion. The product of x ( k ) and this exponential is then filtered in discrete time by using the pulse response h ( k ) . The duration of the pulse response h ( k ) is assumed to be finite and in particular of length no greater than L , an integer. The filter output y ¯ ( k ) is then decimated by a factor of M , yielding the sequence y ( r ) , where the integer r is the decimated time index.

These processing steps are shown in graphical form in [link] . Both sides of the two-sided spectrum of the sampled input signal are seen in [link] (a). For the moment, the input signal is assumed to be real-valued and therefore the spectrum is symmetrical around 0 Hz Even though real-valued inputs are assumed here, all of the ensuing analysis applies to complex-valued signals as well. . A channel of interest in this spectrum has been shaded and its center frequency is noted to be f 0 . Multiplying the input signal by e - j 2 π f 0 k T has the effect of shifting the spectrum to the left (assuming 0 f 0 f s 2 ) and centering the desired channel at 0 Hz. The downconverted signal is now complex-valued, and therefore spectral symmetry around 0 Hz is neither required nor expected. The transfer function of the lowpass filter appears in [link] (c). The filter pulse response h ( k ) is chosen to attain the desired spectral characteristics. In particular, the filter needs to pass the channel of interest without degradation and suppress all others sufficiently. How to design such a pulse response is discussed in Appendix A. In general, the quality of the filter grows with the value of the parameter L . The filter shown here is symmetrical around 0 Hz and its pulse response h ( k ) can therefore be real-valued. This is not required however.

After the application of the shifted signal ρ ( k ) to the filter, the spectrum shown in [link] (d) results. The desired channel is isolated from all others. It is sampled, however, at a rate far faster than required by the Nyquist sampling theorem. The filter output is then decimated by the factor M , resulting in the spectrum shown in [link] (e). The channel's bandwidth is the same as before but now its percentage bandwidth, that is, its bandwidth compared to its final sampling rate, is much higher. In a good digital tuner the percentage bandwidth after decimation usually ranges between 0.5 and 0.9, where unity is the theoretical limit imposed by the sampling theorem.

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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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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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
Joseph
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, An introduction to the fdm-tdm digital transmultiplexer. OpenStax CNX. Nov 16, 2010 Download for free at http://cnx.org/content/col11165/1.2
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