# Tag Info

10

Remember that Wavelet Transforms are nothing but time-localized filtering/correlation operations. The wavelet transforms provide a unified framework for getting around the Heisenberg Uncertainly Principle that the Fourier Transform suffers from. So when you ask "what should my settings be for bandwidth, and center frequency", you are asking for filter ...

9

A sinusoidal signal is represented as $$x(t) = \mathrm{cos}(\omega t) = \mathrm{cos}(2\pi f t)$$ $\omega$ is the angular frequency and $f$ is the frequency. See Frequency definition. Your signal \begin{align} x(t) &= 5 + 30\mathrm{cos}(2000\pi t) + 10\mathrm{cos}(6000\pi t)\\ &= 5\mathrm{cos}(2\pi \times 0 t) + 30\mathrm{cos}(2\pi \times 1000 t) + ...

8

The $BT$ product is the bandwidth-symbol time product where $B$ is the $-3\textrm{ dB}$(half-power) bandwidth of the pulse/filter and $T$ is the symbol duration. For different applications you will find varying recommended values. In GSM telephony for instance, a $BT=0.3$ is recommended. In satellite communications with GMSK, for near-earth missions the ...

7

It's generally not true that a band pass signal can be sampled and recovered without error if $f_s>2B$ is satisfied, where $B=f_h-f_l$ is the signal's bandwidth. This condition is just necessary but not sufficient. You have to make sure that the aliased spectra do not overlap. This results in the following condition on the sampling frequency: $$\frac{... 7 Short answer 10*log(bw/fs) to take into account the oversampling operation because the awgn() function specifies the signal-to-noise ratio per sample, in dB. Longer answer The discrete time AWGN model is$$Y = X+N$$where X is data from continuous time X(t), N is noise sequence from AWGN process N(t) and Y is receive symbols. If X(t) is ... 6 A recording originally at 8kHz and digitally upsampled to 16kHz will have almost no energy in the 4-8kHz range (whatever is here is due to imperfections in the filters used for the upsampling process). I would just use a 4kHz and 5.5kHz high pass; and use a threshold on the signal energy at the output of these filters. ... Unless your recordings are ... 6 An important theorem, known as Weyl's, 1931, is: if function s(t) and related functions ts(t), s'(t) are in L^2 (square integrable) with the related \|\cdot\| L_2 norm symbol then:$$ \| s(t) \|^2 \leq 2\| ts(t) \| \| s'(t) \|$$Equality is attained when s(t) is a modulated Gaussian/Gabor elementary function defined as:$$ s'(t) / s(t) \...

6

In the Nyquist sampling theorem, the bandwidth is usually defined as the largest frequency in the signal; in other words, what the sampling theorem says is that, if you sample this signal at 6.8 kilosamples per second, you will be able to reconstruct the analog signal perfectly. Now, under some conditions, you can do bandpass sampling, where the bandwidth ...

6

It is a common value because at $-3 \ \mathrm{dB}$ the power of the signal is reduced to half its value. I'll show a brief example to make it clearer. Suppose that you have a signal whose amplitude is $1 \ \mathrm{V}$. If we reduce its amplitude by $-3 \ \mathrm{dB}$, the new amplitude would be close to $0.707 \ \mathrm{V}$. The power of a signal is ...

5

Unless I'm missing something here it would just be the difference in frequency between the highest frequency sine wave and the lowest: BW = Fmax - Fmin

5

First a demonstration that the squares of both \begin{align}&[\dots, 0, 0, 1,\hphantom{-}1, 0, 0, \dots] \text{ and}\\ &[\dots, 0, 0, 1, -1, 0, 0, \dots]\end{align} equal $$[\dots, 0, 0, 1, \hphantom{-}1, 0, 0, \dots]\hphantom{\text{ and}}$$ but the squares of their sinc interpolations differ (Fig. 1): Figure 1. Squares of sinc interpolations ...

5

Not at all. The Uncertainty Principle says that a function cannot be both limited in time and limited in frequency. More specifically, the product of the signal's widths in time and in frequency (i.e., its time extension $\Delta_t$ and its bandwidth $\Delta_f$) is bounded from below: $$\Delta_t\cdot\Delta_f\ge C\tag{1}$$ where the constant $C$ depends on ...

5

The final statement is correct, for PSK with proper pulse shaping the baud rate and the bandwidth are the same (the bandwidth will typically be 20-30% higher than the symbol rate, but read on). The baud rate is the symbol rate or the rate of change from one symbol to the next in the modulated constellation. Your earlier figures suggest unfiltered waveforms ...

5

Does this formula mean that these two channels have the same transmission speed? Yes. That's exactly what you should take away from this: Channel capacity has nothing to do with center frequency; only bandwidth and SNR are relevant. Cause I think the one from 1 GHz to 1.02 GHz is faster, since it has a higher frequency. But that's plain wrong. It's not ...

5

A step response test is an easy way to determine the bandwidth. Sum a small step into the control voltage of your oscillator (VCO or NCO), and measure the 90% to 10% fall time of the corrected response at the output of the loop filter as shown in this block diagram. Note that the loop will respond in such a way to completely cancel the injected offset, but ...

4

In order to avoid inter symbol interference (ISI) the baseband transmit signal must meet Nyquist's first criterion. In frequency domain it can be formulated as follows. $$\omega_\mathrm{N} = \frac{v_\mathrm{s}}{2}$$ With $v_\mathrm{s}$ the symbol rate and $\omega_\mathrm{N}$ the 3dB frequency of a low pass filter with symmetric flank. The flank must be ...

4

The need for modulation: Your voice signals generally lies in the frequency range 1kHz to 4kHz and music lies in the range 20Hz to 20Khz. Say your locality has 4-5 AM broadcast stations. Now all the AM broadcast stations cannot transmit their content as is because if all the stations transmit at the same time there will be signal interference. Also, for ...

4

In communications, whether a signal is narrowband (has a small bandwidth) or wideband (has a large bandwidth) is relative to the channel's impulse response. If the channel behaves as a flat filter (constant gain) over the signal's bandwidth, then the signal is narrowband. If the channel introduces significant distortion (that is, if different signal ...

4

Do you perhaps mean small signal model? That is often used in the context of analysing the performance of transistor circuits. The transistors themselves are nonlinear devices, but they can be analysed as linear devices if you choose their operating regime appropriately. Conversely, the large signal model is when the operating regime is such that the ...

4

Interpolated points of the DFT can be computed using a dot-product of a few samples around the peak region with a pre-computed interpolation vector. The interpolation vector is determined by the location of the desired interpolated sample, taking into consideration the amount of zero-padding required, etc., etc. This technique and the methodology to ...

4

To complement my part to this question: Here is a somewhat shorted answer based upon a manual expansion of the odd function $f(x)$ \begin{align*} f(x)&=\ln\left(\arctan\left(\alpha e^x\right)\right)-\ln\left(\arctan\left(\alpha e^{-x}\right)\right)\\ &=f_1x+f_3x^3+O\left(x^5\right)\tag{1} \end{align*} into a series up to the third order. Some ...

4

The noise power continues to be $N_0/2$, independent of bandwidth. The reason is that the noise variance at the output of a filter with frequency response $H(f)$ is $$\sigma_n^2=\int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,df.$$ (This is a direct application of the Wiener-Khinchin theorem). Since you're assuming an orhonormal basis, the noise power is always ...

4

This picture may help clear up the OP's confusion. On the right is the baseband signal of interest, with finite bandwidth centered at DC as shown in the frequency spectrum plot. This is a complex signal, so it has I (in-phase, on the real axis) and Q (quadrature, on the imaginary axis) components. We implement these signals in hardware using two real ...

4

You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92). With a well-behaved signal $f(t)$ as you defined it above (in particular, $f(t)$ is integrable and bandlimited to $B\,\text{Hz}$, and $\text{sup}\,|f(t)| = A$), then $$\left|\frac{\text{d}f(t)}{\text{d}t}\right| \leq 2AB\... 4 Yes, of course, you can sum them. The bandwidth of the resulting signal is simply the min/max of the individual signals. If we assume$$z(t)=x(t)+y(t)$$Then then bandwidth of z(t) will simply be [min(f_{min},g_{min}),max(f_{max},g_{max})], so in general the you will have B_z > B_x Keep in mind that for any real valued signal x(t) the spectrum ... 3 The channel bandwidth and FFT size alone don't provide enough information to describe the entire structure of the OFDM signal. Recall the following relationship:$$ \Delta f = \frac{f_s}{N}  where $\Delta f$ is the subcarrier spacing, $f_s$ is the sample rate used at the modulator input, and $N$ is the FFT size. In your LTE examples, the two channel ...

3

I guess I'll convert my comments above into an answer: apart from the theoretical signal processing concepts, physical devices are made to be used in real, physical systems. 155 Mbps and 622 Mbps are common data rates used on fiber-optic SONET networks, specifically OC-3 and OC-12 lines. Therefore, the diode that you referenced is specifically designed and ...

3

Find the Fourier Transform $F(\omega)$ and perform the following integration.$\frac{1}{2\pi}\int_{-W}^{W}|F(\omega)|^2\mathrm{d}\omega=0.7E$, where $E$ is the total Energy of the function given by: $E=\frac{1}{2\pi}\int_{-\infty}^{+\infty}|F(\omega)|^{2}\mathrm{d}\omega$.The value of $W$ is the 70% Bandwidth.

3

"The telecom people" are the ones providing the internet data to you. Hence, they can easily throttle data at their end, making sure that you do not get faster access than you pay for. If the actual signal processing is adapted based on this I would say depends on the solution.

3

The required bandwidth depends on the symbol rate (or baud rate). If you have a binary system (i.e. two symbols) then the bit rate and the symbol rate are equal. If you use $M>2$ symbols, you can transmit more than 1 bit per symbol and your symbol rate, hence also the required bandwidth, is decreased. The problem with increasing the number of symbols is ...

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