9
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What is a $BT$ (Bandwidth-Time) product with reference to modulation?
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 ...
- 3,366
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How do I calculate the bandwidth from a waveform?
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 \...
- 6,080
8
votes
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Relation between Bandwidth and Baud-Rate for 8-PSK
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 ...
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8
votes
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Correct way to add AWGN to a signal
Short answer
10*log(bw/fs) to take into account the oversampling operation because the awgn() function specifies the signal-to-...
- 6,080
6
votes
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Uncertainty principle - Duration bandwidth principle
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:
$$ \|...
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6
votes
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Bandwidth confusion
In the Nyquist sampling theorem, the bandwidth is usually defined as the largest frequency in the signal; in other words, what the sampling theorem says is that, if you sample this signal at 6.8 ...
- 14.9k
6
votes
Why we measure bandwith at -3dB?
It is a common value because at $-3 \ \mathrm{dB}$ the power of the signal is reduced to half its value. I'll show a brief example to make it clearer.
Suppose that you have a signal whose amplitude ...
- 4,980
6
votes
OFDM use a pulse shaping filter or not?
The purpose of pulse shaping filters is not to overcome ISI as is implied in the OP's question. The only reason for using a pulse shaping filter is spectral efficiency, and in the process ISI can be ...
- 48.8k
5
votes
Interpolation of magnitude of discrete Fourier transform (DFT)
First a demonstration that the squares of both
$$\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, ...
- 13.5k
5
votes
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Value of power spectral density $N_0$ or effect of scaling bandwidth to SNR
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 \...
- 14.9k
5
votes
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Infinite extent of spectrum, but also in time in Oppenheim's Discrete Time Signal Processing?
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.,...
- 88.9k
5
votes
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Why Shannon Theorem has nothing to do with frequency?
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; ...
- 29.6k
5
votes
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Bounds of the derivative of a bounded band-limited function
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 ...
- 14.9k
5
votes
How can I experimentally find the bandwidth of my PLL?
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 ...
- 48.8k
4
votes
What is the definition of small/large signal bandwidth?
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 ...
- 25.2k
4
votes
What is the definition of small/large signal bandwidth?
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 (...
- 14.9k
4
votes
Does "keying on" a sine wave at a zero-crossing reduce its bandwidth?
It is instructive to have a look at the expression for the spectrum of a truncated sine wave to see how the choice of the window length affects the decay of the spectrum. Let's define a windowed ...
- 88.9k
4
votes
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Interpolation of magnitude of discrete Fourier transform (DFT)
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 ...
- 506
4
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Why is the bandwidth doubled with AM?
Bandwidth is defined at positive frequencies. So for the lowpass (baseband) signal in your first figure, the bandwidth equals its upper cut-off frequency, whereas in the bandpass case (your second ...
- 88.9k
4
votes
Mathematical question that comes out of using bilinear transform
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)\...
- 641
4
votes
Spectral Effeciency of BPSK
The spectral efficiency depends on the pulse shape. The basepand BPSK signal can be written as $$s(t)=\sum_k a_k p(t-kT_b),$$ where $a_k$ is equal to either $\sqrt{E_b}$ or $-\sqrt{E_b}$, $E_b$ is the ...
- 14.9k
4
votes
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The necessity of null subcarrier in OFDM?
The only portion of the spectrum that is "occupied" or "transmitted" are the subcarriers that are actually used (i.e. 10 MHz instead of 15 MHz). The null subcarriers referenced in your question are ...
- 24.5k
4
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Why Shannon Theorem has nothing to do with frequency?
This picture may help clear up the OP's confusion.
On the right is the baseband signal of interest, with finite bandwidth centered at DC as shown in the frequency spectrum plot. This is a complex ...
- 48.8k
4
votes
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Passband vs Baseband Bandwidth
You can unify the definitions for baseband and passband bandwidths by saying that bandwidth is only measured at positive frequencies. So for low pass signals, the lower frequency is zero and the upper ...
- 88.9k
4
votes
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How to generate random data with a specific bandwidth
How can I manipulate/control $R_S$?
You usually start with a desired pulse rate $R_S$. Then, the number of samples per symbol is $f_sT_S$, where $f_s$ is the sampling rate and $T_S = 1/R_S$. The ...
- 14.9k
4
votes
sum of 2 signals
Yes, of course, you can sum them. The bandwidth of the resulting signal is simply the min/max of the individual signals. If we assume
$$z(t)=x(t)+y(t)$$
Then then bandwidth of $z(t)$ will simply be $[...
- 42.5k
4
votes
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Why do ATSC channels need 6MHz of bandwidth when they're digital?
A key point is bandwidth is proportional to the symbol rate, or rate of change of the modulation. If the symbol is rectangular shaped, the spectrum is a Sinc function with the first null in Hz at the ...
- 48.8k
4
votes
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What is the intuition explaining the Shannon-Hartley theorem?
Fundamentally, the signal-to-noise-ratio (SNR or $S/N$) is the ratio of signal power to noise power and that power ratio and is usually expressed in $dB$. Shannon and Hartley (likely collaborating at ...
4
votes
What is the intuition explaining the Shannon-Hartley theorem?
Echoing what already answered: you are approaching this backwards. SNR is a concept that's very fundamental and applicable to way more things that just channel capacity.
If you have a signal $y(t)$ ...
- 42.5k
4
votes
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Specify notch bandwidth by pole placement
Summary
Given the equation for the transfer function of the second order notch filter:
$$ H(z) =\frac{1+2\alpha \cos(\omega_n)+\alpha^2}{2+2\cos(\omega_n)}\frac{z^2-2z\cos\omega_n+1}{(z^2-2az\cos\...
- 48.8k
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