# Tag Info

8

Creating the Frequency Vector The arrangement of the output of fft() depends on whether you use an odd or even number of points for your fft. I think this post nicely summarises how the frequencies are arranged. Have a look at it. Since you are using an even number of points, the Nyquist frequency, $F_N = F_s/2$, is present in the output of your fft, and is ...

7

The OP's opening statement is incorrect: $f_s > f_{max}/2$ prevents frequency aliasing for a bandlimited signal, but not amplitude aliasing $f_s > 2 f_{max}$ prevents aliasing. It's as simple as that. There is no such distinction as "amplitude aliasing". Since the OP has stated the signal is band-limited; as long as we can assume that means ...

4

Know that each bin in the FFT is the value for the exponential $e^{j\omega t}$, not sine or cosine. So for the OP's case $5sin(2\pi 3 x-2)$ in terms of exponentials (see Euler's identity) this is: $$\frac{5}{2j}e^{j(6\pi x -2)} - \frac{5}{2j}e^{-j(6\pi x -2)}$$ Each is a DFT bin with that magnitude and phase given by the exponentials above, as long as ...

4

In order to resample without aliasing, the resampling process needs to apply an anti aliasing filter at the new Nyquist Frequency (or thereabouts). So chances are, your resampling process applied a low pass filter at 256 Hz or so and that removed A LOT of the energy.

4

The claim is wrong. Sampling of a pure sinusodial whose frequency is below but arbitrarily close to the Nyquist frequency (half the sampling frequency) is a perfectly valid operation, as long as you can create ideal (zero width transition band) brickwall lowpass filters to be used at the reconstruction interpolation of the continuous waveform from its ...

3

As mentioned in the comments, an impulse response with the shape shown in your question can only be obtained by a system with two real-valued poles. In continuous time, with two distinct poles, the total impulse response is $$h(t)=\frac{e^{-\alpha_1t}-e^{-\alpha_2t}}{\alpha_2-\alpha_1}u(t)\tag{1}$$ It's a basic exercise to determine the location of the ...

3

Below is the analytic result for both the actual max value of $0.901243$ and the maximum value found by the OP of $0.898464$ The reason you are not getting the predicted maximum is your samples of the sine wave are not located exactly at the peak. This is clear if you zoom in on the plot and compare the two peak locations for the number of samples given (as ...

2

I can help with some of your multiple questions. First, a cascade of n buffered RC low pass filters (LPFs), a so-called n-th order synchronous LPF, has the impulse response and step response shown in the screenshot below: This is a screenshot from my paper 1 referenced at the bottom. All R values are the same and all C values are the same. Buffering simply ...

2

Okay, this takes a bit of algebra, Euler's formula, and the geometric series summation formula, and some plugging and chugging, but here is how you can calculate it directly: \begin{aligned} x[m] &= \frac{1}{n}\sum_{k=0}^{n-1} A \cos \left( (m-k) \frac{2\pi}{p} + \phi \right) \\ &= \frac{1}{n}\sum_{k=0}^{n-1} A \left[ \frac{e^{i\left( (m-k) \... 2 Are you familiar with how to obtain the Fourier series of an arbitrary periodic function? Because if you are, I'm confused at your confusion. This is simply the Fourier series expansion of the triangle wave with fundamental frequency f_0=1/T. This can be found by taking the inner product of the function with the basis functions. We note that the DC value (... 2 You're right, to convert from a power density to a power, you integrate over frequency! However: Well, if neither of your transmitter nor receiver are calibrated, then you simply can't find the "real" transmitted or received power. The plots you have are "amplitude relative to maximum ADC input". You don't know what power "maximum ... 2 This is what Fourier series are all about. Under relatively mild conditions, a T-periodic function f(t) can be represented as an infinite sum of complex exponentials:f(t)=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi kt/T}\tag{1}$$Note that if f(t) is real-valued, (1) can equivalently be written as a sum of real-valued sinusoids:$$f(t)=c_0+2\sum_{k=1}^{\...

1

Perhaps standard deviation is what you want. Standard deviation $$\sigma= \sqrt{\frac{1}{n}\sum_{k=1}^n (x_k-x_{ave})^2}$$ (where $x_{ave}$ is the average of your sequence) quantifies the fluctuations of your sequence above and below the sequence's average value. The square of the standard deviation, $\sigma^2$, is the variance of your sequence which is a ...

1

Is amplitude of oscillation a (only) function of initial condition of the system? No. For a system with a transfer function of the form $H(s) = \frac{\cdots}{s^2 + \omega_0^2}$, the amplitude of oscillation will increase any time it is excited by a signal that has a component at $x(t) = \cos \omega_0 t + \phi$. Even if you're not exciting it intentionally,...

1

Regarding your problem it seems that there is a misunderstanding between dB and dBm. When we talk about the power ratios between signals we will express the difference in dB while when we will express the power of the signal we will be interested in the power in dBm which is none other than the power of the signal in question compared to 1mW. Which leads me ...

1

You need to distinguish between: The error rate for the inner symbols - the error is half of the overlapped segments between the symbol to its closest neighbors (by symmetry we consider one and multiply by 2). $$P_{e1} = Pr(|n|>100-G) = 2*Pr(n>100-G) = 2 \frac{100 - G}{200}$$ The error rate for the points 31G and -31G - same as above bu only for one ...

1

I am not sure if you can use $P_{err}=\text{erfc}(G/\sigma)$ because noise is not gaussian distributed. Here is my take on it. Assuming uniform probability of transmission for all 32 symbols $x_i$,the received signal $y=x_i+n$ so given that $x_i$ was transmitted $y$ is also uniformly distributed in the interval $[-100+x_i,100+x_i]$. Suppose say the ...

1

A few points to consider. This all gets a lot easier if you create a sine wave that has an integer number of periods in your time window. You can do this by building your time vector as t = np.linspace(0, tmax*(1-1/npts), npts) If you do this you can approach this as follows: You are using a time window of length of 10s sampled with 100 taps, which ...

1

Puzzle solved, thanks to Cedron Dawg and Dan Boschen! First, I ran a simple N point moving average of a sinewave, using the simulation model below: I used the OP's values: N = 10, P = 40, sinewave amplitude = 1 and a simulation step size, $\Delta t$, equal to unity. The results, shown in the next figure, are the same as those of the OP: The maximum ...

1

The amplitude reduction is simply given as the magnitude of the transfer function of moving average filter. A moving average filter has a rectangular impulse response so the transfer function will be a $sinc()$ function. You need to sample the $sinc()$ function at the frequency or your sign wave

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