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23

Say you're interested in $$M^{j2\pi f_0 t}. \tag{1}$$ Note that $$M = e^{\log M},$$ so $(1)$ can be written as \begin{align} M^{j2\pi f_0 t} &= \left( e^{\log M} \right) ^ {j2\pi f_0 t} \\ &= e^{j2\pi (f_0\log M) t} \\ &= \cos(2\pi (f_0\log M) t) + j \sin(2\pi (f_0\log M) t), \end{align} which is a complex sinusoid with frequency $f_0 \log M$. ...


21

It is actually not distorted, it is sampled at high enough rate. What fools you is the straight lines drawn between sample points, it gives you a false impression of the waveform. It shows you a linear interpolation of the signal. It does not represent how the signal would actually look like. A sampled signal exists only at the sample points, and to convert ...


9

The actual requirement is to sample at GREATER then twice the bandwidth, not at a rate equal to it... So only your 80Hz same set actually meets the requirement, because the 60Hz case is ambiguous in general, consider if you were sampling sin (2PiFt) instead then you would get a flat line at zero amplitude.... And changing the angle between sin and cos would ...


6

in my opinion, either of these two functions of time can be rightly labeled "sinusoidal": $$ x(t) = A \sin(\omega t + \theta) $$ $$ y(t) = A e^{j(\omega t + \theta)} $$ where $A$, $\omega$ can be any non-zero real number (i wouldn't call DC a "sinusoid") and $\theta$ can be any real number.


6

Frequency is more likely decreasing (period is getting longer). This is a sweep sine signal or as some people used to call it - chirp tone. No point in rewriting the literature, so here are some links with very good explanation for both linear and exponential frequency change. For linear sweep you get: $$s(t)=\sin\left(\omega_1t+\frac{\omega_2-\omega_1}{T}\...


6

There is no aliasing as 𝑓 = 30 Hz is less than or equal to the folding frequency, 30 Hz and 40 Hz, respectively. Yes and no. There isn't significant aliasing when you're sampling at 80Hz, because the resulting signal has frequency components at 30Hz and 50Hz. The result is thus unambiguous as long as you take that 50Hz signal into account. There is ...


4

It makes sense after all: your phase estimate is an estimate of the unknown phase angle $\phi$ minus $\pi/2$, because you're using a sine function. So the phase of the FFT at the appropriate frequency index corresponds to $\phi-\pi/2$. So in your example, $\phi\approx -0.98+\pi/2=0.59$, which is a lot closer to the actual phase. If you used a cosine function,...


4

The fundamental property of averaging filters (in swear words, a linear system) is that the output of a sine is a sine of the same frequency, albeit of zero amplitude. So: Which non-trivial moving average weights would you need to find to get a filter that returns the same output as its input? An averaging in phase with the sine: take a filter full of ...


4

The phase of the sinusoid does not matter: A phase shift of a sinusoid is equivalent to shifting it in time, which results in a time shift of both the quantized sinusoid and the quantization error. The power spectrum is invariant to time shifts. We choose to work with sinusoid $A\cos(x)$. Optimality by def. 1 Equivalently to maximizing signal-to-noise ...


4

HINT: $$\frac{\cos(n\pi /6)}{(n+3)\pi}=\frac{\cos[(n+3)\pi/6 -\pi/2]}{(n+3)\pi}=\frac{\sin[(n+3)\pi/6]}{(n+3)\pi}$$


3

The function $$x(t)=e^{j(\omega t+\phi)}\tag{1}$$ is most commonly called a complex exponential (function). Some people call it a complex sinusoid, but I think this is a misnomer, because what is $\sin(z)$, $z\in\mathbb{C}$, then? Note that $(1)$ is also (much less commonly) called cis or cisoid. A sinusoidal function has the form $$x(t)=A\sin(\omega t+\...


3

You're almost there. First, there's a small scaling error in the transform $X_c(\omega)$. Since the Fourier transform of an impulse train is given by $$\mathcal{F}\left\{\sum_{n=-\infty}^{\infty}\delta(t-nT)\right\}=\omega_s\sum_{k=-\infty}^{\infty}\delta(\omega-k\omega_s)\tag{1}$$ with $\omega_s=2\pi/T$, and since multiplication in the time domain ...


3

You need to make sure that the phase of continuous. The easiest way would be something like this. x(n) = cos(phase); phase = phase + 2*pi*current_frequency*sample_time; current_frequency= update_frequency(n); n = n + 1; Note, that this is very inefficient code and only for illustrative purposes.


3

How can we know the used sample frequency fs by looking at y(t)? You can't. The sampling theorem states that any sampling higher than twice the highest signal frequency allows for perfect reconstruction. In you case any sampling frequency higher than 4 would result in the same $y(t)$.


3

Remembering from my 1970 Signal Processing lectures we have ... The crucial thing is the filter used to reconstruct the signal. Let's do the theory first for ideal sampling a perfect sine wave at 2x its frequency and filtering with an ideal low pass filter. The samples are infinitely thin - they are delta functions separated by time t. The filter is an ...


3

There's nothing wrong about this – it's absolutely what you should be seeing. The sine wave is intact, if you applied a reconstruction filter – there's no amplitude fluctuation of the wave, as soon as you limit its bandwidth to the nyquist bandwidth, the analog signal "has" to follow the full sines.


3

The MATLAB / OCTAVE function freqz(b,a) is typically used for displaying the magnitude and the phase of a Frequency Response function $H(e^{j\omega})$ associated with a discrete-time, linear time-invariant (LTI), system with rational transfer function $H(z)$ (having a corresponding LCCDE representation) given by $$H(z) = \frac{ \sum\limits_{k=0}^{M} b_k z^{-...


2

Yes, it is necessary to add an additional pre/suffix to the OFDM symbol that corresponds to the window length when windowing is applied. As the length of the already existing cyclic prefix is usually chosen as the maximum delay spread of the channel no further Inter-symbol interference (ISI) must be introduced or otherwise orthogonality is lost. The ...


2

Any moving average that is of a length different from that of the period of sinusoid (or integer multiple thereof) will output a sinusoid of that same frequency. So you could just make your moving average 1 sample shorter (which would be the same as subtracting some phase of sma(1) from DC).


2

Hint: So you have proved that $$\exp\left({j\frac \pi4 n}\right)+ \exp\left({j\frac{7\pi} 4 n}\right) =\exp\left({-j\frac \pi4 n}\right)+ \exp\left({j\frac \pi4 n}\right)$$ The second derivation does results in the same answer as well. It is not prohibited but one would want to simplify things here. It does not help factorizing with a factor $\exp({j\pi n})...


2

Many terms in mathematics have a long history, and sometimes, they were redefined later, with sounder or more coherent, or simpler, definitions. Sines and cosines are typical examples. I am not professional on math history, so corrections are welcome. From what I know, the present sine and cosine terms originate from (Indian) Sanskrit, through Arabic, to ...


2

Since the OP asked for a real code example, there is one below. It's in Matlab, but optimized for implementation on a real time processor. The idea is to implement the oscillator through a complex phasor multiplication, which only requires one complex multiply per sample and no transcendental functions (which tend to be very expensive). The only potential ...


2

It's an interesting question. Let's see what nonzero complex numbers $w$ have the property that they "act like $e$" in the classical formula, i.e., that $$e^z = w^z$$ for all complex $z=x+iy$. For convenience, suppose we can write $$w=re^{it}$$ The symbol $w^z$ takes the possible multiple values $$w^z =e^{z\log w} = e^{(x+iy)(\overbrace{\ln r + it + 2k\pi i}...


2

signal.windows.cosine is a window function, not a signal, as it says in the docstring: Return a window with a simple cosine shape. You want something like numpy.cos(2*pi*f*t).


2

Those 2D cosine functions are independent of your input image. They are just "cosine waves", or the 64 basis functions that yield 64 coefficients when transforming "$8\times 8$" blocks. Given a $D$ $8\times 8$ matrix for the DCT-II, and a $8\times 8$ image patch $I$, you'll get $64$ coefficients $C$ by: $$C=DID^T$$ [EDIT] If you want to display the 2D DCT-...


2

Since this is a pure sinusoid, it has a bandwidth of 0 Hz. You can multiply it by a carrier signal of the same frequency, pass it through a low pass filter then take only a few samples. What matters is NOT the frequency of the signal, rather the bandwidth. Consider for example a voice signal modulating a 1 GHz carrier. It will be very costly, to sample this ...


2

Every periodic signal can be described as sum of "basis" functions. These basis functions are the sin and cosines at frequencies of integer multiples of a fundamental freqeuency. This fundamental freqeuency is a consequence of the periodicity of the signal. So if a signal is periodic with time $T$, then the fundamental frqeuency is $F_o = \frac{1}{T}$ Hz. ...


2

I would not teach integrals this way. Issues might arise form the definition of terms. In standard calculus, a (note: "a", not "the) primitive function (also called antiderivative or indefinite integral) of a continuous function $f$ is a kind of converse of the concept of derivation. SO: If $F$ is "a" differentiable function whose derivative is equal ...


2

The integration in your question is equivalent to convolution with the unit step function: $$y(t)=\int_{-\infty}^tx(\tau)d\tau=(x\star u)(t)\tag{1}$$ This means that in the Fourier domain we have $$Y(j\omega)=X(j\omega)U(j\omega)\tag{2}$$ With $$U(j\omega)=\frac{1}{j\omega}+\pi\delta(\omega)\tag{3}$$ Eq. $(2)$ becomes $$Y(j\omega)=\frac{X(j\omega)}{j\...


2

First of all, the correct equation for an RRC pulse is given here. You are correct that you need to define the sampling frequency Fs and the symbol interval Ts. The number of samples per symbol interval is then Ts*Fs, which I assume to be an integer. You also need to define beta, and the pulse duration D. For simplicity, let's assume D is given as a multiple ...


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