10

A random process is a collection of random variables, one random variable for each time instant. It is best to write the random process as $$\{X(t)\colon -\infty < t < \infty\} \tag{1}$$ where the $\{$ and $\}$ indicate that a set (or collection) of objects is being defined, and the interior says that a typical member of this set is denoted by $X(t)$ ...


8

Why is the fourier transform a special case of the laplace transform? The Laplace transform produces a 2D surface of complex values, while the Fourier transform produces a 1D line of complex values. The Fourier transform is what you get when you slice the Laplace transform along the jω axis. For instance, a simple lowpass filter $H(s)=\frac{1}{s+1}$ has a ...


8

Here let me show you a simple procedure very similar to pole zero placement which will be helpful for your notch filter design. First, lets analyse the frequency response of a single zero and let $$ H(z) = 1 - b z^{-1} $$ be a first order system with a single zero at $z = b$ where $b$ is a complex constant with a radius $r$ and phase angle $\phi$ radians; ...


8

Abhinav Jain, Welcome to DSP Community. I build for you a proper testing of the run time comparison. Few tips about timing in MATLAB: Never time in a script. Always call a function to do the heavy lifting. When you run something from script it runs in the global scope which means MATLAB can't optimize it as well as it could if it was in a function. When ...


7

A causal system does not need to know the future in order to compute its output. A memoryless system computes the output only from the current input. A memoryless system is always causal (as it doesn't depend on future input values), but a causal system doesn't need to be memoryless (because it may depend on past input or output values). The system $$y[n]=x[...


5

If you have an understanding of Fourier transforms then you probably already have a conceptual model of transforming signals into the frequency domain. The Laplace transform provides an alternative frequency domain representation of the signal - usually referred to as the "S domain" to differentiate it from other frequency domain transforms (such as the Z ...


5

There are a few things going on here. First, one of the fundamental properties of the Fourier transform is that it conserves energy, i.e. the energy of the time domain is exactly equal to the energy of its Fourier transform. What is the power of a sinusoid? Assuming we restrict the calculation to integer multiples of cycles it is as follows (I suggest ...


5

Even though I realize that this is a very late response, I will nevertheless try to answer this question because I find it instructive and also because the number of upvotes suggests that this question is of general interest to the community. As already suggested in the question, let's define two signals $x(t)$ and $w(t)$ as $$x(t)=e^{-kt}u(t),\quad k>0\\...


5

You can achieve this result by using two combs filters : https://en.wikipedia.org/wiki/Comb_filter Put simply, the comb filter consists of adding a delayed version of the signal to itself, causing destructive or constructive interference. For instance, with $K = 20$ and a negative gain value after the delay line, you can significantly decrease or suppress ...


5

Note that option (b) is not correct, and that it is also not equal to what you came up with. Option (b) is just the multiplication of $x(t)$ and $y'(t)$, not the convolution. Your solution and option (c) are both correct, assuming that all derivatives exist and that the convolution integrals converge, because with that assumption the following holds: $$\...


5

Two principles here: When dealing with a differential equation, you define intermediate state variables so everything is in terms of first derivatives. This system is nonlinear, so the state-space equations won't be in terms of matrices. Applying these principles, we define a state vector: $$ \mathbf x = [x_1, x_2]^T, $$ where: $$ x_1 = y \\ x_2 = \dot y $$...


4

The issue is resolved: the butter() function automatically does bilinear transform, you just need to specify the frequencies normalised to the Nyqyist freq.


4

Polyphonic pitch estimation is still a difficult problem. Search on the MIREX forum for current research papers. Simplistic spectrum based approaches usually require component pitches with extremely simple timbres (nearly pure unmodulated sine waves, etc.) rarely found in real-life polyphonic music. More typical polyphonic music can contain a dense set of ...


4

You're probably best off asking your professor for clarification. As you reasoned, it sounds like you're being asked to multiply the image in the frequency domain by either a rectangular or circular mask that eliminates all frequencies outside the mask. This is not typically how you would apply a filter, however, as "ideal" frequency-domain filter masks have ...


4

I'll go through it in the $z$-domain. First, we find the transfer function $H_1(z) = \frac{V(z)}{X(z)}$. As you noted, in the time domain, $x[n]$ and $v[n]$ are related as follows: $$ v[n] = x[n] + g * v[n - M] $$ Take the $z$-transform of the above and you get: $$ V(z) = X(z) + z^{-M}G(z) V(z) $$ taking advantage of the convolution property, which ...


4

I don't feel like it's a very well-posed question, as multi-channel signals are sometimes represented as vector-valued, which would make them multidimensional. However, I assume the answer that is being fished for is: This is a one-dimensional, two-channel setup. The two channels are the child's height and weight. Each signal is one-dimensional, measured ...


4

The best intuitive description of Laplace transform I've ever seen: At first glance, it would appear that the strategy of the Laplace transform is the same as the Fourier transform: correlate the time domain signal with a set of basis functions to decompose the waveform. Not true! Even though the mathematics is much the same, the rationale behind the two ...


4

Hint: define a bound for $|x(t)|$, i.e., $|x(t)|\le A$; now try to find a positive number $B$ such that $|y(t)|\le B$ for any $|x(t)|\le A$ (that's simply the definition of BIBO stability). For the system $y(t)=1/x(t)$ it should be easy to show that the above cannot be satisfied (because $|x(t)|$ can get arbitrarily small).


4

In easy words: A process is stationary if its stochastic properties are independent of the time you look at it. Think of it like this: A stochastic process is just a Random Variable (RV) that, instead of giving you e.g. a real value gives you a function every time you look at it. We call that realizations. If you now take a lot of these realizations, and ...


4

Technically, the impulse $\delta(t)$ is called a distribution, and not a function, but for the purposes of your first course in Fourier transforms, what you need to know is that $\delta(t)$ has the sifting property $$\displaystyle\int_{-\infty}^\infty x(t)\delta(t) \,\mathrm dt = x(0) ~\text{provided that }x(t) ~\text{is continuous at }t=0 \tag 1$$ ...


4

For BIBO stability in the case of discrete time, there is a necessary and sufficient condition given by $\sum |h[n]| < \infty$ that is if the impulse response is absolute summable then the system is BIBO stable. Clearly $$h[n] = \delta[n] - \delta[n-1]$$ and it has a finite support, the impulse response is absolutely summable and therefore BIBO stable.


4

Hint The simplest way is to use the Z transform property "convolution in time domain is multiplication in z domain". See Z transform convolution $$\mathrm{Z}(x[n]*h[n]) = \mathrm{Z}(x[n]) \times \mathrm{Z}(h[n])$$ Then you just need to do inverse Z transform. Some typical Z transforms (including what is useful for you) can be found at Z transform pairs ...


4

First note that: $$ \cos(2\pi 50 t) \longleftrightarrow 0.5 \delta(f+50) + 0.5\delta(f-50) $$ $$\sin(2\pi 150 t) \longleftrightarrow 0.5 j \delta(f+150) -j 0.5\delta(f-150)$$ Hence the baseband spectrum is: $$ X(f) = 0.5 j \delta(f+150) + 0.5 \delta(f+50) + 0.5 \delta(f-50) - 0.5 j\delta(f-150) $$ Then a shift to right by 100 Hz yields (without sampling ...


4

The strategy depends heavily your "realistic scenario", e.g. which estimator, which equalizer, which estimation error, etc. In general scenario, you cannot trust the estimate of $h$, it means that you cannot use coheent detection. Thus try non-coherent detection (detection by energy) : to send bit $0$, use 2 channel uses (2 symbols) $x_0 = [x[0], x[1]] = [1,...


4

The standard meaning of white noise includes an insistence (whether implicit or explicit) that the mean is $0$. Thus, what you want to prove is trivially true: since $$Y[n] = \sum_{k=-\infty}^\infty h[n-k]X[k] = h\star X \big\vert_{n},$$ the linearity of expectation (the notion that $E[aX]=aE[X]$ and that the expectation of a sum is the sum of the ...


4

HINT: Going from your last equation, $$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$ This can be simplified further down by considering the following: \begin{align} e^{j2\pi (f_1T-n)} &= e^{j\pi (f_1T-n)}\cdot e^{j\pi (f_1T-n)}\\ 1 &=e^{j\pi (f_1T-n)}\cdot e^{-j\pi (f_1T-...


4

This is pretty straight forward using the definition of the Discrete Time Fourier Transform (DTFT). The definition of the DTFT: $$ X \left( {e}^{j \omega} \right) = \sum_{m = -\infty}^{\infty} x \left[ m \right] {e}^{-j \omega m} $$ Differentiating with respect to $\omega$: $$\begin{align*} \frac{d}{d \omega} X \left( {e}^{j \omega} \right) & = \...


4

In the following, I suggest that, before using the generic $T(\alpha_1 x_1+\alpha_2 x_2)$ versus $\alpha_1 T( x_1)+\alpha_2T( x_2)$, it can be more informative to try with simpler partial tests, or try counterexamples, based on your intuition. I don't really understand the motivation behind the second group of equations, or why $b$ gets multiplied by $\...


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

It's already answered in the comments so just to close it out: the answers are 3, 7 and 3. The periodicity is basically given by the denominator of the fraction in front of 2*pi*n after the fraction has been simplified (no common divisors between nominator and denominator). The simplified fractions for the cases are 1/3, 5/7 and 7/3 so the periodicity is 3, ...


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