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29

HINT When you sample at below the Nyquist rate, aliasing happens. That means frequencies higher than half the sampling rate get folded back down to below half the sampling rate. Have a read about bandpass sampling. PS: Tell your teacher, that's a really nice question. :-)

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As correctly stated in Peter K.'s answer, this question is about aliasing. Since you can't sample at a rate that is sufficiently high to avoid aliasing - i.e., $f_s>50\textrm{ kHz}$ - you have to take aliasing into account. Now it's your task to figure out the aliased frequencies of the given signals for the different sampling rates. If you understand how ...

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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 $X(t)$ ...

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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 ...

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; ...

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 $$y[n]=x[... 7 The integral doesn't converge in the conventional sense, so you can't solve it with standard methods. Assuming that you know (or can look up) the Fourier transform of the unit step function u(t), it is straightforward to compute the Fourier transform of \cos(\omega_0t)\,u(t) using the modulation property:$$\mathcal{F}\big\{u(t)\big\}=U(\omega)=\pi\delta(...

6

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) & = \... 6 Note that in this case you can see that the system is causal only from the given implementation. It's important to understand that you can't see it from the difference equation (if no initial conditions are given), and in general you can't see it from the transfer function either (if no region of convergence is given). The only case for which the expression ... 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

The property is the Linearity of the DTFT. Linearity means that if your input is a linear combination of signals the output will be the same linear combination of each input by itself: $$\operatorname{DTFT} \left( \alpha x \left[ n \right] + \beta y \left[ n \right] \right) = \alpha \operatorname{DTFT} \left( x \left[ n \right] \right) + \beta \... 5 Two principles here: When dealing with a differential equation, you define intermediate state variables so everything is in terms of first derivatives. This system is nonlinear, so the state-space equations won't be in terms of matrices. Applying these principles, we define a state vector:$$ \mathbf x = [x_1, x_2]^T, $$where:$$ x_1 = y \\ x_2 = \dot y $$... 4 The best intuitive description of Laplace transform I've ever seen: At first glance, it would appear that the strategy of the Laplace transform is the same as the Fourier transform: correlate the time domain signal with a set of basis functions to decompose the waveform. Not true! Even though the mathematics is much the same, the rationale behind the two ... 4 There seems to be an image which is gone now and hence I might be missing something. In order to declare if the system is invariant you should see if a delay of the input yields only a delay in the output. In you case, does the input x_1[n-m] yields y_1[n-m] and etc... As written by @DilipSarwate, By looking at the response for  {x}_{2} - {x}_{1}  ... 4 I'm not sure exactly what you're after, but just to try to add data: Using the separability property of the filter is always the right choice. Given it is separable we now have to apply 1D convolutions. Selecting which method to do it (Frequency Domain / Spatial Domain) depends on the length of the filter and the number of pixels in each dimension of the ... 4 Hint: define a bound for |x(t)|, i.e., |x(t)|\le A; now try to find a positive number B such that |y(t)|\le B for any |x(t)|\le A (that's simply the definition of BIBO stability). For the system y(t)=1/x(t) it should be easy to show that the above cannot be satisfied (because |x(t)| can get arbitrarily small). 4 In easy words: A process is stationary if its stochastic properties are independent of the time you look at it. Think of it like this: A stochastic process is just a Random Variable (RV) that, instead of giving you e.g. a real value gives you a function every time you look at it. We call that realizations. If you now take a lot of these realizations, and ... 4 Technically, the impulse \delta(t) is called a distribution, and not a function, but for the purposes of your first course in Fourier transforms, what you need to know is that \delta(t) has the sifting property$$\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 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 Working with your definitions: v \left( \left( n + 1 \right) {T}_{s} \right) - v \left( n {T}_{s} \right) = \int_{0}^{ \left( n + 1 \right) {T}_{s} } g(u) du - \int_{0}^{ n {T}_{s} } g(u) du = \int_{ n {T}_{s} }^{ \left( n + 1 \right) {T}_{s} } g(u) du $$So basically we have integration (Which is a Low Pass Filter) of White Noise over a Time Interval ... 4 A continuous-time white noise process \{X(t)\colon -\infty < t < \infty\} is a hypothetical construct that we can treat (in the simplified versions that we use on dsp.SE) as a zero-mean wide-sense stationary process with autocorrelation function K\delta(\tau) where \delta(\cdot) is the Dirac delta. More strongly, all the random variables X(t) ... 4 HINT:$$\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}

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