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Q1: 3) Given that x(t) has the Fourier transform X(jw), express the Fourier transforms of the signals listed below in terms of X(jw). You may find useful the Fourier transform properties listed in the attached Table X2(t) = x(3t- 6)See Answer
Q2: Assignment You are going to estimate the bandwidth of FM/PM signals by finding the frequency spectrum of the given FM/PM signals. Compare your estimates with the Carson's rule estimates. Use a mini lab report format to write the report. \text { Assume a signal } m(t)=2 \cos (2 \pi t)+3 \sin (4 \pi t) 1. FM b) Plot the magnitude spectrum of EM(t). Estimate the bandwidth required. c) Compute the bandwidth of pEM(t) by using the Carson's rule. ) Plot the magnitude spectrum of pPM(t). Estimate the bandwidth required. c) Compute the bandwidth of PM(t) by using the Carson's rule.See Answer
Q3: 1. Mystery System. Copyright University of Colorado Denver. This is an exam question. The use of any external resources such as Chegg is not allowed. Download the file black box.p from CANVA Sand save it in the folder where you have your MATLAB code. This file defines a system y=black box (x); Do not try to load this file into MATLAB just have it in the folder where you are running other code from so MATLAB can find the file. a) Use MATLAB to evaluate y=blackbox (x); where x =- [1 2 3 4] c) Use MATLAB to evaluate y blackbox(x); where x =[0 1 2 3 4 d) Is the blackbox system linear? Justify your answer using your answers from (a) - (c) e) Is the blackbox system invariant? Justify your answer using your answers from (a)-(c) Write equation for the blackbox system that gives output y[n] in terms of input x[n]. If you cannot write an equation, describe clearly in words what the black box system is doing. (Hint if the answer is not obvious input other signals into the system and see what comes out). b) Use MATLAB to evaluate y-blackbox (x); where x =2*[1 2 3 4]See Answer
Q4: 2True and False: A Gambler's Ruin (10 pts) [True | False] (1 point) The DTFT is sampling the DFT in the frequency domain. [True | False] (1 point) You cannot guarantee that an FIR filter can achieve a linearphase response. [True | False] (1 point) Zero padding adds new information into the DFT. In fact, zero padding will have the same result as sampling the signal more (i.e., a signal with 10 samples and 90 zeros is the same as a signal with 100 samples). [True | False] (1 point) The frequency mapping H(s) to H(z) is nonlinear, where H(2)is obtained with the bilinear transformation. [True | False] (1 point) IIR filters can be designed stable; however, the actual implementation in hardware can be unstable. [Short Answer] (2 point) What is DFT leakage? [Short Answer] (3 points) In a sentence or two, describe the difference between an analog-to-digital converter and a discrete-to-continuous converter.See Answer
Q5: 1) Consider the rectangular pulse signal x(t)=\left\{\begin{array}{ll} 1, & |t|<T_{1} \\ 0, & |t|>T_{1} \end{array}\right. Find the Fourier transform of this signal and sketch it. Make sure to find the amplitude at the origin and the zeros of X(jw) See Answer
Q6: 2) Consider the following a continuous-time signals: h(t) = 8 (t-to), find y(t)a. dx(t)b. y(t) =find H(jw)dtSee Answer
Q7: Convert each signal to the finite sequence form {a, b, c, d, e}. \text { (a) } u[n]-u[n-4] \text { (b) } u[n]-2 u[n-2]+u[n-4] \text { (c) } n u[n]-2(n-2) u[n-2]+(n-4) u[n-4] \text { (d) } n u[n]-2(n-1) u[n-1]+2(n-3) u[n-3]-(n-4) u[n-4]See Answer
Q8: Consider the following signal Write x(t) in terms of the pulse signal pr (t). ]Determine and express Fourier transformX (w).See Answer
Q9: The figure below shows the Fourier transform of a real bandpass signal, i.e., a signal whose frequencies are not centered around the origin. We want to sample this signal. Let F, in Hz (a) (4 pts) One option is to sample this signal at the Nyquist rate. Then to recover the signal, we pass its sampled version through a low pass filter. What is the Nyquist rate of this signal? (b) (9 pts) Since the signal might have high frequency components, Nyquist rate for this signal can be high. In other words, we need to have a lot of samples of the signal,which means that the sampling scheme is costy. It turns out that for this type of signal,we can sample it at a sampling frequency lower that the Nyquist rate and we can still recover the signal, however in this case, we will use a band pass filter. To see this, we have the following two options for the sampling frequency: • Fs= 0.5 Hz; • Fs = 1 Hz; For each case, draw the spectrum of the signal after sampling it. To recover the signal,which F, can we use? How we should choose the frequencies of the band pass filter?What is the minimum F, we can use and still recover the signal?See Answer
Q10: 4–4 A DSB-SC signal has a carrier frequency of 900 kHz and A. = 10. If this signal is modulated by a waveform that has a spectrum given by Fig. P3–3. Find the magnitude spectrum for this DSB-SC signal.See Answer
Q11: f) At a sampling rate f=15HZ the sampler outputs a signal v2(t). From what you found earlier,and using 5Hz/em as horizontal scale, represent V2(f) graphically on cartesian axes between(2 Marks)-45 and 45HZ.See Answer
Q12: As shown in Figure 1 below, the signal is now passed to a sampler. The sampler samples the signal s(t) at a rate f, and provides the following output: v(t)=s(t) e(t) where e(t) is the Dirac Comb, given by: \mathrm{e}(\mathrm{t})=\sum_{n=-\alpha}^{+\alpha} \delta\left(t-n T_{S}\right) Ts is the sampling interval and n is an integer. Analyse the output sampled signal v(t) by answering the following questions: :) Write the Fourier Transform E(f) of the Dirac Comb e(t). No details required. 1) Using the convolution and duality theorem, find the Fourier Transform V(1) of the sampledsignal v(t) as function of S(f), T, and f. Steps are required.(I Mark)See Answer
Q13: 4. Laplace Transform (20 pts) \text { i. } f(t)=t e^{-a t}\left(\sin \omega_{0} t\right)^{2} u(t) (b) The Laplace transform of a causal signal x(t) is given by X(s)=\frac{1}{s^{2}+2 s+5}, \quad \text { ROC: } \operatorname{Re}\{s\}>-1 Which of the following Fourier transforms can be obtained from X(s) without actu-ally determining the signal x(t)? In each case, either determine the indicated Fouriertransform or explain why it cannot be determined. i. F{r(t)e¬t} ii. F{r(t)e#}See Answer
Q14: Assume a causal LTI system S1 is described by the following differential equation: \frac{d^{2} y(t)}{d t^{2}}+4 \frac{d y(t)}{d t}+3 y(t)=a x(t), \quad y(0)=0, y^{\prime}(0)=0 where a is a constant. Moreover, we know that when the input is e', the output of the system (5 pts) Find the transfer function H1(s) of the system. (The answer should not be in terms of a, i.e., you should find the value of a). (5 pts)Find the output y(t) when the input is x(t) = u(t). (6 pts) The system S1 is linearly cascaded with another causal LTI system S2. Thesystem S2 is given by the following input-output pair: \mathcal{S}_{2} \text { input : } u(t)-u(t-1) \rightarrow \text { output }: \tau(t)-2 \tau(t-1)+\tau(t-2) Find the overall impulse response.See Answer
Q15: a. In this exercise we will look at how to convolve a signal with a system's impulse response function. Given an impulse response h[n] = u[n+ 1] – u[n – 3] and a signal ¤[n] = 2(u[n] – u[n – 11]) lets use MATLAB to convolve them and display the resulting signal y[n] = [n] * h[n]. To do so enter the following lines of code into MATLAB: >> n = [– 20:201: >> X = 0*n: >> x ([0:10] + 21) = 2; >> h = 0*n; >> h([-1:2] + 21) = 1; >> y = conv(h,x,'same '); b. Let's plot the resulting sequences: >> figure ( 2); >> subplot (3,1,1); >> stem (n,x,’filled '); >> axis ([-20 20 0 3]); >> xlabel('$$n$$ ' , 'interpreter ','latex '); >> title ('$$x [n]$$ ' , 'interpreter ','latex '); >> subplot (3,1,2); >> stem (n, h,'filled '); >> axis([– 20 20 0 3]); >> xlabel ('$$n$$ ' , 'interpreter ','latex '); >> title ('$$h[n] $$ ' , 'interpreter ',' latex '); >> subplot (3,1,3); >> stem (n,y,'filled '); >> axis ([-20 20 0 10]); >> xlabel('$$n$$ ' , 'interpreter ','latex '); >> title ('$$y [n]x[n]*h[n] $$ ', 'interpreter ','latex '); Notice that we create the discrete time vector n to have a range from [-20, 20]. We then create x[n] to be the same length as n and set each element to zero. Since MATLAB has a 1-based indexing convention notice that for creation of x[n] we need to set the elements corresponding to n e [0, 10] to 2, but we have to offset the index value by 21 since our n vector begins at -20. We do similar steps to define the impulse response h[n]. Then, we use the conv function to perform the convolution. Note that we use the parameter 'same'. This truncates the convolution to have the same number of samples as the longest input sequence. Normally a convolution results in a sequence of length (M +N – 1) (length of first sequence plus length of second sequence - 1). We also padour sequences with sufficient zeros in order to ensure that the full convolution results are properly captured in our truncated (due to ʻsame'), convolved sequence.See Answer
Q16: When we have a frequency response (or transfer function) for a system we often want to use it to plot the magnitude and phase response of the system. To do so in MATLAB we make use of the abs and angle commands. For a system frequency response given in analytical form, H\left(e^{j \omega}\right)=1+e^{-j \omega} > H = 1 + exp(-j.*w); figure (3); subplot (2,1 ,1); > plot (w, abs (H)); > аxis ([—pi рі 0 2]); > xlabel('$$\omega$$' , ' interpreter ', 'latex '); > title ('$$ | H(e^{j\omega})| $$ ' , ' interpreter ','latex '); subplot (2,1 ,2); > plot (w, angle (H)); ахis ([— pi рі —1.75 1.75]); xlabel('$$\omega$$', 'interpreter ','latex '); title (' $$\angle H(e^{j\omega}) $$ ' , ' interpreter ','latex '); In this case notice that we have used the plot command. Even though these are discrete sequences it is conventionally acceptable to plot the magnitude and phase response as continuous functions. Also notice that we have only plotted the response plots from [-7, 7]. One of the consequences of sampling is that the Discrete-Time Fourier Transform is27-periodic and hence we only need to plot from [-a, 7]. This will become clear later in the semester.See Answer
Q17: Later in the semester it will become useful to determine the frequency response of a signal or system by taking the Fourier Transform empirically (rather than computing it analytically). To do so we make use of the fft and fftshift commands. The fft command is an efficient implementation of the Discrete Fourier Transform (DFT) known as the Fast Fourier Transform (FFT). When the FFT is computed the samples are not ordered properly and thus the fftshift command is called to re-order the samples correctly. Then, to get the magnitude response of the system we again use the abs command (or angle for the phase response). the abs command (or angle for the phase response).Let's compute the Fourier Transform for a system described by a rectangular function, i.e., take h[n] = u[n+10]–u[n-11].For reasons that will become clear later, when we create our discrete sequence we pad the sequence with zeros on each side. In MATLAB enter the following commands to define h[n], take the Fourier Transform of it, and then plot the corresponding magnitude response of the system: n = -100:100; > h = 0*n; h(91:111) = 1;= 1; >> w = linspace(-pi , pi ,l1024); -> H = fftshift (fft (h,1024)); - figure (4); > subplot (2,1,1); - stem (n,h,'filled '); > axis([-100 100 0 1.5]); xlabel('$$n$$ ' , ' interpreter ','latex '); > title ('$$h [n] $$ ', 'interpreter ','latex '); >> subplot ( 2,1,2); > plot (w, abs (H)); > axis ([- pi pi 0 25]); • xlabel('$$\omega$$','interpreter ','latex '); >title ('$$ |H(e^{j\omega})|$$ ', 'interpreter ', 'latex '); A few things to note here are that we explicitly specify that the FFT be computed using 1024 points (rather than the default 201 points that is the length of h). This helps to improve the resolution of the frequency response and we will study why that is later in the semester. We create a plotting vector for w to have the same number of samples between[-7,7] and then use the plot command to display the magnitude response, similar to when we plotted the analytic magnitude response in Exercise 3.See Answer
Q18: Consider the following system \dot{\mathbf{x}}(t)=\left[\begin{array}{rrr} 0 & 0 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -2 \end{array}\right] \mathbf{x}(t)+\left[\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right] u(t), \quad y(t)=\left[\begin{array}{lll} 1 & 1 & 2 \end{array}\right] \mathbf{x}(t) Is it asymptotically stable? Is it BIBO stable?See Answer
Q19: Consider the system \dot{\mathbf{x}}(t)=\left[\begin{array}{rr} -1 & 0 \\ 0 & 0 \end{array}\right] \mathbf{x}(t)+\left[\begin{array}{l} 1 \\ 0 \end{array}\right] u(t) y(t)=\left[\begin{array}{ll} 1 & 1 \end{array}\right] x(t) Not all the eigenvalues have negative real parts, but the system is totally stable. How can we explain this?See Answer
Q20: B) (2 pts) The sun is 93,000,000 miles away from us and has a diameter ofroughly 865,000 miles. Under pinhole projection with a projection screen held 2feet (24 inches) away from the pinhole, what would you expect the diameter of theimage of the sun to be? Give your answer in inches, rounded to one decimalplace. Diameter =inchesSee Answer

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