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. :-)


21

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 ...


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

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 ...


4

Purely by inspection of the block diagram the system is causal, because the output is the sum of the current input sample and stuff that's delayed -- there's no $z$ blocks in there to predict the future, just $z^{-1}$ block to react to the past. Also by your method of finding the transfer function, the system is causal -- with a $3^{rd}$ order numerator and ...


3

I'm not gonna give you a complete answer but I can help you. Your open-loop transfer function has 2 poles at $ -1 ±j \sqrt(3) $ Good news, they are stable. However, they are not damped and they are slow. Your strategy is this 1 - Damp the poles, this will reduce the overshoot. 2 - Try to increase the poles frequency. This will reduce the settling time. 3 - ...


3

First, you are not trying to find the linearity of a signal, but the linearity of a system defined by the transform above between input and output signals. If you suspect your system is not linear (the different expressions depending on the sign of $x(t)$ is a big red flag), then you only need a counterexample rather than a general proof of linearity. In ...


2

The system is causal, provided that the recursion is forward; i.e., it's recursed for increasing $k$. Seeing that you are confused about causality tests, let me elaborate on it. Let's put the definition of causality from Oppenheim's Signals & Systems book : A system is causal if the output at any time depends only on values of the input at the present ...


2

As said by @Juancho, the expression let us suspect that the system could be non-linear. So we could look for a counter-example, yet it is not evident how to find a good one. Anyway, lt us try to better understand the system. Since there is something around sign change. The insight is there to have a first signal that has a sign change, and a second that ...


2

The poles of an FIR filter get many people tripped. The poles are the roots of the pole polynomial which you already have identified as $$P(z) = z^{2D}$$ How many roots does this polynomial have ? Where are the roots all located ?


2

The slope of the lowpass is 20dB/decade or 6dB/octave, that means it's a simple first order lowpass filter. At the corner frequency the gain is -3dB. Same of the phase. It goes from 0 to -90, so it's a first order filter. At the corner frequency, it's 45 degrees. Looking at both phase and level we conclude that it's a first order lowpass with $f_c = 1000 rad/...


2

$\mathcal{F}\{x^*(t)\} = X^*(-f)$ Taken from the table of Fourier transform theorems on slide 6 of this document: https://www.comm.utoronto.ca/~dkundur/course_info/316/KundurFTProperties_handouts.pdf Tables like these are useful in general. Please look it up yourself next time! :)


1

If you have a piecewise function, you need to separate the integral over that function into the same number of (sub-)intervals: $$\begin{align}\int_0^4x(t)e^{-jk\omega_0t}dt&=\int_0^1x(t)e^{-jk\omega_0t}dt+\int_1^2x(t)e^{-jk\omega_0t}dt+\int_2^4x(t)e^{-jk\omega_0t}dt\\&=\int_0^1(-t)e^{-jk\omega_0t}dt+\int_1^2e^{-jk\omega_0t}dt\end{align}$$ with $\...


1

Your intuition that you need initial conditions to fully solve this is correct, so it becomes a problem in test-taking. Were it me, and were the test proctored by the prof, then I would go to the front of the room and ask. Failing that, I would first solve for $y[10]$ in terms of $y[0]$, $y[1]$ and $y[2]$, then I would point out that as they were not given, ...


1

The Nyquist criterion states that the sampling frequency should be minimum twice the signal frequency. In this case it should be 50 kHz. A more precise statement is that the sampling frequency should more than twice the bandwidth. Usually, you want a bandwidth that include zero frequency, making the bandwidth and the maximum frequency the same, but that is ...


1

Your first approach doesn't work because the original signal is simply not the integral of the signal you sketched below it. Just graphically integrate that signal to see this. Your second approach doesn't work for the very same reason. If you integrate the rectangular signal you don't get a signal that is zero for $t>1$. The value of the integral would ...


1

Suppose that you could write $x_.$ as $x_.(t) = a_1x_1(t-n_1)+a_2x_1(t-n_2)$, then by LTI hypothesis, you can derive $y_.(t) = a_1y_1(t-n_1)+a_2y_1(t-n_2)$ with very little computation. You can get a first insight to that method by looking at the shape of $x_2$ which is self-evident, and a little more imagination gives you the structure for $x_3$. I find ...


1

It's hard to give you the "right" hints, since I have no idea what you have covered in class and what you didn't. But should'nt the magnitude be twice as high i both cases? Or what am I missing? The answer graph is pretty bad. The axes are not labelled and there are no units. The continuous spectrum would be $\delta(t)$ distribution so the value ...


1

As mentioned in a comment, just use the fact that $$|1+z|^2=1+2\,\textrm{Re}\{z\}+|z|^2,\qquad z\in\mathbb{C}\tag{1}$$ and take the square root to obtain the amplitude function. There is no simpler expression when $|z|\neq 1$.


1

Your result is correct, but it can be rewritten in an even simpler form, as suggested in the problem statement: $x[n]=u[an+b]$ If you draw the signal then you'll see that it is just a reversed and shifted unit step. From that drawing it should be easy to derive the constants $a$ and $b$.


1

The standard notation for the unit step function is $u(t)$. I've seen the notation $u_k(t)$, with integer $k$ in Oppenheim's Signals and Systems, where $k$ is the order of the derivative of the Dirac delta impulse. I.e., $u_1(t)=\delta'(t)$, etc. Using the same logic, negative values of $k$ indicate integrals of the Dirac delta impulse. So $u_{-2}(t)$ is the ...


1

If you did want to grind through this, you could define $x_3(t) = A x_1(t) + B x_2(t)$, then substitute it into the left side of either of your top two system definitions above. like this: $$x_3(t) = \begin{cases}0 & x_3(t) < 0 \\ x_3(t)+x_3(t-2) & x_3(t) \ge 0\end{cases} = \\ \begin{cases}0 & A x_1(t) + B x_2(t) < 0 \\ A x_1(t) + B x_2(t)+...


1

In terms of linear algebra, functions can be represented as vector with infinite number of components (in the Hilbert's space as they say in physics or $L2$ space in mathematics). The set of function $e^{-st}$ with complex $s$ forms a vector space, and Fourier and Laplace transform X(s) = $\int_{a}^{ \infty} x(t) e^{-st} $ is a projection of $x(t)$ to $e^{-...


1

A 5-point moving average can be performed in different ways. The two principal options consist in: causal: take the current point, and average it with the four most recent past samples, or sum it and divide by the length of the average span (which seems to be your choiice, regarding your for bounds $$ y[n] = \left(\sum^n_{k=n-4} x[k]\right)/5\,,$$ ...


1

HINT (really an answer but not providing code so its a hint): In your setup, a $M$-point moving average filter uses the average of the previous $M-1$ samples and the current sample to compute the current output. So for $n=5$, the output will be $\frac{1}{5}(x[1]+x[2]+x[3]+x[4]+x[5])$. In general, for $n=k$ the output of this $M$-point MA filter is: $$ \frac{...


1

This exercise is meant to help the student appreciate the fact that if the response $y_1(t)$ of an LTI system to an input $x_1(t)$ is known, then the response to an input $$x_2(t)=\sum_{k=1}^Ka_kx_1(t-t_k)\tag{1}$$ is given by $$y_2(t)=\sum_{k=1}^Ka_ky_1(t-t_k)\tag{2}$$ which is a direct consequence of linearity and time-invariance. Consequently, if an input ...


1

It must be added to the problem that $R(\omega)$ is a real-valued, possibly bipolar function. In that case, its inverse discrete-time Fourier transform must be even: $$r[n]=r[-n]\tag{1}$$ From the given relation between $H(e^{j\omega})$ and $R(\omega)$ it is clear that $$h[n]=r[n-25]\tag{2}$$ must hold. I'm sure that you'll manage to combine $(1)$ and $(2)$ ...


1

It is not clear in which part of the problem statement you experience issues. However, a few hints may help you get the gist of such problems The sampling frequency, it is intuitive that in order for a system to monitor the state of another system it must poll it at higher rates than the highest rate in which the monitored signal is able to change its state....


1

I would really recommend that you compute the DTFT of the sequences $x_1[n]=a^nu[n]$ and $x_2[n]=-a^nu[-n-1]$. This is very straightforward if you use the formula for the geometric series: $$\sum_{n=0}^{\infty}q^n=\frac{1}{1-q},\qquad |q|<1\tag{1}$$ It will be important to consider the magnitude of $a$. Note that it's not for nothing that the table in ...


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