# Tag Info

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: $$\mathcal{F}\big\{u(t)\big\}=U(\omega)=\pi\delta(... 6 Note that in this case you can see that the system is causal only from the given implementation. It's important to understand that you can't see it from the difference equation (if no initial conditions are given), and in general you can't see it from the transfer function either (if no region of convergence is given). The only case for which the expression ... 4 @Henrique Luna. Please forgive me. In Part (b) of the problem the words "sequence time" should "time sequence". Sorry for the confusion! Years ago when I created that Part (b) question I was thinking about the answer to the question; "What is the time duration of an N-sample time-domain sequence?" Back then I believed the time ... 4 HINT:$$\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}$$4 Purely by inspection of the block diagram the system is causal, because the output is the sum of the current input sample and stuff that's delayed -- there's no z blocks in there to predict the future, just z^{-1} block to react to the past. Also by your method of finding the transfer function, the system is causal -- with a 3^{rd} order numerator and ... 3 There is a very important point that is being glossed over in this question (which follows how this topic is conventinally taught) which is: The DFT does not care what your sampling rate is. Ultimately, every DFT calculation boils down to these parameters using conventional naming:$$ \begin{aligned} N &= SamplesPerFrame = \frac{Samples}{Frame}\\ k &...

3

It is not. It suffices to find a counter example. Let me tell you that "I am flat". Could you derive my actual value? So, any constant function $f(t) =c$ is differentiable, and have the same derivative, $f'(t)=0$. Only from knowing that $f'(t)=0$, you cannot recover the original (constant) function. Hence the system is not invertible.

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

HINT: There's no need to compute the frequency response by evaluating the transfer functions for $z=e^{j\omega}$. Just compute the poles and zeros and see where they are. Estimate the effect of the poles and zeros on the frequency response by trying to see how they influence the behavior of the transfer function on the unit circle.

2

The system is not invertible because you can always add an arbitrary constant $c$ to any function $x(t)$ and the system will map it to same differentiated function $y(t)$. So, the mapping is not unique or one-to-one and hence not invertible.

2

In complement to Matt L.'s as-usually-excellent-answer, some additional bits on the intuition, a simplification of the problem (to ease resolution) and the construction of a counter-example. They could be useful to understand and solve similar time-invariant/shift-invariant questions. First, on the intuition: the system contains a dilation on the time ...

2

You have a system with the following input-output relation: $$y(t)=\int_{-t}^{\infty}x(-3\tau)d\tau\tag{1}$$ In order to check whether the system is time-invariant or not, we need to compare the shifted output with the output resulting from a shifted input. The shifted output is $$y(t-T)=\int_{-(t-T)}^{\infty}x(-3\tau)d\tau\tag{2}$$ Shifting the input ...

2

These problems are most easily solved by using the $\mathcal{Z}$-transform. But it is also possible to do it in the sample domain. You have two equations for expressing $w[n]$ and $w[n-1]$ in terms of $x[n]$ and $y[n]$: $$x[n]=w[n]-aw[n-1]\tag{1}$$ $$y[n]=w[n]+bw[n-1]\tag{2}$$ From $(1)$ and $(2)$, $w[n]$ is obtained as $$w[n]=\frac{1}{a+b}\big(bx[n]+ay[n]... 2 HINT: The frequency response has the form$$H(e^{j\omega})=a+be^{-j\omega}+ce^{-2j\omega}+be^{-3j\omega}+ae^{-4j\omega}\tag{1}$$which can be rewritten as$$H(e^{j\omega})=e^{-2j\omega}\big[ae^{2j\omega}+be^{j\omega}+c+be^{-j\omega}+ae^{-2j\omega}\big]\tag{2}$$Now note that the term in brackets is purely real-valued. I trust that you can take it from here. 2 HINT: If you rewrite y[n] as$$y[n]=(0.8)^3(0.8)^{n-1}u[n-1]\tag{1}$$does it become easier to find the \mathcal{Z}-transform? 2 HINT: The error energy can be written as$$\begin{align}\sum_k\big|d[k]\big|^2&=\sum_k\left|x[k]-\frac{1}{2\pi}\int_{-W}^W\sum_nx[n]e^{-jn\omega}e^{jk\omega}d\omega\right|^2\\&=\sum_k\left|x[k]-\sum_nx[n]\frac{1}{2\pi}\int_{-W}^We^{j(k-n)\omega}d\omega\right|^2\tag{1}\end{align}$$Now compute the integral$$I(k-n,W)=\frac{1}{2\pi}\int_{-W}^We^{j(k-n)\...

2

As a complement to Matt's answer, on the intuition: $u[n]$ has value $1$ from $n=0$ on. So basically its energy will increase for ever, because it keeps adding ones for $n\ge0$. Then, you build another signal $x[n]$ that grows way faster because you multiply it with the exponential term $4^n$. Therefore, only at the common sense level, one cannot expect it ...

2

HINT: It is based on the fact that $$\sin(x) = \frac{e^{jx} - e^{-jx}}{2j}\quad\text{and}\quad j^2 = -1$$

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 ...

1

It's because $$j \cdot ( a + b) = -\frac{a + b}{j}$$ which stems from the fact that the imaginary unit $j$ has the property : $$j = \frac{-1}{j}$$

1

Let $x_1(t) = t^2$ and $x_2(t) = 1$. We have $$y_1(t) =\sqrt{(t-1)^2(t+1)^2}$$ $$y_2(t) = 1$$ Now let $x_3(t) = x_1(t) + x_2(t) = t^2 + 1$. Then we have $$y_3(t) = \sqrt{((t-1)^2 +1)((t+1)^2 +1)}$$ Obviously $$y_3(t) \not = y_1(t) + y_2(t)=\sqrt{(t-1)^2(t+1)^2} +1$$Therefore the system isn't linear.

1

Given LTI system with transfer function : $$H(s) = \frac{ -2 s}{(s+6)(s+2) } \tag{1}$$ which has the poles $s = \{ -2 , -6 \}$, and zeros $s = 0$. Apply partial fraction expansion : $$H(s) = \frac{1}{s+2} - \frac{3}{s+6} \tag{2}$$ Based on pole locations, there will be three ROCs, ROC1 : -$2 < \mathcal{Re}\{s\} < \infty$ ; stable ROC2 : $-\infty ... 1 Hint Substitute$s = t-1$. That gets you the equations in a more standard form$y(s) = ...$Then go through the same excercise. 1 It's always good to do some plausibility checks on your results. The energy of$x[n]$is computed by summing up the values$|x[n]|^2$. Each one of these values is non-negative, so if you end up with a negative value for the signal energy, you should ask yourself where you went wrong (unless your name is Ramanujan). Check the conditions on the number$a$for ... 1 You correctly figured out that the system must have$3$zeros. So you just have to define a corresponding transfer function$H(z)$that satisfies $$H(z_i)=0,\qquad z_i=e^{j\omega_i},\quad i\in\{0,1,2\}$$ In other words, what is the minimum number of coefficients of a polynomial with$3$zeros? Note that you might need to provide two solutions: one general ... 1 Hint: differentiate from Fourier Transform equation: $$X(e^{j\omega})= \sum\limits_{n=- \infty }^{\infty}x[n]e^{-jn\omega} \Rightarrow \frac{d}{d\omega}X(e^{j\omega})= \frac{d}{d\omega}\sum\limits_{n=- \infty }^{\infty}x[n]e^{-jn\omega}$$ and then find what you want. 1 You're right about the relation between the imaginary part of the frequency response and the odd part of the impulse response: $$\textrm{DTFT}\big\{h_o[n]\big\}=jH_I(e^{j\omega})\tag{1}$$ So from the given$H_I(e^{j\omega})$you can obtain$h_o[n]$. Now note that the odd part of$h[n]$is defined by $$h_o[n]=\frac12\big(h[n]-h[-n]\big)\tag{2}$$ The important ... 1 In such cases it can be useful to introduce auxiliary variables at the input of the integrators. For the given diagram you could use a signal$u(t)$at the input of the second integrator. The equation for its Laplace transform$U(s)$becomes $$U(s)=\frac{1}{s}\big[X(s)-bY(s)\big]-aY(s)\tag{1}$$ You need another equation relating$U(s)$to$Y(s)$, but that ... 1 The question can only be answered if it is clarified what it is that the function$G(z)$describes. If$G(z)$is the system's transfer function then we're done immediately, because only linear time-invariant (LTI) systems can be characterized by a transfer function of that form. If$G(z)$is the transfer function, the output sequence$y[n]\$ is given by the ...

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