# Tag Info

11

According to your definition of autocorrelation, the autocorrelation is simply the covariance of the two random variables $Z(n)$ and $Z(n+\tau)$. This function is also called autocovariance. As an aside, in signal processing, the autocorrelation is usually defined as $$R_{XX}(t_1,t_2)=E\{X(t_1)X^*(t_2)\}$$ i.e., without subtracting the mean. The ...

7

The reason is that if it is true for any $m$, it is also true for $m=kn$. I will sketch the proof in another way. Call $f_s = 1/t_s$ sampling frequency where $t_s$ is sampling period, the two signal $x(t) = \sin(2\pi f_0 t)$ and $x_k(t) = \sin(2\pi (f_0 + k f_s) t)$ have the same values at sampling instants (aliasing), i.e. $x[n] = x_k[n]$. Indeed, \...

6

The signum function is defined by $$\text{sgn}(t)=\begin{cases}-1,&t<0\\0,&t=0\\1,&t>0\end{cases}$$ Using the half-maximum convention, the unit step function is defined by $$u(t)=\begin{cases}0,&t<0\\\frac12,&t=0\\1,&t>0\end{cases}$$ From these two definitions it should be obvious that $$\text{sgn}(t)=2u(t)-1$$ must ...

6

Hint: According to Euler's formula we have $$e^{-j2\pi k}=\cos(2\pi k)-j\sin(2\pi k)=\ldots$$

6

6

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: $$\... 5 In engineering practice, the complex inversion integral is hardly ever used. As an engineer, you will almost exclusively need to invert rational functions, and this can be done by partial fraction expansion and elementary inversions. So first I'll show you how to obtain the inverse Laplace transform by partial fraction expansion, then I'll explain the ... 5 Remember that e^z has a very different meaning than e^x (taking z\in\mathbb{C} and x\in\mathbb{R}). If the exponent was real, then, as you state in your question:$$e^x = 1 \iff x=0$$However, when the exponent is complex, this function acquires a very different meaning. Let z=x+jy, where x,y\in\mathbb{R} and j is the imaginary unit. Then$$...

5

Just use the formula for the geometric series (I use $l=h-k\neq mN$): $$\sum_{n=0}^{N-1}e^{-j\frac{2\pi}{N}nl}=\frac{1-e^{-j\frac{2\pi}{N}Nl}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-e^{-j2\pi l}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-1}{1-e^{-j\frac{2\pi}{N}l}}=0,\quad l\neq mN$$

5

It's possible, in mathematics, to complete real numbers with "infinite" values, with sound topological properties; for instance non-standard analysis or the Extended real number line (discussion at Math StackExchange). However, in this context, for any standard real number $t$, the rule is to set $t\pm \infty = \pm \infty$ (see Arithmetic operations). So, ...

4

The Cauchy Schwarz inequality states that: $$\left|\int_{-\infty}^{\infty}g_1(t)g_2(t) dt\right|^2 \leq \int_{-\infty}^{\infty}|g_1(t)|^2 dt \int_{-\infty}^{\infty}|g_2(t)|^2 dt$$ I'm going to assume that $f(t)$ is real, just to make the math a little easier. From the above we can write: $$\left|\int_{-\infty}^{\infty}f(t)f(t-\tau) dt\right|^2 \leq \int_{... 4 It's a bit contrived, but observe that by the "sifting property" of Dirac-delta function:$$ K(x-X_j) = K(x) * \delta(x-X_j) $$where * denotes convolution and \delta is the Dirac-delta function. Now you can apply Fourier convolution theorem and write:$$ \mathcal{F}\{K(x-X_j)\}(t) = \kappa(t) e^{j t X_j} $$where \mathcal{F}\{g(x)\}(t) denotes ... 4 This is because the impulse response of an integrator is h(t)=u(t). The output which is the convolution with the impulse respoponse is$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$$and with h(t)=u(t) it becomes$$\begin{align} y(t)&=\int_{-\infty}^{\infty}x(\tau)u(t-\tau)d\tau\\ &=\int_{-\infty}^{t}x(\tau)d\tau\tag{1}\end{align}where (... 4 Just do a change of variables: \begin{align} z(t) &= x\star y \big|_{t}\\ &=\int_{\tau=-\infty}^{\tau = \infty} x(\tau)y(t-\tau) \,\mathrm d\tau &\scriptstyle{\text{Set}~\tau=-\lambda, \mathrm d\tau=-\mathrm d\lambda} \tag{1}\\ &=-\int_{-\lambda=-\infty}^{-\lambda = \infty} x(-\lambda)y(t+\lambda) \,\mathrm d\lambda &\scriptstyle{\text{... 4 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: sinceY[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-... 3 The summation on the left side of your equation represents a time domain discrete time periodic signal x[n] whose period is N. x[n] = \sum_{k=-\infty}^{\infty} \delta[n-kN] $$And the summation on the right side of your equation is just the same (periodic) signal x[n] represented via its Discrete Fourier Series DFS synthesis form which is an ... 3 Apart from scale factors, your USB signal is$$f(t)=m(t)\cos(2\pi\nu_ct)-\hat{m}(t)\sin(2\pi\nu_ct)\tag{1}$$(as you've correctly stated in your question). Now if you multiply with a (coherent) carrier 2\cos(2\pi\nu_ct), you get$$\begin{align}g(t)&=2\big[m(t)\cos(2\pi\nu_ct)-\hat{m}(t)\sin(2\pi\nu_ct)\big]\cos(2\pi\nu_ct)\\ &=2m(t)\cos^2(2\pi\...

3

I agree that one of the easiest ways to compute the Hilbert transform in this case is to use the analytic signal. This is most easily obtained via the Fourier transform. Note that the Fourier transform of $$f(t)=\frac{\sin(at)}{at}\tag{1}$$ is $$F(j\omega)=\frac{\pi}{a}\text{rect}\left(\frac{\omega}{2a}\right)\tag{2}$$ i.e., a rectangular function with ...

3

Did you ever think about where $\pi$ came from? Watch out... Let us first draw this weird function $e^{-2j\pi t}$ for several $t\in[0,10]$ (the little blue circles joined by line segments): On one axis, the variable $t$, on the others the real and imaginary parts, respectively. It looks like an everlasting spring. Now draw the function at integers $k\in [... 3 Let$x[n]=a^nu[n], |a|<1$. Autocorrelation is $$\phi_{xx}[n]=\sum_{m=-\infty}^{\infty}x[m]x[m-n]=\sum_{m=-\infty}^{\infty}a^mu[m]a^{m-n}u[m-n]$$ First assume that$n>0. In this case, we have $$u[m]u[m-n]=\begin{cases}0,& \forall m<n\\ 1,& \forall m\ge n\end{cases}$$ Therefore, \begin{align} \phi_{xx}[n]&=\sum_{\color{red}{m=n}}^{\... 3 Doesn't m have to be greater than 0 going from your first line to your second because you aren't taking account of the u(m) term? 3 This may be a lot easier in the frequency domain. Something like this: Even (and real) function in time transforms into a real function in frequency (and vice versa) Convolution in the time domain is equivalent to convolution in the frequency domain Multiplication of two real functions is a real function Inverse transform of a real function is even. 3 Here instead of a general proof method, let us use the method known as the counter example which relies on the fact that a single counter-example that violates the claim is sufficient (enough) to prove that the claim is wrong. Assume that the system was LTI,(the claim is that system is LTI). Then consider the following inputs x_1[n] = \delta[n] and x_2[... 3 Cedron Dawg posted an interesting initial point in this answer. It begins with these steps: \begin{align} U(\omega) &= \sum\limits_{n=0}^{+\infty} e^{-j \omega n} \\ &= \lim_{ N \to \infty } \sum\limits_{n=0}^{N-1} e^{-j \omega n}\\ &= \lim_{ N \to \infty } \left[ \frac{ 1 - e^{-j \omega N} }{ 1 - e^{-j \omega } } \right] \\ &= \... 3 Your result is correct but note that for complex signals, the even and odd parts are defined by $$x_e(t)=\frac12\left[x(t)+x^*(-t)\right]\tag{1}$$ and $$x_o(t)=\frac12\left[x(t)-x^*(-t)\right]\tag{2}$$ where^*$denotes complex conjugation. From$(1)$and$(2)$it follows that the real part of$x_e(t)$is even and its imaginary part is odd, whereas$x_o(...

3

As @MattL points out in a comment, a Gaussian pdf does not imply whiteness. Indeed, it can be argued that the assumption that the process is a continuous-time white noise process is contrary to the belief that the random variables constituting the process even have a pdf of any kind. Continuous-time white noise is a mythical beast that is used to account ...

3

This is not directly related to signal processing, but similar calculations are ubiquitous in DSP. So, here we go: \begin{align}e^{-n/D}\frac{1-e^{(n+1)/D}}{1-e^{1/D}}&=\frac{e^{-n/D}-e^{1/D}}{1-e^{1/D}}\\&=\frac{e^{1/D}}{e^{1/D}}\cdot\frac{e^{-(n+1)/D}-1}{e^{-1/D}-1}\\&=\frac{1-e^{-(n+1)/D}}{1-e^{-1/D}}\end{align}

3

Sample rate conversion systems (expansion or decimation) are time-varying operators. For your example system of decimation by $M$ : $$y[n] = T\{x[n]\} = x[Mn]$$ you can easily see that results of shift in the input and output are not the same; i.e., $$y_1[n] = T\{ x[n-d] \} = x[Mn - d]$$ and $$y_d[n] = y[n-d] = x[M(n - d)] \neq y_1[n]$$

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