0.3 Modelling corruption  (Page 7/11)

 Page 7 / 11
$w\left(t\right)=\frac{A}{2j}\left[{e}^{j2\pi {f}_{0}t},-,{e}^{-j2\pi {f}_{0}t}\right].$

Applying the linearity property [link] and the result of Exercise  [link] gives

$\begin{array}{ccc}\hfill \mathcal{F}\left\{w\left(t\right)\right\}& =& \frac{A}{2j}\left[\mathcal{F},\left\{{e}^{j2\pi {f}_{0}t}\right\},-,\mathcal{F},\left\{{e}^{-j2\pi {f}_{0}t}\right\}\right]\hfill \\ & =& j\frac{A}{2}\left[-,\delta ,\left(f-{f}_{0}\right),+,\delta ,\left(f+{f}_{0}\right)\right].\hfill \end{array}$

Thus, the spectrum of a sine wave is a pair of $\delta$ functions with opposite signs, located symmetrically about zero frequency. The corresponding magnitude spectrum,shown in [link] , is at the heart of one importantinterpretation of the Fourier transform: it shows the frequency content of any signal by displayingwhich frequencies are present (and which frequencies are absent) from the waveform. For example, [link] (a) shows the magnitude spectrum $W\left(f\right)$ of a real-valued signal $w\left(t\right)$ . This can be interpreted as saying that $w\left(t\right)$ contains (or is made up of) “all the frequencies” up to $B$ Hz, and that it contains no sinusoids with higher frequency. Similarly,the modulated signal $s\left(t\right)$ in [link] (b) contains all positive frequencies between ${f}_{c}-B$ and ${f}_{c}+B$ , and no others.

Note that the Fourier transform in [link] is purely imaginary, as it must be because $w\left(t\right)$ is odd (see  [link] ). The phase spectrum is a flat line at $-{90}^{\circ }$ because of the factor  $j$ .

What is the magnitude spectrum of $\mathrm{sin}\left(2\pi {f}_{0}t+\theta \right)$ ? Hint: Use the frequency shift property [link] . Show that the spectrum of $\mathrm{cos}\left(2\pi {f}_{0}t\right)$ is $\frac{1}{2}\left(\delta \left(f-{f}_{0}\right)+\delta \left(f+{f}_{0}\right)\right)$ . Compare this analytical result to the numerical resultsfrom Exercise  [link] .

Let ${w}_{i}\left(t\right)={a}_{i}\mathrm{sin}\left(2\pi {f}_{i}t\right)$ for $i=1,2,3$ . Without doing any calculations, write down the spectrum of $v\left(t\right)={w}_{1}\left(t\right)+{w}_{2}\left(t\right)+{w}_{3}\left(t\right)$ . Hint: Use linearity. Graph the magnitude spectrum of $v\left(t\right)$ in the same manner as in [link] . Verify your results with a simulation mimicking that in  [link] .

Let $W\left(f\right)=\mathrm{sin}\left(2\pi f{t}_{0}\right)$ . What is the corresponding time function?

Convolution in time: it's what linear systems do

Suppose that a system has impulse response $h\left(t\right)$ , and that the input consists of a sum of three impulses occurring at times ${t}_{0}$ , ${t}_{1}$ , and ${t}_{2}$ , with amplitudes ${a}_{0}$ , ${a}_{1}$ , and ${a}_{2}$ (for example, the signal $w\left(t\right)$ of [link] ). By linearity of the Fourier transform, property [link] , the output is a superpositionof the outputs due to each of the input pulses. The output due to the first impulse is ${a}_{0}h\left(t-{t}_{0}\right)$ , which is the impulse response scaled by the size of the input and shifted to beginwhen the first input pulse arrives. Similarly, the outputs to the second and thirdinput impulses are ${a}_{1}h\left(t-{t}_{1}\right)$ and ${a}_{2}h\left(t-{t}_{2}\right)$ , respectively, and the complete output is the sum ${a}_{0}h\left(t-{t}_{0}\right)+{a}_{1}h\left(t-{t}_{1}\right)+{a}_{2}h\left(t-{t}_{2}\right)$ .

Now suppose that the input is a continuous function $x\left(t\right)$ . At any time instant $\lambda$ , the input can be thought of as consisting of an impulse scaled by the amplitude $x\left(\lambda \right)$ , and the corresponding output will be $x\left(\lambda \right)h\left(t-\lambda \right)$ , which is the impulse response scaled by thesize of the input and shifted to begin at time $\lambda$ . The complete output is then given by integrating over all $\lambda$

$y\left(t\right)={\int }_{-\infty }^{\infty }x\left(\lambda \right)h\left(t-\lambda \right)d\lambda \equiv x\left(t\right)*h\left(t\right).$

a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
how did I we'll learn this
f(x)= 2|x+5| find f(-6)
f(n)= 2n + 1
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Got questions? Join the online conversation and get instant answers!