# 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 $\lambda$

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

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