# 0.3 Modelling corruption  (Page 9/11)

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Suppose that a system has an impulse response that is a sinc function, as shown in [link] , and that the input to the system is a white noise (as in specnoise.m ).

1. Mimic convolex.m to numerically find the output.
2. Plot the spectrum of the input and the spectrum of the output (using plotspec.m ). What kind of filter would you call this?

## Convolution $⇔$ Multiplication

While the convolution operator [link] describes mathematically how a linear system acts on a given input, time domain approaches are often notparticularly revealing about the general behavior of the system.Who would guess, for instance in Exercises  [link] and [link] , that convolution with exponentials and sinc functions would act like lowpass filters?By working in the frequency domain, however, the convolution operator is transformed into a simpler point-by-pointmultiplication, and the generic behavior of the system becomes clearer.

The first step is to understand the relationship between convolution intime, and multiplication in frequency. Suppose that the two time signals ${w}_{1}\left(t\right)$ and ${w}_{2}\left(t\right)$ have Fourier transforms ${W}_{1}\left(f\right)$ and ${W}_{2}\left(f\right)$ . Then,

$\mathcal{F}\left\{{w}_{1}\left(t\right)*{w}_{2}\left(t\right)\right\}={W}_{1}\left(f\right){W}_{2}\left(f\right).$

To justify this property, begin with the definition of the Fourier transform [link] and apply the definition of convolution [link] to obtain

$\begin{array}{ccc}\hfill \mathcal{F}\left\{{w}_{1}\left(t\right)*{w}_{2}\left(t\right)\right\}& =& {\int }_{t=-\infty }^{\infty }{w}_{1}\left(t\right)*{w}_{2}\left(t\right){e}^{-j2\pi ft}dt\hfill \\ & =& {\int }_{t=-\infty }^{\infty }\left[{\int }_{\lambda =-\infty }^{\infty },{w}_{1},\left(\lambda \right),{w}_{2},\left(t-\lambda \right),d,\lambda \right]{e}^{-j2\pi ft}dt.\hfill \end{array}$

Reversing the order of integration and using the time shift property [link] produces

$\begin{array}{ccc}\hfill \mathcal{F}\left\{{w}_{1}\left(t\right)*{w}_{2}\left(t\right)\right\}& =& {\int }_{\lambda =-\infty }^{\infty }{w}_{1}\left(\lambda \right)\left[{\int }_{t=-\infty }^{\infty },{w}_{2},\left(t-\lambda \right),{e}^{-j2\pi ft},d,t\right]d\lambda \hfill \\ & =& {\int }_{\lambda =-\infty }^{\infty }{w}_{1}\left(\lambda \right)\left[{W}_{2},\left(f\right),{e}^{-j2\pi f\lambda }\right]d\lambda \hfill \\ & =& {W}_{2}\left(f\right){\int }_{\lambda =-\infty }^{\infty }{w}_{1}\left(\lambda \right){e}^{-j2\pi f\lambda }d\lambda \hfill \\ & =& {W}_{1}\left(f\right){W}_{2}\left(f\right).\hfill \end{array}$

Thus, convolution in the time domain is the same as multiplication in the frequency domain. See [link] .

The companion to the convolution property is the multiplication property, which says that multiplication in the time domainis equivalent to convolution in the frequency domain (see [link] ); that is,

$\begin{array}{ccc}\hfill \mathcal{F}\left\{{w}_{1}\left(t\right){w}_{2}\left(t\right)\right\}& =& {W}_{1}\left(f\right)☆{W}_{2}\left(f\right)\hfill \\ & =& {\int }_{-\infty }^{\infty }{W}_{1}\left(\lambda \right){W}_{2}\left(f-\lambda \right)d\lambda .\hfill \end{array}$

The usefulness of these convolution properties is apparent when applying them to linear systems.Suppose that $H\left(f\right)$ is the Fourier transform of the impulse response $h\left(t\right)$ . Suppose that $X\left(f\right)$ is the Fourier transform of the input $x\left(t\right)$ that is applied to the system. Then [link] and [link] show that the Fourier transform of the output is exactly equal to the product of the transforms of theinput and the impulse response, that is,

$\begin{array}{ccc}\hfill Y\left(f\right)=\mathcal{F}\left\{y\left(t\right)\right\}& =& \mathcal{F}\left\{x\left(t\right)*h\left(t\right)\right\}\hfill \\ & =& \mathcal{F}\left\{h\left(t\right)\right\}\mathcal{F}\left\{x\left(t\right)\right\}=H\left(f\right)X\left(f\right).\hfill \end{array}$

This can be rearranged to solve for

$H\left(f\right)=\frac{Y\left(f\right)}{X\left(f\right)},$

which is called the frequency response of the system because it shows, for each frequency $f$ , how the system responds. For instance, suppose that $H\left({f}_{1}\right)=3$ at some frequency ${f}_{1}$ . Then whenever a sinusoid of frequency ${f}_{1}$ is input into the system, it will be amplified by a factor of 3.Alternatively, suppose that $H\left({f}_{2}\right)=0$ at some frequency ${f}_{2}$ . Then whenever a sinusoid of frequency ${f}_{2}$ is input into the system, it is removed from the output(because it has been multiplied by a factor of 0).

The frequency response shows how the system treats inputs containing various frequencies. In fact, this propertywas already used repeatedly in [link] when drawing curves that describe the behavior of lowpass and bandpassfilters. For example, the filters of Figures  [link] , [link] , and  [link] are used to remove unwanted frequencies from the communications system. In each of these cases, the plotof the frequency response describes concretely and concisely how the system (or filter) affects the input, and how thefrequency content of the output relates to that of the input. Sometimes, the frequency response $H\left(f\right)$ is called the transfer function of the system, since it “transfers” the input $x\left(t\right)$ (with transform $X\left(f\right)$ ) into the output $y\left(t\right)$ (with transform $Y\left(f\right)$ ).

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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