(New page: {| | align="left" style="padding-left: 0em;" | multiplication property |- | <math> \mathcal{Z}(\omega)=\mathcal{X}(\omega)*\mathcal{Y}(\omega) \ </math> |- | <math> Z(f)=X(f)*Y(f) \ </mat...)
 
 
Line 1: Line 1:
{|
+
=How to obtain the multiplication property in terms of f in hertz (from the formula in terms of <math>\omega</math>) =
| align="left" style="padding-left: 0em;" | multiplication property
+
 
|-
+
Denoting
| <math> \mathcal{Z}(\omega)=\mathcal{X}(\omega)*\mathcal{Y}(\omega) \ </math>
+
|-
+
<math> \mathcal{Z}(\omega)=\mathcal{X}(\omega)*\mathcal{Y}(\omega) \ </math>
| <math> Z(f)=X(f)*Y(f) \ </math>
+
 
|-
+
<math> Z(f)=X(f)*Y(f) \ </math>
| <math> Z(f)=\mathcal{Z}(2\pi f)=\frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{X}(\theta)\mathcal{Y}(2\pi f-\theta)d\theta \ </math>
+
 
|-
+
To obtain X(f), use the substitution
| <math>Let\ \varphi =\frac{\theta}{2\pi},\ then\ \theta=2\pi \varphi \ </math>
+
 
|-
+
<math>\omega= 2 \pi f </math>.
|}
+
 
 +
More specifically
 +
 
 +
<math> Z(f)=\mathcal{Z}(2\pi f)=\frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{X}(\theta)\mathcal{Y}(2\pi f-\theta)d\theta \ </math>
 +
 
 +
<math>Let\ \varphi =\frac{\theta}{2\pi},\ then\ \theta=2\pi \varphi \ </math>
 +
 
 
<div align="left" style="padding-left: 0em;">
 
<div align="left" style="padding-left: 0em;">
 
<math>
 
<math>
Line 21: Line 27:
 
</div>
 
</div>
 
<math>Since\ X(\alpha)=\mathcal{X}(2\pi \alpha),Y(\alpha)=\mathcal{Y}(2\pi \alpha) </math>
 
<math>Since\ X(\alpha)=\mathcal{X}(2\pi \alpha),Y(\alpha)=\mathcal{Y}(2\pi \alpha) </math>
 +
 +
----
 +
[[ECE438_HW1_Solution|Back to Table]]

Latest revision as of 12:10, 15 September 2010

How to obtain the multiplication property in terms of f in hertz (from the formula in terms of $ \omega $)

Denoting

$ \mathcal{Z}(\omega)=\mathcal{X}(\omega)*\mathcal{Y}(\omega) \ $

$ Z(f)=X(f)*Y(f) \ $

To obtain X(f), use the substitution

$ \omega= 2 \pi f $.

More specifically

$ Z(f)=\mathcal{Z}(2\pi f)=\frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{X}(\theta)\mathcal{Y}(2\pi f-\theta)d\theta \ $

$ Let\ \varphi =\frac{\theta}{2\pi},\ then\ \theta=2\pi \varphi \ $

$ \begin{align} Z(f) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{X}(2\pi \varphi)\mathcal{Y}(2\pi f-2\pi \varphi)d2\pi \varphi \\ &= \int_{-\infty}^{\infty} \mathcal{X}(2\pi \varphi)\mathcal{Y}(2\pi (f-\varphi))d\varphi \\ &= \int_{-\infty}^{\infty} X(\varphi)Y(f-\varphi)d\varphi \end{align} $

$ Since\ X(\alpha)=\mathcal{X}(2\pi \alpha),Y(\alpha)=\mathcal{Y}(2\pi \alpha) $


Back to Table

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett