Chapter 7: Q34P (page 386)

Show that if (12.2) is written with the factor $\frac{1}{\sqrt{2 π}}$ multiplying each integral, then the corresponding form of Parseval’s (12.24) theorem is $\int_{- \infty}^{\infty} {| f (x) |}^{2} d x = \int_{- \infty}^{\infty} {| g (α) |}^{2} d α$ .

Short Answer

Expert verified

Start from equation 12.20 and change the constant from $\frac{1}{2 π}$ to $\frac{1}{\sqrt{2 π}}$ . Then $\int_{- \infty}^{\infty} {| g (α) |}^{2} d α = \int_{- \infty}^{\infty} {| f (x) |}^{2} d x$ .

Step by step solution

Given Information.

The given function is $\bar{g_{1}} (α) = \int_{- \infty}^{\infty} \bar{f_{1}} (x) e^{i α x} d x$ .It is to verify that the corresponding form of Parseval theorem is $\int_{- \infty}^{\infty} {| g (α) |}^{2} d α = \int_{- \infty}^{\infty} {| f (x) |}^{2} d x$ after multiplying each integral with $\frac{1}{\sqrt{2 π}}$ .

Definition of Parseval’s theorem

Parseval’s theorem is a theorem stating that the energy of a signal can be expressed as its frequency components’ average energy.

Verify Parseval Theorem

$\bar{g_{1}} (α) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} \bar{f_{1}} (x) e^{i α x} d x$

Multiply both sides by $g_{2} (α)$ and integrate over $α$ from $- \infty$ to $\infty$ .

$\begin{matrix} \int_{- \infty}^{\infty} \bar{g_{1}} (α) g_{2} (α) d α = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} [\int_{- \infty}^{\infty} \bar{f_{1}} (x) e^{i α x} d x] g_{2} (α) d α \\ = \int_{- \infty}^{\infty} {\bar{f}}_{1} (x) [\frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{i α x} g_{2} (α) d α] d x \\ = \int_{- \infty}^{\infty} \bar{f_{1}} (x) f_{2} (x) d x \end{matrix}$

If $f_{1} = f_{2} = f$ and $g_{1} = g_{2} = g$ then

$\int_{- \infty}^{\infty} {| g (α) |}^{2} d α = \int_{- \infty}^{\infty} {| f (x) |}^{2} d x$

Thus, Start from equation 12.20 and change the constant from $\frac{1}{2 π}$ to localid="1664271226752" $\frac{1}{\sqrt{2 π}}$ . Then $\int_{- \infty}^{\infty} {| g (α) |}^{2} d α = \int_{- \infty}^{\infty} {| f (x) |}^{2} d x$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Show that if (12.2) is written with the factor $\frac{1}{\sqrt{2 π}}$ multiplying each integral, then the corresponding form of Parseval’s (12.24) theorem is $\int_{- \infty}^{\infty} {| f (x) |}^{2} d x = \int_{- \infty}^{\infty} {| g (α) |}^{2} d α$ .

Short Answer

Step by step solution

Given Information.

Definition of Parseval’s theorem

Verify Parseval Theorem

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Rotational Dynamics

Force

Conservation of Energy and Momentum

Electricity

Waves Physics

Magnetism and Electromagnetic Induction

Study anywhere. Anytime. Across all devices.

Show that if (12.2) is written with the factor 12πmultiplying each integral, then the corresponding form of Parseval’s (12.24) theorem is ∫-∞∞|f(x)|2dx=∫-∞∞|g(α)|2dα.

Short Answer

Step by step solution

Given Information.

Definition of Parseval’s theorem

Verify Parseval Theorem

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Rotational Dynamics

Force

Conservation of Energy and Momentum

Electricity

Waves Physics

Magnetism and Electromagnetic Induction

Study anywhere. Anytime. Across all devices.

Show that if (12.2) is written with the factor $\frac{1}{\sqrt{2 π}}$ multiplying each integral, then the corresponding form of Parseval’s (12.24) theorem is $\int_{- \infty}^{\infty} {| f (x) |}^{2} d x = \int_{- \infty}^{\infty} {| g (α) |}^{2} d α$ .