Modern Physics, 3rd Edition
Modern Physics, 3rd Edition
3rd Edition
ISBN: 9780534493394
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Chapter 5, Problem 34P

(a)

To determine

Sketch thespectral content of the rectangular pulse.

(a)

Expert Solution
Check Mark

Answer to Problem 34P

The spectral content of the rectangular pulse is (2π)1/2V0sinωτω_ and its sketch is drawn here.

Explanation of Solution

Write the expression for the spectral content of the rectangular pulse,

    g(ω)=12V(t)eiωtdt        (I)

Here, g(ω) is the spectral content of the rectangular pulse, V(t) is the strength of a signal as a function of time and ω is the frequency.

Rewrite the above equation,

    g(ω)=12πV(t)(cosωtisinωt)dt=2(2π)1/20τV0cosωtdt=(2π)1/2V0sinωτω

A sketch of g(ω) is drawn here.

Modern Physics, 3rd Edition, Chapter 5, Problem 34P

Conclusion:

In this case, V(t)sinωt is an odd function.

Rewrite the above equation,

    g(ω)=12πV(t)(cosωtisinωt)dt=2(2π)1/20τV0cosωtdt=(2π)1/2V0sinωτω

Therefore, the spectral content of the rectangular pulse is (2π)1/2V0sinωτω_ and its sketch is drawn here.

(b)

To determine

Show that a reciprocity relation ΔωΔtπ.

(b)

Expert Solution
Check Mark

Answer to Problem 34P

A reciprocity relation ΔωΔtπ is shown here.

Explanation of Solution

Write the expression for the change in frequency in this case,

    Δωπτ        (II)

Here, Δω is the change in frequency and τ is the time taken.

Write the expression for the change in time,

    Δt=τ        (III)

Here, Δt is the change in time.

Conclusion:

Multiply (II) and (III),

    ΔωΔt(πτ)τ=π        (IV)

Therefore, a reciprocity relation ΔωΔtπ is shown here.

(c)

To determine

The required range of frequencies to compose a pulse of width 2τ=1μs .

(c)

Expert Solution
Check Mark

Answer to Problem 34P

The required range of frequencies to compose a pulse of width 2τ=1μs is 2×106Hz_ and a pulse of width 2τ=1ns is 2×109Hz_.

Explanation of Solution

Write the expression for the change in frequency in this case,

    ΔωπΔt        (V)

Conclusion:

For 2τ=1μs

Substitute 0.5μs for Δt in (V),

    Δω=πΔt2πΔf=π(0.5μs)(1×106s1μs)2Δf=21×106s=2×106Hz

For 2τ=1ns

Substitute 0.5ns for Δt in (V),

    Δω=πΔt2πΔf=π(0.5ns)(1×109s1ns)2Δf=21×109s=2×109Hz

Therefore, the required range of frequencies to compose a pulse of width 2τ=1μs is 2×106Hz_ and a pulse of width 2τ=1ns is 2×109Hz_.

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