Chapter 5, Problem 34P

(a)

To determine

Sketch thespectral content of the rectangular pulse.

Answer to Problem 34P

The spectral content of the rectangular pulse is $\underset{\_}{{\left(\frac{2}{\pi }\right)}^{1/2}{V}_0\frac{\sin \omega \tau }{\omega }}$ and its sketch is drawn here.

Explanation of Solution

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

    $g\left(\omega \right)=\frac{1}{\sqrt{2}}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}V\left(t\right)e^{-i\omega t}dt}$        (I)

Here, $g\left(\omega \right)$ is the spectral content of the rectangular pulse, $V\left(t\right)$ is the strength of a signal as a function of time and $\omega$ is the frequency.

Rewrite the above equation,

    $\begin{array}{c}g\left(\omega \right)=\frac{1}{\sqrt{2\pi }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}V\left(t\right)\left(\cos \omega t-i\sin \omega t\right)dt}\\ =2{\left(2\pi \right)}^{-1/2}{\displaystyle \underset{0}{\overset{\tau }{\int }}{V}_0\cos \omega tdt}\\ ={\left(\frac{2}{\pi }\right)}^{-1/2}{V}_0\frac{\sin \omega \tau }{\omega }\end{array}$

A sketch of $g\left(\omega \right)$ is drawn here.

Conclusion:

In this case, $V\left(t\right)\sin \omega t$ is an odd function.

Rewrite the above equation,

    $\begin{array}{c}g\left(\omega \right)=\frac{1}{\sqrt{2\pi }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}V\left(t\right)\left(\cos \omega t-i\sin \omega t\right)dt}\\ =2{\left(2\pi \right)}^{-1/2}{\displaystyle \underset{0}{\overset{\tau }{\int }}{V}_0\cos \omega tdt}\\ ={\left(\frac{2}{\pi }\right)}^{-1/2}{V}_0\frac{\sin \omega \tau }{\omega }\end{array}$

Therefore, the spectral content of the rectangular pulse is $\underset{\_}{{\left(\frac{2}{\pi }\right)}^{1/2}{V}_0\frac{\sin \omega \tau }{\omega }}$ and its sketch is drawn here.

(b)

To determine

Show that a reciprocity relation $\Delta \omega \Delta t\approx \pi$.

Answer to Problem 34P

A reciprocity relation $\Delta \omega \Delta t\approx \pi$ is shown here.

Explanation of Solution

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

    $\Delta \omega \approx \frac{\pi }{\tau }$        (II)

Here, $\Delta \omega$ is the change in frequency and $\tau$ is the time taken.

Write the expression for the change in time,

    $\Delta t=\tau$        (III)

Here, $\Delta t$ is the change in time.

Conclusion:

Multiply (II) and (III),

    $\begin{array}{c}\Delta \omega \Delta t\approx \left(\frac{\pi }{\tau }\right)\tau \\ =\pi \end{array}$        (IV)

Therefore, a reciprocity relation $\Delta \omega \Delta t\approx \pi$ is shown here.

(c)

To determine

The required range of frequencies to compose a pulse of width $2\tau =1\text{μs}$ .

Answer to Problem 34P

The required range of frequencies to compose a pulse of width $2\tau =1\text{μs}$ is $\underset{\_}{2\times {10}^6\text{Hz}}$ and a pulse of width $2\tau =1\text{ns}$ is $\underset{\_}{2\times {10}^9\text{Hz}}$.

Explanation of Solution

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

    $\Delta \omega \approx \frac{\pi }{\Delta t}$        (V)

Conclusion:

For $2\tau =1\text{μs}$

Substitute $0.5\text{μs}$ for $\Delta t$ in (V),

    $\begin{array}{c}\Delta \omega =\frac{\pi }{\Delta t}\\ 2\pi \Delta f=\frac{\pi }{\left(0.5\text{μs}\right)\left(\frac{1\times {10}^{-6}\text{s}}{1\text{μs}}\right)}\\ 2\Delta f=\frac{2}{1\times {10}^{-6}\text{s}}\\ =2\times {10}^6\text{Hz}\end{array}$

For $2\tau =1\text{ns}$

Substitute $0.5\text{ns}$ for $\Delta t$ in (V),

    $\begin{array}{c}\Delta \omega =\frac{\pi }{\Delta t}\\ 2\pi \Delta f=\frac{\pi }{\left(0.5\text{ns}\right)\left(\frac{1\times {10}^{-9}\text{s}}{1\text{ns}}\right)}\\ 2\Delta f=\frac{2}{1\times {10}^{-9}\text{s}}\\ =2\times {10}^9\text{Hz}\end{array}$

Therefore, the required range of frequencies to compose a pulse of width $2\tau =1\text{μs}$ is $\underset{\_}{2\times {10}^6\text{Hz}}$ and a pulse of width $2\tau =1\text{ns}$ is $\underset{\_}{2\times {10}^9\text{Hz}}$.