Chapter 6, Problem 6.11P

(a)

To determine

The value of $v_l$, $v_m$ and $v_u$.

Answer to Problem 6.11P

The value of $v_l$ is $1.405\text{GHz}$, value of $v_m$ is $1.43\text{GHz}$ and the value of $v_u$ is $1.455\text{GHz}$.

Explanation of Solution

Write the expression for bandwidth at the center when $v_l\le v\le v_m$.

    $f_v=\frac{v}{v_m-v_l}-\frac{v_l}{v_m-v_l}$        (1)

Here, $f_v$ is the bandwidth and $\nu$ is the frequency.

Write the expression for bandwidth at the center, when $v_m\le v\le v_u$.

    $f_v=-\frac{v}{v_{\text{u}}-v_{\text{m}}}+\frac{v_{\text{u}}}{v_{\text{u}}-v_{\text{m}}}$        (2)

Write the expression for relation between the different frequency.

    $v-v_l=v_m-v_l$        (3)

Conclusion:

Substitute $1$ $f_v$, $1.430\text{GHz}$ for $v$ and $\left(\frac{50\text{MHz}}{2}\right)$ for $v_m-v_l$ in equation (1).

    $\begin{array}{c}1=\frac{1.430\text{GHz}-v_l}{\frac{50\text{MHz}}{2}-v_l}\\ 1.430\text{GHz}-v_l=\left(\frac{50\text{MHz}}{2}\times \frac{1\text{GHz}}{{10}^3\text{MHz}}\right)\\ v_l=1.430\text{GHz}-0.025\text{GHz}\\ =1.405\text{GHz}\end{array}$

Substitute $1$ $f_v$, $1.430\text{GHz}$ for $v$ and $\left(\frac{50\text{MHz}}{2}\right)$ for $v_{\text{u}}-v_{\text{m}}$ in equation (2).

    $\begin{array}{c}1=\frac{v_{\text{u}}-1.430\text{GHz}}{v_{\text{u}}-\frac{50\text{MHz}}{2}}\\ v_{\text{u}}-1.430\text{GHz}=\left(\frac{50\text{MHz}}{2}\times \frac{1\text{GHz}}{{10}^3\text{MHz}}\right)\\ v_{\text{u}}=1.430\text{GHz}+0.025\text{GHz}\\ =1.455\text{GHz}\end{array}$

Substitute $1.430\text{GHz}$ for $v$ in equation (3).

    $\begin{array}{c}v-v_l=v_m-v_l\\ v_{\text{m}}=v-v_l+v_l\\ =1.43\text{GHz}\end{array}$

Thus, the value of $v_l$ is $1.405\text{GHz}$, value of $v_m$ is $1.43\text{GHz}$ and the value of $v_u$ is $1.455\text{GHz}$.

(b)

To determine

The total power measured at the receiver.

Answer to Problem 6.11P

The total power measured at the receiver is $4.908\times {10}^{-18}\text{W}$.

Explanation of Solution

Write the expression for amount of energy detected per second.

    $P={\displaystyle \underset{A}{\int }{\displaystyle \underset{v}{\int }S\left(v\right)f_{v}dvdA}}$        (4)

Here, $S\left(v\right)$ is the spectral flux density and $f_v$ is the efficiency of the detector.

Conclusion:

Substitute functional form of filter function and integrate equation (4).

    $\begin{array}{c}P={\displaystyle \underset{A}{\int }{\displaystyle \underset{v}{\int }S\left(v\right)f_{v}dvdA}}\\ =\pi S_0{\left(\frac{D}{2}\right)}^2{\displaystyle \underset{v_l}{\overset{v_{\text{u}}}{\int }}{f}_{v}dv}\\ =\pi S_0{\left(\frac{D}{2}\right)}^2\left[\frac{1}{2}\left(v_{\text{u}}-v_{\text{l}}\right)\right]\end{array}$        (5)

Substitute $2.5\text{mJy}$ $S_0$, $100\text{m}$ for $D$, $1.455\text{GHz}$ for $v_{\text{u}}$ and $1.405\text{GHz}$ for $v_l$ in equation (5).

    $\begin{array}{c}P=\pi \left(2.5\text{mJy}\right){\left(\frac{100\text{m}}{2}\right)}^2\left[\frac{1}{2}\left(1.455\text{GHz}-1.405\text{GHz}\right)\right]\\ =\pi \left(2.5\text{mJy}\times \frac{{10}^{-29}{\text{Wm}}^{-2}{\text{Hz}}^{-1}}{1\text{mJy}}\right){\left(\frac{100\text{m}}{2}\right)}^2\left[\frac{1}{2}\left(1.455\text{GHz}-1.405\text{GHz}\right)\right]\\ =4.908\times {10}^{-18}\text{W}\end{array}$

Thus, the total power measured at the receiver is $4.908\times {10}^{-18}\text{W}$.

(c)

To determine

The power emitted at the source.

Answer to Problem 6.11P

The power emitted at the source is $7.45\times {10}^{42}\text{W}$.

Explanation of Solution

Write the expression of power of telescope at distance $d$.

    $P_{\text{tec}}=P_{\text{emit}}\left(\frac{\pi {\left(D/2\right)}^2}{4\pi d^2}\right)$        (6)

Here, $P_{\text{tec}}$ is the power of telescope, $P_{\text{emit}}$ is the power emitted at the source and $d$ is the distance from telescope.

Conclusion:

Substitute $4.908\times {10}^{-18}\text{W}$ $P_{\text{tec}}$, $100\text{Mpc}$ for $d$ and $100\text{m}$ for $D$ in equation (6).

    $\begin{array}{c}\left(4.908\times {10}^{-18}\text{W}\right)=P_{\text{emit}}\left(\frac{\pi {\left(\frac{100\text{m}}{2}\right)}^2}{4\pi {\left(100\text{Mpc}\right)}^2}\right)\\ P_{\text{emit}}=\left(4.908\times {10}^{-18}\text{W}\right)\left(\frac{4\pi {\left(100\text{Mpc}\times \frac{3.086\times {10}^{22}\text{m}}{1\text{Mpc}}\right)}^2}{\pi {\left(\frac{100\text{m}}{2}\right)}^2}\right)\\ =7.45\times {10}^{42}\text{W}\end{array}$

Thus, the power emitted at the source is $7.45\times {10}^{42}\text{W}$.