Chapter 17, Problem 1P

(a)

To determine

Calculate the Fourier transform of the function shown in the given Figure.

Answer to Problem 1P

The Fourier transform for the function given in Figure is

$F\left(\omega \right)=j\frac{2A}{\tau }\left[\frac{\omega \tau \cos \left(\frac{\omega \tau }{2}\right)-2\sin \left(\frac{\omega \tau }{2}\right)}{{\omega }^2}\right]$.

Explanation of Solution

Given data:

Refer to Figure given in the textbook.

Formula used:

Write the general expression for the function $f\left(t\right)$,

$f\left(t\right)=mt$ (1)

Write the general expression for definite integral $F\left(\omega \right)$,

$F\left(\omega \right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(t\right)}{e}^{-j\omega t}dt$ (2)

Calculation:

In the given Figure, the function $f\left(t\right)$ is a straight line.

The end points of the function $f\left(t\right)$ is,

$\left(\frac{\tau }{2},A\right)$ and $\left(-\frac{\tau }{2},-A\right)$

The slope of the straight line is calculated as follows,

$\begin{array}{c}m=\frac{A-\left(-A\right)}{\frac{\tau }{2}-\left(\frac{-\tau }{2}\right)}\\ =\frac{2A}{\frac{2\tau }{2}}\\ =\frac{2A}{\tau }\end{array}$

Substitute $\frac{2A}{\tau }$ for m in equation (1) as follows,

$f\left(t\right)=\frac{2A}{\tau }t$

Therefore, for the given Figure, the function $f\left(t\right)$ is,

$f\left(t\right)=\{\begin{array}{l}\frac{2A}{\tau }t,\frac{-\tau }{2}\le t\le \frac{\tau }{2}\\ 0,\text{elsewhere}\end{array}$ (3)

Applying equation (3) in equation (2) as follows,

$F\left(\omega \right)={\displaystyle \underset{\frac{\tau }{2}}{\overset{\frac{\tau }{2}}{\int }}\frac{2A}{\tau }t}{e}^{-j\omega t}dt$

$F\left(\omega \right)=\frac{2A}{\tau }{\displaystyle \underset{\frac{\tau }{2}}{\overset{\frac{\tau }{2}}{\int }}t}{e}^{-j\omega t}dt$ (4)

Consider,

${\displaystyle \int t{e}^{-j\omega t}dt}$ (5)

Write the general expression for integration by parts method as follows,

${\displaystyle \int f{g}^{\prime }}=fg-{\displaystyle \int f^{\prime }g}$

By applying integration by parts to equation (5),

$\begin{array}{l}f=t,g^{\prime }=e^{-j\omega t}\\ f^{\prime }=1,g=-\frac{e^{-j\omega t}}{j\omega }\end{array}$ (6)

Applying equation (6) in equation (5) as follows,

${\displaystyle \int t{e}^{-j\omega t}dt}=-\frac{t{e}^{-j\omega t}}{j\omega }-{\displaystyle \int -}\frac{e^{-j\omega t}}{j\omega }dt$ (7)

Consider,

${\displaystyle \int -}\frac{e^{-j\omega t}}{j\omega }dt$ (8)

Consider,

$\begin{array}{l}u=-j\omega t\\ \frac{du}{dt}=-j\omega \\ dt=\frac{-1}{j\omega }du\end{array}$

Substitute $-j\omega t$ for u and $\frac{-1}{j\omega }du$for dt in equation (8) as follows,

${\displaystyle \int -}\frac{e^{-j\omega t}}{j\omega }dt={\displaystyle \int -}\frac{e^u}{j\omega }\frac{-1}{j\omega }du$

${\displaystyle \int -}\frac{e^{-j\omega t}}{j\omega }dt=\frac{1}{j^2{\omega }^2}{\displaystyle \int e^u}du$ (9)

Substitute equation (9) in equation (7) as follows,

$\begin{array}{c}{\displaystyle \int t{e}^{-j\omega t}dt}=-\frac{t{e}^{-j\omega t}}{j\omega }-\frac{1}{j^2{\omega }^2}{\displaystyle \int e^u}du\\ =-\frac{t{e}^{-j\omega t}}{j\omega }-\frac{1}{j^2{\omega }^2}\left[e^u\right]\\ =-\frac{t{e}^{-j\omega t}}{j\omega }-\frac{e^{-j\omega t}}{j^2{\omega }^2}\left\{\because u=-j\omega t\right\}\\ =\frac{-\left(j\omega t+1\right)e^{-j\omega t}}{j^2{\omega }^2}\end{array}$

The above equation as follows,

${\displaystyle \int t{e}^{-j\omega t}dt}=\frac{e^{-j\omega t}}{-{\omega }^2}\left(-j\omega t-1\right)$ (10)

Substitute equation (10) in equation (4), and applying the limits as follows,

$\begin{array}{c}F\left(\omega \right)=\frac{2A}{\tau }{\left[\frac{e^{-j\omega t}}{-{\omega }^2}\left(-j\omega t-1\right)\right]}_{\frac{-\tau }{2}}^{\frac{\tau }{2}}\\ =\frac{2A}{{\omega }^2\tau }\left[e^{-j\omega \left(\frac{\tau }{2}\right)}\left(\frac{j\omega \tau }{2}+1\right)-e^{j\omega \left(\frac{\tau }{2}\right)}\left(\frac{-j\omega \tau }{2}+1\right)\right]\\ =\frac{2A}{{\omega }^2\tau }\left[e^{-j\omega \left(\frac{\tau }{2}\right)}-e^{j\omega \left(\frac{\tau }{2}\right)}+\frac{j\omega \tau }{2}\left(e^{-j\omega \left(\frac{\tau }{2}\right)}+e^{j\omega \left(\frac{\tau }{2}\right)}\right)\right]\end{array}$

$F\left(\omega \right)=j\frac{2A}{\tau }\left[\frac{\omega \tau \cos \left(\frac{\omega \tau }{2}\right)-2\sin \left(\frac{\omega \tau }{2}\right)}{{\omega }^2}\right]$ (11)

Conclusion:

Thus, the Fourier transform for the function given in Figure is

$F\left(\omega \right)=j\frac{2A}{\tau }\left[\frac{\omega \tau \cos \left(\frac{\omega \tau }{2}\right)-2\sin \left(\frac{\omega \tau }{2}\right)}{{\omega }^2}\right]$

(b)

To determine

Calculate $F\left(\omega \right)$ when $\omega =0$.

Answer to Problem 1P

The function $F\left(\omega \right)=0$ when $\omega =0$.

Explanation of Solution

Given data:

Refer to Part (a).

$F\left(\omega \right)=j\frac{2A}{\tau }\left[\frac{\omega \tau \cos \left(\frac{\omega \tau }{2}\right)-2\sin \left(\frac{\omega \tau }{2}\right)}{{\omega }^2}\right]$

Formula used:

Write the general expression for L’Hospital’s rule as follows,

$\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$ (12)

Calculation:

Applying equation (12) to equation (11) when $\omega =0$ as follows,

$\begin{array}{c}F\left(0\right)=\underset{\omega \to 0}{\lim }2A\left[\frac{\omega \tau \left(\frac{\tau }{2}\right)\left[-\sin \left(\frac{\omega \tau }{2}\right)\right]+\tau \cos \left(\frac{\omega \tau }{2}\right)-2\left(\frac{\tau }{2}\right)\cos \left(\frac{\omega \tau }{2}\right)}{2\omega \tau }\right]\\ =\underset{\omega \to 0}{\lim }2A\left[\frac{-\omega \tau \left(\frac{\tau }{2}\right)\sin \left(\frac{\omega \tau }{2}\right)}{2\omega \tau }\right]\\ =\underset{\omega \to 0}{\lim }2A\left[\frac{-\tau \sin \left(\frac{\omega \tau }{2}\right)}{4}\right]\\ =2A\left[\frac{-\tau \sin \left(\frac{\left(0\right)\tau }{2}\right)}{4}\right]\end{array}$

The above equation becomes,

$\begin{array}{c}F\left(0\right)=2A\left[0\right]\left\{\because \sin 0=0\right\}\\ =0\end{array}$

Conclusion:

Thus, the function $F\left(\omega \right)=0$ when $\omega =0$.

(c)

To determine

Plot $\left|F\left(\omega \right)\right|$ versus $\omega$ when $A=10$ and $\tau =0.1$ where $\left|F\left(\omega \right)\right|$ is an even function of $\omega$.

Answer to Problem 1P

The sketch for $\left|F\left(\omega \right)\right|$ versus $\omega$ is shown in Figure 1.

Explanation of Solution

Given data:

Refer to Part (a),

$F\left(\omega \right)=j\frac{2A}{\tau }\left[\frac{\omega \tau \cos \left(\frac{\omega \tau }{2}\right)-2\sin \left(\frac{\omega \tau }{2}\right)}{{\omega }^2}\right]$ (13)

Calculation:

Appling $A=10$ and $\tau =0.1$ in equation (13) as follows,

$\begin{array}{c}F\left(\omega \right)=j\frac{2\left(10\right)}{0.1}\left[\frac{0.1\omega \cos \left(\frac{0.1\omega }{2}\right)-2\sin \left(\frac{0.1\omega }{2}\right)}{{\omega }^2}\right]\\ =j200\left[\frac{0.1\omega \cos \left(\frac{\omega }{20}\right)-2\sin \left(\frac{\omega }{20}\right)}{{\omega }^2}\right]\\ =j\left[\frac{20\omega \cos \left(\frac{\omega }{20}\right)-400\sin \left(\frac{\omega }{20}\right)}{{\omega }^2}\right]\end{array}$

$\left|F\left(\omega \right)\right|=\left|\frac{20\omega \cos \left(\frac{\omega }{20}\right)-400\sin \left(\frac{\omega }{20}\right)}{{\omega }^2}\right|$ (14)

Create a table as shown in below Table 1.

Table 1

Angular Frequency$\left(\omega \right)$ in $\frac{\text{rad}}{\text{s}}$	Function $\left|F\left(\omega \right)\right|$
–200	0.0785
–190	0.1041
–180	0.1063
–170	0.0819
–160	0.0336
–150	0.0295
–140	0.0943
–130	0.1452
–120	0.1678
–110	0.1522
–100	0.0951
–90	0.0014
–80	0.1161
–70	0.2389
–60	0.3457
–50	0.4162
–40	0.4354
–30	0.3962
–35	0.3012
–15	0.1625
0	5.0000 (Original value is infinity, though for the instance consider one finite value as 5)
10	0.1625
20	0.3012
30	0.3962
40	0.4354
50	0.4162
60	0.3457
70	0.2389
80	0.1161
90	0.0014
100	0.0951
110	0.1522
120	0.1678
130	0.1452
140	0.0943
150	0.0295
160	0.0336
170	0.0819
180	0.1063
190	0.1041
200	0.0785

Sketch the plot for various values of function $\left|F\left(\omega \right)\right|$ is an even function of $\omega$ is shown in Figure 1 using the values in Table 1.

Conclusion:

Thus, the sketch for $\left|F\left(\omega \right)\right|$ versus $\omega$ is shown in Figure 1.