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Three functions $f_1(t), f_2(t)$ and $f_3(t)$, which are zero outside the interval $[0,T]$, are shown in the figure. Which of the following statements are correct?

  • $f_1(t)$ and $f_2(t)$ are orthogonal.

  • $f_1(t)$ and $f_3(t)$ are orthogonal.

  • $f_2(t)$ and $f_3(t)$ are orthogonal.

  • $f_1(t)$ and $f_2(t)$ are orthonormal.

enter image description here

The solution is :

Two functions $f(x)$ & $g(x)$ are said to orthogonal if $\int_{-\infty}^{\infty}f(x)g(x)dx=0$, so we have:

$$\int_{0}^{T}f_2(t)f_3(t)dt=\frac{T}{3}- T +\frac{2T}{3}=0$$

Now my problem is that I can't understand how they did the integral. I have tried to do the integral as:

$$\int_{0}^{T}f_2(t)f_3(t)dt=\int_{0}^{\frac{T}{3}}1\cdot 1dt+\int_{\frac{T}{3}}^{\frac{2T}{3}}(-1)\cdot (-3)dt+\int_{\frac{2T}{3}}^{T}2\cdot 1dt$$

$$\implies \int_{0}^{T}f_2(t)f_3(t)dt= \frac{T}{3}+T+\frac{2T}{3}$$

What am I doing wrong here?

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Your solution is correct and the textbook's solution is not. Clearly, the function $f_2(t)\cdot f_3(t)$ is an everywhere positive function, so its integral should be greater than $0$.

The two functions that are orthogonal are $f_1$ and $f_3$, and your method applied to this pair of functions should show their inner product is $0$.

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  • $\begingroup$ Thank you. Also, I think you have a typo in your first should :-) $\endgroup$
    – paulplusx
    Sep 25, 2018 at 9:25
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    $\begingroup$ @paulplusx Indeed. I fixed it. $\endgroup$
    – 5xum
    Sep 25, 2018 at 9:26

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