Solution for Problem 60
contributed by Jeff Jaso 5/95

1.

Implicit in the notion of causality is a temporal relationship between a system input and a system response. To say that a given input produces a certain system response implies that the input chronologically precedes its effect. A common way of expressing this is that a response must not anticipate its cause or input. Stated in mathematical notation: h(t)=0, for t<0.

2.

For an even function, x(-t) = x(t). By the definition of the Fourier transform,

The righthand equality follows from Euler's Theorem:

$$\exp^{\pm ju}=\cos u\pm j\sin u$$

Substituting -t for t in the expression fo X(f) gives

Comparing the expressions for $F\{x(t)\}$ and $F\{x(-t)\}$, we notice that they differ only by the sign of the imaginary part of the integrand. Since for a given complex number r = a + bi, $r^{\ast} = a - bi$, we can conclude that $X(f) = X^{\ast}(f)$.

3.

For an odd function, y(-t) = -y(t). By the definition of the Fourier transform,

The righthand equality follows from Euler's Theorem:

$$\exp^{\pm ju}=\cos u \pm j\sin u$$

Substituting -t for t in the expression for Y(f) gives

Comparing the expressions for $F\{y(t)\}$ and $F\{y(-t)\}$, we notice that the latter is the negative of the complex conjugate of the former; that is,

$$Y(f) = -Y^{\ast}(f)$$

4.

h(t) = h_e(t) + h_o(t)

where

$$h_{e}(t) = \frac{1}{2}[h(t) + h(-t)]$$

and

$$h_{o}(t) = \frac{1}{2}[h(t) - h(-t)]$$

To show that h_e(t) is even, we must demonstrate that h_e(t) = h_e(-t). Substituting -t for t in the above expression gives

$$h_{e}(-t)= \frac{1}{2}[h(-t) + h(t)]$$

Applying the commutative property of addition gives

$$h_{e}(-t)=\frac{1}{2}[h(t) + h(-t)] = h_{e}(t)$$

Hence, h_e(t) is an even function.

Similarly, to show that h_o(t) is odd, we must demonstrate that h_o(-t) = -h_o(t). Substituting -t for t in the expression for h_o(t) gives

Hence, h_o(t) is an odd function.

Finally, to establish that h(t)=h_e(t) + h_o(t), we simply add the expressions for h_e(t) and h_o(t) and perform some algebra.

5.

Separating the real valued function h(t) into its odd and even components, we make use of the linearity of the Fourier transform to write:

$$F\{h_{e}(t) + h_{o}(t)\} = F\{h_{e}(t)\} + F\{h_{o}(t)\}$$

Invoking the result from part a) that the Fourier transform of an even function is the complex conjugate of the Fourier transform of that function, we can express the Fourier transform of the even part of h(t) as follows:

$$F\{h_{e}(t)\} = H_{e}^{\ast}(f)$$

Recalling the result from part b) that the Fourier transform of an odd function is the negative complex conjugate of the Fourier transform of that function, we can express the Fourier transform of the odd part of h(t) as follows:

$$F\{h_{o}\} = -H_{o}^{\ast}(f)$$

6.

h(t) is a pulse of centered around $\frac{1}{2}$ of width one and height one. To validate the results from parts e) and d), we must rewrite h(t) as the sum of even and odd parts, find the respective Fourier transforms, and take the superposition of the two transforms.

The Fourier transform of a pulse of height and width one centered around $\frac{1}{2}$ is given by

$$H(f) = sincf\exp^{-j\pi f\frac{1}{2}} = sincf\exp^{-j\pi f}$$

The even part of h(t) is given by

$$h_{e}(t) = \frac{1}{2}[h(t) + h(-t)]$$

which graphically represents a pulse centered around $\frac{1}{2}$ of width one and height $\frac{1}{2}$, along with its reflection across the y-axis.

Applying superposition, we can find the Fourier transform of the sum of these two pulses by adding the Fourier transforms of the individual pulses. For the pulse component centered around $\frac{1}{2}$,

$$F\{h_{e_{+\frac{1}{2}}}\} = \frac{1}{2}sincf\exp^{-j\pi f}$$

For the pulse component centered around $-\frac{1}{2}$,

$$F\{h_{e_{-\frac{1}{2}}}\} = \frac{1}{2}sincf\exp^{j\pi f}$$

By superposition,

The odd part of h(t) is given by

$$h_{o}(t) = \frac{1}{2}[h(t) - h(-t)]$$

which graphically represents a pulse centered around $\frac{1}{2}$ of width one and height $\frac{1}{2}$, in addition to its reflection across the origin, centered around $-\frac{1}{2}$ of height $-\frac{1}{2}$ and width one. Using the same procedure as with the even part of h(t), we can find $F\{h_{o}(t)\}$ by taking the superposition of the separate pulses comprising h_o(t).

For the pulse component centered around $\frac{1}{2}$ we have

$$F\{h_{o_{+\frac{1}{2}}}\} = \frac{1}{2}sincf\exp^{-j\pi f}$$

For the pulse component centered around $-\frac{1}{2}$ we have

$$F\{h_{o_{-\frac{1}{2}}}\} = \frac{1}{2}sincf\exp^{j\pi f}$$

Again by superposition,

Thus

Simplification yields

$$H(f) = sincf\exp^{-j\pi f}$$

which agrees with our original result for $F\{h(t)\}$.